ITS HISTORY AND APPLICATIONS

NEČAS CENTER FOR MATHEMATICAL MODELING, Volume 1 HISTORY OF MATHEMATICS, Volume 29 PRODUCT INTEGRATION, ITS HISTORY AND APPLICATIONS Antonín Slvík (I+ A(x)dx)=I+ b A(x)dx+ b x2 A(x 2 )A(x 1 )dx 1 dx 2 + MATFYZPRESS

NEČAS CENTER FOR MATHEMATICAL MODELING, Volume 1 HISTORY OF MATHEMATICS, Volume 29 PRODUCT INTEGRATION, ITS HISTORY AND APPLICATIONS Antonín Slvík PRAGUE 2007

NEČAS CENTER FOR MATHEMATICAL MODELING Editoril bord Michl Beneš, Fculty of Nucler Sciences nd Physicl Engineering, Czech Technicl University Edurd Feireisl, Institute of Mthemtics, Acdemy of Sciences of the Czech Republic Miloslv Feistuer, Fculty of Mthemtics nd Physics, Chrles University Josef Málek, Fculty of Mthemtics nd Physics, Chrles University Jn Mlý, Fculty of Mthemtics nd Physics, Chrles University Šárk Nečsová, Institute of Mthemtics, Acdemy of Sciences of the Czech Republic Jiří Neustup, Institute of Mthemtics, Acdemy of Sciences of the Czech Republic Tomáš Roubíček, Fculty of Mthemtics nd Physics, Chrles University HISTORY OF MATHEMATICS Editoril bord Jindřich Bečvář, Fculty of Mthemtics nd Physics, Chrles University Edurd Fuchs, Fculty of Science, Msryk University Ivn Kolář, Fculty of Science, Msryk University Krel Mčák, Fculty of Eduction, Technicl University Liberec Ivn Netuk, Fculty of Mthemtics nd Physics, Chrles University Štefn Schwbik, Institute of Mthemtics, Acdemy of Sciences of the Czech Republic Emilie Těšínská, Institute of Contemporry History, Acdemy of Sciences of the Czech Republic The publiction ws supported by Nečs Center for Mthemticl Modeling nd by grnt MSM 0021620839 of the Czech Ministery of Eduction. Všechn práv vyhrzen. Tto publikce ni žádná její část nesmí být reprodukován nebo šířen v žádné formě, elektronické nebo mechnické, včetně fotokopií, bez písemného souhlsu vydvtele. c Antonín Slvík, 2007 c MATFYZPRESS, vydvtelství Mtemticko-fyzikální fkulty Univerzity Krlovy v Prze, 2007 ISBN 80-7378-006-2

Tble of contents Prefce 1 Chpter 1. Introduction 3 1.1 Ordinry differentil equtions in the 19th century..... 3 1.2 Motivtion to the definition of product integrl....... 7 1.3 Product integrtion in physics............. 9 1.4 Product integrtion in probbility theory......... 9 Chpter 2. The origins of product integrtion 13 2.1 Product integrtion in the work of Vito Volterr...... 15 2.2 Bsic results of mtrix theory............. 16 2.3 Derivtive of mtrix function............. 19 2.4 Product integrl of mtrix function.......... 22 2.5 Continuous mtrix functions.............. 32 2.6 Multivrible clculus................ 39 2.7 Product integrtion in complex domin.......... 51 2.8 Liner differentil equtions t singulr point....... 59 Chpter 3. Lebesgue product integrtion 65 3.1 Riemnn integrble mtrix functions........... 66 3.2 Mtrix exponentil function.............. 71 3.3 The indefinite product integrl............. 74 3.4 Product integrl inequlities.............. 76 3.5 Lebesgue product integrl.............. 79 3.6 Properties of Lebesgue product integrl......... 83 3.7 Double nd contour product integrls.......... 91 3.8 Generliztion of Schlesinger s definition......... 95 Chpter 4. Opertor-vlued functions 99 4.1 Integrl opertors................. 100 4.2 Product integrl of n opertor-vlued function....... 102 4.3 Generl definition of product integrl.......... 108 Chpter 5. Product integrtion in Bnch lgebrs 111 5.1 Riemnn-Grves integrl............... 112 5.2 Definition of product integrl............. 114 5.3 Useful inequlities................. 116 5.4 Properties of product integrl............. 118 5.5 Integrble nd product integrble functions........ 122 5.6 Additionl properties of product integrl......... 127 Chpter 6. Kurzweil nd McShne product integrls 131 6.1 Kurzweil nd McShne integrls............ 132

6.2 Product integrls nd their properties.......... 133 Chpter 7. Complements 139 7.1 Vrition of constnts................ 139 7.2 Equivlent definitions of product integrl......... 140 7.3 Riemnn-Stieltjes product integrl........... 142 Bibliogrphy 145

Prefce This publiction is devoted to the theory of product integrl, its history nd pplictions. The text represents n English trnsltion of my disserttion with numerous corrections nd severl complements. The definition of product integrl ppered for the first time in the work of Vito Volterr t the end of the 19th century. Although it is rther elementry concept, it is lmost unknown mong mthemticins. Wheres the ordinry integrl of function A provides solution of the eqution y (x) = f(x), the product integrl helps us to find solutions of the eqution y (x) = A(x)y(x). The function A cn be sclr function, but product integrtion is most useful when A is mtrix function; in the ltter cse, y is vector function nd the bove eqution represents in fct system of liner differentil equtions of the first order. Volterr ws trying (on the whole successfully) to crete nlogy of infinitesiml clculus for the product integrl. However, his first ppers didn t meet with gret response. Only the development of Lebesgue integrl nd the birth of functionl nlysis in the 20th century ws followed by the revivl of interest in product integrtion. The ttempts to generlize the notion of product integrl followed two directions: Product integrtion of mtrix functions whose entries re not Riemnn integrble, nd integrtion of more generl objects thn mtrix functions (e.g. opertor-vlued functions). In the 1930 s, the ides of Volterr were tken up by Ludwig Schlesinger, who elborted Volterr s results nd introduced the notion of Lebesgue product integrl. Approximtely t the sme time, Czech mthemticin nd physicist Bohuslv Hostinský proposed definition of product integrl for functions whose vlues re integrl opertors on the spce of continuous functions. New pproches to ordinry integrtion were often followed by similr theories of product integrtion; one of the ims of this work is to document this progress. It cn be lso used s n introductory textbook of product integrtion. Most of the text should be comprehensible to everyone with good knowledge of clculus. Prts of Section 1.1 nd Section 2.8 require bsic knowledge of nlytic functions in complex domin, but both my be skipped. Sections 3.5 to 3.8 ssume tht the reder is fmilir with the bsics of Lebesgue integrtion theory, nd Chpters 4 nd 5 use some elementry fcts from functionl nlysis. Almost every text bout product integrtion contins references to the works of V. Volterr, B. Hostinský, L. Schlesinger nd P. Msni, who hve strongly influenced the present stte of product integrtion theory. The lrgest prt of this 1

publiction is devoted to the discussion of their work. There were lso other pioneers of product integrtion such s G. Rsch nd G. Birkhoff, whose works [GR] nd [GB] didn t hve such gret influence nd will not be treted here. The reders with deeper interest in product integrtion should consult the monogrph [DF], which includes n exhusting list of references. All theorems nd proofs tht were tken over from nother work include reference to the originl source. However, especilly the results of V. Volterr were reformulted in the lnguge of modern mthemtics. Some of his proofs contined gps, which I hve either filled, or suggested different proof. Since the work ws originlly written in Czech, it includes references to severl Czech monogrphs nd rticles; I hve lso provided reference to n equivlent work written in English if possible. I m indebted especilly to professor Štefn Schwbik, who supervised my disserttion on product integrtion, suggested vluble dvices nd devoted lot of time to me. I lso thnk to ing. Tomáš Hostinský, who hs kindly provided photogrph of his grndfther Bohuslv Hostinský. 2

Chpter 1 Introduction This chpter strts with brief look on the prehistory of product integrtion; the first section summrizes some results concerning ordinry differentil equtions tht were obtined prior the discovery of product integrl. The next prt provides motivtion for the definition of product integrl, nd the lst two sections describe simple pplictions of product integrtion in physics nd in probbility theory. 1.1 Ordinry differentil equtions in the 19th century The notion of product integrl ws introduced by Vito Volterr in connection with the differentil eqution of the n-th order y (n) (x) + p 1 (x)y (n 1) (x) + + p n (x)y(x) = q(x). (1.1.1) Such n eqution cn be converted (see Exmple 2.5.5) into system of n liner differentil equtions of the first order y i(x) = n ij (x)y j (x) + b i (x), i = 1,..., n, j=1 which cn be lso written in the vector form y (x) = A(x)y(x) + b(x). (1.1.2) Volterr ws initilly interested in solving this eqution in the rel domin: Given the functions A : [, b] R n n (where R n n denotes the set of ll rel n n mtrices) nd b : [, b] R n, we hve to find ll solutions y : [, b] R n of the system (1.1.2). Lter Volterr considered lso the complex cse, where y : G C n, A : G C n n nd b : G C n, where G C nd C n n denotes the set of ll n n mtrices with complex entries. To be ble to pprecite Volterr s results, let s hve brief look on the theory of Equtions (1.1.1) nd (1.1.2) s developed t the end of the 19th century. A more detiled discussion cn be found e.g. in the book [Kl] (Chpters 21 nd 29). A lrge mount of problems in physics nd in geometry leds to differentil equtions; mthemticins were thus forced to solve differentil equtions lredy since the invention of infinitesiml clculus. The solutions of mny differentil equtions hve been obtined (often in n ingenious wy) in closed form, i.e. expressed s combintions of elementry functions. Leonhrd Euler proposed method for solving Eqution (1.1.1) in cse when the p i re constnts. Substituting y(x) = exp(λx) in the corresponding homogeneous eqution yields the chrcteristic eqution λ n + p 1 λ n 1 + + p n = 0. 3

If the eqution hs n distinct rel roots, then we hve obtined fundmentl system of solutions. Euler knew how to proceed even in the cse of multiple or complex roots nd ws lso ble to solve inhomogeneous equtions. The well-known method of finding prticulr solution using the vrition of constnts (which works even in the cse of non-constnt coefficients) ws introduced by Joseph Louis Lgrnge. More complicted equtions of the form (1.1.1) cn be often solved using the power series method: Assuming tht the solution cn be expressed s y(x) = n=0 n(x x 0 ) n nd substituting to the differentil eqution we obtin recurrence reltion for the coefficients n. Of course, this procedure works only in cse when the solution cn be indeed expressed s power series. Consequently, mthemticins begn to be interested in the problems of existence of solutions. The pioneering result ws due to Augustin Louis Cuchy, who proved in 1820 s the existence of solution of the eqution y (x) = f(x, y(x)) y(x 0 ) = y 0 (1.1.3) under the ssumption tht f nd f y re continuous functions. The sttement is lso true for vector functions y, nd thus the liner Eqution (1.1.2) is specil cse of the Eqution (1.1.3). Rudolf Lipschitz lter replced the ssumption of continuity of f y by weker condition (now known s the Lipschitz condition). f(x, y 1 ) f(x, y 2 ) < K y 1 y 2 Tody, the existence nd uniqueness of solution of Eqution (1.1.3) is usully proved using the Bnch fixed point theorem: We put y 1 (x) = y 0 + y n (x) = y 0 + x x 0 f(t, y 0 ) dt, x x 0 f(t, y n 1 (t)) dt, n 2. If f is continuous nd stisfies the Lipschitz condition, then the successive pproximtions {y n } n=1 converge to function y which solves Eqution (1.1.3). The method of successive pproximtions ws lredy known to Joseph Liouville nd ws used by Émile Picrd. Around 1840 Cuchy proved the existence of solution of Eqution (1.1.3) in complex domin using the so-clled mjornt method (see [VJ, EH]). We re looking for the solution of Eqution (1.1.3) in the neighbourhood of point x 0 C; the solution is holomorphic function nd thus cn be expressed in the form y(x) = c n (x x 0 ) n (1.1.4) n=0 4

in certin neighbourhood of x 0. Suppose tht f is holomorphic for x x 0 nd y y 0 b, i.e. tht f(x, y) = ij (x x 0 ) i (y y 0 ) j. (1.1.5) i,j=0 Substituting the power series (1.1.4) nd (1.1.5) to Eqution (1.1.3) gives n eqution for the unknown coefficients c n ; it is however necessry to prove tht the function (1.1.4) converges in the neighbourhood of x 0. We put nd define F (x, y) = M = sup{ f(x, y), x x 0, y y 0 b} A ij (x x 0 ) i (y y 0 ) j = i,j=0 A ij = M i b j, M (1 (x x 0 )/)(1 (y y 0 )/b). (1.1.6) The coefficients ij cn be expressed using the Cuchy s integrl formul ij = 1 i+j f i!j! x i y j = 1 (2πi) 2 ϕ f(x, y) ϕ b (x x 0 ) i+1 dy dx, (y y 0 ) j+1 where ϕ is circle centered t x 0 with rdius > 0 nd ϕ b is circle centered t y 0 with rdius b > 0. The lst eqution leds to the estimte ij A ij, i.e. the infinite series (1.1.6) is mjornt to the series (1.1.5). Cuchy proved tht there exists solution of the eqution Y (x) = F (x, Y (x)) tht cn be expressed in the form Y (x) = n=0 C n(x x 0 ) n in neighbourhood of x 0 nd such tht c n C n. Consequently the series (1.1.4) is lso convergent in neighbourhood of x 0. In prticulr, for the system of liner equtions (1.1.2) Cuchy rrived t the following result: Theorem 1.1.1. Consider functions ij, b j (i, j = 1,..., n) tht re holomorphic in the disk B(x 0, r) = {x C; x x 0 < r}. Then there exists exctly one system of functions y i (x) = yi 0 + c ij (x x 0 ) i (i = 1,..., n) j=1 defined in B(x 0, r) tht stisfies y i(x) = n ij (x)y j (x) + b i (x), j=1 y i (x 0 ) = y 0 i, 5

where y 0 1,..., y 0 n C re given numbers. As consequence we obtin the following theorem concerning liner differentil equtions of the n-th order: Theorem 1.1.2. Consider functions p 1,..., p n, q tht re holomorphic in the disk B(x 0, r) = {x C; x x 0 < r}. Then there exists exctly one holomorphic function y(x) = c k (x x 0 ) k k=0 defined in B(x 0, r) tht stisfies the differentil eqution nd the initil conditions y (n) (x) + p 1 (x)y (n 1) (x) + + p n (x)y(x) = q(x) y(x 0 ) = y 0, y (x 0 ) = y 0,..., y (n 1) (x 0 ) = y (n 1) 0, where y 0, y 0,..., y (n 1) 0 C re given complex numbers. Thus we see tht the solutions of Eqution (1.1.1), whose coefficients p 1,..., p n, q re holomorphic functions, cn be indeed obtined by the power series method. However, it is often necessry to solve Eqution (1.1.1) in cse when the coefficients p 1,..., p n, q hve n isolted singulrity. For exmple, seprtion of vribles in the wve prtil differentil eqution leds to the Bessel eqution y (x) + 1 ) x y (x) + (1 n2 y(x) = 0, whose coefficients hve singulrity t 0. Similrly, seprtion of vribles in the Lplce eqution gives the Legendre differentil eqution y (x) with singulrities t 1 nd 1. x 2 2x n(n + 1) 1 x 2 y (x) + 1 x 2 y(x) = 0 The behviour of solutions in the neighbourhood of singulrity hs been studied by Bernhrd Riemnn nd fter 1865 lso by Lzrus Fuchs. Consider the homogeneous eqution y (n) (x) + p 1 (x)y (n 1) (x) + + p n (x)y(x) = 0 (1.1.7) in the neighbourhood of n isolted singulrity t x 0 C; we ssume tht the functions p i re holomorphic in the ring P (x 0, R) = {x C; 0 < x x 0 < r}. If we choose n rbitrry P (x 0, R), then the functions p i re holomorphic in U(, r) (where r = x 0 ), nd Eqution (1.1.7) hs n linerly independent holomorphic solutions y 1,..., y n in U(, r). We now continue these functions long the circle ϕ(t) = x 0 + ( x 0 ) exp(it), t [0, 2π] 6

centered t x 0 nd pssing through. We thus obtin different system of solutions Y 1,..., Y n in U(, r). Since both systems re fundmentl, we must hve Y i = n j=1 M ijy j, or in the mtrix nottion Y = My. By clever choice of the system y 1,..., y n it cn be chieved tht M is Jordn mtrix. Using these fcts, Fuchs ws ble to prove the existence of fundmentl system of solutions of Eqution (1.1.7) in P (x 0, R) tht consists of nlytic functions of the form (x x 0 ) λi ( ϕ i 0(x) + ϕ i 1(x) log(x x 0 ) + + ϕ i n i (x) log ni (x x 0 ) ), i = 1,..., n, where ϕ j k re holomorphic functions in P (x 0, R) nd λ i C is such tht exp(2πiλ i ) is n eigenvlue of M with multiplicity n i. Moreover, if p i hs pole of order t most i t x 0 for i {1,..., n}, then the Fuchs theorem gurntees tht ϕ j k hs pole (i.e. not n essentil singulrity) t x 0. This result implies tht Eqution (1.1.7) hs t lest one solution in the form y(x) = (x x 0 ) r k=0 k(x x 0 ) k ; the numbers r nd k cn be clculted by substituting the solution to Eqution (1.1.7) (this is the Frobenius method). We hve now recpitulted some bsic fcts from the theory of ordinry differentil equtions. In lter chpters we will see tht mny of them cn be lso obtined using the theory of product integrtion. 1.2 Motivtion to the definition of product integrl The theory of product integrl is rther unknown mong mthemticins. following text should provide motivtion for the following chpters. We consider the ordinry differentil eqution The y (t) = f(t, y(t)) (1.2.1) y() = y 0 (1.2.2) where f : [, b] R n R n is given function. Thus, we re seeking solution y : [, b] R n tht stisfies (1.2.1) on [, b] (one-sided derivtives re tken t the endpoints of [, b]) s well s the initil condition (1.2.2). An pproximte solution cn be obtined using the Euler method, which is bsed on the observtion tht for smll t, y(t + t). = y(t) + y (t) t = y(t) + f(t, y(t)) t. We choose prtition D : = t 0 < t 1 < < t m = b of intervl [, b] nd put y(t 0 ) = y 0 y(t 1 ) = y(t 0 ) + f(t 0, y(t 0 )) t 1 y(t 2 ) = y(t 1 ) + f(t 1, y(t 1 )) t 2 y(t m ) = y(t m 1 ) + f(t m 1, y(t m 1 )) t m, 7

where t i = t i t i 1, i = 1,..., m. We expect tht the finer prtition D we choose, the better pproximtion we get (provided tht f is well-behved, e.g. continuous, function). We now turn to the specil cse f(t, y(t)) = A(t)y(t), where A(t) R n n is squre mtrix for every t [, b]. The Euler method pplied to the liner eqution yields y(t 0 ) = y 0, y (t) = A(t)y(t) y() = y 0 (1.2.3) y(t 1 ) = (I + A(t 0 ) t 1 )y(t 0 ) = (I + A(t 0 ) t 1 )y 0, y(t 2 ) = (I + A(t 1 ) t 2 )y(t 1 ) = (I + A(t 1 ) t 2 )(I + A(t 0 ) t 1 )y 0, y(t m ) = (I + A(t m 1 ) t m ) (I + A(t 1 ) t 2 )(I + A(t 0 ) t 1 )y 0, where I denotes the identity mtrix. Put P (A, D) = (I + A(t m 1 ) t k ) (I + A(t 1 ) t 2 )(I + A(t 0 ) t 1 ). Provided the entires of A re continuous functions, it is possible to prove (s will be done in the following chpters) tht, if ν(d) 0 (where ν(d) = mx{ t i, i = 1,..., m}), then P (A, D) converges to certin mtrix; this mtrix will be denoted by the symbol (I + A(x) dx) nd will be clled the left product integrl of the mtrix function A over the intervl [, b]. Moreover, the function Y (t) = t (I + A(x) dx) stisfies Y (t) = A(t)Y (t) Y () = I Consequently, the vector function y(t) = t (I + A(x) dx) y 0 is the solution of Eqution (1.2.3). 8

1.3 Product integrtion in physics The following exmple shows tht product integrtion lso finds pplictions outside mthemticl nlysis, prticulrly in fluid mechnics ( more generl tretment is given in [DF]). Consider fluid whose motion is described by function S : [t 0, t 1 ] R 3 R 3 ; the vlue S(t, x) corresponds to the position (t the moment t) of the prticle tht ws t position x t the moment t 0. Thus, for every t [t 0, t 1 ], S cn be viewed s n opertor on R 3 ; we emphsize this fct by writing S(t)(x) insted of S(t, x). If x is position of certin prticle t the moment t, it will move to S(t + t) S(t) 1 (x) (where denotes the composition of two opertors) during the intervl [t, t + t]. Consequently, its instntneous velocity t the moment t is given by S(t + t) S(t) 1 ( (x) x V (t)(x) = lim = lim t 0 t t 0 S(t + t) S(t) 1 ) I (x), t where I denotes the identity opertor. The velocity V (t) is n opertor on R 3 for every t [t 0, t 1 ]; in the following chpters it will be clled the left derivtive of the opertor S. Given the velocity opertor V, how to reconstruct the position opertor S? For smll t we hve S(t + t)(x). = (I + V (t) t) S(t)(x). If we choose sufficiently fine prtition D : t 0 = u 0 < u 1 < < u m = t of intervl [t 0, t], we obtin S(t)(x). = (I + V (u m 1 ) u m ) (I + V (u 0 ) u 1 )(x), where u i = u i u i 1, i = 1,..., m. The bove product (or composition) resembles the product encountered in the previous section. Indeed, pssing to the limit ν(d) 0, we see tht S is the left product integrl of opertor V, i.e. S(t) = t t 0 (I + V (u) du), t [t 0, t 1 ]. In certin sense, the left derivtive nd the left product integrl re inverse opertions. 1.4 Product integrtion in probbility theory Some results of probbility theory cn be elegntly expressed in the lnguge of product integrtion. We present two exmples concerning survivl nlysis nd Mrkov processes; both re inspired by [Gil]. Exmple 1.4.1. Let T be non-negtive continuous rndom vrible with distribution function F (t) = P (T t) nd probbility density function f(t) = F (t). 9

For exmple, T cn be interpreted s the service life of certin component (or the length of life of person etc.). The probbility of filure in the intervl [t, t + t] is P (t T t + t) = F (t + t) F (t). We remind tht the survivl function is defined s S(t) = 1 F (t) = P (T > t) nd the filure rte (or the hzrd rte) is (t) = f(t) S(t) = f(t) 1 F (t) = S (t) S(t) = d log S(t). (1.4.1) dt The nme filure rte stems from the fct tht P (t T t + t T > t) P (t T t + t) lim = lim = t 0 t t 0 P (T > t) t F (t + t) F (t) = lim = f(t) t 0 S(t) t S(t) = (t), i.e. for smll t, the conditionl probbility of filure during the intervl [t, t + t] is pproximtely (t) t. Given the function, Eqution (1.4.1) tells us how to clculte S: ( t ) S(t) = exp (u) du. (1.4.2) We cn lso proceed in different wy: If we choose n rbitrry prtition then D : 0 = t 0 < t 1 < < t m = t, S(t) = P (T > t) = P (T > t 0 )P (T > t 1 T > t 0 ) P (T > t m T > t m 1 ) = m m = P (T > t i T > t i 1 ) = (1 P (T t i T > t i 1 )). i=1 0 i=1 In cse the prtition is sufficiently fine, the lst product is pproximtely equl to m (1 (t i ) t i ). i=1 This product is similr to the one used in the definition of left product integrl, but the fctors re reversed. Its limit for ν(d) 0 is clled the right product integrl of the function on intervl [0, t] nd will be denoted by the symbol S(t) = (1 (u) du) t. (1.4.3) 0 10

Compring Equtions (1.4.2) nd (1.4.3) we obtin the result (1 (u) du) t 0 ( = exp t 0 ) (u) du, which will be proved in Chpter 2 (see Exmple 2.5.6). The product integrl representtion of S hs the dvntge tht it cn be intuitively viewed s the product of probbilities 1 (u) du tht correspond to infinitesiml intervls of length du. The lst exmple corresponds in fct to simple Mrkov process with two sttes s 1 ( the component is operting ) nd s 2 ( the component is broken ). The process strts in the stte s 1 nd goes over to the stte s 2 t time T. We now generlize our clcultion to Mrkov processes with more thn two sttes; before tht we recll the definition of Mrkov process. A stochstic process X on intervl [0, ) is rndom function t X(t), where X(t) is rndom vrible for every t [0, ). We sy tht the process is in the stte X(t) t time t. A Mrkov process is stochstic process such tht the rnge of X is either finite or countbly infinite nd such tht for every choice of numbers n N, n > 1, 0 t 1 < t 2 < < t n, we hve P (X(t n )=x n X(t n 1 )=x n 1,..., X(t 1 )=x 1 ) = P (X(t n )=x n X(t n 1 )=x n 1 ), where x 1,..., x n re rbitrry sttes (i.e. vlues from the rnge of X). The bove condition mens tht the conditionl probbility distribution of the process t time t n depends only on the lst observtion t t n 1 nd not on the whole history. Exmple 1.4.2. Let {X(t); t 0} be Mrkov process with finite number of sttes S = {s 1,..., s n }. For exmple, we cn imgine tht X(t) determines the number of ptients in physicin s witing room (whose cpcity is of course finite). Suppose tht the limit ij (t) = P (X(t + t) = s j X(t) = s i ) lim t 0+ t exists for every i, j = 1,..., n, i j nd for every t [0, ). The number ij (t) is clled the trnsition rte from stte i to stte j t time t. For sufficiently smll t we hve P (X(t + t) = s j X(t) = s i ). = ij (t) t, i j, (1.4.4) P (X(t + t) = s i X(t) = s i ). = 1 j i ij (t) t. (1.4.5) We lso define ii (t) = j i ij (t), i = 1,..., n nd denote A(t) = { ij (t)} n i,j=1. Given the mtrix A, we re interested in clculting the probbilities p i (t) = P (X(t) = s i ), t [0, ), i = 1,..., n, 11

nd p ij (s, t) = P (X(t) = s j X(s) = s i ), 0 s < t, i, j = 1,..., n. The totl probbility theorem gives p j (t) = n p i (0)p ij (0, t). i=1 The probbilities p i (0), i = 1,..., n re usully given nd it is thus sufficient to clculte the probbilities p ij (0, t), or generlly p ij (s, t). Putting P (s, t) = {p ij (s, t)} n i,j=1 we cn rewrite Equtions (1.4.4) nd (1.4.5) to the mtrix form for sufficiently smll t. P (t, t + t). = I + A(t) t (1.4.6) Using the totl probbility theorem once more we obtin p ij (s, u) = n p ik (s, t)p kj (t, u), (1.4.7) for 0 s < t < u, i, j = 1,..., n. This is equivlent to the mtrix eqution P (s, u) = P (s, t)p (t, u). (1.4.8) If we choose sufficiently fine prtition s = u 0 < u 1 < < u m = t of intervl [s, t], then Equtions (1.4.6) nd (1.4.8) imply m P (s, t) = P (u i 1, u i ) =. m (I + A(u i ) u i ). i=1 Pssing to the limit for ν(d) 0 we obtin mtrix which is clled the right product integrl of the function A over intervl [s, t]: i=1 P (s, t) = (I + A(u) du) t. s The lst result cn be gin intuitively interpreted s the product of mtrices I + A(u) du which correspond to trnsition probbilities in the infinitesiml time intervls of length du. 12

