Differentiation Under the Integral Sign with Weak Derivatives

Transcription

1 Differentition Under the Integrl Sign with Wek Derivtives Steve Cheng September 6, 21 Contents 1 Introduction 2 2 Abstrct theory Wek differentition Appliction to integrble functions Relxing the continuity criterion Clssicl differentibility Computtionl exmples Fundmentl Theorem of Clculus Smooth integrnds Another counterexmple Integrnds tht re step functions The Blck-Scholes pricing formul A Differentition under the integrl sign with clssicl derivtives 12 B Bibliogrphy 12 C List of supporting theorems 13 Copyright mtters Permission is grnted to copy, distribute nd/or modify this document under the terms of the GNU Free Documenttion License, Version 1.2 or ny lter version published by the Free Softwre Foundtion; with no Invrint Sections, with no Front-Cover Texts, nd with no Bck-Cover Texts. Prefce In short, in this rticle we discuss the problem of differentiting under the integrl sign. 1

2 We ssume pssing fmilirity of the theory of distributions by Lurent Schwrtz, nd no more. (Even the uthor is just beginner on this theory. However, even reders who do not cre much bout the theory my find the section with worked-out computtionl exmples to be useful. The theory expoused in this rticle is nothing new: nlysts in the nineteenth nd twentieth centuries no doubt recognized the issues tht crop up with exchnging prtil derivtives nd integrtion, nd hd fixes for them though probbly through d-hoc methods before Schwrtz s theory of distributions. But I hope this rticle would serve purpose in explining the ppliction of the concepts t more elementry level thn the stndrd works on distribution theory, while not losing rigor for the mthemticlly-minded udience. Certinly no clculus book nowdys, even non-rigorous ones, dre tlk bout delt functions, even though they re vluble clcultion tools for the engineer nd physicist. I would like to thnk Kmyr Hzveh, who is n engineer, working on the sme problem set s I one dy; I would not hve written this rticle if not for him posing the question on differentition under the integrl; I would lso like to thnk Mtt Towers ( silverfish nd Rymond Puzio t PlnetMth for their encourging comments in my investigtion of the present problem. (Which only confirms how little people seem to know bout differentition under the integrl sign! And I should lso mention tht problem set originted from Prof. Sebstin Jimungl, former physicist, who lso motivted my investigtion with his helpful comments on wht to do when differentiting integrnds tht re not smooth. 1 Introduction Let X be n open subset of R m, nd be mesure spce. Given f : X R, nd consider the integrl with prmeter g(x = f(x, ω dω (ssuming it is well-defined. We wnt to differentite g, nd we hope tht g(x f(x, ω = dω, s often sserted in non-rigorous expositions of clculus without qulifiction. If f is sufficiently nice for exmple, it stisfies the conditions of Theorem A.1 then the swp of the integrtion nd differentition cn be proven to be vlid. However, when we do enough clcultions switching integrtion nd differentition, we soon begin to relize tht the conditions llowing the interchnge re probbly lot more generl thn wht the usul mthemticl theorems would tell us. For exmple, in pplictions, f(x, ω/ my exist except for limited number of singulrities; if we go hed nd differentite under the integrl sign nywy, experimenting nd fudging little with the Dirc delt function nd the like, we seem to lwys obtin the correct result. As the involvement of the Dirc delt function suggests, differentition under the integrl sign cn be more generlly formulted s problem with generlized functions (the distributions of Lurent Schwrtz nd their integrls nd derivtives. In this note, 2

