Rel Anlysis nd Multivrible Clculus: Grdute Level Problems nd Solutions Igor Ynovsky 1
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 2 Disclimer: This hndbook is intended to ssist grdute students with qulifying exmintion preprtion. Plese be wre, however, tht the hndbook might contin, nd lmost certinly contins, typos s well s incorrect or inccurte solutions. I cn not be mde responsible for ny inccurcies contined in this hndbook.
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 3 Contents 1 Countbility 5 2 Unions, Intersections, nd Topology of Sets 7 3 Sequences nd Series 9 4 Notes 13 4.1 Lest Upper Bound Property........................ 13 5 Completeness 14 6 Compctness 16 7 Continuity 17 7.1 Continuity nd Compctness........................ 18 8 Sequences nd Series of Functions 19 8.1 Pointwise nd Uniform Convergence.................... 19 8.2 Normed Vector Spces............................ 19 8.3 Equicontinuity................................ 21 8.3.1 Arzel-Ascoli Theorem........................ 21 9 Connectedness 21 9.1 Reltive Topology.............................. 21 9.2 Connectedness................................ 21 9.3 Pth Connectedness............................. 23 10 Bire Ctegory Theorem 24 11 Integrtion 26 11.1 Riemnn Integrl............................... 26 11.2 Existence of Riemnn Integrl........................ 27 11.3 Fundmentl Theorem of Clculus..................... 27 12 Differentition 30 12.1 R R..................................... 30 12.1.1 The Derivtive of Rel Function................. 30 12.1.2 Rolle s Theorem........................... 30 12.1.3 Men Vlue Theorem........................ 30 12.2 R R m.................................... 31 12.3 R n R m................................... 31 12.3.1 Chin Rule.............................. 34 12.3.2 Men Vlue Theorem........................ 35 12.3.3 x ( f y ) = y ( f x )........................... 36 12.4 Tylor s Theorem............................... 37 12.5 Lgrnge Multipliers............................. 40
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 4 13 Successive Approximtions nd Implicit Functions 41 13.1 Contrction Mppings............................ 41 13.2 Inverse Function Theorem.......................... 41 13.3 Implicit Function Theorem......................... 44 13.4 Differentition Under Integrl Sign..................... 46
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 5 1 Countbility The number of elements in S is the crdinlity of S. S nd T hve the sme crdinlity (S T ) if there exists bijection f : S T. crd S crd T if injective 1 f : S T. crd S crd T if surjective 2 f : S T. S is countble if S is finite, or S N. Theorem. S, T φ. injection f : S T surjection g : T S. Theorem. Q is countble. Proof. Need to show tht there is bijection f : N Q. Since N Q, crd N crd Q, nd therefore, f tht is injective. To show crd N crd Q, construct the following mp. The set of ll rtionl numbers cn be displyed in grid with rows i = 1, 2, 3,... nd columns j = 0, 1, 1, 2, 2, 3, 3,.... Ech ij, the ij th entry in tble, would be represented s j i. Strting from 11, nd ssigning it n = 1, move from ech subsequent row digonlly left-down, updting n. This would give mp g : N Q, which will count ll frctions, some of them more thn once. Therefore, crd N crd Q, nd so g is surjective. Thus, crd N = crd Q, nd Q is countble. Theorem. R is not countble. Proof. It is enough to prove tht [0, 1) R is not countble. Suppose tht the set of ll rel numbers between 0 nd 1 is countble. Then we cn list the deciml representtions of these numbers (use the infinite expnsions) s follows: 1 = 0. 11 12 13... 1n... 2 = 0. 21 22 23... 2n... 3 = 0. 31 32 33... 3n... nd so on. We derive contrdiction by showing there is number x between 0 nd 1 tht is not on the list. For ech positive integer j, we will choose jth digit fter the deciml to be different thn jj : x = 0.x 1 x 2 x 3... x n..., where x j = 1 if jj 1, nd x j = 2 if jj = 1. For ech integer j, x differs in the jth position from the jth number on the list, nd therefore cnnot be tht number. Therefore, x cnnot be on the list. This mens the list s we chose is not bijection, nd so the set of ll rel numbers is uncountble. (Need to worry bout not llowing 9 tils in deciml expnsion: 0.399... = 0.400...). 1 injective = 1-1: f(s 1 ) = f(s 2 ) s 1 = s 2. 2 surjective = onto: t T, s S, s.t. f(s) = t.