Chpter 2 The origins of product integrtion The notion of product integrl hs been introduced by Vito Volterr t the end of the 19th century. We strt with short biogrphy of this eminent Itlin mthemticin nd then proceed to discuss his work on product integrtion. Vito Volterr ws born t Ancon on 3rd My 1860. His fther died two yers lter; Vito moved in with his mother Angelic to Alfonso, Angelic s brother, who supported them nd ws like boy s fther. Becuse of their finncil sitution, Angelic nd Alfonso didn t wnt Vito to study his fvourite subject, mthemtics, t university, but eventully Edordo Almgià, Angelic s cousin nd rilrod engineer, helped to persude them. An importnt role ws lso plyed by Volterr s techer Ròiti, who secured him plce of ssistnt in physicl lbortory. Vito Volterr 1 In 1878 Volterr entered the University of Pis; mong his professors ws the fmous Ulisse Dini. In 1879 he pssed the exmintion to Scuol Normle Superiore of Pis. Under the influence of Enrico Betti, his interest shifted towrds mthemticl physics. In 1882 he offered thesis on hydrodynmics, grduted doctor of physics nd becme Betti s ssistent. Shortly fter, in 1883, the young Volterr won the competition for the vcnt chir of rtionl mechnics nd ws promoted 1 Photo from [McT] 13

to professor of the University of Pis. After Betti s deth he took over his course in mthemticl physics. In 1893 he moved to the University of Turin, but eventully settled in Rome in 1900. The sme yer he mrried Virgini Almgià (the dughter of Edordo Almgià). During the first qurter of the 20th century Volterr not only represented the leding figure of Itlin mthemtics, but lso becme involved in politics nd ws nominted Sentor of the Kingdom in 1905. When Itly entered the world wr in 1915, Volterr volunteered the Army Corps of Engineers nd engged himself in perfecting of irships nd firing from them; he lso promoted the collbortion with French nd English scientists. After the end of the wr he returned to scientific work nd teching t the university. Volterr strongly opposed the Mussolini regime which cme to power in 1922. As one of the few professors who refused to tke n oth of loylty imposed by the fscists in 1931, he ws forced to leve the University of Rome nd other scientific institutions. After then he spent lot of time brod (giving lectures e.g. in Frnce, Spin, Czechoslovki or Romni) nd lso t his country house in Aricci. Volterr, who ws of Jewish descent, ws lso ffected by the ntisemitic rcil lws of 1938. Although he begn to suffer from phlebitis, he still devoted himself ctively to mthemtics. He died in isoltion on 11th October 1940 without greter interest of Itlin scientific community. Despite the fct tht Volterr is best known s mthemticin, he ws mn of universl interests nd devoted himself lso to physics, biology nd economy. His mthemticl reserch often hd origins in physicl problems. Volterr ws lso n enthusistic bibliophile nd his collection, which reched nerly seven thousnd volumes nd is now deposited in the United Sttes, included rre copies of scientific ppers e.g. by Glileo, Brhe, Trtgli, Fermt etc. The monogrph [JG] contins welth of informtion bout Volterr s life nd times. Volterr s nme is closely connected with integrl equtions. He contributed the method of successive pproximtions for solving integrl equtions of the second kind, nd lso noted tht n integrl eqution might be considered s limiting cse of system of lgebric liner equtions; this observtion ws lter utilized by Ivr Fredholm (see lso the introduction to Chpter 4). His investigtions in clculus of vritions led him to the study of functionls (he clled them functions of lines ); in fct he built complete clculus including the definitions of continuity, derivtive nd integrl of functionl. Volterr s pioneering work on integrl equtions nd functionls is often regrded s the dwn of functionl nlysis. An overview of his chievements in this field cn be obtined from the book [VV5]. Volterr ws lso one of the founders of mthemticl biology. The motivtion cme from his son-in-lw Umberto D Ancon, who ws studying the sttistics of Adritic fishery. He posed to Volterr the problem of explining the reltive increse of predtor fishes, s compred with their prey, during the period of First World Wr (see e.g. [MB]). Volterr interpreted this phenomenon with the help of mthemticl models of struggle between two species; from mthemticl point of 14

view, the models were combintions of differentil nd integrl equtions. Volterr s correspondence concerning mthemticl biology ws published in the book [IG]. A more detiled description of Volterr s ctivites (his work on prtil differentil equtions, theory of elsticity) cn be found in the biogrphies [All, JG] nd lso in the books [IG, VV5]. An interesting ccount of Itlin mthemtics nd its intertwining with politics in the first hlf of the 20th century is given in [GN]. 2.1 Product integrtion in the work of Vito Volterr Volterr s first work devoted to product integrtion [VV1] ws published in 1887 nd ws written in Itlin. It introduces the two bsic concepts of the multiplictive clculus, nmely the derivtive of mtrix function nd the product integrl. The topics discussed in [VV1] re essentilly the sme s in Sections 2.3 to 2.6 of the present chpter. The publiction [VV1] ws followed by second prt [VV2] printed in 1902, which is concerned minly with mtrix functions of complex vrible. It includes results which re recpitulted in Sections 2.7 nd 2.8, nd lso tretment of product integrtion on Riemnn surfces. Volterr lso published two short Itlin notes, [VV3] from 1887 nd [VV4] from 1888, which summrize the results of [VV1, VV2] but don t include proofs. Volterr s finl tretment of product integrtion is represented by the book Opértions infinitésimles linéires [VH] written together with Czech mthemticin Bohuslv Hostinský. The publiction ppered in the series Collection de monogrphies sur l théorie des fonctions directed by Émile Borel in 1938. More thn two hundred pges of [VH] re divided into eighteen chpters. The first fifteen chpters represent French trnsltion of [VV1, VV2] with only smll chnges nd complements. The remining three chpters, whose uthor is Bohuslv Hostinský, will be discussed in Chpter 4. As Volterr notes in the book s prefce, the publiction of [VH] ws motivted by the results obtined by Bohuslv Hostinský, s well s by n incresed interest in mtrix theory mong mthemticins nd physicists. As the bibliogrphy of [VH] suggests, Volterr ws lredy wre of the ppers [LS1, LS2] by Ludwig Schlesinger, who linked up to Volterr s first works (see Chpter 3). The book [VH] is rther difficult to red for contemporry mthemticins. One of the resons is somewht cumbersome nottion. For exmple, Volterr uses the sme symbol to denote dditive s well s multiplictive integrtion: The sign pplied to mtrix function denotes the product integrl, while the sme sign pplied to sclr function stnds for the ordinry (dditive) integrl. Clcultions with mtrices re usully written out for individul entries, wheres using the mtrix nottion would hve gretly simplified the proofs. Moreover, Volterr didn t hesitte to clculte with infinitesiml quntities, he interchnges the order of summtion nd integrtion or the order of prtil derivtives without ny justifiction etc. The conditions under which individul theorems hold (e.g. continuity or differentibility of the given functions) re often omitted nd must be deduced from the proof. This is certinly surprising, since the rigorous foundtions of mthemticl 15

nlysis were lredy lid out t the end of the 19th century, nd even Volterr contributed to them during his studies by providing n exmple of function which is not Riemnn integrble but hs n ntiderivtive. The following sections summrize Volterr s chievements in the field of product integrtion. Our discussion is bsed on the text from [VH], but the results re stted in the lnguge of contemporry mthemtics (with occsionl comments on Volterr s nottion). Proofs of most theorems re lso included; they re generlly bsed on Volterr s originl proofs except for few cses where his clcultions with infinitesiml quntities were replced by different, rigorous rgument. 2.2 Bsic results of mtrix theory The first four chpters of the book [VH] recpitulte some bsic results of mtrix theory. Most of them re now tught in liner lgebr courses nd we repet only some of them for reder s convenience, s we will refer to them in subsequent chpters. Volterr refers to mtrices s to substitutions, becuse they cn be used to represent liner chnge of vribles. A composition of two substitutions then corresponds to multipliction of mtrices: If then where x i = n j=1 ij x j nd x i = x i = c ij = n c ij x j, j=1 n b ij x j, j=1 n b ik kj, (2.2.1). We will use the symbol R n n to denote the set of ll squre mtrices with n rows nd n columns. If A = { ij } n i,j=1, B = {b ij } n i,j=1, C = {c ij } n i,j=1, we cn write Eqution (2.2.1) in the form C = B A. A mtrix A = { ij } n i,j=1 is clled regulr if it hs nonzero determinnt. If i {1,..., n}, the theorem on expnsion by minors gives 11 1n det..... = n1 nn n ik A ik, (2.2.2) where A ik = ( 1) i+k M ik 16

is the so-clled cofctor corresponding to minor M ik ; the minor is defined s the determinnt of mtrix obtined from A by deleting the i-th row nd k-th column. Since the determinnt of mtrix with two or more identicl rows is zero, it follows tht n jk A ik = δ ij det A. (2.2.3) for ech pir of numbers i, j {1,..., n}; recll tht δ ij is the Kronecker symbol { 1 if i = j, δ ij = 0 if i j. If we thus define the mtrix A 1 = { } n Aji, det A i,j=1 then Eqution (2.2.3) yields where AA 1 = I = A 1 A, 1 0 0 0 1 0 I =...... 0 0 1 is the identity mtrix. The mtrix A 1 is clled the inverse of A. The definition of mtrix multipliction gives the following rule for multipliction of block mtrices: Theorem 2.2.1. 1 Consider mtrix tht is prtitioned into m 2 squre blocks A ij nd mtrix prtitioned into m 2 squre blocks B ij such tht A ij hs the sme dimensions s B ij for every i, j {1,..., m}. Then A 11 A 1m B 11 B 1m C 11 C 1m.......... =....., A m1 A mm B m1 B mm C m1 C mm where C ik = j A ijb jk. We will be often deling with block digonl mtrices, i.e. with mtrices of the form 1 [VH], p. 27 A 1 0 0 0 A 2 0...... 0 0 A m 17

composed of smller squre mtrices A 1, A 2,..., A m ; Volterr denotes such mtrix by the symbols { } m A1 A 2 A m or A i, but we don t follow his nottion. The following theorem expresses the fct tht every squre mtrix cn be trnsformed to certin cnonicl form clled the Jordn norml form. Volterr proves the theorem by induction on the dimension of the mtrix; we refer the reder to ny good liner lgebr textbook. Theorem 2.2.2. 1 To every mtrix A R n n there exist mtrices C, J R n n such tht A = C 1 JC nd J hs the form J 1 0 0 0 J J = 2 0...... 0 0 J m, where J i = i=1 λ i 0 0 0 1 λ i 0 0....... 0 0 1 λ i nd {λ 1,..., λ m } re (not necessrily distinct) eigenvlues of A. Recll tht if A = C 1 BC for some regulr mtrix C, then the mtrices A, B re clled similr. Thus the previous theorem sys tht every squre mtrix is similr to certin Jordn mtrix. The next two theorems concern the properties of block digonl mtrices nd re simple consequences of Theorem 2.2.1. Theorem 2.2.3. If A i is squre mtrix of the sme dimensions s B i for every i {1,..., m}, then A 1 0 0 B 1 0 0 A 1 B 1 0 0 0 A 2 0 0 B. 2 0........... = 0 A 2 B 2 0....... 0 0 A m 0 0 B m 0 0 A m B m Theorem 2.2.4. The inverse of block digonl mtrix composed of invertible mtrices A 1,..., A m is equl to 1 [VH], p. 20 24 1 A 1 0 0 0 A 2 0...... = 0 0 A m A 1 1 0 0 0 A 1 2 0...... 0 0 A 1 m. 18

2.3 Derivtive of mtrix function In this section we focus on first of the bsic opertions of Volterr s mtrix clculus, which is the derivtive of mtrix function. A mtrix function A : [, b] R n n will be clled differentible t point x (, b) if ll the entries ij, i, j {1,..., n} of A re differentible t x; in this cse we denote A (x) = { ij(x)} n i,j=1. We lso define A (x) for the endpoints x = nd x = b s the mtrix of the corresponding one-sided derivtives (provided they exist). Definition 2.3.1. Let A : [, b] R n n be mtrix function tht is differentible nd regulr t point x [, b]. We define the left derivtive of A t x s d dx A(x) = A (x)a 1 A(x + x)a 1 (x) I (x) = lim x 0 x nd the right derivtive of A t x s A(x) d dx = A 1 (x)a (x) = lim x 0 A 1 (x)a(x + x) I. x Volterr doesn t use the mtrix nottion nd insted writes out the individul entries: { } d n dx { ik (x + x) ik (x) ij} = lim A jk (x), x 0 x { { ik } d dx = lim n x 0 } A ki (x) kj(x + x) kj (x), x where {A ji } n i,j=1 denote the entries of A 1. He lso defines the left differentil s d{ ij } = A(x + dx)a(x) 1 = I + A (x)a(x) 1 dx = nd the right differentil s { ij }d = A(x) 1 A(x + dx) = I + A(x) 1 A (x) dx = { { δ ij + δ ij + } n ik(x)a jk (x) dx } n A ki (x) kj(x) dx, where dx is n infinitesiml quntity. Both differentils re considered s mtrices tht differ infinitesimlly from the identity mtrix. 19

Volterr uses infinitesiml quntities without ny scruples, which sometimes leds to very unredble proofs. This is lso the cse of the following theorem; Volterr s justifiction hs been replced by rigorous proof. Theorem 2.3.2. 1 If A, B : [, b] R n n re differentible nd regulr mtrix functions t x [, b], then d dx (AB) = d ( ) ( d dx A + A dx B A 1 = A A d dx + d ) dx B A 1, (AB) d dx = B d ( dx + B 1 A d ) ( B = B 1 A d dx dx + d ) dx B B, where ll derivtives re tken t the given point x. Proof. The definition of the left derivtive gives d dx (AB) = (AB) (AB) 1 = (A B + AB )B 1 A 1 = A A 1 + AB B 1 A 1, where the expression on the right hnd side is equl to ( ) d d dx A + A dx B A 1, but cn be lso trnsformed to the form AA 1 A A 1 + AB B 1 A 1 = A The second prt is proved in similr wy. ( A d dx + d ) dx B A 1. Corollry 2.3.3. 2 Consider mtrix function A : [, b] R n n tht is differentible nd regulr on [, b]. Then for n rbitrry regulr mtrix C R n n we hve d dx (AC) = d dx A. The corollry cn be expressed like this: The left derivtive of mtrix function doesn t chnge, if the function is multiplied by constnt mtrix from right. It is lso esy to prove dul sttement: The right derivtive of mtrix function doesn t chnge, if the function is multiplied by constnt mtrix from left. Symboliclly written, (CA) d dx = A d dx. As Volterr notes, this is generl principle: Ech sttement concerning mtrix functions remins true, if we replce ll occurences of the word left by the word 1 [VH], p. 43 2 [VH], p. 39 20

right nd vice vers. A precise formultion nd justifiction of this dulity principle is due to P. R. Msni nd will be given in Chpter 5, Remrk 5.2.2. Theorem 2.3.4. 1 If A : [, b] R n n is differentible nd regulr t x [, b], then d dx (A 1 ) = A d dx, (A 1 ) d dx = d dx A, where ll derivtives re tken t the given point x. Proof. Differentiting the eqution AA 1 = I yields A A 1 + A(A 1 ) = 0, nd consequently (A 1 ) = A 1 A A 1. The sttement follows esily. Corollry 2.3.5. 2 If A, B : [, b] R n n re differentible nd regulr mtrix functions t x [, b], then ( d d dx (A 1 B) = A 1 dx B d ) dx A A, (AB 1 ) d ( dx = B A d dx B d ) B 1, dx where ll derivtives re tken t the given point x. Proof. A simple consequence of Theorems 2.3.2 nd 2.3.4. Theorem 2.3.6. 3 Consider functions A, B : [, b] R n n tht re differentible nd regulr on [, b]. If d dx A = d dx B on [, b], then there exists mtrix C R n n such tht B(x) = A(x)C for every x [, b]. Proof. Define C(x) = A 1 (x)b(x) for x [, b]. Corollry 2.3.5 gives ( d d dx C = A 1 dx B d ) dx A A = 0, which implies tht 0 = C (x) for every x [, b]. This mens tht C is constnt function. A combintion of Theorem 2.3.6 nd Corollry 2.3.3 leds to the following sttement: Two mtrix functions hve the sme left derivtive on given intervl, if nd only if one of the functions is obtined by multiplying the other by constnt mtrix from right. This is the fundmentl theorem of Volterr s differentil clculus; dul sttement is gin obtined by interchnging the words left nd 1 [VH], p. 41 2 [VH], p. 44 3 [VH], p. 46 21

right. Both sttements represent n nlogy of the well-known theorem: Two functions hve the sme derivtive if nd only if they differ by constnt. Theorem 2.3.7. 1 Consider mtrix function A : [, b] R n n tht is differentible nd regulr on [, b]. Then for n rbitrry regulr mtrix C R n n we hve ( ) d d dx (CA) = C dx A C 1. Proof. A simple consequence of Theorem 2.3.2. Theorem 2.3.8. Let A : [, b] R n n be mtrix function of the form A 1 (x) 0 0 0 A 2 (x) 0 A(x) =.....,. 0 0 A k (x) where A 1,..., A k re squre mtrix functions. If d dx A i(x) = B i (x), i = 1,..., k, then B 1 (x) 0 0 d dx A(x) = 0 B 2 (x) 0...... 0 0 B k (x). Proof. The sttement follows from the definition of left derivtive nd from Theorems 2.2.3 nd 2.2.4. 2.4 Product integrl of mtrix function Consider mtrix function A : [, b] R n n with entries { ij } n i,j=1. For every tgged prtition D : = t 0 ξ 1 t 1 ξ 2 t m 1 ξ m t m = b of intervl [, b] with division points t i nd tgs ξ i we denote 1 [VH], p. 41 t i = t i t i 1, i = 1,..., m, ν(d) = mx 1 i m t i. 22

We lso put P (A, D) = P (A, D) = 1 (I + A(ξ i ) t i ) = (I + A(ξ m ) t m ) (I + A(ξ 1 ) t 1 ), i=m m (I + A(ξ i ) t i ) = (I + A(ξ 1 ) t 1 ) (I + A(ξ m ) t m ). i=1 Volterr now defines the left integrl of A s the mtrix b { ij } = lim P (A, D) ν(d) 0 (in cse the limit exists) nd the right integrl s b { ij } = lim P (A, D) ν(d) 0 (gin if the limit exists). Volterr isn t very precise bout the mening of the limit tken with respect to prtitions; we mke the following greement: If M(D) is mtrix which is dependent on the choice of tgged prtition D of intervl [, b], then the equlity lim M(D) = M ν(d) 0 mens tht for every ε > 0 there exists δ > 0 such tht M(D) ij M ij < ε for every tgged prtition D of intervl [, b] stisfying ν(d) < δ nd for i, j = 1,..., n. The following definition lso introduces different nottion to better distinguish between ordinry nd product integrls. Definition 2.4.1. Consider function A : [, b] R n n. If the limit lim P (A, D) or lim P (A, D) ν(d) 0 ν(d) 0 exists, it is clled the left (or right) product integrl of A over the intervl [, b]. We use the nottion for the left product integrl nd (I + A(t) dt) = lim P (A, D) ν(d) 0 (I + A(t) dt) = lim ν(d) 0 P (A, D) 23

for the right product integrl. We note tht in the cse when the upper limit of integrtion coincides with the lower limit, then (I + A(t) dt) = (I + A(t) dt) = I. In the subsequent text we use the following convention: A function A : [, b] R n n is clled Riemnn integrble, if its entries ij re Riemnn integrble functions on [, b]. In this cse we put b { b A(t) dt = ij (t) dt We will often encounter integrls of the type b xk x2 } n. i,j=1 A(x k ) A(x 1 ) dx 1 dx k. These integrls should be interpreted s iterted integrls, where x k [, b] nd x i [, x i+1 ] for i {1,..., k 1}. Lemm 2.4.2. Let A : [, b] R n n be Riemnn integrble function such tht A(x)A(y) = A(y)A(x) for every x, y [, b]. Then for every k N. b xk = 1 k! x2 b b A(x k ) A(x 1 ) dx 1 dx k = b A(x k ) A(x 1 ) dx 1 dx k Proof. If P (k) denotes ll permuttions of the set {1,..., k} nd for every π P (k), then M π = {(x 1,..., x k ) [, b] k ; x π(1) x π(2) x π(k) } b b = π P (k) b The ssumption of commuttivity implies A(x k ) A(x 1 ) dx 1 dx k = M π A(x k ) A(x 1 ) dx 1 dx k = A(x k ) A(x 1 ) dx 1 dx k. M π b xk 24 x2 A(x k ) A(x 1 ) dx 1 dx k

for every permuttion π P (k), which completes the proof. Theorem 2.4.3. 1 If A : [, b] R n n is Riemnn integrble function, then both product integrls exist nd (I + A(x) dx) = I + b xk x2 A(x k ) A(x 1 ) dx 1 dx k, (I + A(x) dx) = I + b xk x2 A(x 1 ) A(x k ) dx 1 dx k. Proof. Volterr s proof goes s follows: Expnding the product P (A, D) gives 1 m (I + A(ξ i ) t i ) = I + A(ξ ik ) A(ξ i1 ) t i1 t ik. i=m 1 i 1< <i k m Volterr now rgues tht for ν(d) 0 we obtin nd generlly 1 i 1< <i k m 1 i 1<i 2 m 1 i 1 m A(ξ i1 ) t i1 A(ξ i2 )A(ξ i1 ) t i1 t i2 A(ξ ik ) A(ξ i1 ) t i1 t ik for every k {1,..., m}. rrived t the result (I + A(x) dx) = I + b b x2 A(x 1 ) dx 1, b xk A(x 2 )A(x 1 ) dx 1 dx 2, x2 A(x k ) A(x 1 ) dx 1 dx k Using the fct tht m for ν(d) 0, Volterr b xk x2 A(x k ) A(x 1 ) dx 1 dx k. The proof for the right product integrl is crried out in similr wy. The infinite series expressing the vlue of the product integrls re often referred to s the Peno series. They were discussed by Giuseppe Peno in his pper [GP] from 1888 deling with systems of liner differentil equtions. Remrk 2.4.4. The proof of Theorem 2.4.3 given by Volterr is somewht unstisfctory. First, he didn t justify tht 1 i 1< <i k m 1 [VH], p. 49 52 A(ξ ik ) A(ξ i1 ) t i1 t ik b xk x2 A(x k ) A(x 1 ) dx 1 dx k 25

for ν(d) 0. This cn be done s follows: Define X k = {(x 1,..., x k ) R k ; x 1 < x 2 < < x k b} nd let χ k be the chrcteristic function of the set X k. Then = lim lim ν(d) 0 1 i 1< <i k m m ν(d) 0 i 1,...,i b b b = = b xk A(ξ ik ) A(ξ i1 ) t i1 t ik = A(ξ ik ) A(ξ i1 )χ(ξ i1,..., ξ ik ) t i1 t ik = A(x k ) A(x 1 )χ(x 1,..., x k ) dx 1 dx k = x2 A(x k ) A(x 1 ) dx 1 dx k. The second problem is tht Volterr didn t explin the equlity lim ν(d) 0 I + I + m 1 i 1< <i k m lim ν(d) 0 1 i 1< <i k m A(ξ ik ) A(ξ i1 ) t i1 t ik = A(ξ ik ) A(ξ i1 ) t i1 t ik. We postpone its justifiction to Chpter 5, Lemm 5.5.9. Theorem 2.4.5. 1 If A : [, b] R n n is Riemnn integrble function, then the infinite series x (I + A(t) dt) = I + x (I + A(t) dt) = I + x xk x xk x2 x2 converge bsolutely nd uniformly for x [, b]. A(x k ) A(x 1 ) dx 1 dx k, A(x 1 ) A(x k ) dx 1 dx k Proof. We give only the proof for the first series: Its sum is mtrix whose (i, j)-th entry is the number 1 [VH], p. 51 52 n l 1,...,l k 1=1 x xk x2 i,l1 (x k ) lk 1,j(x 1 ) dx 1 dx k. (2.4.1) 26

The functions ij re Riemnn integrble, therefore bounded: There exists positive number M R such tht ij (t) M for i, j {1,..., n} nd t [, b]. Using Lemm 2.4.2 we obtin the estimte n x xk x2 i,l1 (x k ) lk 1,j(x 1 ) dx 1 dx k l 1,...,l k 1=1 n k 1 M k b for every x [, b]. Since xk x2 dx 1 dx k = 1 (nm(b )) k n k! 1 (nm(b )) k = 1 n k! n enm(b ), we see tht (ccording to the Weierstrss M-test) the infinite series (2.4.1) converges uniformly nd bsolutely on [, b]. Theorem 2.4.6. 1 If A : [, c] R n n is Riemnn integrble function, then c (I + A(x) dx) = c (I + A(x) dx) b (I + A(x) dx) nd for every c [, b]. Proof. {D 2 k } c c (I + A(x) dx) = (I + A(x) dx) (I + A(x) dx) Tke two sequences of tgged prtitions {Dk 1} of intervl [, b] nd of intervl [b, c] such tht lim k ν(d1 k) = lim k ν(d2 k) = 0. If we put D k = D 1 k D2 k, we obtin sequence of tgged prtitions {D k} of intervl [, c] such tht lim k ν(d k ) = 0. Consequently c (I + A(x) dx) = lim P (A, D k) = lim P (A, k k D2 k) lim P (A, k D1 k) = = c (I + A(x) dx) (I + A(x) dx). The second prt is proved in the sme wy. 1 [VH], p. 54 56 b b 27

Remrk 2.4.7. Volterr lso offers different proof of Theorem 2.4.6, which goes s follows 1 : Denote D(i) = {x R i ; x 1 x i c}, D(j, i) = {x R i ; x 1 x j b x j+1 x i c} for ech pir of numbers i N nd j {0,..., i}. Clerly i 1 D(i) = D(0, i) D(i j, i j) D(0, j) D(i, i) (2.4.2) for every i N. We hve j=1 (I + A(x) dx) = I + b i=1 c (I + A(x) dx) = I + i=1 D(i,i) D(0,i) A(x i ) A(x 1 ) dx 1 dx i, A(x i ) A(x 1 ) dx 1 dx i. Since both infinite series converge bsolutely, their product is equl to the Cuchy product: c (I + A(x) dx) (I + A(x) dx) = I + b + + i=1 i=1 D(0,i) ( i 1 ( j=1 i=1 D(i,i) A(x i ) A(x 1 ) dx 1 dx i + ) A(x j ) A(x 1 ) dx 1 dx j D(0,j) A(x i ) A(x 1 ) dx 1 dx i + ( )) A(x i j ) A(x 1 ) dx 1 dx i j = D(i j,i j) c = I + A(x i ) A(x 1 ) dx 1 dx i = (I + A(x) dx) i=1 D(i) (we hve used Eqution (2.4.2)). If nd b re two rel numbers such tht < b, we usully define 1 [VH], p. 54 56 b f = b f. 28

The following definition ssigns mening to product integrl whose lower limit is greter thn its upper limit. Definition 2.4.8. For ny function A : [, b] R n n we define (I + A(t) dt) = lim b ν(d) 0 i=1 m (I A(ξ i ) t i ) = (I A(t) dt) nd (I + A(t) dt) = lim b ν(d) 0 i=m 1 (I A(ξ i ) t i ) = (I A(t) dt), provided tht the integrls on the right hnd sides exist. Corollry 2.4.9. If A : [, b] R n n is Riemnn integrble function, then b (I + A(t) dt) = I + ( 1) k xk x2 b A(x 1 ) A(x k ) dx 1 dx k, (I + A(t) dt) b = I + ( 1) k xk x2 b A(x k ) A(x 1 ) dx 1 dx k. The following sttement represents generlized version of Theorem 2.4.6. Theorem 2.4.10. 1 If A : [p, q] R n n is Riemnn integrble function, then c (I + A(x) dx) = c (I + A(x) dx) b (I + A(x) dx), for every, b, c [p, q]. c c (I + A(x) dx) = (I + A(x) dx) (I + A(x) dx) Proof. If b c, then the sttement reduces to Theorem 2.4.6. Let s hve look t the cse b < = c: Denote E(j, i) = {x R i ; b x 1 x j nd x j+1 x i b} for ech pir of numbers i N nd j {0,..., i}. A simple observtion revels tht E(j, i) = E(j, j) E(0, i j) (2.4.3) for every i N nd j {1,..., i 1}. We lso ssert tht 1 [VH], p. 56 58 E(0, i) E(2, i) = E(1, i) E(3, i) (2.4.4) b 29

for every i N. Indeed, if x R i is member of the union on the left side, then x E(2k, i) for some k. If 2k < i nd x 2k x 2k+1, then x E(2k +1, i). If 2k = i, or 2k < i nd x 2k+1 < x 2k, then x E(2k 1, i). In ny cse, x is member of the union on the right side; the reverse inclusion is proved similrly. Now, the Peno series expnsions might be written s (I + A(x) dx) = I + A(x 1 ) A(x i ) dx 1 dx i, b i=1 E(i,i) (I + A(x) dx) = I + ( 1) i i=1 E(0,i) A(x 1 ) A(x i ) dx 1 dx i (we hve used Corollry 2.4.9). Since both infinite series converge bsolutely, their product is equl to the Cuchy product: (I + A(x) dx) (I + A(x) dx) = I + b + + ( 1) i i=1 ( i 1 i=1 ( j=1 E(0,i) i=1 E(i,i) A(x 1 ) A(x i ) dx 1 dx i + ) A(x 1 ) A(x k ) dx 1 dx k E(j,j) A(x 1 ) A(x i ) dx 1 dx i + )) (( 1) i j A(x 1 ) A(x i j ) dx 1 dx i j = E(0,i j) = I + i=1 j=0 i ( 1) i j E(j,i) A(x 1 ) A(x i ) dx 1 dx i, where the lst equlity is consequence of Eqution (2.4.3). Eqution (2.4.4) implies i ( 1) i j A(x 1 ) A(x i ) dx 1 dx i = 0 j=0 E(j,i) for every positive number i, which proves tht (I + A(x) dx) (I + A(x) dx) = I. b We see tht our sttement is true is even in the cse b > = c. The remining cses re now simple to check: For exmple, if < c < b, then c (I + A(x) dx) = c (I + A(x) dx) b (I + A(x) dx) c c (I + A(x) dx) = 30

= (I + A(x) dx) (I + A(x) dx) The prove the second prt, we clculte c. b (I + A(x) dx) c = (I A(x) dx) = c (I A(x) dx) b (I A(x) dx) = c = (I + A(x) dx) (I + A(x) dx) c. b Corollry 2.4.11. If A : [, b] R n n is Riemnn integrble function, then ( 1 (I + A(x) dx) = (I + A(x) dx)), b ( ) 1 (I + A(x) dx) = (I + A(x) dx). b Theorem 2.4.12. 1 If A : [, b] R n n is Riemnn integrble function, then the functions x Y (x) = (I + A(t) dt), stisfy the integrl equtions x Z(x) = (I + A(t) dt) Y (x) = I + Z(x) = I + for every x [, b]. Proof. Theorem 2.4.3 implies x x A(t)Y (t) dt, Z(t)A(t) dt A(t)Y (t) = A(t) + t xk x2 A(t)A(x k ) A(x 1 ) dx 1 dx k. (2.4.5) 1 [VH], p. 52 53 31

The Peno series converges uniformly nd the entries of A re bounded, therefore the series (2.4.5) lso converges uniformly for t [, b] nd might be integrted term by term to obtin + x t xk x x2 A(t)Y (t) dt = x The other integrl eqution is deduced similrly. A(t) dt+ A(t)A(x k ) A(x 1 ) dx 1 dx k dt = Y (x) I. 2.5 Continuous mtrix functions Volterr is now redy to stte nd prove the fundmentl theorem of clculus for product integrl. Recll tht the ordinry fundmentl theorem hs two prts: 1) If f is continuous function on [, b], then the function F (x) = x f(t) dt stisfies F (x) = f(x) for every x [, b]. 2) If f is continuous function on [, b] nd F its ntiderivtive, then b f(t) dt = F (b) F (). The function x f(t) dt is usully referred to s the indefinite integrl of f; similrly, the functions x (I + A(t) dt) nd (I + A(t) dt) x re clled the indefinite product integrls of A. Before proceeding to the fundmentl theorem we mke the following greement: A mtrix function A : [, b] R n n is clled continuous, if the entries ij of A re continuous functions on [, b]. Theorem 2.5.1. 1 If A : [, b] R n n is continuous mtrix function, then the indefinite product integrls Y (x) = x (I + A(t) dt) nd Z(x) = (I + A(t) dt) x stisfy the equtions Y (x) = A(x)Y (x), Z (x) = Z(x)A(x) for every x [, b]. Proof. The required sttement is esily deduced by differentiting the integrl equtions obtined in Theorem 2.4.12. 1 [VH], p. 60 61 32