3 we justify, using generlized derivtives, differentition under the integrl sign, in the cses when f does not stisfy the prerequisite bsic conditions, or even when f(x, ω/ fils to exist for certin vlues of x nd ω. Although the theory concerns the wek derivtives of generlized functions, in prcticl clcultions, the finl results obtined will often be seen to hve strong (usul/clssicl derivtives tht gree with their wek derivtives except t certin singulrities. Thus wek derivtives will intervene s clcultion tool, much s complex nlysis intermedites results bout rel functions, nd fundmentl solutions/green s functions intermedite results bout ordinry solutions of prtil differentil equtions. 2 Abstrct theory For n ordinry function f : X R, the mening of the expression f(x, ω dω is immedite. But to obtin the utmost generlity, nd to simplify the proofs of the fundmentl result, we need to extend the notion of integrting function with respect to one vrible while holding the other fixed, to rbitrry generlized functions f. This leds us nturlly to the following definition. Definition 2.1. Let f(x, ω be generlized function of x X, for ech fixed ω. Let φ denote ny function from some suitble clss of test functions on X defining the spce of generlized functions. Suppose tht the Lebesgue integrl ( f(x, ω φ(x dx dω := f(, ω, φ dω (1 X exists s finite quntity (converges bsolutely for ll test functions φ. Then we define the generlized function g(x = f(x, ω dω by X ( f(x, ω dω φ(x dx := g, φ := ( X f(x, ω φ(x dx dω. (2 However, without dditionl ssumptions on f, the liner functionl defined by eqution (2 my not be continuous liner functionl, nd hence will not be genuine generlized function. To fix this problem, we cn impose this strightforwrd hypothesis: Criterion 2.2 (Continuity. For every convergent 1 sequence {φ n } of test functions, we stipulte tht sup f(, ω, φ n dω <. (3 n Proof tht the criterion suffices. For ny sequence of test functions {φ n } converging to φ, we hve f(, ω, φ n converging to f(, ω, φ s n, for ech ω. Then the bound (3 gives the dominting fctor for the Lebesgue Dominted Convergence Theorem tht llows us to deduce g, φ n g, φ. 1 The convergence is in the topology of the spce of test functions for the generlized functions. 3

4 2.1 Wek differentition Proposition 2.3. Let f(x, ω be generlized function of x X, for ech ω. If f(x, ω dω exists, then so does f(x, ω dω. Proof. First, we hve to show tht hypothesis (1 holds for f/ : tht f, φ dω exists for every test function φ. But the integrnd is, by the definition of the derivtive of generlized function, equl to f(, ω, φ/. And the integrl over of this quntity exists becuse f(, ω dω exists nd φ/ is lso test function. Criterion 2.2 is verified likewise: sup f, φ n dω = sup f(, ω, φ n dω <, n n noting tht φ n / φ/ whenever φ n φ. Theorem 2.4 (Differentition under the integrl sign. Let f(x, ω be generlized function of x, for ech ω, such tht f(x, ω dω exists. Then f(x, ω dω = f(x, ω dω. Proof. The integrl on the right-hnd side of the eqution bove mkes sense in light of Proposition 2.3. Set g(x = f(x, ω dω. For every test function φ, g, φ = g, φ = f(, ω, φ dω = f, φ dω := f dω, φ. More generlly, the opertor / in the theorem nd its proof cn in fct be replced by ny continuous liner opertor T on generlized functions with dul T : T g, φ = g, T φ = f(, ω, T φ dω = T f, φ dω := T f dω, φ. The dul T of T will be gurnteed to exist if the underlying spce of test functions is reflexive, such s tht of the compctly-supported smooth functions, D = C c (X. For exmple, T could be tken to be the Fourier trnsform opertor. 4

5 3 Appliction to integrble functions The deceptively esy result of Theorem 2.4 might pper to be tutology: the interchnge of the ction, nd the integrl is, fter ll, built into the bstrct definition (2. Fortuntely, the theorem does hve rel content for functions f L 1 (X integrble under Lebesgue mesure. The bstrct definition of f(x, ω dω, s generlized function, when interpreted s n ordinry function, coincides with ctul integrl of f over, by virtue of Fubini s theorem, which is estblished for integrble functions. Furthermore, integrble functions lwys stisfy Criterion 2.2, for convergent sequence of test functions must be uniformly bounded on ll of X. If the spce of test functions is D = C c (X (the infinitely-differentible functions with compct support, even just locl integrbility suffices. Tht is: Criterion 3.1 (Locl integrbility. f : X R is mesurble function, nd f(x, ω dω dx <, for ny compct set K X. K An importnt cse of this sitution is when x f(x, ω dω is continuous: it is then, of course, loclly integrble. These simple observtions lredy cover most pplictions, which do not involve nsty Wierstrss-like functions, but only functions tht re nice (nd smooth perhps except t finite or countble number of points. On the other hnd, this does not men the theory just developed is useless; indeed, s the exmples in the next section show, even seemingly minor singulrities cn turn out to be significnt in the finl result. 3.1 Relxing the continuity criterion If we re unsure whether the ssumption of Criterion 2.2 holds, we could try to ignore it, with the penlty tht we would hve to del with discontinuous functionls. But if the derivtive of discontinuous functionl g is defined formlly by g, φ = g, φ for every test function φ, just s it is for continuous functionls, then we redily see tht the proof of Theorem 2.4 survives. If functionl g is not known to be continuous, then the consequence is tht we would not be ble to find sequence of generlized functions g n converging to g (in the sense tht g n, φ g, φ for every test function φ. However, if we do mnge to find such sequence, then we cn deduce tht g is in fct continuous (by sequentil completeness of the spce of distributions. Alterntively, we might be ble to compute g(x = f(x, ω dω nd discover directly tht it is continuous functionl without verifying Criterion 2.2 priori. It even works to first clculte g/ formlly by differentition under the integrl sign, nd discover tht it is continuous. Then we cn lso deduce fter-the-fct tht g must be continuous too, by the theorem of distribution theory tht nti-derivtives of ny generlized function exist s generlized functions. 5