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 6 Problem (F 01, #4). The set of ll sequences whose elements re the digits 0 nd 1 is not countble. Let S be the set of ll binry sequences. We wnt to show tht there does not exist one-to-one mpping from the set N onto the set S. Proof. 1) Let A be countble subset of S, nd let A consist of the sequences s 1, s 2,.... We construct the sequence s s follows. If the nth digit in s n is 1, let the nth digit of s be 0, nd vice vers. Then the sequence s differs from every member of A in t lest one plce; thus s / A. However, s S, so tht A is proper 3 subset of S. Thus, every countble subset of S is proper subset of S, nd therefore, S is not countble. Proof. 2) Suppose there exists f : N S tht is injective. We cn lwys exhibit n injective mp f : N S by lwys picking different sequence from the set of sequences tht re lredy listed. (One wy to do tht is to choose binry representtion for ech n N). Suppose f : N S is surjective. Then, ll sequences in S could be listed s s 1, s 2,.... We construct the sequence s s follows. If the nth digit in s n is 1, let the nth digit of s be 0, nd vice vers. Then the sequence s differs from every member of the list. Therefore, s is not on the list, nd our ssumption bout f being surjective is flse. Thus, there does not exist f : N S surjective. Theorem. crd (A) < crd (P (A)) 4. Proof. crd (A) crd (P (A)), since A cn be injectively mpped to the set of oneelement sets of A, which is subset of P (A). We need to show there is no onto mp between A nd P (A). So we would like to find thing in P (A) which is not reched by f. In other words, we wnt to describe subset of A which cnnot be of the form f() for ny A. Suppose A = P (A). Then there is 1-1 correspondence f : A P (A). We obtin contrdiction to the fct tht f is onto by exhibiting subset X of A such tht X f() for ny A. For every A, either f(), or / f(). Let X = { A : / f()}. Consider A. If f(), then / X, so f() X. If / f(), thn X, so f() X. Therefore, X f(), A, contrdiction. Therefore, crd (A) < crd (P (A)). Theorem. Suppose tht f : [0, 1] R is n incresing function. Show tht f cn hve t most countble number of discontinuities. Proof. Let E = {x [0, 1] : f is discontinuous t x}. Given ny x E, we know tht lim t x f(t) < lim t x + f(t) nd, using this fct, we choose r(x) Q such tht lim t x f(t) < r(x) < lim t x + f(t). We hve defined 1 1 function r : E Q. 3 A is proper subset of B if every element of A is n element of B, nd there is n element of B which is not in A. 4 Power set of set S is the set whose elements re ll possible subsets of S, i.e. S = {1, 2}, P (S) = 2 S = {φ, {1}, {2}, {1, 2}}. P (A) = 2 n, if A = n.
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 7 2 Unions, Intersections, nd Topology of Sets Theorem. Let E α be collection of sets. Then ( α E α ) c = α (E c α). Proof. Let A = ( E α ) c nd B = ( E c α). If x A, then x / E α, hence x / E α for ny α, hence x E c α for every α, so tht x E c α. Thus A B. Conversely, if x B, then x E c α for every α, hence x / E α for ny α, hence x / E α, so tht x ( E α ) c. Thus B A. Theorem. ) For ny collection G α of open sets, α G α is open. b) For ny collection F α of closed sets, α F α is closed. c) For ny finite collection G 1,..., G n of open sets, n i=1 G i is open. d) For ny finite collection F 1,..., F n of closed sets, n i=1 F i is closed. Proof. ) Put G = α G α. If x G, then x G α for some α. Since x is n interior point of G α, x is lso n interior point of G, nd G is open. b) By theorem bove, ( α F α ) c = α (F c α) ( α F α ) c = α (F c α), (2.1) nd F c α is open. Hence ) implies tht the right eqution of (2.1) is open so tht α F α is closed. c) Put H = n i=1 G i. For ny x H, there exists neighborhoods N ri of x, such tht N ri G i (i = 1,..., n). Put r = min(r 1,..., r n ). Then N r (x) G i for i = 1,..., n, so tht N r (x) H, nd H is open. d) By tking complements, d) follows from c): ( n i=1 F i) c = n i=1 (F c i ). Theorem. S M X. S open reltive to M open U X such tht S = U M. Proof. S open reltive to M. To ech x S, r x, x y < r x, y M y S. Define U = x S N rx (x) U X open. It is cler tht S U M. By our choice of N rx (x), we hve N rx (x) M S, x S so tht U M S S = U M. If U is open in X nd S = U M, every x S hs neighborhood N rx (x) U. Then N rx (x) M S so tht S is open reltive to M.