The differentil equtions from the previous theorem cn be rewritten in the form d dx x (I + A(t) dt) = A(x), (I + A(t) dt) x d dx = A(x), which closely resembles the first prt of the ordinry fundmentl theorem. We see tht the left (or right) derivtive is in certin sense inverse opertion to the left (or right) product integrl. Remrk 2.5.2. A function Y : [, b] R n n is solution of the eqution Y (x) = A(x)Y (x), x [, b] nd stisfies Y () = I if nd only if Y solves the integrl eqution Y (x) = I + x This is specil type of eqution of the form y(x) = f(x) + A(t)Y (t) dt, x [, b]. (2.5.1) x K(x, t)y(t) dt, which is tody clled the Volterr s integrl eqution of the second kind. Volterr proved (see e.g. [Kl, VV4]) tht such equtions my be solved by the method of successive pproximtions; in cse of Eqution (2.5.1) we obtin the solution Y (x) = I + b xk x2 A(x k ) A(x 1 ) dx 1 dx k, which is exctly the Peno series. Theorem 2.5.3. 1 Consider continuous mtrix function A : [, b] R n n. If there exists function Y : [, b] R n n such tht for every x [, b], then d Y (x) = A(x) dx (I + A(x) dx) = Y (b)y () 1. Similrly, if there exists function Z : [, b] R n n such tht 1 [VH], p. 62 63 Z(x) d dx = A(x) 33

for every x [, b], then (I + A(x) dx) = Z() 1 Z(b). Proof. We prove the first prt: The functions x (I + A(t) dt) nd Y (x) hve the sme left derivtive for every x [, b]. Theorem 2.3.6 implies the existence of mtrix C such tht x (I + A(t) dt) = Y (x)c for every x [, b]. Substituting x = yields C = Y () 1. Theorem 2.5.4. 1 If A : [, b] R n n is continuous function, then the function x Y (x) = (I + A(t) dt) is the fundmentl mtrix of the system of differentil equtions n y i(x) = ij (x)y j (x), i = 1,..., n. (2.5.2) j=1 Proof. Let y k denote the k-th column of Y, i.e. x y k (x) = (I + A(t) dt) e k, where e k is the k-th vector from the cnonicl bsis of R n. Theorem 2.5.1 implies tht ech of the vector functions y k, k = 1,..., n yields solution of the system (2.5.2). Since y k () = e k, the system of functions {y k } n is linerly independent nd represents fundmentl set of solutions of the system (2.5.2). Exmple 2.5.5. 2 Volterr now shows the fmilir method of converting liner differentil eqution of the n-th order y (n) (x) = p 1 (x)y n 1 (x) + p 2 (x)y n 2 (x) + + p n (x)y(x) to system of equtions of the first order. If we introduce the functions z 0 = y, z 1 = z 0, z 2 = z 1,..., z n 1 = z n 2, then the bove given n-th order eqution is equivlent to the system of equtions written in mtrix form s 1 [VH], p. 69 2 [VH], p. 70 z 0 z 1. z n 2 z n 1 0 1 0 0 0 0 1 0 =....... 0 0 0 1 p n p n 1 p n 2 p 1 z 0 z 1. z n 2 z n 1. 34

The fundmentl mtrix of this system cn be clculted using the product integrl nd the solution of the originl eqution (which corresponds to the function z 0 ) is represented by the first column of the mtrix (we obtin set of n linerly independent solutions). Exmple 2.5.6. If n = 1, then the function A : [, b] R n n is in fct sclr function, nd we usully write b (1 + A(t) dt) insted of b (I + A(t) dt). Using Theorem 2.4.3 nd Lemm 2.4.2 we obtin x ( 1 x k ( x ) y(x) = (1 + A(t) dt) = 1 + A(t) dt) = exp A(t) dt, k! which is indeed solution of the differentil eqution y (x) = A(x)y(x) nd stisfies y() = 1. Exmple 2.5.7. 1 Recll tht if A R n n, then the exponentil of A is defined s exp A = k=0 A k k! The fundmentl mtrix of the system of equtions is given by y i(x) = n ij y j (x), j=1 x Y (x) = (I + A dt) = I + I + b xk i = 1,..., n x2 (x ) k A k = e (x )A k! A k dx 1 dx k = (2.5.3). (we hve used Theorem 2.4.3 nd Lemm 2.4.2), which is well-known result from the theory of differentil equtions. We lso remrk tht similr clcultion leds to the reltion x (I + A dt) = e (x )A, x [, b]. Exmple 2.5.8. 2 Volterr is lso interested in ctully clculting the mtrix e A(x ). Convert A to the Jordn norml form J 1 0 0 λ i 0 0 0 0 J A = C 1 2 0...... C, where J 1 λ i = i 0 0....... 0 0 J k 0 0 1 λ i 1 [VH], p. 70 71 2 [VH], p. 66 68 35

for i {1,..., k}. If e λix 0 0 0 0 x 1 1! eλix e λix 0 0 0 S i (x) = x 2 x1 2! eλix 1! eλix e λix 0 0....... is squre mtrix which hs the sme dimensions s J i, it is esily verified tht S i (x) 1 = S i ( x), d dx S i(x) = S i (x) S i (x) 1 = J i. Applying Theorem 2.3.8 to mtrix we obtin nd Theorem 2.3.7 gives d dx S 1 (x) 0 0 0 S 2 (x) 0 S(x) =...... 0 0 S k (x) J 1 0 0 d dx S(x) = 0 J 2 0...... 0 0 J k J 1 0 0 ( C 1 S(x) ) = C 0 J 1 2 0...... 0 0 J k C = A. Theorem 2.3.6 implies the existence of mtrix D R n n such tht e (x )A = C 1 S(x)D for every x [, b]; substituting x = yields D = S() 1 C. Remrk 2.5.9. Volterr gives no indiction how to guess the clcultion in the previous exmple. We my proceed s follows: Let gin A = C 1 JC, where J 1 0 0 0 J J = 2 0...... 0 0 J k, J i = The definition of mtrix exponentil implies λ i 0 0 0 1 λ i 0 0....... 0 0 1 λ i exp(ax) = exp(c 1 JxC) = C 1 exp(jx)c 36.

for every x R nd it suffices to clculte exp(jx). We see tht λ i 0 0 0 0 0 0 0 0 λ J i x = x i 0 0....... + x 1 0 0 0........ 0 0 0 λ i 0 0 1 0 It is esy to clculte n rbitrry power of the mtrices on the right hnd side (the second one is nilpotent mtrix); the definition of mtrix exponentil then gives e λix 0 0 0 0 x 1 1! eλix e λix 0 0 0 exp(j i x) = x 2 x1 2! eλix 1! eλix e λix 0 0,....... exp(j 1 x) 0 0 0 exp(j 2 x) 0 exp(jx) =....... 0 0 exp(j k x) Theorem 2.5.10. 1 If A : [, b] R n n is continuous function nd ϕ : [c, d] [, b] continuously differentible function such tht ϕ(c) = nd ϕ(d) = b, then (I + A(x) dx) = d (I + A(ϕ(t))ϕ (t) dt). c Proof. Define for every x [, b]. Then Y (x) = x (I + A(t) dt) d dt (Y ϕ) = Y (ϕ(t))ϕ (t)y (ϕ(t)) 1 = A(ϕ(t))ϕ (t) for every t [c, d]. The fundmentl theorem for product integrl gives (I + A(x) dx) = Y (b)y () 1 = Y (ϕ(d))y (ϕ(c)) 1 = d (I + A(ϕ(t))ϕ (t) dt). c 1 [VH], p. 65 37

Theorem 2.5.11. 1 If A : [, b] R n n is continuous function, then Proof. Denote ( ) ( b det (I + A(x) dx) = exp Y (x) = ) n ii (x) dx. i=1 x (I + A(t) dt), x [, b]. The determinnt of Y (x) might be interpreted s function of the n 2 entries y ij (x), i, j {1,..., n}. The chin rule therefore gives (det Y ) (x) = n i,j=1 (det Y ) y y ij(x). ij Formul (2.2.2) for the expnsion of determinnt by minors implies nd consequently (det Y ) (x) = = n i,j, (det Y ) y ij = Y ij, n y ij(x)y ij (x) = i,j=1 n i,j, ik (x)y kj (x)y ij (x) = ( n ) ik (x)δ ik (x) det Y (x) = ii (x) det Y (x) (we hve used Theorem 2.5.1 nd Eqution (2.2.3)). However, the differentil eqution ( n ) (det Y ) (x) = ii (x) det Y (x) hs unique solution tht stisfies i=1 i=1 det Y () = det I = 1. It is given by ( ) x n det Y (x) = exp ii (t) dt, x [, b]. i=1 1 [VH], p. 61 62 38

Theorem 2.5.12. 1 If A : [, b] R n n is continuous function nd C R n n regulr mtrix, then (I + C 1 A(x)C dx) = C 1 (I + A(x) dx)c. Proof. Define Y (x) = x (I + A(t) dt) for every x [, b]. Theorem 2.3.7 gives ( ) d d dx (C 1 Y ) = C 1 dx Y C = C 1 AC, nd therefore (I + C 1 A(x)C dx) = C 1 Y (b)(c 1 Y ()) 1 = C 1 (I + A(x) dx)c. 2.6 Multivrible clculus In this section we turn our ttention to mtrix functions of severl vribles, i.e. to functions A : R m R n n, where m, n N. We introduce the nottion { } n A ij (x) = (x), x k x k i,j=1 provided the necessry prtil derivtives exist. Definition 2.6.1. Let G be domin in R m nd x G. Consider function A : G R n n tht is regulr t x nd such tht A x k (x) exists. We define the left prtil derivtive of A t x with respect to the k-th vrible s d A(x) = A (x)a 1 (x). dx k x k Remrk 2.6.2. Volterr lso introduces the left differentil of A s the mtrix m ( ) d da = A(x 1 + dx 1,..., x m + dx m )A 1 (x 1,..., x m ) = I + A(x) dx k, dx k 1 [VH], p. 63 (2.6.1) 39

which differs infinitesimlly from the identity mtrix. He lso clims tht da = m ( ( ) ) d I + A(x) dx k, dx k since the product of infinitesiml quntities cn be neglected. Recll the following well-known theorem of multivrible clculus: If f 1,..., f m : R m R re functions tht hve continuous prtil derivtives with respect to ll vribles, then the following sttements re equivlent: (1) There exists function F : R m R such tht F x i = f i for i = 1,..., m. (2) f i x j f j x i = 0 for i, j = 1,..., m, i j. Volterr proceeds to the formultion of similr theorem concerning left derivtives. Definition 2.6.3. Let A, B : R m R n n be mtrix functions tht possess prtil derivtives with respect to the i-th nd j-th vrible. We define (A, B) xi,x j = B x i A x j + BA AB. Volterr s proof of the following lemm hs been slightly modified to mke it more redble. We lso require the equlity of mixed prtil derivtives, wheres Volterr supposes tht the mixed derivtives cn be interchnged without ny comment. Lemm 2.6.4. 1 Let m N, i, j {1,..., m}, i j. Let G be n open set in R m nd x G. Consider pir of mtrix functions X, Y : G R n n tht possess prtil derivtives with respect to x i nd x j t x, nd function S : G R n n tht stisfies d dx i S(x) = X(x), (2.6.2) Then the equlity x i holds t the point x. 2 S (x) = 2 S (x). x i x j x j x i ( (S 1 Y d ) ) S S = S 1 (X, Y ) xi,x dx j S j Proof. Using the formul for the derivtive of n inverse mtrix nd the ssumption (2.6.2) we clculte 1 [VH], p. 81 S 1 x i ( Y d ) ( 1 S S S = S S 1 Y dx j x i d ) S S = dx j 40

further nd finlly ( = S 1 X Y d ) ( ( )) d S S = S 1 XY + X S S, dx j dx j S 1 S 1 ( Y ( Y d ) ( Y S S = S 1 x i dx j x i x i ( )) S S 1 S = x j ( Y = S 1 2 S S 1 S S 1 ) S = x i x i x j x j x i ( Y = S 1 (XS) S 1 S ) 1 S S S 1 S = x i x j x j x i ( Y = S 1 X ( ) ( ) ) d d X S S X S = x i x j dx j dx j ( Y = S 1 X ( ) ( ) ) d d X S S X S, x i x j dx j dx j d ) ( S S = S 1 Y dx j x i d ) ( ( ) ) d S XS = S 1 Y X S X S. dx j dx j The product rule for differentition gives (using the previous three equtions) x i ( (S 1 Y d ) ) ( ( ) d S S = S 1 XY + X S + Y X dx j dx j x i x j ( ) ( ) ( ) ) d d d X S S X + Y X S X S = S 1 (X, Y ) xi,x dx j dx j dx j S. j Theorem 2.6.5. 1 If B 1,..., B m : R m R n n re continuously differentible with respect to ll vribles, then the following sttements re equivlent: (1) There exists function A : R m R n n such tht B k = d dx k A for k = 1,..., m. (2) (B i, B j ) xi,x j = 0 for i, j = 1,..., m, i j. Proof. We strt with the impliction (1) (2): = x j 1 [VH], p. 78 85 B i B j = ( ) A A 1 x j x i x j x i ( A x i ) A 1 + A x i A 1 x j x i x i ( A x j ( ) A A 1 = x j ) A 1 A x j A 1 x i = 41

= A A 1 A A 1 = A 1 A A A 1 + A 1 A A A 1 = B j B i B i B j x i x j x j x i x i x j x j x i (sttement (1) implies tht the mixed prtil derivtives of A re continuous, nd therefore interchngeble). The reverse impliction (2) (1) is first proved for m = 2: Suppose tht the function A : R 2 R n n from (2) exists. Choose x 0 R nd define S(x, y) = x x 0 (I + B 1 (t, y) dt). Then d dx S = B 1 = d dx A, which implies the existence of mtrix function T : R R n n such tht A(x, y) = S(x, y)t (y) (T is independent on x). We clculte d dy T = d ( d dy (S 1 A) = S 1 dy A d ) dy S S = S (B 1 2 ddy ) S S. We now relx the ssumption tht the function A exists; the function on the right hnd side of the lst eqution is nevertheless independent on x, becuse Lemm 2.6.4 gives (S (B 1 2 ddy ) ) x S S = S 1 (B 1, B 2 ) x,y S = 0. Thus we define T (y) = y ( I + S 1 (x, t) (B 2 (x, t) ddy ) ) S(x, t) S(x, t) dt y 0 (where x is rbitrry) nd A = ST. Since nd d dx A = d dx (ST ) = d dx S = B 1 d dy A = d dy (ST ) = d ( ) d dy S + S dy T S 1 = B 2, the proof is finished; we now proceed to the cse m > 2 by induction: Choose x 0 R nd define S(x 1,..., x m ) = x 1 x 0 (I + B 1 (t, x 2,..., x m ) dt). 42

If the function A : R m R n n exists, we must hve nd consequently d S = B 1 = d A dx 1 dx 1 d dx k T = A(x 1,..., x m ) = S(x 1,..., x m )T (x 2,..., x m ) for some mtrix function T : R m 1 R n n. Then ( B k d dx k (S 1 A) = S 1 d dx k S ) S, k = 2,..., m. We now relx the ssumption tht the function A exists nd define ( U k = S 1 B k d ) S S, k = 2,..., m. dx k Ech of these functions U k is indeed independent on x 1, becuse Lemm 2.6.4 gives Since U k x 1 = S 1 (B 1, B k ) x1,x k S = 0. (U i, U j ) xi,x j = S 1 (B i, B j ) xi,x j S = 0, i, j = 2,..., m, the induction hypothesis implies the existence of function T of m 1 vribles x 2,..., x m such tht d dx k T = U k, k = 2,..., m. We now let A = ST nd obtin nd d dx k A = d dx 1 A = d dx k (ST ) = d dx 1 (ST ) = for k = 2,..., m, which completes the proof. d dx 1 S = B 1 d ( ) d S + S T S 1 = B k dx k dx k Remrk 2.6.6. Volterr s proof of Theorem 2.6.5 contins deficiency: We hve pplied Lemm 2.6.4 to the function without verifying tht S(x 1,..., x m ) = x 1 x 0 (I + B 1 (t, x 2,..., x m ) dt) 2 S (x) = 2 S (x), x i x 1 x 1 x i i {2,..., m}. 43

This equlity follows from the well-known theorem of multivrible clculus provided the derivtives S x i exist in some neighbourhood of x for every i {1,..., m}, nd the derivtives 2 S (x) x 1 x i re continuous t x for every i {2,..., m}. We hve nd consequently S x 1 = B 1 2 S x 1 x i = B 1 x i, which is continuous function for every i {2,..., m}. The existence of the derivtives S x i for i {2,..., m} is certinly not obvious but follows from Theorem 3.6.14 on differentiting the product integrl with respect to prmeter, which will be proved in Chpter 3. Remrk 2.6.7. An nlogy of Theorem 2.6.5 holds lso for right derivtives; the condition (B i, B j ) xi,x j = 0 must be replced by (B i, B j ) xi,x j = 0, where (A, B) xi,x j = B x i A x j + AB BA. The second fundmentl notion of multivrible clculus is the contour integrl. While Volterr introduces only product integrls long contour ϕ in R 2, which he denotes by X dx Y dy, ϕ we give generl definition for curves in R m ; we lso use different nottion. We will lwys consider curves tht re given using prmetriztion ϕ : [, b] R m tht is piecewise continuously differentible, which mens tht ϕ (x) exists for x (, b], ϕ +(x) exists for x [, b), nd ϕ (x) = ϕ +(x) except finite number of points in (, b). The imge of the curve is then defined s ϕ = ϕ([, b]) = {ϕ(t); t [, b]}. Definition 2.6.8. Consider piecewise continuously differentible function ϕ : [, b] R m nd system of m mtrix functions B 1,..., B m : ϕ R n n. The contour product integrl of these functions long ϕ is defined s (I+B 1 dx 1 + +B m dx m ) = ϕ (I+(B 1 (ϕ(t))ϕ 1(t) + + B m (ϕ(t))ϕ m(t)) dt). 44

Given n rbitrry curve ϕ : [, b] R m, we define the curve ϕ s ( ϕ)(t) = ϕ( t), t [ b, ]. This curve hs the sme imge s the originl curve, but is trversed in the opposite direction. For ny pir of curves ϕ 1 : [ 1, b 1 ] R m, ϕ 2 : [ 2, b 2 ] R m such tht ϕ 1 (b 1 ) = ϕ 2 ( 2 ) we define the composite curve ϕ 1 + ϕ 2 by (ϕ 1 + ϕ 2 )(t) = { ϕ1 (t), t [ 1, b 1 ], ϕ 2 (t b 1 + 2 ), t [b 1, b 1 + b 2 2 ]. Theorem 2.6.9. 1 Contour product integrl hs the following properties: (1) If ϕ 1 + ϕ 2 is curve obtined by joining two curves ϕ 1 nd ϕ 2, then ϕ 1+ϕ 2 (I + B 1 dx 1 + + B m dx m ) = = ϕ 2 (I + B 1 dx 1 + + B m dx m ) ϕ 1 (I + B 1 dx 1 + + B m dx m ). (2) If ϕ is curve obtined by reversing the orienttion of ϕ, then ( 1 (I + B 1 dx 1 + + B m dx m ) = (I + B 1 dx 1 + + B m dx m )). ϕ ϕ Proof. Let ϕ 1 : [ 1, b 1 ] R m, ϕ 2 : [ 2, b 2 ] R m. Then b 1+b 2 2 b 1 (I + B 1 dx 1 + + B m dx m ) = ϕ (I + (B 1 (ϕ(t b 1 + 2 ))ϕ 1(t) + + B m (ϕ(t b 1 + 2 ))ϕ m(t)) dt) b 1 1 (I + (B 1 (ϕ(t))ϕ 1(t) + + B m (ϕ(t))ϕ m(t)) dt). The chnge of vribles Theorem 2.5.10 gives b 1+b 2 2 1 [VH], p. 91 b 1 (I + (B 1 (ϕ(t b 1 + 2 ))ϕ 1(t) + + B m (ϕ(t b 1 + 2 ))ϕ m(t)) dt) = 45

= b 2 2 (I + (B 1 (ϕ(t))ϕ 1(t) + + B m (ϕ(t))ϕ m(t)) dt), which proves the first sttement. The second one is lso direct consequence of Theorem 2.5.10. Note tht the chnge of vribles theorem ws proved only for continuously differentible functions, while our contours re piecewise continuously differentible. It is however lwys possible to prtition the integrtion intervl in such wy tht the integrted functions re continuously differentible on every subintervl. Definition 2.6.10. Let G be subset of R m nd B 1,..., B m : G R n n. The contour product integrl (I + B 1 dx 1 + + B m dx m ) is clled pth-independent in G if (I + B 1 dx 1 + + B m dx m ) = (I + B 1 dx 1 + + B m dx m ) ϕ ψ for ech pir of curves ϕ, ψ : [, b] G such tht ϕ() = ψ() nd ϕ(b) = ψ(b). Using Theorem 2.6.9 it is esy to see tht the contour product integrl is pthindependent in G if nd only if for every closed curve ϕ in G. (I + B 1 dx 1 + + B m dx m ) = I ϕ As lredy mentioned, Volterr is concerned especilly with curves in R 2. His effort is directed towrds proving the following theorem: Theorem 2.6.11. 1 Let G be simply connected domin in R 2. Consider pir of functions A, B : G R n n such tht (A, B) x,y = 0 t every point of G. Then (I + A dx + B dy) = I ϕ for every piecewise continuously differentible closed curve ϕ in G. Although Volterr s proof is somewht incomplete, we try to indicte its min steps in the rest of the section. Theorem 2.6.11 is of gret importnce for Volterr s he uses it to prove n nlogue of Cuchy theorem for product integrl in complex domin; this topic will be discussed in the next section. Definition 2.6.12. A set S in R 2 is clled simple in the x-direction, if the set S {(x, y 0 ); x R} is connected for every y 0 R. Similrly, S is simple in the y-direction, if the set S {(x 0, y); y R} is connected for every x 0 R. 1 [VH], p. 95 46

Equivlently sid, the intersection of S nd line prllel to the x-xis (or the y-xis) is either n intervl (possibly degenerted to single point), or n empty set. Definition 2.6.13. Let F be closed bounded subset of R 2 tht is simple in the x-direction. For every y R denote π F = {y R; there exists x R such tht (x, y) G}. (2.6.3) Further, for every y π F let x A (y) = inf{x; (x, y) F }, x B (y) = sup{x; (x, y) F }. The mening of these symbols is illustrted by the following figure. Note tht the segment [x A (y), x B (y)] {y} is contined in F for every y π F, i.e. the set F is enclosed between the grphs of the functions y x A (y) nd y x B (y), y π F. y π F (x, y) F x A (y) x B (y) x Definition 2.6.14. Let F be closed bounded subset of R 2 tht is simple in the x-direction. The double product integrl of continuous function A : F R n n is defined s (I + A(x, y) dx dy) = F sup π F inf π F ( ( ) ) xb(y) I + A(x, y) dx dy. x A(y) Before proceeding to the next theorem we recll tht Jordn curve is closed curve with no self-intersections. Formlly written, it is curve with prmetriztion ϕ : [, b] R 2 tht is injective on [, b) nd ϕ() = ϕ(b). It is known tht Jordn 47

curve divides the plne in two components the interior nd the exterior of ϕ. In the following text we denote the interior of ϕ by Int ϕ. Theorem 2.6.15. 1 Consider piecewise continuously differentible Jordn curve ϕ : [, b] R 2 such tht the set F = ϕ Int ϕ is simple in the x-direction. Assume tht ϕ strts t point C = (c x, c y ) such tht c y = inf π F nd c x = x A (c y ). Denote S(x, y) = x x A(y) (x A(y),y) (I + X(t, y) dt) (I + X dx + Y dy), where the second integrl is tken over tht prt of ϕ tht joins the points C nd (x A (y), y) (see the figure below). C (x A (y), y) (x, y) ϕ Let G be n open neighbourhood of the set F. Then the eqution C (I + X dx + Y dy) = (I + S 1 (X, Y ) x,y S dx dy) ϕ F holds for ech pir of continuously differentible functions X, Y : G R n n. Theorem 2.6.15 might be regrded s n nlogy of Green s theorem, since it provides reltionship between the double product integrl over F nd the contour product integrl over the boundry of F. The proof in [VH] is somewht obscure (minly becuse of Volterr s clcultions with infinitesiml quntities) nd will not be reproduced here. A sttement similr to Theorem 2.6.15 will be proved in Chpter 3, Theorem 3.7.4. Theorem 2.6.16. 2 Consider piecewise continuously differentible Jordn curve ϕ : [, b] R 2 such tht the set F = ϕ Int ϕ is simple in the x-direction. Let G be n open neighbourhood of the set F. If A, B : G R n n is pir of continuously differentible functions such tht (A, B) x,y = 0 t every point of G, then (I + A dx + B dy) = I. ϕ 1 [VH], p. 92 94 2 [VH], p. 95 48

Proof. Let C = (c x, c y ) be the point such tht c y = inf π F nd c x = x A (c y ). If A is the strting point of ϕ, we my write ϕ = ϕ 1 + ϕ 2, where ϕ 1 is the prt of the curve between the points A, C nd ϕ 2 is the prt between C, A provided we trvel long ϕ in direction of its orienttion. ϕ 1 A ϕ 2 C Theorem 2.6.15 gives (I + A dx + B dy) = I, ϕ 2+ϕ 1 nd consequently (I + A dx + B dy) = (I + A dx + B dy) = ϕ ϕ 1+ϕ 2 = (I + A dx + B dy) ( ) 1 (I + A dx + B dy) (I + A dx + B dy) = I. ϕ 2 ϕ 2+ϕ 1 ϕ 2 Remrk 2.6.17. In cse the set G in sttement of the lst theorem is simple both in the x direction nd in the y direction, there is simpler lterntive proof of Theorem 2.6.16. It is bsed on Theorem 2.6.5, which we proved for G = R 2, but the proof is exctly the sme lso for sets G which re simple in x s well s in y direction. Consequently, the ssumption (A, B) x,y = 0 nd Theorem 2.6.5 imply the existence of function T : G R 2 such tht A(x, y) = d dx T (x, y), B(x, y) = d T (x, y) dy for every (x, y) G. Thus for rbitrry closed curve ϕ : [, b] G we hve (I + A dx + B dy) = ϕ ϕ ( I + d dx T dx + d ) dy T dy = = ( ( T I + x (ϕ(t)) T (ϕ(t)) 1 ϕ 1(t) + T ) ) y (ϕ(t)) T (ϕ(t)) 1 ϕ 2(t) dt = 49

= (I + ddt ) (T ϕ)(t) dt = T (ϕ(b))t (ϕ()) 1 = I. Both the sttement nd its proof is esily generlized to the cse of curves in R m, m > 2. Volterr now comes to the justifiction of Theorem 2.6.12: Let G be simply connected domin in R 2 nd A, B : G R n n such tht (A, B) x,y = 0 t every point of G. We hve to verify tht (I + A dx + B dy) = I (2.6.4) ϕ for every piecewise continuously differentible closed curve ϕ in G. Theorem 2.6.16 ensures tht the sttement is true, if the set F = ϕ Int ϕ is simple in the x- direction. Volterr first notes tht it remins true, if F cn be split by curve in two prts ech of which is simple in the x-direction. ϕ 2d T ϕ 1 ϕ 2c ϕ 2 ϕ 1d ϕ 1b S ϕ 2b ϕ 1c Indeed, using the nottion from the bove figure, if ϕ 1 = ϕ 1 + ϕ 1b + ϕ 1c + ϕ 1d nd ϕ 2 = ϕ 2 + ϕ 2b + ϕ 2c + ϕ 2d, then ϕ 1 (I + A dx + B dy) = I, ϕ 2 (I + A dx + B dy) = I, nd thus (I + A dx + B dy) = (I + A dx + B dy) (I + A dx + B dy) = I. ϕ 1+ϕ 2 ϕ 2 ϕ 1 Now if S denotes the initil point of ϕ, then (I + A dx + B dy) = ϕ = T (I + A dx + B dy) S (I + A dx + B dy) ϕ 1+ϕ 2 50 S (I + A dx + B dy) = I, T

where T S denotes the contour product integrl tken long the prt of ϕ tht connects the points S, T, nd S T is tken long the sme curve with reversed orienttion (it is thus the inverse mtrix of T S ). By induction it follows tht (2.6.4) holds if the set F = ϕ Int ϕ cn be decomposed into finite number of subsets which re simple in the x-direction. Volterr now sttes tht this is possible for every curve ϕ in considertion, nd so Theorem 2.6.12 is proved. He however gve no justifiction of the lst sttement, so his proof remins incomplete. 2.7 Product integrtion in complex domin So fr we hve been concerned with rel mtrix functions defined on rel intervl, i.e. with functions A : [, b] R n n. Most of our results cn be, without greter effort, generlized to complex-vlued mtrix functions, i.e. to functions A : [, b] C n n. However, in the following two sections, we will focus our interest to mtrix functions of complex vrible, i.e. A : G C n n, where G is subset of the complex plne. A mtrix function A = { jk } n j, will be clled differentible t point z C, if its entries jk re differentible t tht point. We will use the nottion A (z) = { jk(z)} n j,. The function A is clled holomorphic in n open domin G C, if it is differentible everywhere in G. Definition 2.7.1. The left derivtive of complex mtrix function A t point z C is defined s d dz A = A (z)a 1 (z), provided tht A is differentible nd regulr t the point z. Ech mtrix function A of complex vrible z might be interpreted s function of two rel vribles x, y, where z = x + iy. The Cuchy-Riemnn eqution sttes tht thus the left derivtive stisfies A (z) = A x (x + iy) = 1 i d dz A = d dx A = 1 d i dy A, A (x + iy), y provided ll the derivtives exist. We now proceed to the definition of product integrl long contour in the complex domin. We gin restrict ourselves to contours with piecewise continuously differentible prmetriztion ϕ : [, b] C, i.e. ϕ (x) exists for ll x (, b], 51