6 3.2 Clssicl differentibility Finlly, we hve to question whether the derivtive of generlized functions obtined in Theorem 2.4 hs nything to do with usul notion of differentition. The nswer is ffirmtive: Theorem 3.2 (Differentition under the integrl sign. Let f : X R be n ordinry function such tht the generlized derivtives f/ re represented by ordinry functions (for ech ω, nd both f nd f/ re loclly integrble s in Criterion 3.1. Then lmost everywhere on X, f(x, ω dω = f(x, ω dω where clssicl differentition is used on both sides of the eqution. Of course, if both sides re continuous in x, then equlity holds for every x X. Proof. To void mbiguity, we temporrily distinguish generlized derivtives from ordinry derivtives by the ddition of brckets: [/ ]. Set g(x = f(x, ω dω, nd let φ be test function supported on compct set K. Then we hve [ ] [ ] g f, φ =, φ dω (Theorem 2.4 f(x, ω = φ(x dx dω (Theorems C.2, C.4 K ( f(x, ω = dω φ(x dx (used locl integrbility. Thus K [ ] g = h := f(, ω dω, in the sense tht the loclly-integrble function h represents the generlized function [g/ ] by integrtion. Now pply Theorem C.4 to g nd h. Compre with Theorem A.1. Loosely speking, tht theorem requires the integrl f/ dω to be loclly bounded function of the prmeter; here we only demnd the integrl to be loclly integrble, which is slightly weker requirement. (For instnce, the function f(x, y = x y for x, y [, 1] fils the first requirement but stisfies the second. A nice dvntge of our theory though, from the point of the lzy physicist or engineer, is tht it llows us to neglect checking the integrbility conditions beforehnd, nd to just go hed nd differentite under the integrl sign formlly. If the result of the clcultion mkes sense (i.e. the relevnt integrls converge, then the opertions will be justified. 6

7 4 Computtionl exmples 4.1 Fundmentl Theorem of Clculus To wrm-up, let us look t the following musing, if hopelessly roundbout, demonstrtion of the Fundmentl Theorem of Clculus. For f continuous, nd x b, we hve d dx x f(t dt = d dx b 1(t xf(t dt = = b b f(t 1(x t dt x f(t δ(x t dt = f(x, where δ, is of course, the (infmous Dirc delt function, nd 1( denotes n indictor. (In effect, 1(x t reduces to H(x t, where H is the Heviside step function, so its derivtive is δ(x t. Actully, our derivtion is not quite rigorous, though the method does seem to work. However, rther thn obsessing over trivil clcultion, we defer the explntion of the rigorous method to the next exmples, where we cn mke serious blunders when we fil to understnd the theory correctly or t lest when we re not creful. 4.2 Smooth integrnds Consider the well-known Gmm function: Γ(x = e t t x 1 dt for x >. Since Γ is continuous (nd the integrnd is non-negtive, it stisfies Criterion 3.1, nd therefore differentition under the integrl sign is llowed: d k dx k Γ(x = e t k x k tx 1 dt. More generlly, let f : X R be n integrnd tht is smooth in the prmeter, stisfying Criterion 3.1; then we cn differentite (wekly under the integrl sign s mny times s we wnt. But there is no gurntee tht f/ hs to be integrble over in the Lebesgue sense condition in Theorem 3.2 for clssicl differentibility of f(x, ω dω. Such nomlies hppen, for exmple, with the function f(x, y = y 2 (1 cos xy for y > nd x R. The integrl f(x, y dy converges (nd is continuous function of x see Theorem A.2, but f x dy = sin xy s dy =? sin xy lim dy = π y s y 2 sgn(x does not exist s rel Lebesgue integrl. But it is not hrd to see tht the interprettion of the divergent integrl s the improper limit, s written bove, is correct. Here re the gory detils: Let φ be smooth function supported on compct intervl [, b]; then f dy, φ := x b f(x, y dφ dx dy = lim dx s s b f(x, y dφ dx dy, dx 7