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 8 Theorem. K Y X. K compct reltive to X K compct reltive to Y. Proof. Suppose K is compct reltive to X, nd let {V α } be collection of sets, open reltive to Y, such tht K α V α. By the bove theorem, U α open reltive to X, such tht V α = Y U α, α; nd since K is compct reltive to X, we hve K U α1 Uαn Since K Y K V α1 Vαn. Suppose K is compct reltive to Y. Let {U α } be collection of open subsets of X which covers K, nd put V α = Y U α. Then will hold for some choice of α 1,..., α n ; nd since V α U α, implies. Theorem. Compct subsets of metric spces re closed. Proof. Let K X be compct subset. We prove K c is open. Suppose x X, x / K. If y K, let V y nd W y be neighborhoods of x nd y, respectively, of rdius less thn 1 2 d(x, y). Since K is compct, K W y 1 Wyn = W. If V = V y1 Vyn, then V is neighborhood of x which does not intersect W. Hence V K c, so tht x is n interior point of K c. Theorem. Closed subsets of compct sets re compct. Proof. Suppose F K X, F is closed (reltive to X), nd K is compct. Let {V α } be n open cover of F. If F c is djoined to {V α }, we obtin n open cover Ω of K. Since K is compct, there is finite subcollection Φ of Ω which covers K, nd hence F. If F c is member of Φ, we my remove it from Φ nd still retin n open cover of F. We hve thus shown tht finite subcollection of {V α } covers F. Corollry. If F is closed nd K is compct, then F K is compct. Proof. K is closed (by theorem bove), nd thus F K is closed. Since F K K, the bove theorem shows tht F K is compct. Theorem. If {K α } is collection of compct subsets of metric spce X such tht the intersection of every finite subcollection of {K α } is nonempty, then K α is nonempty. Proof. Fix member K 1 of {K α }. Assume tht no point of K 1 belongs to every K α. Then the sets Kα c form n open cover of K 1 ; nd since K 1 is compct, K 1 Kα c 1 K c αn. But this mens tht K 1 Kα1 Kαn is empty. contrdiction.
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 9 3 Sequences nd Series A sequence {p n } converges to p X if: ɛ > 0, N: n N p n p < ɛ lim n p n = p. A sequence {p n } is Cuchy if: ɛ > 0, N: n, m N p n p m < ɛ. Cuchy criterion: A sequence converges in R k it is Cuchy sequence. Cuchy criterion: n converges ɛ > 0, N : m k=n k ɛ if m n N. The series n is sid to converge bsolutely if n converges. Problem (S 03, #2). If 1, 2, 3,... is sequence of rel numbers with j=1 j <, then lim N N j=1 j exists. ( n converges bsolutely n converges). Proof. If s n = n j=1 j is prtil sum, then for m n we hve s n s m = n j j=1 m j = j=1 n j j=m n j Since j converges, given ɛ > 0, N, s.t. m, n N (m n), then n j=m j < ɛ. Thus {s n } is Cuchy sequence in R, nd converges. lim N N j=1 j exists. Problem (F 01, #2). Let N denote the positive integers, let n = ( 1) n 1 n, nd let α be ny rel number. Prove there is one-to-one nd onto mpping σ : N N such tht σ(n) = α. n=1 j=m Proof. n = ( 1) n 1 n 1 = 1, 2 = 1 2, 3 = 1 3, 4 = 1 4,.... 2n > 0, 2n 1 < 0, n = 1, 2,... (positive terms) = n=1 2n = 0. (negtive terms) = n=1 2n 1 n=1 1 2n, nd lim n 1 2n 1 = 0. n=1 1 2n = 1 2 = n=1 n=1 1 n diverges, nd lim n 1 2n = 1 2n 1 diverges by comprison with Clim: α R, there is one-to-one nd onto mpping σ : N N such tht n=1 σ(n) = α, where σ(n) is the rerrngement of indices of the originl series. Given α, choose positive terms in sequentil order until their sum exceeds α. At this switch point, choose negtive terms until their sum is less thn α. Repet the process. Note: This process never stops becuse no mtter how mny positive nd negtive terms re tken, there re still infinitely mny both positive nd negtive terms left; the sum of positive terms is, the sum of negtive terms is. Let the sum of terms t the Nth step be denoted by S N, S N = N n=1 σ(n). At switch point, α S N is bounded by the size of the term dded: α S N 0, N All terms { n } will eventully be dded to the [ sum (σ : N N is surjective (onto)) t different steps (σ : N N is injective (1-1)). σ : {1, 2, 3,...} {n1, n 2, n 3,...} ].