ϕ +(x) exists for ll x [, b), nd ϕ (x) = ϕ +(x) except finite number of points in (, b). Definition 2.7.2. Let ϕ : [, b] C be piecewise continuously differentible contour in the complex plne nd A mtrix function which is defined nd continuous on ϕ. The left product integrl long ϕ is defined s (I + A(z) dz) = ϕ (I + A(ϕ(t))ϕ (t) dt). (2.7.1) As Volterr remrks, the left contour product integrl is equl to the limit lim P (A, D), ν(d) 0 where D is tgged prtition of [, b] with division points t i, tgs ξ i [t i 1, t i ] nd P (A, D) = 1 (I + A(ϕ(ξ i ))(ϕ(t i ) ϕ(t i 1 ))). (2.7.2) i=m Insted of our ϕ (I + A(z) dz) he uses the nottion A(z) dz. ϕ The product integrl ϕ (I+A(z) dz) cn be converted to contour product integrl tken long contour ϕ in R 2 with the prmetriztion ϕ(t) = (Re ϕ(t), Im ϕ(t)), t [, b]. Indeed, define A 1 (x, y) = A(x + iy) nd A 2 (x, y) = ia(x + iy). Then (I + A(z) dz) = ϕ (I + A(ϕ(t))ϕ (t) dt) = thus (I + (A 1 (ϕ(t))re ϕ (t) + A 2 (ϕ(t))im ϕ (t)) dt), (I + A(z) dz) = ϕ (I + A(x + iy) dx + ia(x + iy) dy). (2.7.3) ϕ The following theorem is n nlogy of Theorem 2.6.9. It cn be proved directly in the sme wy s Theorem 2.6.9, or lterntively by using the reltion (2.7.3) nd pplying Theorem 2.6.9. Theorem 2.7.3. 1 The left contour product integrl hs the following properties: 1 [VH], p. 107 52

(1) If ϕ 1 + ϕ 2 is curve obtined by joining two curves ϕ 1 nd ϕ 2, then (I + A(z) dz) = (I + A(z) dz) (I + A(z) dz). ϕ 1+ϕ 2 ϕ 2 ϕ 1 (2) If ϕ is curve obtined by reversing the orienttion of ϕ, then ( 1 (I + A(z) dz) = (I + A(z) dz)). ϕ ϕ Our interest in product integrl of mtrix function A : [, b] R n n stems from the fct tht it provides solution of the differentil eqution (or system of equtions) y (x) = A(x)y(x), where y : [, b] R n. The sitution is similr in the complex domin: Since the contour product integrl is limit of the products (2.7.2), we expect tht the solution of the differentil eqution y (z) = A(z)y(z) tht stisfies y(z 0 ) = y 0 will be given by ( ) y(z) = (I + A(w) dw) y 0, ϕ where ϕ : [, b] C is contour connecting the points z 0 nd z. However, this definition of y is correct only if the product integrl is independent on the choice of prticulr contour, i.e. if (I + A(z) dz) = (I + A(z) dz), ϕ ψ whenever ϕ nd ψ re two curves with the sme initil points nd the sme endpoints. From Theorem 2.7.3 we see tht ϕ+( ψ) (I + A(z) dz) should be the identity mtrix. Equivlently sid, (I + A(z) dz) = I ϕ should hold for every closed contour ϕ. Volterr proves tht the lst condition is stisfied in every simply connected domin G provided tht the function A is holomorphic in G. He first uses the formul (2.7.3) to convert the integrl in complex domin to n integrl in R 2. Then, since (A, ia) x,y = ia x A + iaa AiA = 0, y 53

Theorem 2.6.11 implies tht the contour product integrl long ny closed curve in G is equl to the identity mtrix. Becuse we didn t prove Theorem 2.6.11, we offer different justifiction tken over from [DF]. Theorem 2.7.4. 1 If G C is simply connected domin nd A : G C n n holomorphic function in G, then the contour product integrl of A is pthindependent in G. Proof. Let ϕ : [, b] G be curve in G. We expnd the product integrl of A long ϕ to the Peno series (I + A(z) dz) = ϕ (I + A(ϕ(t))ϕ (t) dt) = = I + b A(ϕ(t))ϕ (t) dt + b t2 This infinite series might be written s (I + A(z) dz) = I + ϕ ϕ(b) ϕ() A(z) dz + A(t 2 )A(t 1 )ϕ (t 2 )ϕ (t 1 ) dt 1 dt 2 + ϕ(b) z2 ϕ() ϕ() A(z 2 )A(z 1 ) dz 1 dz 2 + where the contour integrls re ll tken long ϕ (or its initil segment). However, since ordinry contour integrls of holomorphic functions re pth-independent in G, the sum of the infinite series depends only on the endpoints of ϕ. In cse the product integrl is pth-independent in given domin G, we will occsionlly use the symbol z 2 z 1 (I + A(z) dz) to denote product integrl tken long n rbitrry curve in G with initil point z 1 nd endpoint z 2. Volterr now clims tht if G is simply connected domin nd A is holomorphic mtrix function in G, then the function Y (z) = z z 0 (I + A(w) dw) provides solution of the differentil eqution Y (z) = A(z)Y (z) in G. Theorem 2.7.5. If G C is simply connected domin nd A : G C n n holomorphic function in G, then the function ( z ) Y (z) = (I + A(w) dw) z 0 1 [DF], p. 62 63 54

stisfies Y (z) = A(z)Y (z) in G. Proof. The sttement is obtined by differentiting the series z z 0 (I + A(w) dw) = I + with respect to z. z z 0 A(w) dw + z w2 z 0 z 0 A(w 2 )A(w 1 ) dw 1 dw 2 + Corollry 2.7.6. Let G C be simply connected domin nd A : G C n n holomorphic function. If z 0 G nd y 0 C n, then the function y : G C n defined by ( z ) y(z) = (I + A(w) dw) y 0 z 0 stisfies y (z) = A(z)y(z) in G nd y(z 0 ) = y 0. Theorem 2.7.7. 1 Let G C be domin nd A : G C n n holomorphic mtrix function in G. If ϕ 1, ϕ 2 : [, b] G re two positively oriented Jordn curves such tht ϕ 1 Int ϕ 2 nd Int ϕ 2 \ Int ϕ 1 G, then re similr mtrices. ϕ 1 (I + A(z) dz) nd ϕ 2 (I + A(z) dz) Proof. We introduce two disjoint uxiliry segments ψ 1, ψ 2 tht connect the curves ϕ 1, ϕ 2 (see the figure). ψ 1 ϕ 2b ϕ 1b ϕ 1 ϕ 1 = ϕ 1 + ϕ 1b, ϕ 2 = ϕ 2 + ϕ 2b ϕ 2 ψ 2 Theorem 2.7.4 gives ( ) 1 ( (I + A(z) dz) (I + A(z) dz) (I + A(z) dz) = (I + A(z) dz) ϕ 2 ψ 1 ϕ 1 ψ 2 1 [VH], p. 114 116 ) 1 55

nd ( ) 1 ( ) 1 (I + A(z) dz) (I + A(z) dz) (I + A(z) dz) = (I + A(z) dz). ϕ 1b ψ 1 ϕ 2b ψ 2 Multiplying the first equlity by the second from left yields ( ) 1 ( ) 1 (I + A(z) dz) (I + A(z) dz) (I + A(z) dz) (I + A(z) dz) ϕ 1b ψ 1 ϕ 2b ϕ 2 ( ) 1 (I + A(z) dz) (I + A(z) dz) = I, ψ 1 ϕ 1 which cn be simplified to ( ) 1 (I +A(z) dz) (I +A(z) dz) (I +A(z) dz) = (I +A(z) dz). (2.7.4) ψ 1 ϕ 2 ψ 1 ϕ 1 Remrk 2.7.8. Volterr offers slightly different proof of the previous theorem: From Theorem 2.7.4 he deduces tht ( ) 1 ( ) 1 (I +A(z) dz) (I +A(z) dz) (I +A(z) dz) (I +A(z) dz) = I, ϕ 2 ψ 1 ϕ 2 ψ 1 which implies (2.7.4). This rgument is however incorrect, becuse the domin bounded by ψ 1 + ϕ 2 ψ 1 ϕ 2 need not be simply connected. Definition 2.7.9. Let R > 0, G = {z C; 0 < z z 0 < R}. Suppose A : G C n n is holomorphic in G. Let ϕ : [, b] G be positively oriented Jordn curve, z 0 Int ϕ. Then (I + A(z) dz) = CJC 1, ϕ where J is certin Jordn mtrix, which is, ccording to Theorem 2.7.7, independent on the choice of ϕ. This Jordn mtrix is clled the residue of A t the point z 0. Exmple 2.7.10. 1 We clculte the residue of mtrix function 1 [VH], p. 117 120 T (z) = A z z 0 + B(z) 56

t the point z 0, where A C n n nd B is mtrix function holomorphic in the neighbourhood U(z 0, R) of point z 0. We tke the product integrl long the circle ϕ r (t) = z 0 + re it, t [0, 2π], r < R nd obtin ϕ r (I + T (z) dz) = 2π 0 (I + ia + ire it B(z 0 + re it ) dt). Volterr suggests the following procedure (which is however not fully correct, see Remrk 2.7.11): Since ia + ire it B(z 0 + re it ) ia for r 0, we hve ϕ r (I + T (z) dz) 2π 0 (I + ia dt). The integrls ϕ r (I +T (z) dz), r (0, R), re ll similr to single Jordn mtrix. Their limit 2π 0 (I + ia dt) is thus similr to the sme mtrix nd it is sufficient to find its Jordn norml form. By the wy, this integrl is equl to e 2πiA, giving n nlogy of the residue theorem: The mtrix ϕ r (I + T (z) dz) is similr to e 2πiA. (2.7.5) Consider the Jordn norml form of A: A 1 0 0 0 A A = C 1 2 0...... 0 0 A k C, where A j = Using the result of Exmple 2.5.8 we obtin C 1 where 2π 0 λ j 0 0 0 1 λ j 0 0....... 0 0 1 λ j S 1 (2π) 0 0 0 S 2 (2π) 0 (I + ia dt)= C 1...... 0 0 S k (2π). S 1 (0) 0 0 1 S 1 (2π) 0 0 0 S 2 (0) 0 0 S 2 (2π) 0........ = C 1........ C,. 0 0 S k (0) 0 0 S k (2π) e iλjx 0 0 0 0 x 1 1! eiλjx e iλjx 0 0 0 S j (x) = x 2 x1 2! eiλjx 1! eiλjx e iλjx 0 0....... 57

is squre mtrix of the sme dimensions s A j. The Jordn norml form of the mtrix S 1 (2π) 0 0 0 S 2 (2π) 0.....,. 0 0 S k (2π) nd therefore lso the demnded residue, is V 1 0 0 0 V 2 0...... 0 0 V k, where V j = is squre mtrix of the sme dimensions s S j. e 2πiλj 0 0 0 1 e 2πiλj 0 0....... 0 0 1 e 2πiλj Remrk 2.7.11. The clcultion from the previous exmple contins two deficiencies: First, Volterr interchnges the order of limit nd product integrl to obtin 2π 2π lim (I + ia + ire it B(z 0 + re it ) dt) = (I + ia dt) r 0 0 0 without ny further comment. However, the convergence ia+ire it B(z 0 +re it ) ia for r 0 is uniform nd in this cse, s we will prove in Chpter 5, Theorem 5.6.4, the sttement is in fct true. The second deficiency is more serious. Volterr seems to hve ssumed tht if some mtrices S(r), r (0, R) (in our cse S(r) is the product integrl tken long ϕ r ), re ll similr to single Jordn mtrix J, then lim r 0 S(r) is lso similr to J. This sttement is incorrect, s demonstrted by the exmple S(r) = ( ) 1 0, r 1 where S(r) is similr to for r > 0, but ( ) 1 0 1 1 lim S(r) = r 0 ( ) 1 0 0 1 isn t. The mentioned sttement cn be proved only under dditionl ssumptions on S(r). For exmple, if the mtrices S(r), r > 0, hve n distinct eigenvlues λ 1,..., λ n, then the limit mtrix lim r 0 S(r) hs the sme eigenvlues, becuse det(s(0) λi) = lim r 0 det(s(r) λi) = (λ λ 1 ) (λ λ n ). 58

This mens tht ll the mtrices S(r), r 0, re similr to single Jordn mtrix λ 1 0 0 0 0 λ 2 0 0....... 0 0 0 λ n A more detiled discussion of the residue theorem for product integrl cn be found in the book [DF]; for exmple, if the set σ(a) σ(a) = {λ 1 λ 2 ; λ 1 nd λ 2 re eigenvlues of mtrix A} doesn t contin ny positive integers, then the sttement (2.7.5) is shown to be true. 2.8 Liner differentil equtions t singulr point In this section we ssume tht the reder is fmilir with the bsics of the theory of nlytic functions (see e.g. [EH] or [VJ]). We re interested in studying the differentil eqution Y (z) = A(z)Y (z), (2.8.1) where the function A is holomorphic in the ring P (z 0, R) = {z C; 0 < z z 0 < R} nd R > 0. If we choose z 1 P (z 0, R) nd denote r = min( z 1 z 0, R z 1 z 0 ), then the function z Y 1 (z) = (I + A(w) dw) z 1 provides solution of (2.8.1) in B(z 1, r) = {z C; z z 1 < r}; the product integrl is pth-independent, becuse A is holomorphic in U(z 1, r). The holomorphic function Y 1 cn be continued long n rbitrry curve ϕ v P (z 0, R); this procedure leds to (multiple-vlued) nlytic function Y, which will be denoted by Y(z) = z z 1 (I + A(w) dw), z P (z 0, R). If the element (z 1, Y 1 ) Y is continued long curve ϕ in P (z 0, R) to n element (z 2, Y 2 ) Y (we write this s (z 1, Y 1 ) ϕ (z 2, Y 2 )), then (using Y 1 (z 1 ) = I) Y 2 (z 2 ) = ϕ (I + A(w) dw) Y 1 (z 1 ) = ϕ (I + A(w) dw). Let ϕ be the circle with center z 0 which psses through the point z 1, i.e. ϕ(t) = z 0 + (z 1 z 0 ) exp(it), t [0, 2π]. 59

If (z 1, Y 2 ) Y is the element such tht (z 1, Y 1 ) ϕ (z 1, Y 2 ), then d dz Y 1 = d dz Y 2 = A(z) for z B(z 1, r). Consequently, there is mtrix C C n n such tht Y 2 (z) = Y 1 (z) C. Substituting z = z 1 gives C = ϕ (I + A(w) dw). Volterr refers 1 to the point z 0 s point de rmifiction bélien of the nlytic function Y; this mens tht it is the brnch point of Y, but not of its derivtive A, which is single-vlued function. Volterr proceeds to prove tht Y cn be written in the form Y = S 1 S 2, where S 1 is single-vlued in P (z 0, R) nd S 2 is n nlytic function tht is uniquely determined by the mtrix C = ϕ (I + A(w) dw). Here is the proof 2 : We write C = M 1 T M, where T is Jordn mtrix. Then T 1 0 0 0 T T = 2 0...... 0 0 T k, kde T j = e 2πiλj 0 0 0 1 e 2πiλj 0 0......., 0 0 1 e 2πiλj where we hve expressed the eigenvlues of T in the form exp(2πiλ j ); this is certinly possible s the mtrices C nd consequently lso T re regulr, nd thus hve nonzero eigenvlues. We now define the nlytic function V(z) = z z 1 ( I + U ) dw, z P (z 0, R), w z 0 where U 1 0 0 0 U U = 2 0...... 0 0 U k nd U j = λ j 0 0 0 1 λ j 0 0....... 0 0 1 λ j nd U j hs the sme dimensions s T j for j {1,..., k}. Consider function element (z 1, V 1 ) of V; wht hppens if we continue it long the circle ϕ? As in the cse of function Y we obtin the result 1 [VH], p. 121 2 [VH], p. 122 124 (z 1, V 1 ) ϕ (z 1, V 1 D), 60

where D = ϕ ( I + U ) 2π dw = (I + iu dw). w z 0 0 In Exmple 2.7.10 we hve clculted the result where S 1 (2π) 0 0 0 S 2 (2π) 0 D = S(2π) S(0) 1 = S(2π) =.....,. 0 0 S k (2π) e iλjx 0 0 0 0 x 1 1! eiλjx e iλjx 0 0 0 S j (x) = x 2 x1 2! eiλjx 1! eiλjx e iλjx 0 0........ The mtrix D is similr to the Jordn mtrix T ; thus D = N 1 T N, nd consequently (z 1, V 1 ) ϕ (z 1, V 1 N 1 T N), We now consider the nlytic function nd continue its element long ϕ: (z 1, Y 1 ) ϕ (z 1, Y 1 M 1 T M). S 1 (z) = Y(z)M 1 NV(z) 1 (z 1, Y 1 M 1 NV 1 1 ) ϕ (z 1, Y 1 M 1 T MM 1 N(V 1 N 1 T N) 1 ) = (z 1, Y 1 M 1 NV 1 1 ). Thus the nlytic function S 1 is in fct single-vlued in P (z 0, R). finished by putting S 2 (z) = V(z)N 1 M. The proof is Consequently Y = S 1 S 2 nd S 2 is uniquely determined by the mtrix C. We now briefly turn our ttention to the nlytic function V. Assume tht (z 1, V 1 ) ψ (z, V 2 ), 61

where ψ : [, b] P (z 0, R), ψ() = z 1, ψ(b) = z. It is known from complex nlysis tht given the curve ϕ, we cn find function function g : [, b] C such tht exp(g(t)) = ψ(t) z 0, for every t [, b]; g is continuous brnch of logrithm of the function ψ z 0. For convenience we will use the nottion g(t) = log(ψ(t) z 0 ) with the understnding tht g is defined s bove. We lso hve for every t [, b]. We now clculte g (t) = ψ (t) ψ(t) z 0 V 2 (z) = ψ ( I + U ) dz = z z 0 ( ) U I + ψ (t) dt. ψ(t) z 0 Substituting v = g(t) gives where g(b) V 2 (z) = (I + U dv) = S(g(b))S(g()) 1, g() S 1 (z) 0 0 0 S 2 (z) 0 S(z) =...... 0 0 S k (z) is block digonl mtrix composed of the mtrices e λjz 0 0 0 0 z 1 1! eλjz e λjz 0 0 0 S j (z) = z 2 z1 2! eλjz 1! eλjz e λjz 0 0........ Consequently, the solution of Eqution (2.8.1), i.e. the nlytic function Y, cn be expressed s Y(z) = S 1 (z)s 2 (z) = S 1 (z)s(g(b))s(g()) 1 N 1 M, (2.8.2) where S(g(b)) = S(log(z z 0 )), 62

S j (log(z z 0 )) = (z z 0 ) λj 0 0 0 0 (z z 0 ) λj log(z z 0 ) (z z 0 ) λj 0 0 0 = (z z 0 ) λj log2 (z z 0) 2! (z z 0 ) λj log(z z 0 ) (z z 0 ) λj 0 0........ The bove result cn be pplied to obtin the generl form of solution of the differentil eqution y (n) (z) + p 1 (z)y (n 1) (z) + + p n (z)y(z) = 0, (2.8.3) where the functions p i re holomorphic in P (z 0, R). We hve seen tht this eqution of the n-th order is esily converted to system of liner differentil equtions of first order, which cn be written in the vector form s y (z) = A(z)y(z), where A is holomorphic mtrix function in P (z 0, R). The fundmentl mtrix of this system is given by (2.8.2); the first row of this mtrix then yields the fundmentl system of solutions (composed of n nlytic functions) of Eqution (2.8.3). From the form of Eqution (2.8.2) we infer tht every solution of Eqution (2.8.3) cn be expressed s liner combintion of nlytic functions of the form ) (z z 0 ) (ϕ λj j 0 (z) + ϕj 1 (z) log(z z 0) + + ϕ j n j (z) log nj (z z 0 ), where ϕ j k re holomorphic functions in P (z 0, R). Thus we see tht Volterr ws ble to obtin the result of Lzrus Fuchs (see Chpter 1) using the theory of product integrtion. The next chpters of Volterr s book [VH] re concerned with the study of nlytic functions on Riemnn surfces; the topic is rther specil nd we don t follow Volterr s tretment here. 63

64

Chpter 3 Lebesgue product integrtion While it is sufficient to use the Riemnn integrl in pplictions, it is rther unstisfctory from the viewpoint of theoreticl mthemtics. The generliztion of Riemnn integrl due to Henri Lebesgue is bsed on the notion of mesure. The problem of extending Volterr s definition of product integrl in similr wy hs been solved by Ludwig Schlesinger. Volterr s nd Schlesinger s works differ in yet nother wy: Volterr did not worry bout using infinitesiml quntities, nd it is not lwys esy to trnslte his ides into the lnguge of modern mthemtics. Schlesinger s proofs re rther precise nd cn be red without greter effort except for occsionlly strnge nottion. The foundtions of mthemticl nlysis in 1930 s were firmer thn in 1887; moreover, Schlesinger inclined towrds theoreticl mthemtics, s opposed to Volterr, who lwys kept pplictions in mind. Ludwig Schlesinger 1 Schlesinger s biogrphies cn be found in [Lex, McT]: Ludwig (Ljos in Hungrin) Schlesinger ws born on the 1st November 1864 in Hungrin town Trnv (Ngyszombt), which now belongs to Slovki. He studied mthemtics nd physics t the universities of Heidelberg nd Berlin, where he received doctorte in 1887. 1 Photo from [McT] 65

The dvisors of his thesis (which ws concerned with homogeneous liner differentil equtions of the fourth order) were Lzrus Fuchs (who lter becme his fther-in-lw) nd Leopold Kronecker. Two yers lter Schlesinger becme n ssocite professor in Berlin nd in 1897 n invited professor t the University of Bonn. During the yers 1897 to 1911 he served s n ordinry professor nd lso s the hed of the deprtment of higher mthemtics t the University of Kolozsvár (now Cluj in Romni). In 1911 he moved to Giessen in Germny where he continued to tech until his retirement in 1930. Ludwig Schlesinger died on the 16th December 1933. Schlesinger devoted himself especilly to complex function theory nd liner differentil equtions; he lso mde vluble contributions to the history of mthemtics. He trnslted Descrtes Geometrie into Germn, nd ws one of the orgnizers of the centenry festivities dedicted to the hundredth nniversry of János Bolyi, one of the pioneers of non-eucliden geometry. The most importnt works of Schlesinger include Hndbuch der Theorie der lineren Differentilgleichungen (1895 98), J. Bolyi in Memorim (1902), Vorlesungen über linere Differentilgleichungen (1908) nd Rum, Zeit und Reltivitätstheorie (1920). Schlesinger s pper on product integrtion clled Neue Grundlgen für einen Infinitesimlklkul der Mtrizen [LS1] ws published in 1931. The uthor links up to Volterr s theory of product integrl. He strts with the Riemnn-type definition nd estblishes the bsic properties of the product integrl. His proofs re nevertheless originl while Volterr proved most of his sttements using the Peno series expnsion, Schlesinger prefers the ε δ proofs. He then proceeds to define the Lebesgue product integrl (s limit of product integrls of step functions) nd explores its properties. A continution of this pper ppered in 1932 under the title Weitere Beiträge zum Infinitesimlklkul der Mtrizen [LS2]. Schlesinger gin studies the properties of Lebesgue product integrl nd is lso concerned with contour product integrtion in R 2 nd in C. This chpter summrizes the most importnt results from both Schlesinger s ppers; the finl section then presents generliztion of Schlesinger s definition of the Lebesgue product integrl. 3.1 Riemnn integrble mtrix functions When deling with product integrl we need to work with sequences of mtrices nd their limits. Volterr ws minly working with the individul entries of the mtrices nd convergence of sequence of mtrices ws for him equivlent to convergence of ll entries. Schlesinger chooses different pproch: He defines the norm of mtrix A = { ij } n i,j=1 by [A] = n mx 1 i,j n ij. He lso mentions nother norm Ω A = mx{ λ ; λ is n eigenvlue of A} 66

nd sttes tht n Ω A ij 2 [A]. i,j=1 The second inequlity is obvious, the first is proved in [LS1] 1. Schlesinger s norm [A] hs the nice property tht [A B] [A] [B] for every A, B R n n, but its disdvntge is tht [I] = n. In the following text we will use the opertor norm A = sup{ Ax ; x 1}, where Ax nd x denote the Eucliden norms of vectors Ax, x R n. This simplifies Schlesinger s proofs slightly, becuse I = 1 nd A B A B still holds for every A, B R n n. It should be noted tht the spce R n n is finite-dimensionl, therefore it doesn t mtter which norm we choose since they re ll equivlent. The convergence of sequence of mtrices nd the limit of mtrix function is now defined in stndrd wy using the norm introduced bove. For n rbitrry mtrix function A : [, b] R n n nd tgged prtition D : = t 0 ξ 1 t 1 ξ 2 t m 1 ξ m t m = b of intervl [, b] with division points t i nd tgs ξ i we denote where t k = t k t k 1. P (A, D) = m (I + A(ξ k ) t k ), Schlesinger is now interested in the limit vlue of P (A, D) s the lengths of the intervls [t k 1, t k ] pproch zero (if the limit exists independently on the choice of ξ k [t k 1, t k ]). Clerly, the limit is nothing else thn Volterr s right product integrl. Definition 3.1.1. Consider function A : [, b] R n n. In cse the limit lim P (A, D) ν(d) 0 exists, it is clled the product integrl of function A on intervl [, b] nd denoted by the symbol (I + A(t) dt). 1 [LS1], p. 34 35 67

Remrk 3.1.2. Schlesinger in fct defines the product integrl s the limit of the products P (A, D) = Y 0 m (I + A(ξ k ) t k ), where Y 0 is n rbitrry regulr mtrix (which plys the role of n integrtion constnt ). In the following text we ssume for simplicity tht Y 0 = I. Also, insted of Schlesinger s nottion b {{ (I + A(x) dx) we use the symbol (I + A(x) dx) b to denote the product integrl. Lemm 3.1.3. 1 Let A 1, A 2,..., A m R n n be rbitrry mtrices. Then ( m ) (I + A 1 )(I + A 2 ) (I + A m ) exp A k. Proof. A simple consequence of the inequlities I + A k 1 + A k exp A k. Corollry 3.1.4. 2 If A(x) M for every x [, b], then P (A, D) e M(b ) for every tgged prtition D of intervl [, b]. Corollry 3.1.5. If the function A : [, b] R n n is product integrble nd A(x) M for every x [, b], then (I + A(x) dx) em(b ). Schlesinger s first tsk is to prove the existence of product integrl for Riemnn integrble mtrix functions, i.e. functions A : [, b] R n n whose entries ij re Riemnn integrble on [, b]. The proof is substntilly different from the proof given by Volterr; the technique is similr to Cuchy s proof of the existence of b f for continuous function f (see [CE, SŠ]). Definition 3.1.6. Consider function A : [, b] R n n nd let [c, d] [, b]. The oscilltion of A on intervl [c, d] is the number 1 [LS1], p. 37 2 [LS1], p. 38 osc(a, [c, d]) = sup{ A(ξ 1 ) A(ξ 2 ) ; ξ 1, ξ 2 [c, d]}. 68

The following chrcteriztion of Riemnn integrble function will be needed in subsequent proofs: Lemm 3.1.7. If A : [, b] R n n is Riemnn integrble function, then lim ν(d) 0 m osc(a, [t k 1, t k ]) t k = 0. Proof. The sttement follows esily from Drboux s definition of the Riemnn integrl which is bsed on upper nd lower sums; it is in fct equivlent to Riemnn integrbility of the given function (see e.g. [Sch2]). Definition 3.1.8. We sy tht tgged prtition D is refinement of tgged prtition D (we write D D), if every division point of D is lso division point of D (no condition being imposed on the tgs). Lemm 3.1.9. 1 Let the function A : [, b] R n n be such tht A(x) M for every x [, b]. Then for every pir of tgged prtitions D, D of intervl [, b] such tht D D we hve P (A, D) P (A, D ) e M(b ) m (osc(a, [t k 1, t k ]) t k + (M t k ) 2 e M tk ), where t i, i = 0,..., m re division points of the prtition D. Proof. Let the prtition D consist of division points nd tgs D : = t 0 ξ 1 t 1 ξ 2 t m 1 ξ m t m = b. First, we refine it only on the subintervl [t k 1, t k ], i.e. we consider prtition D which contins division points nd tgs t k 1 = u 0 η 1 u 1 u l 1 η l u l = t k nd coincides with the prtition D on the rest of intervl [, b]. Then k 1 P (A, D ) P (A, D) (I + A(ξ i ) t i ) We estimte i=1 l (I + A(η j ) u j ) I A(ξ k ) t k j=1 1 [LS1], p. 39 41 k 1 (I + A(ξ i ) t i ) i=1 m i=k+1 m i=k+1 (I + A(ξ i ) t i ). (I + A(ξ i ) t i ) em(b ) 69

nd l (I + A(η j ) u j ) I A(ξ k ) t k l (A(η j ) A(ξ k )) u j + j=1 l + p=2 1 r 1< <r p l j=1 A(η r1 ) A(η rp ) u r1 u rp osc(a, [t k 1, t k ]) t k + l p=2 1 r 1< <r p l M p u r1 u rp = = osc(a, [t k 1, t k ]) t k + l (1 + M u j ) 1 j=1 l M u j osc(a, [t k 1, t k ]) t k +e M tk 1 M t k osc(a, [t k 1, t k ]) t k +(M t k ) 2 e M tk. Therefore we conclude tht P (A, D) P (A, D ) e M(b ) (osc(a, [t k 1, t k ]) t k + (M t k ) 2 e M tk ). Now, since the given prtition D cn be obtined from D by successively refining the subintervls [t 0, t 1 ],..., [t m 1, t m ], we obtin P (A, D) P (A, D ) e M(b ) m j=1 (osc(a, [t k 1, t k ]) t k + (M t k ) 2 e M tk ). Corollry 3.1.10. 1 Consider Riemnn integrble function A : [, b] R n n. Then for every ε > 0 there exists δ > 0 such tht whenever ν(d) < δ nd D D. P (A, D) P (A, D ) < ε Proof. The sttement follows from the previous lemm, Lemm 3.1.7 nd the estimte m m (M t k ) 2 e M tk ν(d)m 2 e Mν(D) t k = (b )ν(d)m 2 e Mν(D). 1 [LS1], p. 39 41 70