8 nd we cn chnge the outer integrl to limit of integrls since f dφ/dx is L 1 ([, b] (, function. Next, we perform n integrtion by prts, then switch the order of integrtion, which is llowed since (f/x φ is continuous on the compct set [, b] [, s] (nd so hs finite integrl there: = lim s s b f φ(x dx dy = lim x s b ( s Finlly, we cn move the limit to inside the first integrl since s bounded for ll x [, b] nd s (, : = b ( lim s s f x dy φ(x dx. f x dy φ(x dx. sin xy y dy is uniformly This verifies the clim tht f s sin xy x dy is the loclly-integrble function lim s y dy. We leve it to the reder to ponder the proper interprettion of this pprent nonsense: π δ(x = d ( π dx 2 sgn(x = d dx 4.3 Another counterexmple This exmple comes from [Gelbum]: let sin xy y dy = sin xy dy = x y f(x, y = x3 2 y 2 e x /y, x R, y (, 1] f(x, y = 3x2 2 x y 2 e x /y 2x4 2 y 3 e x /y Theorem 3.2 pplies. Computing directly, we find d dx 1 f(x, y dy = d 2 dx (xe x = e x2 (1 2x 2 = 1 f(x, y x cos(xy dy. for ll x, but the left- nd right-hnd sides do not gree t the single point x =. Of course, there is no contrdiction, for Theorem 3.2 only sserts the equlity of the derivtives lmost everywhere. In the context of Theorem A.1, observe tht f/x is integrble (with respect to y when x, but is not integrble when x =. Moreover, the dominting fctor Θ(y cn be found for this integrnd if x is restricted to compct set not including the origin, but not if x cn get rbitrrily close to the origin. 4.4 Integrnds tht re step functions Next, we come to detiled exmple involving singulrities described by delt functions. Let = [, 1] (with the elements denoted by the vrible y, nd c be constnt. Define f(x, y = y 1(xy > c, g(x = 1 dy f(x, y dy, for x R nd y [, 1]. 8

9 For comprison, we first compute g nd its derivtive directly, without differentition under the integrl sign: g(x = 1 ( ( c 2 1 1(x > c, 2 x dg(x dx = 1 ( c 2 1(x > c (except t x = c. x x Next follows tempting, but unrigorous nd totlly wrong computtion. (I illustrte these exmples becuse I mde the sme mistkes before I fully bsorbed the proper theory. Blunder: We try to pply the chin rule to the Heviside step function: dg 1 dx = y H(xy c dy = x 1 y 2 δ(xy c dy = y 2 ( c 2 xy=c =, x for < c x < 1. Even without knowing the true solution for dg/dx, we know tht the nswer just obtined must be wrong becuse it hs the wrong order in x nd fils dimensionl nlysis: if y is unitless, but both x nd c hve units of distnce, then g is lso unitless, nd thus dg/dx must hve units of inverse distnce. It turns out tht the use of the chin rule is correct, but the substitution 1 ( c γ(y δ(xy c dy γ(y = γ xy c= x is incorrect. To understnd this intuitively, recll tht δ function ppered from prtil differentition with respect to x; tht mens the stndrd properties of the delt function pply only with respect to the vrible x: e.g. γ(x δ(x dx = γ( for continuous functions γ. But we were integrting with respect to the vrible y, the integrnd γ(y δ(xy c. Indeed, if we write δ(xy c δ(x c/y, we see tht 1 δ(xy c dy = 1 ( δ x c dy c y x, becuse now y 1 in the rgument of the delt function cuses the integrl to ccumulte vlues differently thn if the rgument hd been y itself. The correct vlue of the integrl is, insted, suggested by differentil chnge of vribles: 1 ( y δ x c c c dy = δ(x u y c u u 2 du = c2 u 3 u=x if x c, zero otherwise, substitute u = c y, dy = c u 2 du yielding the desired nswer (with the dimensionl inconsistency fixed. 9