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 10 Root Test. Given n, put α = lim sup n n n. Then ) if α < 1, n converges; b) if α > 1, n diverges; c) if α = 1, the test gives no informtion. Rtio Test. The series n ) converges if lim sup n n+1 n < 1, b) diverges if n+1 n 1 for ll n n 0, where n 0 is some fixed integer. Alternting Series. Suppose () 1 2... ; (b) 2m 1 0, 2m 0, m = 1, 2, 3,... ; (c) lim n n = 0. Then n converges. Geometric Series. x < 1 : k=0 x k = 1 1 x. P roof : S n = 1 + x + x 2 + + x n xs n = x + x 2 + x 3 + + x n+1 = (1 x)s n = 1 x n+1 S n = n x k = 1 xn+1 1 x k=0
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 11 Problem (F 02, #4). By integrting the series prove tht 1 1 + x 2 = 1 x 2 + x 4 x 6 + x 8 π 4 = 1 1 3 + 1 5 1 7 + 1 9. Justify crefully ll the steps (especilly tking the limit s x 1 from below). Proof. Geometric Series: 1 1 + x 2 = 1 x 2 + x 4 x 6 + x 8 ; x < 1. dx 1 + x 2 = tn 1 (x) = x x3 3 + x5 5 x7 7 + x9 9. dx 1 + x 2 = [tn 1 (x)] 1 0 = tn 1 1 tn 1 0 = π 4 0 = π 4. 1 0 1 0 1 ɛ 0 dx 1 ɛ 1 + x 2 = lim ɛ 0 0 dx 1 + x 2 = f n f 0 f(x)dx S(x) S N (x) = 1 ɛ 0 dx 1 + x 2 { S }} { [1 x 2 + x 4 ] } {{ } converges uniformly for x 1 ɛ dx = lim f n (x)dx f(x)dx f n (x)dx f(x) f n (x) dx S is uniform limit of S N = N ( 1) n x 2n. n=1 ( 1) n x 2n, x [0, 1 ɛ] x 2n N+1 N+1 1 ɛ 0 S N (x)dx f f n dx = (b ) f f n. (1 ɛ) 2n = N+1 Above clcultions show tn 1 x = ( 1) n x 2n+1, x < 1. 2n + 1 Alternting series test right side converges. π 4 = ( 1) n 2n + 1 = 1 1 3 + 1 5 1 7 + 1 9. (1 ɛ)2(n+1) (1 ɛ) 2 0.
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 12 Logrithm. x < 1 : log(1 + x) = x 0 1 1 + x = 1 x + x2 x 3 + = dt 1 + t = x x2 2 + x3 3 x4 ( 1) n x n n=0 4 + = n=1 n+1 xn ( 1) log(1 + x) is not vlid for x > 1, or x < 1. Clim: If x = 1, the series converges to log 2. Proof: n=1 ( 1)n 1 xn n is uniformly convergent for x [0, 1], since the sum of ny number of consecutive terms strting with the n th hs bsolute vlue t most, since for 0 < x < 1 we hve lternting series. x n n 1 n Binomil Series. x < 1 : ( (1 + x) α α(α 1) (α n + 1) = 1 + x n α = 1 + αx + n! 2 n=1 = 1 + αx + α(α 1) x 2 + 2! α(α 1)(α 2) x 3 + 3! Problem (F 03, #3). The sequence 1, 2,... with n = n. Proof. By Binomil Series Theorem with α = n, x = 1 n, we get: ( 1 + 1 ) n = 1 + n 1 n(n 1) 1 n(n 1)(n 2) + + n n 2! n2 3! 1 + 1 + 1 2! + 1 3! + + 1 n! = n k=0 1 k! e, s n. n ) ( x 2 α + + n ( 1 + 1 n) n converges s ) x n + 1 n(n 1)(n 2) 1 + + n3 n! 1 n n
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 13 4 Notes 4.1 Lest Upper Bound Property An ordered set S is sid to hve the lest upper bound 5 property if: E S, E is not empty, nd E is bounded bove, then sup E exists in S. Completeness xiom: If E is nonempty subset of R tht is bounded bove, then E hs lest upper bound. Problem (F 02, #2). Show why the Lest Upper Bound Property (every set bounded bove hs lest upper bound) implies the Cuchy Completeness Property (every