Theorem 3.1.11. 1 The product integrl (I+A(x) dx) b exists for every Riemnn integrble function A : [, b] R n n. Proof. Tke ε > 0. Corollry 3.1.10 gurntees the existence of δ > 0 such tht P (A, D) P (A, D ) < ε/2 whenever ν(d) < δ nd D D. Consider pir of tgged prtitions D 1, D 2 of intervl [, b] stisfying ν(d 1 ) < δ nd ν(d 2 ) < δ. These prtitions hve common refinement, i.e. prtition D such tht D D 1, D D 2 (the tgs in D cn be chosen rbitrrily). Then P (A, D 1 ) P (A, D 2 ) P (A, D 1 ) P (A, D) + P (A, D) P (A, D 2 ) < ε. We hve proved tht every Riemnn integrble function A : [, b] R n n stisfies certin Cuchy condition nd this is lso the end of Schlesinger s proof; the existence of product integrl follows from the Cuchy condition in the sme wy s in the nloguous theorem for the ordinry Riemnn integrl (see e.g. [Sch2]). Theorem 3.1.12. 2 Consider Riemnn integrble function A : [, b] R n n. If c [, b], then (I + A(x) dx) = (I + A(x) dx) c (I + A(x) dx). c Proof. As Schlesinger remrks, the proof follows directly from the definition of product integrl (see the proof in Chpter 2). 3.2 Mtrix exponentil function Let A : [, b] R n n be constnt function. If D m is prtition of [, b] to m subintervls of length (b )/m, then ( P (A, D m ) = I + b ) m m A. Since ν(d m ) 0 s m, we hve (I + A(x) dx) ( = lim I + b ) m m m A = e (b )A. The lst equlity follows from the fct tht e A = lim m (I + A/m) m for every A R n n ; recll tht the mtrix exponentil ws defined in Chpter 2 using the series e A A m = m!. (3.2.1) 1 [LS1], p. 41 2 [LS1], p. 41 m=0 71

Lemm 3.2.1. If A 1,..., A m R n n nd B 1,..., B m R n n, then m m m i 1 m A i B i = B j (A i B i ). i=1 i=1 i=1 j=1 j=i+1 A j Proof. m m m A i B i = (B 1 B i 1 A i A m B 1 B i A i+1 A m ) = i=1 i=1 i=1 m i 1 m = B j (A i B i ) i=1 j=1 j=i+1 A j. Theorem 3.2.2. 1 Consider Riemnn integrble function A : [, b] R n n. Then lim ν(d) 0 m e A(ξk) tk = lim ν(d) 0 m (I + A(ξ k ) t k ) = (I + A(t) dt). Proof. Since every Riemnn integrble function is bounded, we hve A(x) M for some M R nd for every x [, b]. The definition of mtrix exponentil (3.2.1) implies e A(ξk) tk (I + A(ξ k ) t k ) ( A(ξ k ) t k ) 2 e A(ξk) tk (M t k ) 2 e M tk for k = 1,..., m. According to Lemm 3.2.1, m m e A(ξk) tk (I + A(ξ k ) t k ) = m j 1 = (I + A(ξ k ) t k ) (e A(ξj) tj I A(ξ j ) t j ) j=1 e M(b ) 1 [LS1], p. 42 m j=1 e A(ξj) tj I A(ξ j ) t j e M(b ) M 2 e M(b ) M 2 ν(d)e Mν(D) m j=1 m k=j+1 m j=1 e A(ξk) tk ( t j ) 2 e M tj t j = (b )e M(b ) M 2 ν(d)e Mν(D). 72

By choosing sufficiently fine prtition D of [, b], the lst expression cn be mde rbitrrily smll. Definition 3.2.3. The trce of mtrix A = { ij } n i,j=1 is the number Tr A = n ii. i=1 Theorem 3.2.4. 1 If A : [, b] R n n is Riemnn integrble function, then ( ) ( ) b det (I + A(x) dx) = exp Tr A(x) dx. Proof. det ( = lim (I + A(x) dx) m ν(d) 0 ) ( = det lim ν(d) 0 e Tr A(ξk) tk = lim ν(d) 0 exp ) m e A(ξk) tk = lim ν(d) 0 m det e A(ξk) tk = ( m ) ( ) b Tr A(ξ k ) t k = exp Tr A(x) dx (we hve used theorem from liner lgebr: det exp A = exp Tr A). Remrk 3.2.5. This formul (sometimes clled the Jcobi formul) ppered lredy in Volterr s work. Schlesinger employs different proof nd his sttement is lso more generl it requires only the Riemnn integrbility of A, in contrst to Volterr s ssumption tht A is continuous. Corollry 3.2.6. If A : [, b] R n n is Riemnn integrble function, then the product integrl (I + A(x) dx) b is regulr mtrix. Recll tht Volterr hs lso ssigned mening to product integrls whose lower limit is greter thn the upper limit; his definition for the right integrl ws (I + A(t) dt) = lim b ν(d) 0 k=m 1 (I A(ξ k ) t k ). If A is Riemnn integrble, we know tht this is equivlent to (I + A(t) dt) = lim b ν(d) 0 k=m 1 e A(ξk) tk. 1 [LS1], p. 43 44 73

Thus I = = lim lim ν(d) 0 ν(d) 0 ( m e A(ξk) tk m e A(ξk) tk lim ) 1 e A(ξk) tk = k=m ν(d) 0 k=m 1 e A(ξk) tk = (I + A(t) dt) (I + A(t) dt), b which proves tht (I +A(t) dt) b is the inverse mtrix of (I +A(t) dt) b ; compre with Volterr s proof of Theorem 2.4.10. 3.3 The indefinite product integrl Schlesinger now proceeds to study the properties of the indefinite product integrl, i.e. of the function Y (x) = (I + A(t) dt) x Theorem 3.3.1. 1 If A : [, b] R n n is Riemnn integrble, then the function Y (x) = (I + A(t) dt) x is continuous on [, b]. Proof. We prove the right-continuity of Y t x 0 [, b); continuity from left is proved similrly. Let x 0 x 0 + h b. The function A is bounded: A(x) M for some M R. We now employ the inequlity from Lemm 3.1.9. Let D be prtition of intervl [x 0, x 0 + h]. Then I + A(x 0 )h P (A, D ) e Mh (osc(a, [x 0, x 0 + h])h + (Mh) 2 e Mh ). Pssing to the limit ν(d ) 0 we obtin x I + A(x 0+h 0)h (I + A(t) dt) emh (osc(a, [x 0, x 0 + h])h + (Mh) 2 e Mh ), which implies Therefore x 0. x0+h lim (I + A(t) dt) = I. h 0+ x 0 lim (Y (x 0 + h) Y (x 0 )) = Y (x 0 ) h 0+ ( (I + A(t) dt) x 0+h x 0 I ) = 0. 1 [LS1], p. 44 46 74

Theorem 3.3.2. 1 If A : [, b] R n n is Riemnn integrble, then the function Y (x) = (I + A(t) dt) x, stisfies the integrl eqution Y (x) = I + x Y (t)a(t) dt, x [, b]. Proof. It is sufficient to prove the sttement for x = b. Let D : = t 0 ξ 1 t 1 ξ 2 t m 1 ξ m t m = b be tgged prtition of intervl [, b]. We define k Y k = (I + A(ξ i ) t i ), k = 0,..., m. i=1 Then Y k Y k 1 = Y k 1 A(ξ k ) t k, k = 1,..., m. (3.3.1) Since Y 0 = I nd Y m = P (A, D), dding the equlities (3.3.1) for k = 1,..., m yields m P (A, D) I = Y k 1 A(ξ k ) t k. The function A is bounded: A(x) M for some M R. We estimte Y (b) I b + P (A, D) I Y (t)a(t) dt Y (b) P (A, D) + b Y (t)a(t) dt Y (b) P (A, D) + m + (Y k 1 m Y (t k 1 ))A(ξ k ) t k + (Y (t k 1 ) Y (ξ k ))A(ξ k ) t k + 1 [LS1], p. 46 47 m + Y (ξ k )A(ξ k ) t k b Y (t)a(t) dt. (3.3.2) 75

Using the inequlities m m (Y k 1 Y (t k 1 ))A(ξ k ) t k M Y k 1 Y (t k 1 ) t k M m m e M(b ) t k (osc(a, [t j 1, t j ]) t j + (M t j ) 2 e M tj ) j=1 m Me M(b ) (b ) osc(a, [t j 1, t j ]) t j + M 2 ν(d)e Mν(D) j=1 (we hve used Lemm 3.1.9) nd m m (Y (t k 1 ) Y (ξ k ))A(ξ k ) t k M osc(y, [t k 1, t k ]) t k, we see tht ll terms on the right-hnd side of (3.3.2) cn be mde rbitrrily smll if the prtition D is sufficiently fine. Corollry 3.3.3. 1 If A : [, b] R n n is continuous, then the function Y (x) = (I + A(t) dt) x provides solution of the differentil eqution Y (x) = Y (x)a(x), x [, b] nd stisfies the initil condition Y () = I. Remrk 3.3.4. The function Y is therefore the fundmentl mtrix of the system n y i(x) = ji (x)y j (x), i = 1,..., n. j=1 Schlesinger uses the nottion D x Y (x) = A(x), where D x Y = Y 1 Y, i.e. D x is exctly Volterr s right derivtive of mtrix function. 3.4 Product integrl inequlities In this section we summrize vrious inequlities tht will be useful lter. 1 [LS1], p. 47 48 76

Lemm 3.4.1. 1 If A : [, b] R n n is Riemnn integrble function, then ( ) b (I + A(x) dx) exp A(x) dx. Proof. Lemm 3.1.3 implies tht ( m m ) (I + A(ξ i ) t i ) exp A(ξ i ) t i i=1 for every tgged prtition D of intervl [, b]; the proof is completed by pssing to the limit ν(d) 0. Lemm 3.4.2. 2 Let m N, A k, B k R n n for every k = 1,..., m. Then ( m m m ) ( ( (I + B k ) (I + A k ) exp m ) ) A k exp B k A k 1. i=1 Proof. Define Y k = k (I + A i ), Z k = i=1 k (I + B i ), k = 0,..., m (where the empty product for k = 0 equls the identity mtrix). Then for k = 1,..., m. This implies where i=1 Y k Y k 1 = Y k 1 A k, Z k Z k 1 = Z k 1 B k, Z k Y k = (Z k 1 Y k 1 )(I + B k ) + E k, (3.4.1) E k = Y k 1 (B k A k ). Applying the equlity (3.4.1) m times on the difference Z m Y m we obtin 1 [LS1], p. 51 2 [LS1], p. 52 53 Z m Y m = m 1 E k (I + B k+1 ) (I + B m ) + E m. 77

We lso estimte E k exp i=1 ( k 1 ) A i B k A k i=1 (the empty sum for k = 0 equls zero), ( m 1 k 1 ) ( m ) Z m Y m exp A i B k A k exp ( B i A i + A i ) + Since = m 1 + exp ( m 1 i=1 A i ) exp A i B k A k exp i k + exp i=1 ( m 1 i=1 A i ) i=k+1 B m A m = ( m i=k+1 B m A m i=k+1 B i A i ( m m ) ( m ) exp A i B k A k exp B i A i. B k A k exp ( B k A k ) 1, we conclude tht m m (I + B k ) (I + A k ) = Zm Y m i=1 ( m ) m ( ( m )) exp A i (exp ( B k A k ) 1) exp B i A i = i=1 i=k i=k+1 ( m ) m ( ( m ) ( m )) = exp A i exp B i A i exp B i A i = i=k+1 ( m ) ( ( m ) ) = exp A i exp B i A i 1. i=1 i=1 ) + Corollry 3.4.3. 1 If A, B : [, b] R n n re Riemnn integrble functions, then (I + B(x) dx) (I + A(x) dx) 1 [LS1], p. 53 78

( ) ( ( b ) ) b exp A(x) dx exp B(x) A(x) dx 1. Proof. The previous lemm ensures tht for every tgged prtition D of intervl [, b] we hve m m P (B, D) P (A, D) = (I + B(ξ k ) t k ) (I + A(ξ k ) t k ) ( m ) ( m ) ) exp A(ξ k ) t k (exp B(ξ k ) A(ξ k ) t k 1. The proof is completed by pssing to the limit ν(d) 0. Remrk 3.4.4. Lemm 3.4.2 is not present in Schlesinger s work, he proves directly the Corollry 3.4.3; our presenttion is perhps more redble. 3.5 Lebesgue product integrl The most vluble contribution of Schlesinger s pper is his generlized definition of product integrl which is pplicble to ll mtrix functions with bounded nd mesurble (i.e. bounded Lebesgue integrble) entries. From historicl point of view, such generliztion certinly wsn t strightforwrd one. Recll the originl Lebesgue s definition: To compute the integrl b f of bounded mesurble function f : [, b] [m, M], we choose prtition then form the sets D : m = m 0 < m 1 < < m p = M, E 0 = {x [, b]; f(x) = m}, E j = {x [, b]; m j 1 < f(x) m j }, j = 1,..., p, nd compute the lower nd upper sums s(f, D) = m 0 µ 0 + p m j 1 µ j, S(f, D) = m 0 µ 0 + j=1 p m j µ j, (3.5.1) j=1 where µ j = µ(e j ) is the Lebesgue mesure of the set E j. Since S(f, D) s(f, D) = p (m j m j 1 )µ j ν(d)(b ), j=1 79

the sums in (3.5.1) pproch common limit s ν(d) 0 nd we define b f(x) dx = lim s(f, D) = lim S(f, D). ν(d) 0 ν(d) 0 Similr procedure cnnot be used to define product integrl of mtrix function A : [, b] R n n, becuse R n n is not n ordered set. Schlesinger ws insted inspired by n equivlent definition of Lebesgue integrl which is due to Friedrich Riesz (see [FR, KZ]): A bounded function f : [, b] R is integrble, if nd only if there exists uniformly bounded sequence of step (i.e. piecewise-constnt) functions {f n } n=1 such tht f n f lmost everywhere on [, b]; in this cse, b f(x) dx = lim n b f n (x) dx. To proceed to the definition of product integrl we first recll tht (see Theorem 3.2.2) m (I + A(x) dx) = lim e A(ξk) tk ν(d) 0 for every Riemnn integrble function A : [, b] R n n. The product on the right side might be interpreted s m e A(ξk) tk where A D is step function defined by = (I + A D (t) dt), A D (t) = A(ξ k ), t (t k 1, t k ) (the vlues A(t k ), k = 0,..., m, might be chosen rbitrrily). If {D k } is sequence of tgged prtitions of [, b] such tht lim k ν(d k ) = 0, it is esily proved tht lim A D k (t) = A(t) (3.5.2) k t every point t [, b] t which A is continuous. Since Riemnn integrble functions re continuous lmost everywhere, the Eqution (3.5.2) holds.e. on [, b]. We re therefore led to the following generlized definition of product integrl: (I + A(x) dx) = lim k (I + A k(x) dx) where {A k } is suitbly chosen sequence of mtrix step functions tht converge to A lmost everywhere. Definition 3.5.1. A function A : [, b] R n n is clled step function if there exist numbers = t 0 < t 1 < < t m = b 80,

such tht A is constnt function on every intervl (t k 1, t k ), k = 1,..., m. Clerly, mtrix function A = { ij } n i,j=1 is step function if nd only if ll the entries ij re step functions. Definition 3.5.2. A sequence of functions A k : [, b] R n n, k N, is clled uniformly bounded if there exists number M R such tht A k (x) M for every k N nd every x [, b]. Definition 3.5.3. A function A : [, b] R n n is clled mesurble if ll the entries ij re mesurble functions. Lemm 3.5.4. Let A k : [, b] R n n, k N, be uniformly bounded sequence of mesurble functions such tht lim A k(x) = A(x) k.e. on [, b]. Then A k A in the norm of the spce L 1, i.e. b lim k A k (x) A(x) dx = 0. Proof. Choose ε > 0. As A k (x) M for every k N nd every x [, b], we cn estimte b A k (x) A(x) dx ε(b ) + 2Mµ({x; A k (x) A(x) ε}). The convergence A k A.e. implies convergence in mesure 1, i.e. for every ε > 0 we hve lim k µ({x; A(x) A k(x) ε}) = 0. Therefore for every ε > 0. b lim k A k (x) A(x) dx ε(b ) Theorem 3.5.5. 2 Let A k : [, b] R n n, k N, be sequence of step functions such tht Then the limit lim k 1 [IR], Proposition 8.3.3, p. 256 2 [LS1], p. 55 56 b A k (x) A(x) dx = 0. lim (I + A k(x) dx) k 81

exists nd is independent on the choice of the sequence {A k }. Proof. We verify tht (I + A k (x) dx) b is Cuchy sequence. Corollry 3.4.3 we hve (I + A l(x) dx) (I + A m (x) dx) ( ) ( ( b ) ) b exp A m (x) dx exp A l (x) A m (x) dx 1. According to The ssumption of our theorem implies tht the sequence of numbers b A m(x) dx is bounded nd tht b lim l,m A l (x) A m (x) dx = 0, which proves the existence of the limit. To verify the uniqueness consider two sequences of step functions {A k }, {B k } tht stisfy the ssumption of the theorem. We construct sequence {C k }, where C 2k 1 = A k nd C 2k = B k. Then C k A.e. nd lim k b C k (x) A(x) dx = 0, which mens tht lim k (I + C k (x) dx) b exists. Every subsequence of {C k} must hve the sme limit, therefore lim (I + A k(x) dx) k = lim k (I + B k(x) dx). Definition 3.5.6. Consider function A : [, b] R n n. Assume there exists uniformly bounded sequence of step functions A k : [, b] R n n such tht lim A k(x) = A(x) k.e. on [, b]. Then the function A is clled product integrble nd we define (I + A(x) dx) = lim k (I + A k(x) dx). We use the symbol L ([, b], R n n ) to denote the set of ll product integrble functions. Remrk 3.5.7. The correctness of the previous definition is gurnteed by Lemm 3.5.4 nd Theorem 3.5.5. Every function A L ([, b], R n n ) is clerly bounded 82

nd mesurble (step functions re mesurble nd the limit of mesurble functions is gin mesurble). Assume on the contrry tht A : [, b] R n n is mesurble function on [, b] such tht ij (x) M, x [, b], i, j = 1,..., n. There exists 1 sequence of step functions {A k } which converge to A in the L 1 norm. This sequence contins 2 subsequence {B k } of mtrix functions B k = {b k ij }n i,j=1 such tht B k A.e. on [, b]. Without loss of generlity we cn ssume tht the sequence {B k } is uniformly bounded (otherwise consider the functions min(mx( M, b k ij ), M)). We hve thus found uniformly bounded sequence of step functions which converge to A.e. on [, b]. This mens tht L ([, b], R n n ) = { A : [, b] R n n ; A is mesurble nd bounded }. Schlesinger remrks tht it is possible to further extend the definition of product integrl to encompss ll mtrix functions with Lebesgue integrble (not necessrily bounded) entries, but he doesn t give ny detils. We return to this question t the end of the chpter. 3.6 Properties of Lebesgue product integrl After hving defined the Lebesgue product integrl in [LS1], Schlesinger crefully studies its properties. Interesting results my be found lso in [LS2]. Lemm 3.6.1. Assume tht {A k } is uniformly bounded sequence of functions from L ([, b], R n n ), nd tht A k A. e. on [, b]. Then b A(x) dx = lim k b A k (x) dx. Proof. According to the Lebesgue s dominted convergence theorem, b lim k A k (x) dx = b (we hve used continuity of the norm). lim A k(x) dx = k b A(x) dx Corollry 3.6.2. Inequlities 3.4.1 nd 3.4.3 re stisfied for ll step functions. As consequence of the previous lemm we see they re vlid even for functions from L ([, b], R n n ). The next sttement represents dominted convergence theorem for the Lebesgue product integrl. 1 [RG], Corollry 3.29, p. 47 2 [IR], Theorem 8.4.14, p. 267, nd Theorem 8.3.6, p. 257 83

Theorem 3.6.3. 1 Assume tht {A k } is uniformly bounded sequence of functions from L ([, b], R n n ) such tht A k A. e. on [, b]. Then (I + A(x) dx) = lim k (I + A k(x) dx). Proof. The function A is mesurble nd bounded, therefore A L ([, b], R n n ). To complete the proof we use Corollry 3.4.3 in the form (I + A(x) dx) (I + A k (x) dx) nd Lemm 3.5.4. ( ) ( ( b ) ) b exp A(x) dx exp A k (x) A(x) dx 1 Remrk 3.6.4. The previous theorem holds lso for Riemnn product integrl in cse we dd n extr ssumption tht the limit function A is Riemnn product integrble. Definition 3.6.5. If M is mesurble subset of [, b] nd A L ([, b], R n n ), we define (I + A(x) dx) = (I + χ M (x)a(x) dx) M (where χ M is the chrcteristic function of the set M). The previous definition is correct, becuse the product χ M A is obviously mesurble bounded function. Remrk 3.6.6. The following theorem is proved in the theory of Lebesgue integrl 2 : For every f L 1 ([, b]) nd every ε > 0 there exists δ > 0 such tht f(x) dx < ε M whenever M is mesurble subset of [, b] nd µ(m) < δ. Schlesinger proceeds to prove n nloguous theorem for the product integrl (he speks bout totl continuity ). Theorem 3.6.7. 3 δ > 0 such tht 1 [LS1], p. 57 58 2 [RG], theorem 3.26, p. 46 3 [LS1], p. 59 For every A L ([, b], R n n ) nd every ε > 0 there exists (I + A(x) dx) I < ε M 84

whenever M is mesurble subset of [, b] nd µ(m) < δ. Proof. Substituting B = 0 to Corollry 3.4.3 we obtin (I + A(x) dx) ( ) ( ( ) ) I exp A(x) dx exp A(x) dx 1, M M M which completes the proof (see Remrk 3.6.6). Schlesinger now turns his ttention to the indefinite product integrl. Recll tht if f L 1 ([, b]), then the indefinite integrl F (x) = x f(t) dt, x [, b] is n bsolutely continuous function nd F (x) = f(x). e. on [, b]. Before looking t product nlogy of this theorem we stte the following lemm. Lemm 3.6.8. If A L ([, b], R n n ), then for lmost ll x (, b). 1 lim h 0 h Proof. If f L 1 ([, b]), then 1 1 lim h 0 h x+h x h 0 A(t) A(x) dt = 0 f(x + t) f(x) dt = 0 for lmost ll x (, b) (every such x is clled the Lebesgue point of f). Applying this equlity to the entries of A we obtin 1 lim h 0 h x+h for lmost ll x (, b). x 1 A(t) A(x) dt = lim h 0 h h 0 A(x + t) A(x) dt = 0 Theorem 3.6.9. 2 If A L ([, b], R n n ), then the indefinite integrl Y (x) = (I + A(t) dt) x. stisfies Y 1 (x)y (x) = A(x) for lmost ll x [, b]. Proof. According to the definition of derivtive, 1 [IR], Theorem 6.3.2, p. 194 2 [LS1], p. 60 61 Y 1 (x)y Y 1 (x)y (x + h) I (x) = lim. h 0 h 85

We now prove tht ( ) Y 1 x+h (x)y (x + h) I 1 lim = lim (I + A(t) dt) I = A(x) (3.6.1) h 0+ h h 0+ h for lmost ll x [, b]; the procedure is similr for the limit from left. We estimte ( ) ( ) x+h 1 x+h (I + A(t) dt) I A(x) h 1 (I + A(t) dt) e A(x)h + h ( 1 A k (x)h k + h k! 1 (I + A(t) dt) h k=2 x x+h x x x e A(x)h ) + A(x) 2 h e A(x) h (3.6.2) Since A(x) M for some M R, the Corollry 3.4.3 yields ( ) ( ) x+h 1 x+h (I + A(t) dt) e A(x)h 1 x+h = (I + A(t) dt) (I + A(x) dt) h h x ( ) ( ( 1 x+h ) ) x+h h exp A(x) dt exp A(t) A(x) dt 1 x x ( = exp( A(x) h) 1 k 1 x+h A(t) A(x) dt) h k! x exp(mh) ( 1 h x+h x A(t) A(x) dt + (2M) 2 h exp(2mh) Equtions (3.6.2), (3.6.3), nd Lemm 3.6.8 imply Eqution (3.6.1). x ) x =. (3.6.3) Remrk 3.6.10. In the previous theorem we hve tcitly ssumed tht the mtrix Y (x) = (I + A(t) dt) x is regulr for every x [, b]. Schlesinger proved it only for A R([, b], R n n ) (see Corollry 3.2.6), but the proof is esily djusted to A L ([, b], R n n ): If {A k } is uniformly bounded sequence of step functions such tht A k A. e. on [, b], then (using 3.2.4 nd Lebesgue s dominted convergence theorem) det(i + A(t) dt) = det lim k (I + A k(t) dt) = lim k det(i + A k(t) dt) ( ) ( b ) b = lim exp Tr A k (t) dt = exp Tr A(t) dt > 0. k 86 =

Theorem 3.6.11. 1 If A L ([, b], R n n ), then (I + A(x) dx) = I + b xk x2 A(x 1 ) A(x k ) dx 1 dx k (where the integrls on the right side re tken in the sense of Lebesgue). Proof. Let {A k } be uniformly bounded sequence of step functions such tht A k A. e. on [, b]. Every function A k is ssocited with prtition D k : = t k 0 < t k 1 < < t k m(k) = b such tht A k (x) = A k j, x (t k j 1, t k j ). According to the definition of Lebesgue product integrl, (I + A(x) dx) = lim k (I + A k(x) dx) m(k) = lim exp(a k j t k j ). k j=1 Schlesinger proves 2 first tht the product integrl might be lso clculted s (I + A(x) dx) m(k) = lim (I + A k i t k i ), (3.6.4) k i=1 provided tht lim ν(d k) = 0 (3.6.5) k (which cn be ssumed without loss of generlity); note tht if (3.6.5) is not stisfied, (3.6.4) need not hold (consider A = A k = I nd the prtitions = t k 0 < t k 1 = b for every k N). Schlesinger s proof of (3.6.4) seems too complicted nd even fulty; we insted rgue similrly s in the proof of Theorem 3.2.2: Tke positive number M such A k (x) M for every k N nd x [, b]. Then exp(a k j t k j ) I A k j t k j (M t k j ) 2 e M tk j for every k N nd j = 1,..., m(k). According to Lemm 3.2.1, m(k) m(k) exp(a k j t k j ) (I + A k j t k j ) = j=1 j=1 m(k) j 1 = (I + A k l t k l ) (exp(a k j t k j ) I A k j t k j ) j=1 l=1 1 [LS2], p. 487 2 [LS2], p. 485 486 m(k) l=j+1 exp(a k l t k l ) 87

m(k) m(k) e M(b ) exp(a k j t k j ) I A k j t k j e M(b ) M 2 ( t k j ) 2 e M tk j j=1 j=1 m(k) e M(b ) M 2 ν(d k )e Mν(Dk) j=1 t k j = e M(b ) M 2 ν(d k )e Mν(Dk) (b ). This completes the proof of (3.6.4). Schlesinger now sttes tht m(k) (I + A k i t k i ) = I + i=1 m(k) nd concludes the proof sying tht lim k 1 i 1< <i s m(k) s=1 1 i 1< <i s m(k) A k i 1 A k i s t k i 1 t k i s A k i 1 A k i s t k i 1 t k i s = b xs x2 = A(x 1 ) A(x s ) dx 1 dx s The lst step perhps deserves better explntion: Denote nd X s = {(x 1,..., x s ) R s ; x 1 < x 2 < < x s b}, X s k = 1 i 1< <i s m(k) [t i1 1, t i1 ] [t i2 1, t i2 ] [t is 1, t is ], where s nd k re rbitrry positive integers. If χ s nd χ s k denote the chrcteristic functions of X s nd Xk s, then χs k χs for k. Consequently lim A k i 1 A k i s t k i 1 t k i s = = lim k = b b b b k 1 i 1< <i s m(k) = b b b xs A k (x 1 ) A k (x s )χ s k(x 1,..., x s ) dx 1 dx s = A(x 1 ) A(x s )χ s (x 1,..., x s ) dx 1 dx s = x2 A(x 1 ) A(x s ) dx 1 dx s (we hve used the dominted convergence theorem). Remrk 3.6.12. The deficiency in the previous proof is tht Schlesinger didn t justify the equlity m(k) lim I + A k i 1 A k i s t k i 1 t k i s = k s=1 1 i 1< <i s m(k) 88

I + lim k s=1 1 i 1< <i s m(k) A k i 1 A k i s t k i 1 t k i s. We hve lredy encountered similr inccurcy when discussing Volterr s proof of the Peno series expnsion theorem for product integrl; see lso Msni s proof of Theorem 5.5.10. Remrk 3.6.13. Recll tht, ccording to Theorem 2.3.5, the right derivtive of mtrix function stisfies (CD 1 ) d ( dx = D C d dx D d ) D 1. dx Consider two continuous mtrix functions A, B defined on [, b]. Using the previous formul nd lso the convention tht ( ) 1 x y (I + A(t) dt) = (I + A(t) dt) for y > x, we infer the equlity ( ) x d (I + B(t) dt) (I + A(t) dt) dx = b y b x = (I + A(t) dt) (B(x) A(x))(I + A(t) dt) for every x [, b]. Denoting S(x) = (I + A(t) dt) x b we obtin ( ) x d (I + B(t) dt) (I + A(t) dt) dx = S(x)(B(x) A(x))S 1 (x), b nd consequently (since the left hnd side is equl to I for x = b) x x x x x (I + B(t) dt) (I + A(t) dt) = (I + S(t)(B(t) A(t))S 1 (t) dt) b x Substituting x = nd inverting both sides of the eqution yields x. b (I + A(t) dt) (I + B(t) dt) = (I + S(t)(B(t) A(t))S 1 (t) dt). (3.6.6) b A similr theorem (concerning the left product integrl) ws lredy present in Volterr s work 1. Schlesinger proves 2 tht the sttement remins true even if A, B L ([, b], R n n ). The proof is rther technicl nd we don t reproduce it here. 1 [VH], p. 85 86 2 [LS2], p. 488 489 89