10 Hving illustrted this pitfll, we proceed with the rigorous computtion. Firstly, f stisfies Criterion 3.1, so differentition under the integrl is llowed. Secondly, to dispel ny doubts bout the vlidity of f/x = y δ(x c/y, we compute it formlly. For ny test function φ (with compct support: we hve, indeed, f dφ, φ = f, x dx = y 1(xy > c dφ dx = y dx c/y dφ ( c dx dx = y φ. y Now, pplying Theorem 2.4 nd performing the sme differentil chnge of vribles u = c/y it is legitimte now tht the integrls re relly Lebesgue integrls we find: dg dx, φ = 1 f, φ dy = x 1 ( c y φ dy = y c c 2 φ(u du. u3 But we recognize the lst integrl s the functionl, evluted t φ, corresponding to the integrble function x c 2 x 3 1(x > c. It is gin instructive to exmine wht Theorem A.1 would sy in this sitution. Clerly, tht theorem cnnot possibly pply here, becuse f/x = whenever the derivtive exists in the clssicl sense, nd so integrting this nïvely would yield zero identiclly. Theorem A.1 sserts tht it pplies if f is everywhere differentible with respect to x nd lmost every ω, but here we relly hve f being lmost everywhere differentible with respect to x for every ω, These two conditions re not the sme. In fct, lthough both X nd hppened to be subsets of R in the current exmple, the roles they ply re not symmetric. Theorem A.1 uses no mesure theory on X-spce, but only on - spce, so nturlly, only in -spce do the notions of lmost everywhere enter into the hypotheses nd the conclusions of the theorem. Even if we rbitrrily redefine the prtil derivtives to be zero whenever they do not exist, it is esy to see, in this exmple, tht the difference quotient see the remrks fter Theorem A.1 cnnot be dominted. Another wy to understnd this sitution, is if we decompose f/ into sum f(x, ω = p(x, ω + q(x, ω, of n ordinry L 1 (X function p nd generlized function q, then f(x, ω dω = p(x, ω dω + q(x, ω dω, where the fr right term is exctly the contribution to the derivtive tht would be missed out when differentiting clssiclly under the integrl sign. 4.5 The Blck-Scholes pricing formul Here we present n pplied clcultion from the field of mthemticl finnce. (This is one of the problems tht I hd been working on, leding me to investigte more thoroughly differentition under the integrl. Consider the function P (x = e rτ E[mx(xY K, ], x >, Y = e (r 1 2 σ2 τ+σ τz, Z Norml(, 1. 1