Cuchy sequence hs limit) of the rel numbers. Proof. Suppose {x n } Cuchy. The problem is to show tht {x n } converges. We first show tht {x n } is bounded. Fix ɛ > 0 nd let N be such tht x n x m < ɛ if n, m > N. Then for ny fixed n > N, the entire sequence is contined in the closed bll of center x n nd rdius mx{d(x n, x 1 ), d(x n, x 2 ),..., d(x n, x N ), ɛ}. Thus {x n } is bounded. Define z n = sup{x k } k n. Since {x n } is bounded, ech z n is finite rel number nd is bounded bove in bsolute vlue by M. If m > n, then z m is obtined by tking the sup of smller set thn is z n ; hence {z n } is decresing. By the gretest lower bound property, Z = {z n n N} hs n infimum. Let x = inf Z. We clim tht x n x. 6 For ech ɛ > 0 there is corresponding integer N such tht x z N x+ɛ. Since {x n } is Cuchy, by tking lrger N if necessry, we know tht k N x k [x N ɛ, x N +ɛ]. It follows tht z N [x N ɛ, x N + ɛ]. Hence for k N, x k x x k x N + x N z N + z N x ɛ + ɛ + ɛ = 3ɛ. 5 lest upper bound of E sup E. 6 Ide: Since {x n} is Cuchy, the terms of this sequence would pproch one nother. {z n} lso pproches {x n }. Since z n x, {z n } pproches x. It follows tht {x n } pproches x.
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 14 5 Completeness A metric spce X is complete if every Cuchy sequence of elements of X converges to n element of X. Lemm. A convergent sequence is Cuchy sequence. Proof. x n x mens ɛ > 0, N, such tht n N, x x n < ɛ. Hence x n x m x n x + x x m 2ɛ when n, m N. Thus, x n is Cuchy sequence. Lemm. If x n is Cuchy, then x n is bounded. Proof. If the sequence is x 1, x 2, x 3,..., ɛ > 0 nd N is such tht x n x m < ɛ if n, m > N, then for ny fixed m > N, the entire sequence is contined in the closed bll of center x m nd rdius mx{d(x m, x 1 ), d(x m, x 2 ),..., d(x m, x N ), ɛ}. Lemm. If x n is Cuchy nd x nk x, then x n x. Proof. Let ɛ > 0. Since x n is Cuchy choose k > 0 so lrge tht x n x m < ɛ whenever n, m k. Since x nk x choose l > 0 so lrge (i.e. n l lrge) tht x nj x < ɛ whenever j l. Set N = mx(k, n l ). If m, n j > N, then x m x x m x nj + x nj x < ɛ + ɛ = 2ɛ. Theorem. [, b] is complete. Proof. Let x n be Cuchy sequence in [, b]. Let x nk be monotone subsequence. Since x nk b, x nk converges (by the Lest Upper Bound). x nk c. Since [, b] is closed, c [, b]. Any x n tht is Cuchy in [, b], converges in [, b]. [, b] is complete. The bove theorem is specific cse of the following Lemm: Theorem. Let x n be Cuchy sequence in compct metric spce X. Then x n converges to some point of X. Proof. Since X is (sequentilly) compct, then for ny sequence x n X, there is subsequence x nk c, c X. Using the bove theorem (x n Cuchy nd x nk c x n c), we see tht x n c X. Theorem. R is complete. Proof. Let x n be Cuchy sequence in R. x n is bounded (by the Lemm bove). {x n } [, b], nd see bove. A direct consequence of the bove theorem is the following: In R k, every Cuchy sequence converges. Theorem. [0, 1) is not complete.