Theorem 3.6.14. 1 Let A : [, b] [c, d] R n n be such tht the integrl P (t) = (I + A(x, t) dx) exists for every t [c, d] nd tht A (x, t) t M, x [, b], t [c, d], for some M R. Then P d dt = P 1 (t)p (t) = b S(x, t) A t (x, t)s 1 (x, t) dx, where S(x, t) = (I + A(u, t) du) x b. Proof. The definition of derivtive gives ( ) P 1 (t)p 1 (t) = lim (I + A(x, t) dx) (I + A(x, t + h) dx) I. h 0 h Using Eqution (3.6.6) we convert the bove limit to ( 1 lim (I + S(x, t)(a(x, t + h) A(x, t))s 1 (x, t) dx) h 0 h b ) I. Expnding the product integrl to Peno series (see Theorem 3.6.11) we obtin 1 lim h 0 h b xk x2 (x 1, t, h) (x k, t, h) dx 1 dx k, (3.6.7) where (x, t, h) = S(x, t)(a(x, t + h) A(x, t))s 1 (x, t). As the Peno series converges uniformly (the Weierstrss M-test, see Theorem 2.4.5), we cn interchnge the order of limit nd summtion. According to the men vlue theorem there is ξ(h) [t, t + h] such tht A(x, t + h) A(x, t) h = A (x, ξ(h)) t M. The dominted convergence theorem therefore implies 1 [LS2], p. 490 491 1 lim h 0 h b (x 1, t, h) dx 1 = b lim h 0 (x 1, t, h) h dx 1 = 90

nd for k 2 = 1 lim h 0 h b xk = b xk x2 b S(x 1, t) A t (x 1, t)s 1 (x 1, t) dx 1, x2 (x 1, t, h) (x k, t, h) dx 1 dx k = ) lim h 0 which completes the proof. ( h k 1 (x 1, t, h) (x k, t, h) h h dx 1 dx k = 0, The following sttement generlizes Theorem 2.5.12; Schlesinger replces Volterr s ssumption A C([, b], R n n ) by weker condition A L ([, b], R n n ). Theorem 3.6.15. 1 If A L ([, b], R n n ) nd C R n n is regulr mtrix, then (I + C 1 A(x)C dx) = C 1 (I + A(x) dx) C. Proof. Since (C 1 AC) k = C 1 A k C for every k N, we hve If A is step function, then exp(c 1 AC) = C 1 exp(a)c. (I + C 1 A(x)C dx) m = e C 1 A(ξ i)c t i = i=1 m = C 1 e A(ξi) ti C = C 1 (I + A(x) dx) C. i=1 In the generl cse when A L ([, b], R n n ), we rewrite the bove eqution with simple functions A k in plce of A, nd then pss to the limit k. 3.7 Double nd contour product integrls A considerble prt of the pper [LS2] is devoted to double nd contour product integrls (in R 2 s well s in C). Probbly the most remrkble chievement is Schlesinger s proof of the Green s theorem for product integrl, which is reproduced in the following text. Definition 3.7.1. Let G be the rectngle [, b] [c, d] in R 2 nd A : G R n n mtrix function on G. The double product integrl of A over G is defined s (I + A(x, y) dx dy) ( ( ) ) b d = I + A(x, y) dx dy, G c 1 [LS2], p. 489 91

provided both integrls on the right hnd side exist (in the sense of Lebesgue). Definition 3.7.2. Let G be the rectngle [, b] [c, d] in R 2 nd P, Q : G R n n continuous functions on G. We denote U(x, y) = (I + P (t, c) dt) x (I + Q(x, t) dt) y, c y x T (x, y) = (I + Q(, t) dt) (I + P (t, y) dt) for every x [, b], y [c, d]. The contour product integrl over the boundry of rectngle G is defined s the mtrix c (I + P (x, y) dx + Q(x, y) dy) G = U(b, d)t (b, d) 1. (3.7.1) Remrk 3.7.3. Schlesinger refers to the mtrices U(b, d) nd T (b, d) s to the integrl over the lower step nd integrl over the upper step of the rectngle G. They re clerly specil cse of the contour product integrl s defined by Volterr (see definition 2.6.8); the mtrix (3.7.1) corresponds to the vlue of contour product integrl long the (nticlockwise oriented) boundry of G. Theorem 3.7.4. 1 Let G be the rectngle [, b] [c, d] in R 2 nd P, Q : G R n n continuous mtrix functions on G. Assume tht the derivtives P y, Q x exist nd re continuous on G. Then (I + P (x, y) dx + Q(x, y) dy) =(I + T (P, Q) T 1 dx dy), G G where (P, Q) = Q x P y + P Q QP, T (x, y) = (I + Q(, t) dt) y (I + P (t, y) dt) c x. Proof. A simple clcultion revels tht (compre to Lemm 2.6.4) 1 [LS2], p. 496 497 T (P, Q) T 1 = ( ( T Q T d ) ) T 1. x dy 92

Tking the product integrl over G we obtin (I + T (P, Q) T 1 dx dy) = G ( [ ( = I + T (x, y) Q(x, y) T (x, y) d ) ] ) b d T (x, y) 1 dy. (3.7.2) dy c According to the rules for differentiting product of functions (see Theorem 2.3.2), ( ) T d dy = (I + P (t, y) dt) x x d Q(, y)(i + P (t, y) dt) + (I + P (t, y) dt) dy. x Theorem 3.6.14 on differentiting the product integrl with respect to prmeter yields ( ) x d lim (I + P (t, y) dt) x dy = 0, nd consequently The equlities (3.7.2) nd (3.7.3) imply lim T d = Q(, y). (3.7.3) x dy (I + T (P, Q) T 1 dx dy) G = ( ( ( = I + lim T (x, y) Q(x, y) T (x, y) d ) ) ) d T (x, y) 1 dy = x b dy c ( ( = lim I + T (x, y) Q(x, y) T (x, y) d ) ) d T (x, y) 1 dy x b dy c (3.7.4) (we hve used Theorem 3.6.3 on interchnging the order of limit nd integrl). For every x [, b] we hve nd lso T (x, d) 1 T (x, y) = T (x, y) d dy = ( T (x, d) 1 T (x, y) ) d dy ( I + T (x, u) d ) y du du Using Theorem 3.6.15 nd Eqution (3.6.6) we rrive t d = ( I + ( T (x, d) 1 T (x, u) ) ) y d du du. ( ( I + T (x, y) Q(x, y) T (x, y) d ) ) d T (x, y) 1 dy = T (x, d) dy c 93 d

( ( I + T (x, d) 1 T (x, y) Q(x, y) ( T (x, d) 1 T (x, y) ) ) ) d d T (x, y) 1 T (x, d) dy dy c ( T (x, d) 1 = T (x, d) I + ( T (x, d) 1 T (x, y) ) ) c d dy dy d d (I + Q(x, y) dy) T (x, d) 1 = T (x, d) T (x, d) 1 T (x, c)(i + Q(x, y) dy) Finlly, Eqution (3.7.4) gives c T (x, d) 1 = T (x, c)(i + Q(x, y) dy) (I + T (P, Q) T 1 dx dy) G = lim x b ( d T (x, d) 1. T (x, c)(i + Q(x, y) dy) c d ) d T (x, d) 1 = c c = T (b, c)(i + Q(b, y) dy) d T (b, d) 1 = (I + P (x, y) dx + Q(x, y) dy). c G Remrk 3.7.5. The previous theorem represents n nlogy of Green s theorem for the product integrl; we hve lredy encountered similr sttement when discussing Volterr s work. Volterr s nlogy of the curl opertor ws (P, Q) = Q x P y while Schlesinger s curl hs the form + QP P Q, (P, Q) = Q x P y + P Q QP. The reson is tht Volterr stted his theorem for the left product integrl, while Schlesinger ws concerned with the right product integrl (see Theorem 2.6.15 nd Remrk 2.6.7). Wheres Volterr worked with simply connected domin G (see definition 2.6.12), Schlesinger considers only rectngles. Consider functions P, Q tht stisfy ssumptions of Theorem 3.7.4 nd such tht everywhere in G. Then (P, Q) = 0 (3.7.5) (I + P (x, y) dx + Q(x, y) dy) G = I, 94

which in consequence mens tht the vlues of contour product integrl over the lower step nd over the upper step re the sme. Schlesinger then denotes the common vlue of the mtrices U(x, y) nd T (x, y) (see definition 3.7.2) by the symbol (b,d) (I + P (x, y) dx + Q(x, y) dy). Clerly (x,y) (I + P (u, v) du + Q(u, v) dv) d dx (,c) (,c) d = T (x, y) = P (x, y), dx (x,y) (I + P (u, v) du + Q(u, v) dv) d dy = U(x, y) d = Q(x, y). dy (,c) Schlesinger now proceeds to define product integrl long contour nd shows tht (in simply connected domin) the condition (3.7.5) implies tht the vlue of product integrl depends only on the endpoints of the contour. His method is lmost the sme s Volterr s nd we don t repet it here. At the end of pper [LS2] Schlesinger trets mtrix functions of complex vrible. He defines the contour product integrl in complex domin nd recpitultes the results proved erlier by Volterr (theorems 2.7.4, 2.7.7, nd 2.7.6). 3.8 Generliztion of Schlesinger s definition Thnks to the definition proposed by Ludwig Schlesinger it is possible to extend the clss of product integrble functions nd to work with bounded mesurble functions insted of Riemnn integrble functions. At this plce we remind the nottion L ([, b], R n n ) = { A : [, b] R n n ; A is mesurble nd bounded }. Schlesinger ws wre tht his definition might be extended to ll mtrix functions with Lebesgue integrble (not necessrily bounded) entries, i. e. to the clss { } L([, b], R n n ) = A : [, b] R n n ; (L) b A(t) dt < where the symbol (L) emphsizes tht we re deling with the Lebesgue integrl. Clerly L L. If {A k } is uniformly bounded sequence of functions which converge to A lmost everywhere, then ccording to lemm 3.5.4 lim A k A 1 = lim k k b 95 A k (x) A(x) dx = 0,,

i.e. A k converge to A lso in the norm of spce L([, b], R n n ). Tking ccount of Theorem 3.5.5 it is nturl to stte the following definition. Definition 3.8.1. A function A : [, b] R n n is clled product integrble if there exists sequence of step functions {A k } such tht lim A k A 1 = 0. k We define (I + A(t) dt) = lim k (I + A k(t) dt). exists iff A L([, b], R n n ), i. e. iff the integrl (L) b A(t) dt exists. Interested reders re referred to the book [DF] for more detils bout the theory of product integrl bsed on definition 3.8.1. As n interesting exmple we present the proof of theorem on differentiting the product integrl with respect to the upper bound of integrtion. We strt with preliminry lemm (which follows lso from Theorem 3.3.2, but we don t wnt to use it s we re seeking nother wy to prove it). Remrk 3.8.2. The correctness of the previous definition is ensured by theorem 3.5.5. Since step functions belong to the spce L([, b], R n n ), which is complete, every product integrble function lso belongs to this spce. Moreover, step functions form dense subset in this spce 1, nd therefore (I + A(t) dt) b Lemm 3.8.3. If A : [, b] R n n is step function, then Y (x) = I + x Y (t)a(t) dt, x [, b]. Proof. There exist prtition = t 0 < t 1 < < t m = b nd mtrices A 1,..., A m R n n such tht Then A(x) = A k, x (t k 1, t k ). Y (x) = (I + A(t) dt) x = e A1(t2 t1) e Ak 1(tk 1 tk 2) e Ak(x tk 1) for every x [t k 1, t k ]. The function Y is continuous on [, b] nd differentible except finite number of points; we hve 1 [RG], Corollry 3.29, p. 47 Y (x) = Y (x)a(x) (3.8.1) 96

for x [, b]\{t 0, t 1,..., t m }. This implies Y (x) = I + x Y (t)a(t) dt, x [, b]. Theorem 3.8.4. 1 Consider function A L([, b], R n n ). For every x [, b] the integrl x Y (x) = (I + A(t) dt) (3.8.2) exists nd the function Y stisfies the eqution Y (x) = I + x Y (t)a(t) dt, x [, b]. (3.8.3) Proof. Let A k : [, b] R n n, k N be sequence of step functions such tht Then clerly lim A k A 1 = lim k k x lim k b A k (t) A(t) dt = 0, i. e. the definition (3.8.2) is correct. Denote Y k (x) = (I + A k (t) dt) Becuse A k re step functions, Lemm 3.8.3 implies Y k (x) = I + x A k (t) A(t) dt = 0. (3.8.4) x. x [, b], Y k (t)a k (t) dt, x [, b]. (3.8.5) According to Corollry 3.4.3, ( b ) ( ( b ) ) Y l (x) Y m (x) exp A l (t) dt exp A l (t) A m (t) dt 1 = = exp A l 1 (exp A l (t) A m (t) 1 1). From Eqution (3.8.4) we see tht {A k } is bounded nd Cuchy sequence with respect to the norm 1. The previous inequlity therefore implies tht Y k converge uniformly to Y, i. e. 1 [DF], p. 54 55 Y k Y = sup Y k (x) Y (x) 0 pro k. x [,b] 97

We now estimte x Y k (t)a k (t) dt x Y (t)a(t) dt Y ka k Y A 1 (Y k Y )A k 1 + Y (A k A) 1 A k 1 Y k Y + (A k A) 1 Y, nd consequently x lim k Y k (t)a k (t) dt = x Y (t)a(t) dt. The equlity (3.8.3) is obtined by pssing to the limit in eqution (3.8.5). Corollry 3.8.5. If A L([, b], R n n ), then the function x Y (x) = (I + A(t) dt), x [, b], is bsolutely continuous on [, b] nd lmost everywhere on [, b]. Y (x) 1 Y (x) = A(x) Remrk 3.8.6. In our proof of the previous theorem we hve employed Schlesinger s estimte from Corollry 3.4.3, whose proof is somewht lborious. The uthors of [DF] insted mke use of different inequlity, which is esier to demonstrte. Let A, B : [, b] R n n be two step functions. Denoting x x Y (x) = (I + A(t) dt), Z(x) = (I + B(t) dt), we see tht the function Y Z 1 is continuous on [, b] nd differentible except finite number of points. Using Eqution (3.8.1) we clculte (Y Z 1 ) = Y Z 1 + Y (Z 1 ) = Y Y 1 Y Z 1 Y Z 1 Z Z 1 = Y (A B)Z 1, nd consequently Y (x)z 1 (x) = I + x (Y Z 1 ) (t) dt = I + x Y (t)(a(t) B(t))Z 1 (t) dt. Multiplying this eqution by Z from right nd substituting x = b we obtin (I + A(t) dt) (I + B(t) dt) = Y (b) Z(b) b Y (t) A(t) B(t) Z 1 (t) dt Z(b) e 2 B 1 e A 1 A B 1 (we hve used Lemm 3.4.1 to estimte Y (t), Z 1 (t) nd Z(b) ). The mening of the lst inequlity is similr to the mening of inequlity from Corollry 3.4.3: If two step functions A, B re close with respect to the norm 1, then the vlues of their product integrls re lso close to ech other. 98

Chpter 4 Opertor-vlued functions In the previous chpters we hve encountered vrious definitions of product integrl of mtrix function A : [, b] R n n. It is known tht every n n mtrix represents liner trnsformtion on the spce R n, nd tht the composition of two liner trnsformtions corresponds to multipliction of their mtrices. It is thus nturl to sk whether it is possible to define the product integrl of function A defined on [, b] whose vlues re opertors on certin (possibly infinitedimensionl) vector spce X. This pioneering ide (lthough in less generl scope) cn be lredy found in the second prt of the book [VH] written by Bohuslv Hostinský. He studies certin specil liner opertors on the spce of continuous functions (he clls them liner functionl trnsformtions ) nd clcultes their product integrls s well s their left nd right derivtives. Hostinský imgines function s vector with infinitely mny coordintes; the liner opertor on R n given by y i = n ij x j then goes over to the opertor on continuous functions given by y(t) = q p j=1 A(t, u)x(u) du. This ide ws not new one lredy Volterr noted 1 tht integrl equtions cn be treted s limiting cses of systems of liner lgebric equtions. His observtion ws lso lter used by Ivr Fredholm, who obtined the solution of n integrl eqution s limit of the solutions of liner lgebric equtions. Bohuslv Hostinský ws born on the 5th December 1884 in Prgue. He studied mthemtics nd physics t the philosophicl fculty of Chrles University in Prgue nd obtined his doctorl degree in 1907 (his disserttion thesis ws devoted to geometry). He spent short time s high school techer nd visited Pris in 1908 09 (he cooperted especilly with Gston Drboux). Since 1912 he gve lectures s privtdozent t the philosophicl fculty in Prgue nd ws promoted to professor of theoreticl physics t the fculty of nturl sciences in Brno in 1920; he held this position until his deth on the 12th April 1951. 1 [Kl], Chpter 45 99

Bohuslv Hostinský 1 Hostinský initilly devoted himself to pure mthemtics, especilly to differentil geometry. However, his interest grdully moved to theoreticl physics, in prticulr to the kinetic theory of gses (being influenced minly by Borel, Ehrenfest nd Poincré). Of course, this discipline requires good knowledge of probbility theory; Hostinský is considered one of the pioneers of the theory of Mrkov processes nd their ppliction in physics. Aprt from tht he lso worked on differentil nd integrl equtions nd is known for his criticl ttitude towrds the theory of reltivity. A good overview of the life nd work of Bohuslv Hostinský is given in [Ber]. 4.1 Integrl opertors In the following discussion we focus our ttention to the spce C([p, q]) of continuous functions defined on [p, q]. Every mpping T : C([p, q]) C([p, q]) is clled n opertor on C([p, q]); we will often write T f insted of T (f). The opertor is clled liner if T (f 1 + bf 2 ) = T (f 1 ) + bt (f 2 ) for ech pir of functions f 1, f 2 C([p, q]) nd ech pir of numbers, b R. The inverse opertor of T, which is denoted by T 1, stisfies T 1 (T (f)) = T (T 1 (f)) = f (4.1.1) for every function f C([p, q]); note tht the inverse opertor need not lwys exist. The lst eqution cn be shortened to 1 Photo provided by Tomáš Hostinský T 1 T = T T 1 = I, 100

where denotes the composition of opertors nd I is the identity opertor, which stisfies If = f for every function f. Bohuslv Hostinský ws concerned especilly with integrl opertors of the first kind T f(x) = q nd with the opertors of the second kind p T f(x) = f(x) + K(x, y)f(y) dy q p K(x, y)f(y) dy, where the function K (the so-clled kernel) is continuous on [p, q] [p, q]. Thus if f C([p, q]), then lso T f C([p, q]). These opertors ply n importnt role in the theory of integrl equtions nd hve been studied by Vito Volterr, Ivr Fredholm, Dvid Hilbert nd others since the end of the 19th century (see [Kl], Chpter 45). Hostinský strts 1 with recpitultion of the bsic properties of the integrl opertors. If K(x, y) = 0 for y > x, we obtin either the Volterr opertor of the first kind T f(x) = x or the Volterr opertor of the second kind p T f(x) = f(x) + K(x, y)f(y) dy, x p K(x, y)f(y) dy. Composition of two integrl opertors of the second kind T 1 f(x) = f(x) + T 2 f(x) = f(x) + q p q produces nother opertor of the second kind whose kernel is 1 [VH], p. 182 186 (T 2 T 1 )f(x) = f(x) + J(x, y) = K 1 (x, y) + K 2 (x, y) + p K 1 (x, y)f(y) dy, K 2 (x, y)f(y) dy, q p q J(x, y)f(y) dy, p K 2 (x, z)k 1 (z, y) dz. (4.1.2) 101

An importnt question is the existence of inverse opertors (which corresponds to solvbility of the corresponding integrl equtions). The Volterr opertor of the second kind lwys hs n inverse opertor T 1 f(x) = f(x) + x p N(x, y)f(y) dy, which is gin Volterr opertor of the second kind whose kernel N(x, y) (usully clled the resolvent kernel) cn be found using the method of successive pproximtions. Severl sufficient conditions re known for the existence of the inverse opertor to the Fredholm opertor of the second kind; we do not dwell into the detils nd only mention tht the inverse opertor is gin Fredholm opertor of the second kind T 1 f(x) = f(x) + q p N(x, y)f(y) dy, with kernel N(x, y). Equtions (4.1.1) nd (4.1.2) imply tht the kernel stisfies K(x, y) + N(x, y) = q p K(x, t)n(t, y) dt = q p N(x, t)k(t, y) dt. (4.1.3) Generlly, the opertors of the first kind need not hve n inverse opertor. 4.2 Product integrl of n opertor-vlued function We now ssume tht the integrl opertor kernel depends on prmeter u [, b]: T (u)f(x) = f(x) + q p K(x, y, u)f(y) dy Thus T is function defined on [, b] whose vlues re opertors on (C([p, q])). Hostinský now proceeds 1 to clculte its left derivtive. He doesn t stte ny definition, but his clcultion follows Volterr s definition of the left derivtive of mtrix function. Assume tht K is continuous on [p, q] [p, q] [, b] nd tht the derivtive K (x, y, u) u exists nd is continuous for every u [, b] nd x, y [p, q]. We choose prticulr u [, b] nd ssume tht the inverse opertor T 1 (u) exists. According to Eqution (4.1.2), the kernel of the opertor T (u + u) T 1 (u) is J(x, y, u, u) = K(x, y, u + u) + N(x, y, u) + 1 [VH], p. 186 188 q p K(x, t, u + u)n(t, y, u) dt = = K(x, y, u + u) K(x, y, u) + K(x, y, u) + N(x, y, u)+ 102

+ q = K(x, y, u + u) K(x, y, u) + p K(x, t, u + u)n(t, y, u) dt = q (we hve used Eqution (4.1.3)). Denote p (K(x, t, u + u) K(x, t, u)) N(t, y, u) dt J(x, y, u, u) M(x, y, u) = lim = u 0 u = K q (x, y, u) + u p K (x, t, u)n(t, y, u) dt. (4.2.1) u According to Bohuslv Hostinský, the left derivtive of T t u is n opertor of the second kind whose kernel is M(x, y, u). By similr method he deduces tht the right derivtive of T t u is n opertor of the second kind with the kernel M (x, y, u) = K (x, y, u) + u q p N(x, t, u) K (t, y, u) dt. u These sttements re somewht confusing, becuse the left derivtive should be the opertor T (u + u) T 1 (u) I lim, u 0 u which is n opertor of the first kind with kernel M(x, y, u). Similrly, the right derivtive should be rther lim u 0 T 1 (u) T (u + u) I, u i.e. n opertor of the first kind with kernel M (x, y, u). The next problem tckled by Hostinský is the clcultion of the left integrl of the opertor-vlued function T (u)f(x) = f(x) + q p K(x, y, u)f(y) dy. We gin ssume tht the function K is continuous on [p, q] [p, q] [, b]. As in the cse of derivtives, Hostinský provides no definition nd strts 1 directly with the clcultion, which is gin somewht strnge: He chooses tgged prtition D : = t 0 < t 1 < < t m = b, ξ i [t i 1, t i ], nd forms the opertor 1 [VH], p. 188 191 T (ξ m ) t m T (ξ 1 ) t 1, 103

where t i = t i t i 1. He now clims it is n opertor of the second kind, which is not true. The correct procedure tht he probbly hs in mind is to tke the opertor-vlued function nd to form the opertor S(u)f(x) = q p K(x, y, u)f(y) dy P (S, D) = (I + S(ξ m ) t m ) (I + S(ξ 1 ) t 1 ), which is n opertor of the second kind with the kernel m s=1 1 i 1< <i s m q p q p K(x, z 1, ξ is ) K(z s 1, y, ξ i1 ) t is t i1 dz 1 dz s 1 (we hve used Eqution (4.1.2)). Pssing to the limit for ν(d) 0 we clculte tht the left product integrl of the function S is n opertor of the second kind with the kernel L(x, y,, b) = (4.2.2) q q b ts t2 = K(x, z 1, t s ) K(z s 1, y, t 1 ) dt 1 dt s dz 1 dz s 1. s=1 p p In similr wy cn clculte the right product integrl of S, which is obtined s the limit of opertors P (S, D) = (I + S(ξ 1 ) t 1 ) (I + S(ξ m ) t m ) for ν(d) 0; the result is n opertor of the second kind with the kernel = s=1 q p q b ts p R(x, y,, b) = (4.2.3) t2 K(x, z 1, t 1 ) K(z s 1, y, t s ) dt 1 dt s dz 1 dz s 1. In nlogy with product integrls of mtrix functions we use the symbols (I + S(t) dt) nd (I + S(t) dt) to denote the left nd right product integrls, respectively. Let us briefly summrize their properties: If the opertor-vlued function S is defined on [, c] nd b [, c], then 1 c (I + S(t) dt) = c (I + S(t) dt) b (I + S(t) dt). 1 [VH], p. 193 104

The left side of the lst equlity represents n opertor of the second kind whose kernel is L(x, y,, c), while the right side is composition of two second-kind opertors with kernels L(x, y,, b) nd L(x, y, b, c). Thus ccording to Eqution (4.1.2) we hve L(x, y,, c) = L(x, y,, b) + L(x, y, b, c) + Similrly, the right integrl stisfies nd we consequently q p L(x, z, b, c)l(z, y,, b) dz. (4.2.4) c c (I + S(t) dt) = (I + S(t) dt) (I + S(t) dt), R(x, y,, c) = R(x, y,, b) + R(x, y, b, c) + q p b R(x, z,, b)r(z, y, b, c) dz. (4.2.5) Hostinský next demonstrtes 1 tht the derivtive nd the integrl re reverse opertions; this mens tht if we consider the left (right) integrl s function of its upper bound, then its left (right) derivtive is the originl function. If f is n rbitrry continuous function, then lim c b c b f = f(b). Using this result nd the definition of L we conclude L(x, y, b, c) lim = K(x, y, b) c b c b (the series in the definition of L is uniformly convergent nd we cn interchnge the order of limit nd summtion). Consequently, Eqution (4.2.4) gives L(x, y,, b) L(x, y,, c) L(x, y,, b) = lim = b c b c b ( 1 q ) = lim L(x, y, b, c) + L(x, z, b, c)l(z, y,, b) dz = c b c b = K(x, y, b) + q p p K(x, z, b)l(z, y,, b) dz. If N L (x, y,, b) is the resolvent kernel corresponding to L(x, y,, b), then, ccording to Eqution (4.2.1), the left derivtive of the left integrl of function S is n opertor of the first kind with kernel 1 [VH], p. 199 200 L (x, y,, b) + b q p L b (x, t,, b)n L(t, y,, b) dt = 105

+ q p ( K(x, t, b) + = K(x, y, b) + q p q p K(x, z, b)l(z, y,, b) dz+ ) K(x, z, b)l(z, t,, b) dz N L (t, y,, b) dt = K(x, y, b) (we hve pplied Eqution (4.1.3) to functions L nd N L ), which completes the proof. Finlly let us note tht in cse when S is constnt function, i.e. if the kernel K does not depend on u, we obtin where K s (x, y) = L(x, y,, b) = R(x, y,, b) = q p q p (b ) s K s (x, y), (4.2.6) s! s=1 K(x, z 1 )K(z 1, z 2 ) K(z s 1, y) dz 1 dz s 1. Remrk 4.2.1. There exists close reltionship between the formuls derived by Hostinský nd those obtined by Volterr. Recll tht if A : [, b] R n n is continuous mtrix function, then (I + A(t) dt) = I + L(, b), where L(, b) is mtrix with components L ij (, b) = s=1 n z 1,...,z s 1=1 b ts t2 i,z1 (t s ) zs 1,j(t 1 ) dt 1 dt s. This resembles Eqution (4.2.2) the only difference is tht in (4.2.2) we integrte over intervl [p, q] insted of tking sum over {1,..., n}. Similrly, the right integrl of mtrix function stisfies (I + A(t) dt) = I + R(, b), where n R ij (, b) = b ts t2 s=1 z 1,...,z s 1=1 which is n nlogy of Eqution (4.2.3). 106 i,z1 (t 1 ) zs 1,j(t s ) dt 1 dt s,

Also the formul c (I + A(t) dt) = c (I + A(t) dt) b (I + A(t) dt) implies i.e. the components stisfy I + L(, c) = (I + L(b, c))(i + L(, b)), L ij (, c) = L ij (, b) + L ij (b, c) + n L i,z (b, c)l z,j (, b). In similr wy we obtin tht the components of the right integrl stisfy R ij (, c) = R ij (, b) + R ij (b, c) + z=1 n R i,z (, b)r z,j (b, c). z=1 These reltions resemble Equtions (4.2.4) nd (4.2.5). Remrk 4.2.2. Product integrtion of opertor-vlued functions cn be used to solve certin integro-differentil equtions. Hostinský considers 1 the eqution f (x, t) = t q p K(x, y)f(y, t) dt, whose kernel is independent of t; s he remrks, its solution ws found by Volterr (without using product integrtion) in 1914. More generlly, we cn consider the integro-differentil eqution with the initil condition f (x, t) = t q p K(x, y, t)f(y, t) dt f(x, t 0 ) = f 0 (x), x [p, q]. This eqution cn be rewritten using the opertor nottion to f (t) = S(t)f(t), t where S(t) is n integrl opertor of the first kind for every t. Thus, for smll t we hve f(t + t) = f(t) + ts(t)f(t) = (I + S(t) t)f(t). 1 [VH, p. 192] 107