11 which gives the rbitrge-free price of Europen cll option for stock, given the current price of the stock x, modelled s geometric Brownin motion. (The prmeter r is the interest rte, σ is the voltility of the stock, τ is the time from the present to the mturity of the Europen cll option, nd K is its strike price. The expecttion cn of course be rewritten s n integrl over R of the integrnd multiplied by the density of stndrd norml distribution, but for the ske of vriety, let us stick with the form given bove. (Thus, here is tken to be some probbility spce. The function P hs the nlyticl form given by P (x = xφ(d + Ke rτ Φ(d, d ± (x = (r ± 1 2 σ2 τ + log(x/k σ τ. Clculting dp/dx is somewht messy from this formul; we strt insted with differentition under the expecttion sign: [ ] dp dx = e rτ E mx(xy K, = e rτ E [ Y 1(xY > K ] x [ ] = e 1 2 σ2τ E e σ τz 1(Z > d = Φ(d +. The exchnge of differentition nd expecttion is justified by the following estimte: b E [ mx(xy K, ] dx b E[xY ] dx <, < < b, nd Criterion 3.1 is stisfied 2. To obtin d 2 P/dx 2 = Φ(d + /(xσ τ, it is ctully much esier to differentite dp/dx directly. But to illustrte the techniques, we do differentition under the expecttion sign gin: d 2 P dx 2 = d [ ] dp dx dx = e rτ E Y 1(xY > K = e rτ E [ Y δ(x K/Y ] x = e rτ E [ KU 1 δ(x U ], U = K/Y. At this point we hve no choice but to express the expecttion s n integrl over the rels, becuse of the involvement of delt function. More formlly, if we suppose tht d 2 P/dx 2 is loclly-integrble function h, then we must show tht h(xψ(x dx = Ke rτ E [ U 1 ψ(u ] for ll test functions ψ, nd tht obviously necessittes expnding the right-hnd side s Lebesgue integrl. The clcultion is not too hrd if we observe tht log U hs the distribution Norml(log K (r 1 2 σ2 τ, σ 2 τ, so tht [ ] δ(x U Ke rτ E = Ke rτ U δ(x u u density for log-norml { }} { exp( 1 2 d (u 2 u 2π σ τ du = Ke rτ x 2 Φ (d (x. 2 We cn lso verify tht the hypotheses of Theorem 3.2 indeed hold for the first derivtive, but the conclusions from tht theorem, nmely tht the clssicl derivtive exists nd equls the generlized derivtive, re lredy immedite from our clculted results. But Theorem 3.2 does not pply for the second derivtive, becuse Y 1(xY > K/s not n ordinry function. 11

12 The ltter expression looks different from the direct formul for d 2 P/dx 2, but they re in fct equl. We hve plyed fst-nd-loose with delt functions in the lst clcultion, but clerly it bbrevites the rigorous rguments we hve lredy mde. The reder who is still unsure bout the method is invited to repet the explicit clcultion with functionls. A Differentition under the integrl sign with clssicl derivtives Theorem A.1 (Differentition under the integrl sign. Let X be n open subset of R m, nd be mesure spce. Suppose tht the function f : X R stisfies the following conditions: Then 1. f(x, ω is n integrble function of ω for ech x X. 2. For lmost ll ω, the derivtive f(x, ω/ exists for ll x X. 3. There is n integrble function Θ: R such tht f(x, ω/ Θ(ω for ll x X. f(x, ω dω = f(x, ω dω. This result is proved by (wht else? the Lebesgue Dominted Convergence Theorem. A more precise condition, in plce of (3 bove, tht comes out of the proof of the theorem, is tht the difference quotient (f(x + te i, ω f(x, ω/t must be dominted by some Θ(ω for ll x nd t. For completeness, we lso include the following nlogous result for continuity: Theorem A.2 (Continous dependence on integrl prmeter. Let X be n open subset of R m (or ny metric spce, nd be mesure spce. Suppose tht the function f : X R stisfies the following conditions: 1. f(x, ω is mesurble function of ω for ech x X. 2. For lmost ll ω, f(x, ω is continuous in x. 3. There is n integrble function Θ: R such tht f(x, ω Θ(ω for ll x X. Then f(x, ω dω is continuous function of x. B Bibliogrphy References [Follnd] [Gelbum] Gerld B. Follnd, Rel Anlysis: Modern Techniques nd Their Applictions, second ed. Wiley-Interscience, Bernrd R. Gelbum nd John M. H. Olmsted. Counterexmples in Anlysis. Dover,