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 15 Bolzno-Weierstrss. Every bounded, infinite subset S R hs limit point. Proof. If I 0 is closed intervl contining S, denote by I 1 one of the closed hlf-intervls of I 0 tht contins infinitely mny points of S. Continuing in this wy, we define nested sequence of intervls {I n }, ech of which contins infinitely mny points of S. If c = n=1 I n, then it is cler tht c is limit point of S. Lemm. Every bounded sequence of R hs convergent subsequence. Proof. If sequence x n contins only finitely mny distinct points, the conclusion is trivil nd obvious. Otherwise we re deling with bounded infinite set, to which the Bolzno-Weierstrss theorem pplies, giving us limit point x. If, for ech integer k 1, x nk is point of the sequence such tht x nk x 1/k, then it is cler tht is convergent subsequence. x nk Theorem. A closed subspce Y of complete metrix spce X is complete. Proof. Let y n be Cuchy sequence in Y. Then y n is lso Cuchy sequence in X. Since X is complete, x X such tht y n x in X. Since Y is closed, x Y. Consequently, y n x in Y. Theorem. A complete subspce Y of metric spce X is closed in X. Proof. Suppose x X is limit point of Y. y n in Y tht converges to x. y n is Cuchy sequence in X; hence it is lso Cuchy sequence in Y. Since Y is complete, y n y Y. Since limits of sequences re unique, y = x nd x belongs to Y. Hence Y is closed. Theorem. V is normed spce. If we hve the following impliction v n < n=1 v n <, n=1 then V is complete (V is Bnch spce). Proof. Sy v n is Cuchy. Choose n k such tht m, n n k v m v n < 2 k. We my ssume tht n 1 < n 2 <.... v nk+1 v nk < 2 k k=1 v n k+1 v nk <. It follows tht n=1 (v n k+1 v nk ) <. s s K = K (v nk+1 v nk ) = v nk+1 v n1 v nk+1 converges v n converges. k=1
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 16 6 Compctness M is (sequentilly) compct if for ny sequence x n M, there is subsequence x nk c, c M. M is (topologiclly) compct if ny open cover of M, M G α, G α open, contins finite subcover. Problem (W 02, #2). [, b] is compct. Proof. Let x n be sequence in [, b]. Let x nk be monotone subsequence x nk b x nk c. Since [, b] is closed, c [, b]. [, b] is (sequentilly) compct. Lemm. If M is compct, every open cover of M hs countble subcover. Theorem. If M is sequentilly compct, then it is topologiclly compct. Proof. Sy tht M G 1 G2 hs countble subcover. Need to show tht there is finite subcover, i.e. M n k=1 G k for some n. Suppose tht fils for every n; then for every n = 1, 2,..., there would exist n x n M \ G k. k=1 Tht sequence would hve convergent subsequence {x nk }. Let x be its limit, x nk x. Then x would be contined in G m for some m, nd thus x nk G m for ll n k sufficiently lrge, which is impossible for n k > m (since x nk M\G 1 Gnk ). We hve reched contrdiction. So there must be finite subcovering. Problem (S 02, #3). If M is topologiclly compct, then it is sequentilly compct. Proof. Let x n M nd E be the rnge of {x n }. If E is finite, then there is x E nd sequence {n i }, with n 1 < n 2 <..., such tht x n1 = x n2 = = x The subsequence {x ni } converges to x. If E is infinite, E hs limit point x in M (s n infinite subset of compct set). Every neighborhood of x contins infinitely mny points of M. For ech k, B 1 (x) k contins infinitely mny x n s. Select one nd cll it x nk, such tht, n k > n k 1 >.... We hve subsequence {x nk } so tht d(x, x nk ) < 1 k 0. x n k x. Problem (F 02, #1). Let K be compct subset nd F be closed subset in the metric spce X. Suppose K F = φ. Prove tht 0 < inf{d(x, y) : x K, y F }. Proof. Given x K, x / F, d x = d(x, F ) > 0. Then, the bll centered t tht x with rdius d x /2, i.e. B dx 2 (x), stisfies B dx 2 (x) F = φ. Since x ws tken rbitrry, this is true x K, x / F. K x K B d x (x). Since K is compct, x 1,..., x n K, n <, such tht 2 K n k=1 B d x (x k ), nd B dx k (x k ) F = φ. Since min k {d xk } > 0, we hve 2 2 0 < inf{d(x, y) : x K, y F }.