Arguing similrly s in the cse of mtrix functions, we conclude tht f(t) = t t 0 (I + S(u) du)f 0, i.e. f(x, t) = f 0 (x) + q p L(x, y, t 0, t)f 0 (y) dt. However, the effort of Bohuslv Hostinský is not primrily directed towrds integrodifferentil equtions. He shows tht the right product integrl kernel R(x, y,, b) cn be used to obtin the solution of certin differentil eqution known from the work of Jcques Hdmrd. The lst chpter of [VH] is devoted to integrl opertors of the first kind. By n pproprite choice of the kernel, Hostinský is ble to produce infinitesiml opertors of the first kind, i.e. opertors tht differ infinitesimlly from the identity opertor (thus the kernel is something like the Dirc δ-function). The composition of such infinitesiml opertors leds to n opertor whose kernel K provides solution of the eqution K(x, y, u, v) = q p K(x, z, u, w)k(z, y, w, v) dz. This eqution is known s the Chpmn or Chpmn-Kolmogorov eqution nd is often encountered in the theory of stochstic processes (see lso Section 1.4 nd Eqution (1.4.7), which represents discrete version of the Chpmn eqution). The topic is rther specil nd we don t discuss it here. 4.3 Generl definition of product integrl Although Hostinský is interested only in integrl opertors on the spce C([p, q]), it is possible to work more generlly with liner opertors on n rbitrry Bnch spce X. Before proceeding to the corresponding definitions we recll tht the norm of liner opertor T on the spce X is defined s T = sup{ T (x) ; x = 1}. Let L(X) denote the spce of ll bounded liner opertors on X, i.e. opertors whose norm is finite; L(X) is normed vector spce. Definition 4.3.1. Let X be Bnch spce, A : [, b] L(X), t [, b]. Assume tht A(t) 1 exists. We define the left nd right derivtives of A t t s d A(t) = lim dt h 0 A(t) d dt = lim h 0 A(t + h)a(t) 1 I, h A(t) 1 A(t + h) I, h 108

provided the limits exist. Of course, t the endpoints of [, b] we require only the existence of the corresponding one-sided limits. To every function A : [, b] L(X) nd every tgged prtition we ssign the opertors P (A, D) = P (A, D) = D : = t 0 < t 1 < < t m = b, ξ i [t i 1, t i ], 1 (I + A(ξ i ) t i ) = (I + A(ξ m ) t m ) (I + A(ξ 1 ) t 1 ), i=m m (I + A(ξ i ) t i ) = (I + A(ξ 1 ) t 1 ) (I + A(ξ m ) t m ). i=1 Definition 4.3.2. Consider function A : [, b] L(X). The left nd right product integrls of A re the opertors provided the limits exist. (I + A(t) dt) = lim P (A, D), ν(d) 0 (I + A(t) dt) = lim ν(d) 0 P (A, D), Exmple 4.3.3. Recll tht for every A L(X) we define the exponentil e A L(X) by e A A n = n!, n=0 where A 0 = I is the identity opertor, A 1 = A nd A n+1 = A n A for every n N. It cn be proved (see Theorem 5.5.11) tht (I + A dt) = (I + A dt) = e A(b ) (the specil cse for integrl opertors with constnt kernel follows lso from Eqution (4.2.6)). We stte the next theorem without proof; more generl version will be proved in Chpter 5 (see Theorem 5.6.2). Theorem 4.3.4. Let A : [, b] L(X) be continuous function. Then the functions t Y (t) = (I + A(u) du), 109

stisfy for every t [, b]. t Z(t) = (I + A(u) du) d Y (t) = A(t), dt Z(t) d dt = A(t) Definition 4.3.2 represents strightforwrd generliztion of the Riemnn product integrl s defined by Volterr. In the following chpter we provide n even more generl definition pplicble to functions A : [, b] X, where X is Bnch lgebr. It is lso possible to proceed in different wy nd try to generlize the Lebesgue product integrl from Chpter 3; the following prgrphs outline this possibility. Definition 4.3.5. A function A : [, b] L(X) is clled step function if there exists prtition = t 0 < t 1 < < t m = b nd opertors A 1,..., A m L(X) such tht A(t) = A i for t (t i 1, t i ), i = 1,..., m. For such step function we define 1 (I + A(t) dt) = e Ai(ti ti 1), i=m m (I + A(t) dt) = e Ai(ti ti 1). Definition 4.3.6. A function A : [, b] L(X) is clled product integrble if there exists sequence of step functions A n : [, b] L(X), n N such tht i=1 b lim (L) A n (t) A(t) dt = 0 n (where (L) denotes the Lebesgue integrl of rel function). We then define (I + A(t) dt) = lim (I + A n (t) dt), n (I + A(t) dt) = lim n (I + A n(t) dt). The method from Chpter 3 cn be gin used to show tht the vlue of product integrl does not depend on the choice of prticulr sequence of step functions {A n } n=1. The bove defined integrl is clled the Lebesgue product integrl or the Bochner product integrl; the clss of integrble functions is lrger s compred to Definition 4.3.2. More informtion bout this type of product integrl cn be found in [DF, Sch1]. 110

Chpter 5 Product integrtion in Bnch lgebrs A finl tretment of Riemnn product integrtion is given in the rticle [Ms] by Pesi Rustom Msni; it ws published in 1947. Consider mtrix-vlued function f : [, b] R n n nd recll tht Volterr defined the product integrl of f s the limit of products 1 P (f, D) = (I + f(ξ k ) x k ) k=m corresponding to tgged prtitions D of intervl [, b]. This definition is lso pplicble to opertor-vlued functions f : [, b] L(X), where L(X) is the spce of ll bounded liner opertors on Bnch spce X. It is just sufficient to replce multipliction by composition of opertors in the definition of P (f, D); the role of identity mtrix is now plyed by the identity opertor I. Msni s intent ws to define the product integrl of function f : [, b] X for the most generl spce X possible. Let X be normed vector spce equipped with the opertion of multipliction. Assuming there is vector 1 X such tht 1 x = x 1 = x for every x X nd 1 = 1, we let P (f, D) = 1 (1 + f(ξ k ) x k ), k=m where D is n rbitrry tgged prtition of [, b]. product integrl s the limit We would like to define the (1 + f(t) dt) = lim P (f, D). ν(d) 0 To obtin resonble theory it is necessry tht the spce X is complete, i.e. it is Bnch spce. Before giving n overview of Msni s result let s strt with short biogrphy (see lso [PRM, IMS]). Pesi Rustom Msni ws born in Bomby, 1919. He obtined his doctorl degree t Hrvrd in 1946; the thesis concerned product integrtion in Bnch lgebrs nd it ws supervised by Grrett Birkhoff 1. During the yers 1948 58 Msni held the chirs of professor of mthemtics nd science resercher in Bomby nd then he returned to the United Sttes. In the 1970 s he ccepted the position t the University of Pittsburgh. Msni ws ctive in mthemtics even fter his retirement in 1989. He died in Pittsburgh on the 15th October 1999. 1 G. Birkhoff lso devoted himself to product integrtion, see [GB]. 111

Pesi R. Msni 1 Msni contributed to the development of integrtion theory, functionl nlysis, theory of probbility nd mthemticl sttistics. The ppendix in [DF] written by Msni lso concerns product integrtion. He collborted with Norbert Wiener nd edited his collected works fter Wiener s deth. Msni ws lso interested in history, philosophy, theology nd politics. 5.1 Riemnn-Grves integrl We begin with brief recpitultion of fcts concerning integrtion of vector-vlued functions (see lso [Ms]). The notion of Grves integrl is direct generliztion of Riemnn integrl nd ws presented by Lwrence M. Grves in 1927. Let X be Bnch spce, f : [, b] X. To every tgged prtition D : = t 0 < t 1 < < t m = b of intervl [, b] with tgs ξ i [t i 1, t i ], i = 1,..., m we ssign the sum m S(f, D) = f(ξ i ) t i, i=1 where t i = t i t i 1. We recll tht if T (D) X is vector dependent on the choice of tgged prtition D, then lim T (D) = T ν(d) 0 mens tht to every ε > 0 there is δ > 0 such tht T (D) T < ε for every prtition D of [, b] such tht ν(d) < δ. Definition 5.1.1. A function f : [, b] X is clled integrble if lim ν(d) 0 S(f, D) = S f 1 Photo from http://www.york.c.uk/depts/mths/histstt/people/ 112

for some S f X. We spek of Riemnn-Grves integrl of function f on [, b] nd denote S f = b f(t) dt. The following theorem provides dditionl two equivlent chrcteriztions of integrble functions; recll tht the nottion D D mens tht the prtition D is refinement of prtition D (see Definition 3.1.8). Theorem 5.1.2. Let f : [, b] X. The following sttements re equivlent: 1) f is integrble nd b f(t) dt = S f. 2) Every sequence of prtitions {D n } n=1 of [, b] such tht ν(d n ) 0 stisfies lim n S(f, D n ) = S f. 3) For every ε > 0 there is prtition D ε of [, b] such tht S(f, D) S f < ε for every D D ε. The proof proceeds in the sme wy s in the cse when f is rel function. The rest of this section summrizes the bsic results concerning the Riemnn-Grves integrl; gin, the proofs cn be crried out in the clssicl wy. Theorem 5.1.3. Let f : [, b] X. Then the following sttements re equivlent: 1) f is integrble. 2) For every ε > 0 there exists δ > 0 such tht S(f, D 1 ) S(f, D 2 ) < ε whenever D 1 nd D 2 re tgged prtitions of [, b] stisfying ν(d 1 ) < δ, ν(d 2 ) < δ. Theorem 5.1.4. If f : [, b] X is n integrble function, then it is bounded nd b f(t) dt (b ) sup f(t). t [,b] Theorem 5.1.5. Let f : [, b] X. If the integrl b f(t) dt exists nd if [c, d] [, b], then the integrl d f(t) dt exists s well. c Theorem 5.1.6. Let f : [, c] X, < b < c. Suppose tht the integrls b f(t) dt nd c b f(t) dt exists. Then the integrl c f(t) dt lso exists nd c f(t) dt = b f(t) dt + c b f(t) dt. The following two sttements generlize the fundmentl theorem of clculus to the cse of vector-vlued functions f : [, b] X. The derivtive of such function is of course defined s f f(x) f(x 0 ) (x 0 ) = lim x x 0 x x 0 provided the limit exists (in the cse when x 0 is one of the boundry points of [, b] we require only existence of the corresponding one-sided limit). Since X is normed spce, the lst eqution mens tht lim f(x) f(x 0 ) x x 0 f (x 0 ) x x 0 = 0. 113

Theorem 5.1.7. Let f : [, b] X be integrble nd put F (x) = x f(t) dt. If f is continuous t x 0 [, b], then F (x 0 ) = f(x 0 ). Theorem 5.1.8. Let F : [, b] X nd F (t) = f(t) for every t [, b]. If f is n integrble function, then b f(t) dt = F (b) F (). Theorem 5.1.9. Let f, g : [, b] X be integrble functions, α, β R. Then b (αf(t) + βg(t)) dt = α b f(t) dt + β b g(t) dt. The set of ll integrble functions is thus vector spce; it is interesting to note tht if the spce X is equipped with the opertion of multipliction (i.e. it is Bnch lgebr, see the next section), then product of two integrble functions need not be n integrble function. Another surprising fct concerning the Riemnn-Grves integrl is tht every bounded function which is lmost everywhere continuous is lso integrble, but the converse sttement is no longer true (it holds only in finitedimensionl spces X). 5.2 Definition of product integrl Msni turns his ttention to the product nlogy of the Riemnn-Grves integrl. In the sequel we ssume tht X is Bnch lgebr (Msni uses the term normed ring), i.e. tht 1) X is Bnch spce, 2) X is n ssocitive lgebr with unit vector 1 X, 1 = 1, 3) x y x y for every x, y X. The second condition mens tht for every pir x, y X the product x y X is defined, tht the multipliction is ssocitive nd tht there exists vector 1 X such tht 1 x = x 1 = x for every x X nd 1 = 1; we use the sme symbol 1 to denote the unit vector of X s well s the number 1 R; the mening should be lwys cler from the context. Let f : [, b] X. To every prtition D of [, b] we ssign the product P (f, D) = 1 (1 + f(ξ i ) t i ) = (1 + f(ξ m ) t m ) (1 + f(ξ 1 ) t 1 ). i=m Definition 5.2.1. A function f : [, b] X is clled product integrble if there is vector P f X such tht for every ε > 0 there exists prtition D ε of [, b] such tht P (f, D) P f < ε 114

whenever D D ε. The vector P f is clled the (left) product integrl of f nd we use the nottion b (1 + f(t) dt) = P f. Remrk 5.2.2. Msni lso defines the right product integrl s the limit of the products P (f, D) = m (1 + f(ξ i ) t i ) = (1 + f(ξ 1 ) t 1 ) (1 + f(ξ m ) t m ), i=1 which re obtined by reversing the order of fctors in P (f, D). Msni uses the symbols b {{ b }} (1 + f(t) dt), (1 + f(t) dt) to denote left nd right product integrls. As he remrks, it is sufficient to study either the left integrl or the right integrl, respectively. This is becuse the following principle of dulity holds: To every Bnch lgebr X there is dul lgebr X which is identicl with X except the opertion of multipliction: We define the product x y in X s the vector y x, where the lst multipliction is crried out in X. Every sttement C bout Bnch lgebr X hs corresponding dul sttement C, which is obtined by reversing the order of ll products in C. Consequently, every occurence of the term left product integrl must be replced by right product integrl nd vice vers. A dul sttement C is true in X if nd only if C is true in X. In cse C is true in every Bnch lgebr, the sme cn be sid of C. Theorem 5.2.3. 1 Let f : [, b] X be bounded function. The following sttements re equivlent: 1) f is product integrble nd b (1 + f(t) dt) = P f. 2) Every sequence of prtitions {D n } n=1 of intervl [, b] such tht ν(d n ) 0 stisfies lim n P (f, D n ) = P f. 3) lim ν(d) 0 P (f, D) = P f. Proof. The equivlence of sttements 2) nd 3) is proved in the sme wy s in the cse of ordinry integrl. Assume tht 3) holds, i.e. to every ε > 0 there exists δ > 0 such tht P (f, D) P f < ε for every prtition D of intervl [, b] which stisfies ν(d) < δ. Let D ε be such prtition. Then for every D D ε we hve ν(d) ν(d ε ) < δ, nd therefore P (f, D) P f < ε; thus we hve proved the impliction 3) 1). Msni gives only brief indiction of the proof of 1) 3), detils re left to the reder; boundedness of f is importnt here. The following theorem represents Cuchy condition for the existence of product integrl. Theorem 5.2.4. Let f : [, b] X be bounded. The following sttements re equivlent: 1 [Ms], p. 157 159 115

1) f is product integrble. 2) To every ε > 0 there is prtition D ε such tht P (f, D) P (f, D ε ) < ε whenever D D ε. 3) To every ε > 0 there exists δ > 0 such tht P (f, D 1 ) P (f, D 2 ) < ε whenever D 1, D 2 re prtitions of [, b] stisfying ν(d 1 ) < δ, ν(d 2 ) < δ. Proof. The equivlence of sttements 1) nd 2) is proved in the sme wy s in the cse of ordinry integrl. The sttement 3) is clerly equivlent to the sttement 3) of the previous theorem. 5.3 Useful inequlities We now present five inequlities which will be useful lter. Msni didn t prove the first three; we hve however met the first two in Chpter 3 see the Lemms 3.1.3 nd 3.4.2. Although we hve proved them only for mtrices, the proofs re vlid even for elements of n rbitrry Bnch lgebr X. Lemm 5.3.1. 1 Let x k X for k = 1,..., m. Then ( m m ) (1 + x k ) exp x k. Lemm 5.3.2. 2 Let x k, y k X for k = 1,..., m. Then ( m m m ) ( ) (1 + x k ) (1 + y k ) exp m x k exp x k y k 1. Lemm 5.3.3. 3 Let x k X for k = 1,..., m. Then ( m m ) (1 + x k ) 1 exp x k 1. Proof. Elementry clcultion yields m (1 + x k ) 1 = m m j=1 1 [Ms], p. 153 2 [Ms], p. 154 3 [Ms], p. 153 1 i 1< <i j m j=1 1 i 1< <i j m m x i1 x ij 1 j! j=1 x i1 x ij m i 1,...,i j=1 x i1 x ij = 116

( m m ) 1 = j! ( x 1 + + x m ) j exp x k 1. j=1 Lemm 5.3.4. 1 Let x k X for k = 1,..., m. Then ( ) ( ) m m m m (1 + x k ) 1 + x k exp x k 1 x k. Proof. The sttement is simple consequence of the inequlity ( ) m m m m (1 + x k ) 1 + x k = (1 + x j ) 1 j=k+1 nd Lemm 5.3.3. Lemm 5.3.5. 2 Let m, n N, u, v, x j, y k X, x j, y k 1/2 for every j = 1,..., m nd k = 1,..., n. Then m n m n (1 + x j ) (u v) (1 + y k ) exp 2 x j + y k u v. j=1 x k j=1 Proof. Define f(t) = e 2t te 2t 1. Then nd therefore We get Now let x, w X, x 1/2. Then which implies f (t) = e 2t (1 2t) 0, t [0, 1/2], e 2t te 2t 1 = f(t) f(0) = 0, t [0, 1/2]. 1 t e 2t, t [0, 1/2]. w w + x w + x w (1 + x) w + x w, (1 + x) w w (1 x ) w exp( 2 x ). (5.3.1) For y X, y 1/2 we obtin in similr wy 1 [Ms], p. 153 2 [Ms], p. 152 153 w w + w y + w y w (1 + y) + y w, 117

w (1 + y) w (1 y ) w exp( 2 y ). (5.3.2) To complete the proof it is sufficient to use m times the Inequlity (5.3.1) nd n times the Inequlity (5.3.2). 5.4 Properties of product integrl This section summrizes the bsic properties of product integrble functions. We first prove tht every product integrble function is necessrily bounded. Lemm 5.4.1. 1 To every : [, b] (0, ) there exists tgged prtition D : = t 0 < t 1 < < t m = b, ξ i [t i 1, t i ], such tht t i t i 1 (ξ i ). Proof. The system of intervls {(t (t)/2, t + (t)/2), t [, b]} forms n open covering of [, b] nd the result follows from the Heine Borel theorem. It is lso simple consequence of Cousin s lemm (see [Sch2], p. 55 or [RG], Lemm 9.2). Theorem 5.4.2. 2 Every product integrble function f is bounded nd ( ) (1 + f(t) dt) exp (b ) sup f(t). t [,b] Proof. Assume tht f is not bounded. Choose N N nd δ > 0. Define { min(δ, (2 f(x) ) (x) = 1 ) if f(x) > 0, δ if f(x) = 0. According to Lemm 5.4.1 there exists tgged prtition D : = t 0 < t 1 < < t m = b, ξ i [t i 1, t i ], such tht t i t i 1 (ξ i ). (5.4.1) Clerly ν(d) δ. Since f is not bounded, we cn find sequence of points {x n } n=1 from [, b] such tht x n x [, b] nd f(x n ) n. There must be point y {x n } n=1, which lies in the sme intervl [t j 1, t j ] s the point x nd such tht ( ) 1 f(y) f(x) f(y) f(x) N exp( m) min (t i t i 1 ). 1 i m Let D 1 nd D 2 be tgged prtitions tht re obtined from D by replcing the tg ξ j by x nd y, respectively. Then, ccording to Lemm 5.3.5 nd Inequlity (5.4.1), P (f, D 1 ) P (f, D 2 ) exp 2 f(ξ i ) (t i t i 1 ) f(x) f(y) (t j t j 1 ) i j 1 [Ms], p. 162 2 [Ms], p. 163 118

exp( m) f(x) f(y) (t j t j 1 ) N. Since ν(d 1 ) = ν(d 2 ) = ν(d) δ, the number δ cn be rbitrrily smll nd N rbitrrily lrge, we rrive t contrdiction with Theorem 5.2.4. The second prt of the theorem is esily proved using Lemm 5.3.1, which gurntees tht ( m ) ( ) P (f, D) exp f(ξ i ) (t i t i 1 ) exp (b ) sup f(t) t [,b] i=1 for every tgged prtition D of [, b]. Theorem 5.4.3. 1 Assume tht b (1 + f(t) dt) exists. (1 + f(t) dt) exists s well. d c If [c, d] [, b], then Proof. Denote M = sup t [,b] f(t) <. Let D 1, D 2 be tgged prtitions of [c, d], D A tgged prtition of [, c] stisfying ν(d A ) < 1/(2M) nd D B tgged prtition of [d, b] stisfying ν(d B ) < 1/(2M). Letting we obtin (using Lemm 5.3.5) therefore D 1 = D A D 1 D B, D 2 = D A D 2 D B, P (f, D 1) P (f, D 2) = P (f, D B ) (P (f, D 1 ) P (f, D 2 )) P (f, D A ) exp ( 2 D A D B f(ξ i )(t i t i 1 ) ) P (f, D 1 ) P (f, D 2 ) exp( 2M(b )) P (f, D 1 ) P (f, D 2 ), P (f, D 1 ) P (f, D 2 ) exp(2m(b )) P (f, D 1) P (f, D 2). Becuse f is product integrble, to every ε > 0 there is tgged prtition D ε of intervl [, b] such tht P (f, D ) P (f, D ε) < ε exp(2m(b )) whenever D D ε. Without loss of generlity ssume tht D ε = D A D ε D B, where D A is prtition of [, c] stisfying ν(d A ) < 1/(2M), D ε is prtition of [c, d] nd D B is prtition of [d, b] stisfying ν(d B ) < 1/(2M). If D D ε, we construct the prtition D = D A D D B. Then P (f, D) P (f, D ε ) exp(2m(b )) P (f, D ) P (f, D ε) < ε. 1 [Ms], p. 163 165 119

Theorem 5.4.4. 1 If < b < c nd the integrls b (1+f(t) dt) nd c b (1+f(t) dt) exist, then the integrl c (1 + f(t) dt) exists s well nd c (1 + f(t) dt) = c (1 + f(t) dt) b (1 + f(t) dt). Proof. Msni s proof is somewht incomplete; we present modified version. The ssumptions imply the existence of tgged prtition Dε 1 of [, b] nd tgged prtition Dε 2 of [b, c] such tht P (f, D1 ) (1 + f(t) dt) < ε, c P (f, D2 ) (1 + f(t) dt) < ε b whenever D 1 D 1 ε nd D 2 D 2 ε. Let D ε = D 1 ε D 2 ε. Then every tgged prtition D D ε cn be written s D = D 1 D 2, where D 1 D 1 ε nd D 2 D 2 ε. We hve P (f, D) = P (f, D 2 ) P (f, D 1 ) nd c P (f, D) (1 + f(t) dt) (1 + f(t) dt) b ( P (f, D2 ) P (f, D 1 ) (1 + f(t) dt)) + ( ) c + P (f, D 2 ) (1 + f(t) dt) (1 + f(t) dt) ( which completes the proof. b ) c (1 + f(t) dt) + ε ε + ε (1 + f(t) dt), b Sttements similr to Theorem 5.4.4 hve lredy ppered in the work of Volterr nd Schlesinger. Their versions re however less generl: They ssume tht f is Riemnn integrble on [, c], which implies the existence of product integrl on [, c] nd the rest of the proof is trivil. Msni on the other hnd proves tht the existence of product integrl on [, b] nd on [b, c] implies the existence of product integrl on [, c]. The sme remrk lso pplies to Theorem 5.4.3. In the following section we prove tht the product integrl exists if nd only if the function is 1 [Ms], p. 165 120

(Riemnn-Grves) integrble; the proof of this fct is nevertheless bsed on the use of Theorem 5.4.3. Lemm 5.4.5. Every x X such tht x 1 < 1 hs n inverse element nd x 1 = (1 x) n. n=0 Proof. The condition x 1 < 1 implies tht the infinite series given bove is bsolutely convergent; let x 1 be defined s the sum of tht series. If s k = k (1 x) n, n=0 then s k+1 = 1 + (1 x) s k = 1 + s k (1 x). Pssing to the limit k we obtin i.e. x 1 x = x x 1 = 1. x 1 = 1 + (1 x) x 1 = 1 + x 1 (1 x), Theorem 5.4.6. 1 If f : [, b] X is product integrble function, then b (1 + f(t) dt) is n invertible element of the Bnch lgebr X. Proof. Denote M = sup t [,b] f(t) <. Choose δ > 0 such tht exp(mδ) < 2 nd prtition D : = t 0 < t 1 < < t m = b such tht ν(d) δ. Then (1 + f(t) dt) = 1 i=m t i t i 1 (1 + f(t) dt) (5.4.2) Lemm 5.3.3 implies tht for every i = 1,..., m t i (1 + f(t) dt) 1 t i 1 exp(m(t i t i 1 )) 1 < 1, ti i.e. t i 1 (1 + f(t) dt) is (ccording to Lemm 5.4.5) n invertible element of the lgebr X. As consequence of (5.4.2) we obtin ( 1 m t i (1 + f(t) dt)) = (1 + f(t) dt) t i 1 i=1 1. 1 [Ms], p. 165 166 121

5.5 Integrble nd product integrble functions Msni now proceeds to prove n importnt theorem which sttes tht the clsses of integrble nd product integrble functions coincide. The fct tht the existence of Riemnn integrl implies the existence of product integrl ws lredy known to Volterr; the reverse impliction ppers for the first time in Msni s pper. Lemm 5.5.1. Let f : [, b] X be bounded function. For every ε > 0 there exists δ > 0 such tht if [c, d] [, b], d c < δ nd D is tgged prtition of [c, d], then P (f, D) (1 + S(f, D)) ε(d c). Proof. Denote M = sup t [,b] f(t). Choose δ > 0 such tht (exp(mδ) 1) < ε/m. Then ccording to Lemm 5.3.4 P (f, D) (1 + S(f, D)) (exp(m(d c)) 1)M(d c) ε(d c). Definition 5.5.2. Let Y X. The dimeter of the set Y is the number dim Y = sup{ y 1 y 2 ; y 1, y 2 Y }. The convex closure of Y is the set { k } k conv Y = α i y i ; k N, y i Y, α i 0, α i = 1. i=1 i=1 Theorem 5.5.3. 1 If Y X, then dim conv Y = dim Y. Proof. The proof is not difficult, lthough it s not included in Msni s pper. Since Y conv Y, it is sufficient to prove tht dim conv Y dim Y. Let y 1, y 2 conv Y, l m y 1 = α i yi 1, y 2 = β j yj 2, i=1 j=1 1 [Ms], p. 159 122

where yi 1, y2 j Y, i = 1,..., l, j = 1,..., m, l m α i = β j = 1. i=1 Then ( m l ) y 1 y 2 = β j α i yi 1 j=1 i=1 l i=1 j=1 l i=1 α i m α i β j yi 1 yj 2 j=1 m β j yj 2 = l m α i β j (yi 1 yj 2 ) l j=1 i=1 j=1 i=1 j=1 m α i β j dim Y = dim Y. Lemm 5.5.4. 1 Let f : [, b] X be product integrble function. Then for every ε > 0 there is prtition D : = t 0 < t 1 < < t m 1 < t m = b such tht 1 1 (1 + f(ξ i ) t i ) (1 + f(η i ) t i ) < ε i=m k=m for every choice of ξ i, η i [t i 1, t i ], i = 1,..., m. Proof. Follows from Theorem 5.2.4. Remrk 5.5.5. Msni notes tht the reverse impliction is not vlid; his counterexmple is 0 if x = 0, f(x) = 1/x if x (0, 1/2), 2 if x [1/2, 1]. Tking the prtition t 0 = 0 < t 1 = 1/2 < t 2 = 1 we obtin 1 (1 + f(ξ i ) t i ) = 0 i=2 for every choice of ξ i [t i 1, t i ], but f is not product integrble (becuse it is not bounded). Lemm 5.5.6. 2 Consider function f : [, b] X. Assume tht for every ε > 0 there is prtition D ε : = t 0 < t 1 < < t n 1 < t n = b such tht n n f(ξ i ) t i f(η i ) t i < ε 1 [Ms], p. 160 161 2 [Ms], p. 159 160 i=1 i=1 123

for every choice of ξ i, η i [t i 1, t i ]. Then f is n integrble function. Proof. If we introduce the nottion { n n } f([t i 1, t i ]) t i = f(ξ i ) t i ; ξ i [t i 1, t i ], i=1 i=1 then the ssumption of the lemm might be written s ( n ) dim f([t i 1, t i ]) t i < ε. i=1 To prove tht f is integrble it is sufficient to verify tht for every prtition D D ε which consists of division points t i 1 = t i 0 < t i 1 < < t i m(i) = t i, i = 1,..., n nd for every choice of ξj i [ti j 1, ti j ], η i [t i 1, t i ] we hve m(i) n n P (f, D) P (f, D ε ) = f(ξj) t i i j f(η i ) t i < ε. i=1 j=1 i=1 But m(i) n f(ξj) t i i j = i=1 j=1 m(i) n i=1 j=1 t i j t i f(ξ i j) t i conv ( n ) f([t i 1, t i ]) t i, i=1 nd the proof is completed by using Theorem 5.5.3. Theorem 5.5.7. 1 Every product integrble function f : [, b] X is integrble. Proof. We verify tht the ssumption of Theorem 5.5.6 is fulfilled. According to Lemm 5.5.1 it is possible to choose numbers = s 0 < s 1 < < s n 1 < s n = b in such wy tht P (f, D k ) (1 + S(f, D k )) ε (s k s k 1 ) 3 (b ) for every tgged prtition D k of intervl [s k 1, s k ]. Since f is product integrble on [s k 1, s k ], there exists (ccording to Lemm 5.5.4) prtition s k 1 = t k 0 < t k 1 < < t k m(k) = s k such tht 1 i=m(k) 1 [Ms], p. 167 169 (1 + f(ξ k i ) t k i ) 1 i=m(k) (1 + f(ηi k ) t k i ) < ε 3 (s k s k 1 ) (b ) 124

for every choice of ξi k, ηk i [tk i 1, tk i ]. For such ξk i, ηk i we hve m(k) m(k) m(k) f(ξi k ) t k i f(ηi k ) t k i 1 1 + f(ξi k ) t k i (1 + f(ξi k ) t k i ) + i=1 + 1 i=m(k) + 1 i=m(k) (1 + f(ξ k i ) t k i ) (1 + f(ηi k ) t k i ) 1 + i=1 i=m(k) m(k) i=m(k) 1 (1 + f(ηi k ) t k i ) + f(ηi k ) t k i < ε(s k s k 1). (b ) Adding these inequlities for k = 1,..., n nd using the tringle inequlity leds to m(k) m(k) n n f(ξi k ) t k i f(ηi k ) t k i < ε. i=1 This mens tht the prtition D cn be chosen s = t 1 0 < t 1 1 < < t 1 m(1) = t2 0 < < t n 1 m(n 1) = tn 0 < t n 1 < < t n m(n) = b. We now follow Msni s proof of the reverse impliction which sys tht every integrble function is lso product integrble nd tht the product integrl might be expressed using the Peno series. As we know, the history of the theorem cn be trced bck to Volterr (in the cse X = R n n ). Msni ws probbly the first one to give rigorous proof. We will be working with tgged prtitions D : = t 0 < t 1 < < t m(d) = b, ξ i [t i 1, t i ]. For every n m(d) we define T n (f, D) = f(ξ i1 ) f(ξ in ) t 1 t n nd m(d) i 1>i 2> >i n 1 T (f, D) = T 1 (f, D) + + T m(d) (f, D). We stte the following lemm without proof; see Remrk 2.4.4 for the proof in the finite-dimensionl cse (the difficulty in the generl cse is hidden in the fct tht the product of two integrble functions need not be integrble). Lemm 5.5.8. 1 Let f : [, b] X be n integrble function, n N. Then the limit T n (f) = lim ν(d) 0 T n (f, D) exists nd 1 [Ms], p. 174 176 T n (f) = b t1 tn 1 f(t 1 ) f(t n ) dt n dt 1. 125