13 [Hoskins] [Jones] [Lützen] Roy F. Hoskins nd J. Sous Pinto. Theories of Generlized Functions. Horwood Publishing, 25. D. S. Jones. The Theory of Generlized Functions, second ed. Cmbridge University Press, Jesper Lützen. The Prehistory of the Theory of Distributions. Springer- Verlg, [Schwrtz] Lurent Schwrtz. Théorie des Distributions, volume I. Hermnn, [Tlvil] Erik Tlvil. Necessry nd Sufficient Conditions for Differentiting Under the Integrl Sign. Americn Mthemticl Monthly, 18 (June-July Found on-line t: [Lützen] does not develop ny mthemticl theory, but I hppened upon it reserching the current problem. It is n interesting look into the intuition nd motivtion behind the theory of generlized functions. [Tlvil] is n interesting, light-going pper giving the exct conditions for exchnging clssicl differentition nd integrtion, in terms of Henstock integrls tht cn integrte more derivtives. The results of tht pper re essentilly souped-up versions of our Theorem 3.2, nd, remrkbly, the techniques used there re quite nlogous to our distributionl methods. On the other hnd, the restriction to clssicl differentition in tht pper mens tht, the theory cnnot mke sense of discontinuities in the integrnd, or divergent integrls, which I think, re probbly more importnt in pplictions thn derivtives tht re not Lebesgue-integrble. C List of supporting theorems Here is list of some of the theorems we hve ppeled to implicitly but my not be fmilir to the reder. (Actully, I too ws not too fmilir with them before I strted writing this rticle. Theorem C.1 (Fundmentl Theorem of Clculus for Lebesgue integrls. Let f be Lebesgue-integrble function on compct intervl [, b]. If G(x = x f(t dt, for x [, b], then G is bsolutely continuous, nd G = f lmost everywhere. Moreover, if f = g for n bsolutely continous g, then G = g + c for some constnt c. Proof. See, for exmple, Theorem 3.35 in [Follnd]. Theorem C.2 (Integrtion by Prts. Let X be n open set in R m. The usul integrtionby-prts formul is vlid for loclly-integrble functions f : X R bsolutely continuous 3 3 Here nd elsewhere, bsolutely continuous in x i relly mens bsolutely continuous in x i when the other vribles x j, j i, re held fixed, fter some modifiction of the function on set of mesure zero. 13

14 in x i, with f/ loclly integrble, nd smooth test functions φ: X R vnishing outside compct set in X: f f(x, φ = φ(x dx = f(x φ(x dx = f, φ. X X Thus the generlized derivtive grees with the ordinry derivtive for such f. Proof. This sttement ppers in [Schwrtz], Ch. 2, Sect. 5, Theorem V, prt (1. It cn be proven by pplying Theorem C.1 nd the integrtion by prts formul for Lebesgue-Stieljes integrls, Theorem 3.36 in [Follnd]. Wrning. the condition f is bsolutely continuous cnnot be replced by the weker condition tht f exists lmost everywhere nd is loclly integrble. The sme sort of restriction enters into the second prt of Theorem C.1, where we need to mke sure tht g is bsolutely continuous before concluding G = g + c. We do not even hve to try hrd to find pthologicl counterexmples to show this necessity: if H is the Heviside step function, then H = clssiclly, but of course we lredy know tht its generlized derivtive is the Dirc delt. In terms of Theorem C.1: H is not bsolutely continuous not even continuous nd cnnot be expressed s the integrl of its derivtive. (Even if function is continuous, Theorem C.1 cn fil: the stndrd counterexmple being the Cntor function (lso known s the Devil s stircse with zero clssicl derivtive lmost everywhere. Theorem C.3. If f is ny generlized function on X, n open set in R m, then the prtil differentil eqution u/ = f hs infinitely mny generlized solutions u, ech differing by some generlized function independent of x i. Proof. See [Schwrtz], Ch. 2, Sect. 5, Theorem IV. Theorem C.4. Let X be n open set in R m. If loclly-integrble function f : X R hs generlized derivtive represented by loclly-integrble function h: X R, then f is bsolutely continuous in x 1, nd it dmits n ordinry derivtive of h lmost everywhere. (This result is converse to Theorem C.2. Proof. See [Schwrtz], Sect. 5, Theorem V, prt (2. We reproduce the proof here: To void mbiguity, we temporrily distinguish generlized derivtives from ordinry derivtives by the ddition of brckets: [/ ]. Define n indefinite integrl G(x = h(x dx i. Then G(x is bsolutely continuous in x i, nd G/ = h lmost everywhere. It is lso true tht [G/ ] = h. From the ltter sttement, we know from Theorem C.3 tht G f = c where c is some generlized function independent of x i. Since G f = c is loclly-integrble function, we must hve c/ = using ordinry derivtives. It follows tht f/ exists lmost everywhere, nd f/ = G/ = h. The function f is bsolutely continuous in x i since f = G c nd both G nd c re bsolutely continuous in x i. Theorem C.5. The differentition opertor / is continuous mpping from generlized functions to generlized functions. Proof. See [Schwrtz], Ch. 3, Sect. 5. Theorem C.6. The spces D nd D re reflexive. 14

15 Proof. See [Schwrtz], Ch. 3, Sect. 3, Theorem XIV. 15