Msni refers to the following lemm s to the extension of Tnnery s theorem. Lemm 5.5.9. 1 Consider function f : [, b] X nd ssume tht the following conditions re stisfied: 1) There exists T n (f) = lim ν(d) 0 T n (f, D) for every n N. 2) M n = sup D T n (f, D) < for every n N, where the supremum is tken over ll prtitions D of intervl [, b] which consist of t lest n division points. 3) The series n=1 M n is convergent. Then T (f) = lim T (f, D) = T n (f). ν(d) 0 Proof. The series T (f) = n=1 T n(f) is convergent, becuse T n (f) M n for every n N. We will prove tht T (f) = lim ν(d) 0 T (f, D). Choose ε > 0. There exists number n(ε) N such tht k=n(ε)+1 n=1 M k < ε/3. According to the first ssumption, there exists δ > 0 such tht T k (f, D) T k (f) < ε 3n(ε), k = 1,..., n(ε), for every tgged prtition D of [, b] tht stisfies ν(d) < δ. Without loss of generlity we ssume tht δ is so smll tht D consists of t lest n(ε) division points, i.e. T 1 (f, D),..., T n(ε) (f, D) re well-defined. Now for every tgged prtition D tht stisfies ν(d) < δ we estimte m(d) n(ε) T (f, D) T (f) = T k (f, D) T k (f) T k (f, D) T k (f) + + m(d) k=n(ε)+1 T k (f, D) + k=n(ε)+1 T k (f) < n(ε) ε 3n(ε) + 2 k=n(ε)+1 M k < ε. Theorem 5.5.10. 2 Let f : [, b] X be n integrble function. Then f is lso product integrble nd 1 [Ms], p. 189 191 2 [Ms], p. 176 177 (1 + f(t) dt) = 1 + T n (f). n=1 126

Proof. Denote M = sup t [,b] f(t) <. For every prtition D of intervl [, b] we hve P (f, D) = 1 + T (f, D), (1 + f(t) dt) = 1 + lim T (f, D), ν(d) 0 T n (f, D) (b )n M n, n! (b ) n M n = exp(m(b )) 1 <. n! n=1 The sttement of the theorem is therefore consequence of the preceding two lemms. We hve proved tht function is product integrble if nd only if it is integrble. Thus, in the rest of this chpter we use the terms integrble nd product integrble s synonyms. Theorem 5.5.11. 1 Let f : [, b] X be n integrble function. Suppose tht f(x) f(y) = f(y) f(x) for ech pir x, y X. Then ( ) b (1 + f(t) dt) = exp f(t) dt. Proof. A simple consequence of Theorem 5.5.10 nd the equlity b t1 (see Lemm 2.4.2). ( tn 1 ) n f(t 1 ) f(t n ) dt n dt 1 = 1 b f(t) dt n! 5.6 Additionl properties of product integrl This section is devoted to Msni s versions of the fundmentl theorem of clculus, the uniform convergence theorem, nd the chnge of vribles theorem. Theorem 5.6.1. 2 Let f : [, b] X be n integrble function. Denote Y (x) = x (1 + f(t) dt), x [, b]. 1 [Ms], p. 179 2 [Ms], p. 178 127

Then Y (x) = 1 + x f(t)y (t) dt, x [, b]. (5.6.1) Proof. Using Theorem 5.5.10 we obtin Since Y (t) = 1 + t t1 t f(t 1 ) dt 1 + tn 1 t t1 f(t 1 )f(t 2 ) dt 2 dt 1 +. (5.6.2) f(t 1 ) f(t n ) dt n dt 1 (b )n M n, n! the series (5.6.2) is uniformly convergent. Becuse f is bounded, the series f(t)y (t) = f(t) + t f(t)f(t 1 ) dt 1 + x t1 f(t)f(t 1 )f(t 2 ) dt 2 dt 1 + is lso uniformly convergent nd might be integrted term by term on [, x]; performing this step leds to Eqution (5.6.1). Corollry 5.6.2. 1 If f : [, b] X is continuous function, then for every x [, b]. Y (x)y (x) 1 = f(x) The previous corollry represents n nlogy of the formul d dx x f(t) dt = f(x) (see Theorem 5.1.7). Also the Newton-Leibniz formul b f (x) dt = f(b) f() (see Theorem 5.1.8) hs the following product nlogy (whose proof we omit). Theorem 5.6.3. 2 Assume tht Z : [, b] X stisfies Z (x)z(x) 1 = f(x) for every x [, b]. Then (1 + f(t) dt) = Z(b)Z() 1 provided the function f is integrble. 1 [Ms], p. 181 2 [Ms], p. 182 128

The next theorem estblishes criterion for interchnging the order of limit nd product integrl, i.e. for the formul lim n (1 + f n (t) dt) = (1 + lim f n(t) dt). n We hve lredy encountered such criterion in Chpter 3 when discussing the Lebesgue product integrl; Schlesinger s sttement represented in fct product nlogy of the Lebesgue dominted convergence theorem. Msni s theorem concerns the Riemnn product integrl nd requires uniform convergence to perform the interchnge of limit nd integrl. Theorem 5.6.4. 1 Let {f n } n=1 be sequence of integrble functions which converge uniformly to function f on intervl [, b]. Then (1 + f(t) dt) = lim (1 + f n (t) dt). n Proof. The existence of b (1+f(t) dt) follows from the fct tht the limit of uniformly convergent sequence of integrble functions is gin n integrble function. For n rbitrry tgged prtition D we cn use Lemm 5.3.2 to estimte ( ( ) ) P (f, D) P (f n, D) exp(m(b )) exp f(ξ i ) f n (ξ i ) t i 1, where M = sup t [,b] f(t). Choose ε > 0 nd find corresponding ε 0 > 0 such tht exp(m(b )) (exp(ε 0 (b )) 1) < ε/3. Let n 0 N be such tht f(t) f n (t) < ε 0 for every t [, b] nd n n 0. The prtition D cn be chosen so tht the inequlities P (f, D) (1 + f(t) dt) < ε/3, hold. Then for every n n 0 we hve i P (f n, D) (1 + f n (t) dt) < ε/3 (1 + f n (t) dt) (1 + f(t) dt) (1 + f(t) dt) P (f, D) + + P (f, D) P (f n, D) + P (f n, D) (1 + f n (t) dt) < ε. 1 [Ms], p. 171 129

Msni lso proved generlized version of the chnge of vribles theorem for the product integrl (compre to Theorem 2.5.10); we stte it without proof. Theorem 5.6.5. 1 Let f : [, b] X be n integrble function, ϕ : [α, β] [, b] incresing, ϕ(α) =, ϕ(β) = b. If ϕ exists nd is integrble on [, b], then (1 + f(t) dt) = β (1 + f(ϕ(u))ϕ (u) du). α 1 [Ms], p. 187 188 130

Chpter 6 Kurzweil nd McShne product integrls The introduction of Lebesgue integrtion signified revolution in mthemticl nlysis: Every Riemnn integrble function is lso Lebesgue integrble, but the clss of functions hving Lebesgue integrl is considerbly lrger. However, there exist functions f which re Newton integrble nd (N) b f(t) dt = F (b) F (), where F is n ntiderivtive of f, but the Lebesgue integrl (L) b f(t) dt does not exist. Consider for exmple the function F (x) = { x 2 sin(1/x 2 ) if x (0, 1], 0 if x = 0. This function hs derivtive F (x) = f(x) for every x [0, 1] nd f(x) = { 2x sin(1/x 2 ) (2/x) cos(1/x 2 ) if x (0, 1], 0 if x = 0. The function f is therefore Newton integrble nd (N) 1 0 f(t) dt = F (1) F (0). If we denote k = 1, b k = 1 (k + 1/2)π kπ for every k N, then bk k which implies tht f(t) dt 1 0 bk k f(t) dt = F (b 1 k) F ( k ) = (k + 1/2)π, f(t) dt 1 (k + 1/2)π =. The Lebesgue integrl (L) 1 f(t) dt therefore does not exist. 0 Jroslv Kurzweil (nd lter independently Rlph Henstock) introduced new definition of integrl which voids the bove mentioned drwbck of the Lebesgue 131

integrl. The Kurzweil integrl (lso known s the guge integrl or the Henstock- Kurzweil integrl) encompsses the Newton nd Lebesgue (nd consequently lso Riemnn) integrls. Another benefit is tht the definition of Kurzweil integrl is obtined by gentle modifiction of Riemnn s definition nd is considerbly simpler thn Lebesgue s definition. Jroslv Kurzweil 1 Edwrd J. McShne 2 The definition of integrl due to E. J. McShne is similr to Kurzweil s definition nd in fct represents n equivlent definition of Lebesgue integrl. In this chpter we first summrize the definitions of Kurzweil nd McShne integrls; in the second prt we turn our ttention to product nlogies of these integrls. The proofs in this chpter re often omitted nd my be found in the originl ppers (the references re included). 6.1 Kurzweil nd McShne integrls A finite collection of point-intervl pirs D = {([t i 1, t i ], ξ i )} m i=1 prtition of intervl [, b] if is clled n M- = t 0 < t 1 < < t m = b, ξ i [, b], i = 1,..., m. A K-prtition is M-prtition which moreover stisfies ξ i [t i 1, t i ], i = 1,..., m. Given function : [, b] (0, ) (the so-clled guge on [, b]), prtition D is clled -fine if 1 Photo tken by Š. Schwbik 2 Photo from [McT] [t i 1, t i ] (ξ i (ξ i ), ξ i + (ξ i )), i = 1,..., m. 132

For given function f on [, b] nd M-prtition D of [, b] denote S(f, D) = m f(ξ i )(t i t i 1 ). i=1 Definition 6.1.1. Let X be Bnch spce. A vector S f X is clled the Kurzweil (McShne) integrl of function f : [, b] X if for every ε > 0 there is guge : [, b] (0, ) such tht S(f, D) S f < ε for every -fine K-prtition (M-prtition) D of intervl [, b]. We define (K) b f(t) dt = S f or (M) b f(t) dt = S f, respectively. We stte the following theorems without proofs; they cn be found (together with more informtion bout the Kurzweil, McShne nd Bochner integrls) in the book [SY]; other good sources re [Sch2, RG]. Theorem 6.1.2. Let X be Bnch spce. Then every McShne integrble function f : [, b] X is lso Kurzweil integrble (but not vice vers) nd (K) b f(t) dt = (M) b f(t) dt. Theorem 6.1.3. Let X be Bnch spce. Then every Lebesgue (Bochner) integrble function f : [, b] X is lso McShne integrble nd (M) b f(t) dt = (L) b f(t) dt. The converse sttement holds if nd only if X is finite-dimensionl spce. 6.2 Product integrls nd their properties We now proceed to the definitions of Kurzweil nd McShne product integrls. The definition of Kurzweil product integrl ppered for the first time in the pper [JK]; the uthors spek bout the Perron product integrl nd use the nottion (P P ) b (I + A(t) dt). The McShne product integrl ws studied in [Sch1, SS]. 133

For the ske of simplicity we confine our exposition only to mtrix functions A : [, b] R n n insted of working with opertor-vlued functions A : [, b] L(X) or even with functions A : [, b] X with vlues in Bnch lgebr X. For n rbitrry M-prtition D of [, b] nd mtrix function A : [, b] R n n denote P (A, D) = 1 (I + A(ξ i )(t i t i 1 )). i=m Definition 6.2.1. Consider function A : [, b] R n n. A mtrix P A R n n is clled the Kurzweil (McShne) product integrl of A if for every ε > 0 there is guge : [, b] (0, ) such tht P (A, D) P A < ε for every -fine K-prtition (M-prtition) D of intervl [, b]. We define (K) (I + A(t) dt) = P A, or (M) (I + A(t) dt) = P A, respectively. We lso denote { KP ([, b], R n n ) = A : [, b] R n n ; (K) { MP ([, b], R n n ) = A : [, b] R n n ; (M) } (I + A(t) dt) exists, } (I + A(t) dt) exists. The right product integrls cn be introduced using the products m P (A, D) = (I + A(ξ i )(t i t i 1 )), i=1 but we limit our discussion to the left integrls. Exmple 6.2.2. Assume tht the Riemnn product integrl (R) b (I + A(t) dt) exists, i.e. for every ε > 0 we cn find δ > 0 such tht P (A, D) (I + A(t) dt) < ε for every prtition D of [, b] which such tht ν(d) < δ. If we put (x) = δ 2, x [, b], 134

then every -fine K-prtition D of [, b] stisfies ν(d) < δ. This mens tht the Kurzweil product integrl of A exists nd (K) (I + A(t) dt) = (R) (I + A(t) dt). Exmple 6.2.3. Consider the function f(x) = It cn be proved (see [SS]) tht { 1/x if x (0, 1], 0 if x = 0. (M) 1 (1 + f(x) dx) = 0. 0 It is worth noting tht neither the Riemnn integrl (R) 1 0 (1 + f(x) dx) nor the Lebesgue integrl (L) 1 0 (1 + f(x) dx) exist; this follows e.g. from Theorem 6.2.10. Theorem 6.2.4. 1 Consider function A : [, b] R n n. Then the following conditions re equivlent: 1) The integrl (K) b (I + A(t) dt) exists nd is invertible. 2) There exists n invertible mtrix P A such tht for every ε > 0 there is guge : [, b] (0, ) such tht 1 e A(ξi)(ti ti 1) P A < ε i=m whenever D = {([t i 1, t i ], ξ i )} m i=1 is -fine K-prtition of [, b]. If one of these conditions is fulfilled, then (K) (I + A(t) dt) = P A. A similr sttement holds lso for McShne product integrl. Theorem 6.2.5. Consider function A : [, b] R. The integrl (K) b A(t) dt exists if nd only if the integrl (K) b (1 + A(t) dt) exists nd is different from zero. In this cse the following equlity holds: (K) 1 [JK], p. 651, nd [SS] ( (1 + A(t) dt) = exp (K) b A(t) dt ). 135

A similr sttement holds lso for McShne product integrl. Proof. Assume tht (K) b A(t) dt = S A exists nd choose ε > 0. Since the exponentil function is continuous t the point S A, there is δ > 0 such tht e x e SA < ε, x (SA δ, S A + δ). (6.2.1) Let : [, b] (0, ) be guge such tht S(A, D) S A < δ for every -fine K-prtition of intervl [, b]. Ech of these prtitions stisfies 1 e A(ξi)(ti ti 1) e SA = e S(A,D) e SA < ε i=m nd using Theorem 6.2.4 we obtin (K) ( (1 + A(t) dt) = exp (K) b A(t) dt The reverse impliction is proved in similr wy using the equlity ( 1 ) S(A, D) = log e A(ξi)(ti ti 1). i=m ). Remrk 6.2.6. The previous theorem no longer holds for mtrix functions A : [, b] R n n. Jroslv Kurzweil nd Jiří Jrník constructed 1 two functions A, B : [ 1, 1] R 2 2 such tht 1 (K) A(t) dt exists, 1 1 (K) B(t) dt doesn t exist, 1 1 (K) (I + A(t) dt) doesn t exist, 1 1 (K) (I + B(t) dt) exists. 1 Theorem 6.2.7. Every McShne product integrble function is lso Kurzweil product integrble (but not vice vers) nd (K) (I + A(t) dt) = (M) (I + A(t) dt). 1 [JK], p. 658 136

Proof. The inclusion M P KP follows from the fct tht every K-prtition is lso M-prtition; the equlity of the two product integrls is then obvious. We only hve to prove tht MP KP. For n rbitrry function f : [, b] R denote A f (t) = I f(t), where I is the identity mtrix of order n (A f is therefore mtrix-vlued function on [, b]). Then evidently P (A, D) = I P (f, D) for every prtition D of [, b] nd A f is product integrble (in the Kurzweil or McShne sense) if nd only if f is product integrble. Theorem 6.1.2 gurntees the existence of function f : [, b] R tht is Kurzweil integrble, but not McShne integrble; then (ccording to Theorem 6.2.5) the corresponding function A f : [, b] R n n is Kurzweil product integrble, but not McShne product integrble. Theorem 6.2.8. 1 Consider function A : [, b] R n n. Suppose tht the integrl (M) b (I + A(t) dt) exists nd is invertible. Then for every x (, b) the integrl Y (x) = (M) x (I + A(t) dt) exists s well nd the function Y stisfies lmost everywhere on [, b]. Y (x) = A(x)Y (x) Remrk 6.2.9. In Chpter 3 we hve defined the Lebesgue (or Bochner) product integrl (L) b (I+A(t) dt); the definition ws bsed on the pproximtion of A by sequence of step functions which converge to A in the norm of spce L([, b], R n n ). The following theorem describes the reltionship between McShne nd Lebesgue product integrls. Theorem 6.2.10. 2 Consider function A : [, b] R n n. The following conditions re equivlent: 1) A is Lebesgue (Bochner) integrble. 2) The McShne product integrl (M) b (I + A(t) dt) exists nd is invertible. If one of these conditions is fulfilled, then (M) (I + A(t) dt) = (L) (I + A(t) dt). 1 [JK], p. 652 656 nd [SS] 2 [Sch1], p. 329 334 nd [SS] 137

Remrk 6.2.11. We conclude this chpter by compring the clsses of functions which re integrble ccording to different definitions presented in the previous text. Let R, L, M nd K be the clsses of ll functions A : [, b] R n n which re integrble in the sense of Riemnn, Lebesgue, McShne nd Kurzweil, respectively. In similr wy let RP nd LP denote the clsses of Riemnn product integrble nd Lebesgue product integrble functions. Insted of working with the clsses KP nd MP it is more convenient to concentrte on the clsses { } KP = A : [, b] R n n ; (K) (I + A(t) dt) exists nd is invertible, MP = { A : [, b] R n n ; (M) } (I + A(t) dt) exists nd is invertible. The following digrm shows the inclusions between the bove mentioned clsses. R L = M K = = = RP LP = MP KP 138

Chpter 7 Complements This finl chpter contins dditionl remrks on product integrtion theory. The topics discussed here complement the previous chpters; however, most proofs re omitted nd the text is intended only to rouse reder s interest (references to other works re included). 7.1 Vrition of constnts Product integrl enbles us to express solution of the differentil eqution y (x) = A(x)y(x), x [, b], where A : [, b] R n n, y : [, b] R n. The fundmentl mtrix of this system is z x 1(x) 1 zn(x) 1 Z(x) = (I + A(t) dt) =..... z1 n (x) zn(x) n nd its columns z i (x) = zi 1 (x). z n i (x) thus provide fundmentl system of solutions. We now focus our ttention to the inhomogeneous eqution, i = 1,..., n (7.1.1) y (x) = A(x)y(x) + f(x), y() = y 0. x [, b], (7.1.2) A method for solving this system using product integrl (bsed on the well-known method of vrition of constnts) ws first proposed by G. Rsch in the pper [GR]; it cn be lso found in the monogrph [DF]. We ssume tht the functions A : [, b] R n n nd f : [, b] R n re continuous, nd we try to find the solution of (7.1.2) in the form y(x) = n z i (x)c i (x), (7.1.3) i=1 where c i : [, b] R, i = 1,..., n re certin unknown functions. If we denote c 1 (x) c(x) =. c n (x), 139

then the equtions (7.1.1) nd (7.1.3) imply We obtin y(x) = Z(x)c(x). y (x) = Z (x)c(x)+z(x)c (x) = A(x)Z(x)c(x)+Z(x)c (x) = A(x)y(x)+Z(x)c (x), nd using Eqution (7.1.2) Consequently which implies f(x) = Z(x)c (x). c (x) = Z(x) 1 f(x), c() = Z() 1 y() = y 0, c(x) = y 0 + x Z(t) 1 f(t) dt. The solution of the system (7.1.2) is thus given by the explicit formul y(x) = Z(x)c(x) = Z(x)y 0 + Z(x) x x = (I + A(t) dt)y 0 + (I + A(t) dt) = x (I + A(t) dt)y 0 + x x x Z(t) 1 f(t) dt = ( ) (I + A(s) ds)f(t) dt = t ( x ) (I + A(s) ds)f(t) dt. We summrize the result: The solution of the inhomogeneous system (7.1.2) hs the form n y(x) = z i (x)c i (x), i=1 where z 1,..., z n : [, b] R n is the fundmentl system of solutions of the corresponding homogeneous eqution, the functions c i : [, b] R, i = 1,..., n re continuously differentible nd stisfy t n c i(x)z i (x) = f(x), i=1 x [, b]. 7.2 Equivlent definitions of product integrl Consider tgged prtition D : = t 0 < t 1 < < t m = b, ξ i [t i 1, t i ], i = 1,..., m. Ludwig Schlesinger proved (see Theorem 3.2.2) tht the product 140

integrl of Riemnn integrble function A : [, b] R n n cn be clculted not only s ( 1 ) (I + A(t) dt) = lim (I + A(ξ k ) t k ), ν(d) 0 but lso s k=m ( 1 ) (I + A(t) dt) = lim e A(ξk) tk. ν(d) 0 The equivlence of these definitions cn be intuitively explined using the fct tht k=m e A(ξk) tk = 1 + A(ξ k ) t k + O(( t k ) 2 ), nd the terms of order ( t k ) 2 nd higher do not chnge the vlue of the integrl. We hve lso encountered similr theorem pplicble to the Kurzweil nd McShne integrls (see Theorem 6.2.4). We now proceed to more generl theorem concerning equivlent definitions of product integrl, which ws given in [DF]. Definition 7.2.1. A function f(z) = c k z k (7.2.1) k=0 is clled dmissible, if the series (7.2.1) hs positive rdius of convergence r > 0 nd f(0) = c 0 = 1, f (0) = c 1 = 1. For exmple, the functions z exp z, z 1 + z nd z (1 z) 1 re dmissible. For every mtrix A R n n such tht A < r we put f(a) = c k A k. k=0 Theorem 7.2.2. 1 If f is n dmissible function nd A : [, b] R n n continuous mtrix function, then ( 1 ) (I + A(t) dt) = lim f(a(ξ k ) t k ). ν(d) 0 According to the previous theorem, the product integrl of function A cn be defined s the limit ( 1 ) f(a(ξ k ) t k ), 1 [DF], p. 50 53 lim ν(d) 0 k=m k=m 141

where f is n dmissible function. Product integrl defined in this wy is usully denoted by the symbol b f(a(t) dt), e.g. (I + A(t) dt), e A(t) dt, (I A(t) dt) 1 etc. The integrl b ea(t) dt is tken s primry definition in the monogrph [DF]. We note tht it is possible to prove n nlogy of Theorem 7.2.2 even for the Kurzweil nd McShne product integrls (see [JK, Sch1]). 7.3 Riemnn-Stieltjes product integrl Consider two functions f, g : [, b] R. integrl is defined s the limit Then the ordinry Riemnn-Stieltjes b f(x) dg(x) = lim ν(d) 0 m f(ξ i )(g(t i ) g(t i 1 )), (7.3.1) i=1 where D : = t 0 < t 1 < < t m = b is tgged prtition of [, b] with tgs ξ i [t i 1, t i ], i = 1,..., m (provided the limit exists). This integrl ws introduced in 1894 by Thoms Jn Stieltjes (see [Kl], Chpters 44 nd 47), who ws working with continuous functions f nd non-decresing functions g. Lter in 1909 Friedrich Riesz discovered tht the Stieltjes integrl cn be used to represent continuous liner functionls on the spce C([, b]). Also, if g(x) = x, we obtin the ordinry Riemnn integrl. Assume tht the function g is of bounded vrition, i.e. tht { m } sup g(t i ) g(t i 1 ) <, i=1 where the supremum is tken over ll prtitions = t 0 < t 1 < < t m = b of intervl [, b]. Then (see e.g. [RG]) the Riemnn-Stieltjes integrl exists for every continuous function f. In prticulr, if f is continuous nd g is step function defined s g = g 1 χ [t0,t 1) + g 2 χ [t1,t 2) + + g m 1 χ [tm 2,t m 1) + g m χ [tm 1,t m], where = t 0 < t 1 < < t m = b, g 1,..., g m R nd χ M denotes the chrcteristic function of set M, then b f(x) dg(x) = f(t 1 )(g 2 g 1 ) + + f(t m 1 )(g m g m 1 ). 142

Now consider mtrix function A : [, b] R n n. Riemnn-Stieltjes integrl cn be defined s The product nlogy of (I + da(t)) = lim ν(d) 0 i=m (see e.g. [Sch3, GJ, Gil, DN]), or even more generlly s (I + f(t)da(t)) = lim ν(d) 0 i=m 1 (I + A(t i ) A(t i 1 )) (7.3.2) 1 (I + f(ξ i )(A(t i ) A(t i 1 ))), where f : [, b] R (see the entry Product integrl in [EM]). We now present some bsic sttements concerning the Riemnn-Stieltjes product integrl (7.3.2). Product integrls of the type (7.3.2) re encountered in survivl nlysis (when working with the cumultive hzrd A(t) = t (s) ds insted of the hzrd rte 0 (t); see Exmple 1.4.1) nd in the theory of Mrkov processes (when working with cumultive intensities A ij (t) = t 0 ij(s) ds for i j nd A ii (t) = j i A ij(t) insted of the trnsition rtes ij (t); see Exmple 1.4.2). A sufficient condition for the existence of the limit (7.3.2) is gin tht the vrition of A is finite. A different sufficient condition (see [DN]) sys tht the product integrl exists provided A is continuous nd its p-vrition is finite for some p (0, 2), i.e. { m } sup A(t i ) A(t i 1 ) p <, i=1 where the supremum is gin tken over ll prtitions = t 0 < t 1 < < t m = b of intervl [, b]. If A : [, b] R n n is step function defined s A = A 1 χ [t0,t 1) + A 2 χ [t1,t 2) + + A m 1 χ [tm 2,t m 1) + A m χ [tm 1,t m], where = t 0 < t 1 < < t m = b nd A 1,..., A m R n n, then (I + da(t)) = (I + A m A m 1 ) (I + A 2 A 1 ). (7.3.3) Thus, if A k 1 A k = I for some k = 2,..., m, then (I + da(t)) = 0, i.e. the product integrl need not be regulr mtrix. Eqution (7.3.3) lso suggests tht the indefinite product integrl Y (x) = x (I + da(t)), x [, b], 143

need not be continuous function. If A : [, b] R n n is continuously differentible function, it cn be proved tht (I + da(t)) = (I + A (t) dt). As we hve seen in the previous chpters, the Riemnn product integrl provides solution of the differentil eqution or the equivlent integrl eqution y (x) = A(x)y(x), y() = y 0, y(x) y 0 = x A(t)y(t) dt. Similrly, the Riemnn-Stieltjes product integrl cn be used s tool for solving the generlized differentil eqution or the equivlent integrl eqution dy(x) = da(x)y(x), y() = y 0, y(x) y 0 = x da(t)y(t). The detils re given in the pper [Sch3]. There exists definition of product integrl (see [JK, Sch1, Sch3]) tht generlizes both the Riemnn nd Riemnn-Stieltjes product integrls: Consider mpping V tht ssigns squre mtrix of order n to every point-intervl pir (ξ, [x, y]), where [x, y] [, b] nd ξ [x, y]. We define V (t, dt) = provided the limit exists. The choice lim ν(d) 0 i=m 1 V (ξ i, [t i 1, t i ]), V (ξ, [x, y]) = I + A(ξ)(y x) leds to the Riemnn product integrl, wheres V (ξ, [x, y]) = I + A(y) A(x) gives the Riemnn-Stieltjes product integrl. We note tht it is lso possible to define the Kurzweil-Stieltjes nd McShne- Stieltjes product integrls (see [Sch3]), whose definitions re bsed on the notion of -fine M-prtitions nd K-prtitions (see Chpter 6); these integrls generlize the notion of Riemnn-Stieltjes product integrl. 144

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Previous Volumes of the Series HISTORY OF MATHEMATICS 1. J.Bečvář,E.Fuchs(ed.):HistoryofMthemticsI,1994 2. J. Bečvář et l.: Edurd Weyr(1852 1903), 1995 3. J. Bečvář, E. Fuchs(ed.): Mthemtics of the 19th Century, 1996 4. J.Bečvář,E.Fuchs(ed.):Mn Art Mthemtics,1996 5. J. Bečvář(ed.): Jn Vilém Pexider(1874 1914), 1997 6. P.Šrmnová,Š.Schwbik:ABriefGuidetotheHistory of the Integrl, 1996 7. J. Bečvář, E. Fuchs(ed.): History of Mthemtics II, 1997 8. P. Šišm: Grph Theory 1736 1963, 1997 9. K.Mčák:OntheOriginoftheProbbilityClculus,1997 10. M. Němcová: Frntišek Josef Studničk(1836 1903), 1998 11. J. Bečvář, E. Fuchs(ed.): Mthemtics throughout the Ages I, 1998 12. J.Bečvář,E.Fuchs(ed.):Mthemticsinthe16thnd17th Century, 1999 13. M. Bečvářová: On the History of the Assocition(1862 1869), 1999 14. K. Lepk: History of Fermt Quotients(Fermt Lerch), 2000 15. K. Mčák: Three Medievl Collections of Mthemticl Problems, 2001 16. J. Bečvář, E. Fuchs(ed.): Mthemtics throughout the Ages II, 2001 17. E. Fuchs(ed.): Mthemtics throughout the Ages I, 2001(Engl.) 18. K. Mčák, G. Schuppener: Mthemtics t the Jesuit Clementinum in the Yers 1600 1740, 2001 19. J. Bečvář et l.: Mthemtics in Medievl Europe, 2001 20. M. Bečvářová: Euclid s Elements, its Editions nd Trnsltions, 2002

21. P. Šišm: Mthemtics t the Brno Germn Technicl University, 2002 22. M. Hykšová: Krel Rychlík(1885 1968), 2003 23. J. Bečvář, M. Bečvářová, H. Vymzlová: Mthemtics in Ancient Egypt nd Mesopotmi, 2003 24. J. Bečvář, E. Fuchs(ed.): Mthemtics throughout the Ages III, 2004 25. E. Fuchs(ed.): Mthemtics throughout the Ages II, 2004(Engl.) 26. K. Mčák: The Development of Probbility Theory in the Czech Lnds until 1938, 2005 27. Z. Kohoutová, J. Bečvář: Vldimír Kořínek(1899 1981), 2005 28. J.Bečvář,M.Bečvářová,J.Škod:EmilWeyrndhisItlinSty in the Yer 1870/71, 2006 29. A. Slvík: Product Integrtion, its History nd Applictions, 2007(Engl.)

Antonín Slvík PRODUCT INTEGRATION, ITS HISTORY AND APPLICATIONS Nečs Center for Mthemticl Modeling, Volume 1 History of Mthemtics, Volume 29 Vydl MATFYZPRESS vydvtelství Mtemticko-fyzikální fkulty Univerzity Krlovy v Prze Sokolovská83,18575Prh8 jko svou 200. publikci Vydání publikce bylo podpořeno výzkumným záměrem MŠMT č. MSM 0021620839 Z předloh připrvených v systému TEX vytisklo Reprostředisko UK MFF Sokolovská83,18675Prh8 První vydání Prh 2007 ISBN 80-7378-006-2