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 17 7 Continuity Limits of Functions: lim x p f(x) = q if: ɛ > 0, δ such tht x E 0 < x p < δ f(x) q < ɛ. A function f is continuous t p: lim x p f(x) = f(p) if: ɛ > 0, δ such tht x E x p < δ f(x) f(p) < ɛ. Negtion: f is not continuous t p if: ɛ > 0, δ such tht x E x p < δ f(x) f(p) > ɛ. f is uniformly continuous on X if: ɛ > 0, δ such tht x, z X x z < δ f(x) f(z) < ɛ. Negtion: f is not uniformly continuous on X if: ɛ > 0, δ such tht x, z X x z < δ f(x) f(z) > ɛ. Exmples: f(x) = 1 x on (0, 1] nd f(x) = x2 on [1, ) re not uniformly continuous. Theorem. f : X Y is continuous f 1 (V ) is open in X for every open set V in Y. Proof. 7 Suppose f is continuous on X. Let V be n open set in Y. We hve to show tht every point of f 1 (V ) is n interior point of f 1 (V ). Let x f 1 (V ). Choose ɛ such tht B ɛ (f(x)) V. Since f is continuous 8, δ > 0 such tht f(b δ (x)) B ɛ (f(x)) V. Hence, B δ (x) f 1 (V ). Since f 1 (V ) contins n open bll bout ech of its points, f 1 (V ) is open. Suppose f 1 (V ) is open in X for every open set V in Y. Let x X nd let ɛ > 0. Then f 1 (B ɛ (f(x))) is open in X. Hence, δ such tht B δ (x) f 1 (B ɛ (f(x))). Applying f, we obtin f(b δ (x)) B ɛ (f(x)), nd so f is continuous. Problem (S 02, #4; S 03, #1). A function f : (0, 1) R is the restriction to (0, 1) of continuous function F : [0, 1] R f is uniformly continuous on (0, 1). Proof. We show tht if f : (0, 1) R is uniformly continuous, then there is continuous F : [0, 1] R with F (x) = f(x) for ll x (0, 1). Let x n be sequence in (0, 1) converging to 0. Since f is uniformly continuous, given ɛ > 0, δ, s.t. x y < δ f(x) f(y) < ɛ. Therefore, we hve f(x n ) f(x m ) < ɛ for n, m lrge enough. f(x n ) is Cuchy sequence, so it converges to some ξ. Define F (0) = lim n f(x n ) = ξ. We wnt to show tht this limit is well defined. Let y n be nother sequence, s.t. y n 0, so f(y n ) is Cuchy by the sme rgument. Since the sequence (f(x 1 ), f(y 1 ), f(x 2 ), f(y 2 ),...) is Cuchy by still the sme rgument, nd tht there is subsequence f(x n ) ξ, then the entire sequence converges to ξ. Thus, F (0) = lim n f(x n ) = ξ is well defined. By the sme set of rguments, F (1) = η. The function F : [0, 1] R given by f(x) for x (0, 1), F (x) = ξ for x = 0, η for x = 1. is the unique continuous extension of f to [0, 1]. F : [0, 1] R is continuous, nd [0, 1] is compct. Therefore, F is uniformly continuous on [0, 1]. Thus, f = F (0,1) : (0, 1) R is uniformly continuous. 7 Gmelin, Green, p. 26; Edwrds, p. 51. 8 f continuous t x ɛ > 0, δ such tht z B δ (x) f(z) B ɛ (f(x)), or f(b δ (x)) B ɛ (f(x)).
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 18 7.1 Continuity nd Compctness Theorem. Let f : X Y is continuous, where X is compct. Then f(x) is compct. Proof. 1) Let {V α } be n open cover of f(x), (V α Y ). Since f is continuous, f 1 (V α ) is open. Since X is compct, there re finitely mny α 1,..., α n, such tht X f 1 (V α1 )... f 1 (V αn ). Since f(f 1 (E)) E for every E Y, then f(x) V α1... Vαn. f(x) is compct. Proof. 2) Let {y n } be sequence in the imge of f. Thus we cn find x n X, such tht y n = f(x n ). Since X is compct the sequence {x n } hs convergent subsequence {x nk } with limit s X. Since f is continuous, lim y n k = lim f(x n k ) = f(s) k k Hence, the given sequence {y n } hs convergent subsequence which converges in f(x). f(x) is compct. Problem (F 01, #1). Let K R be compct nd f(x) continuous on K. Then f hs mximum on K (i.e. there exists x 0 K, such tht f(x) f(x 0 ) for ll x K). Proof. By theorem bove, the imge f(k) is closed nd bounded. Let b be its lest upper bound. Then b is dherent to f(k). Since f(k) is closed b f(k), tht is x 0 K, such tht b = f(x 0 ), nd thus f(x 0 ) f(x), x K. Theorem. Let f : [, b] R be continuous function. Then: 1) f is bounded; 2) f ssumes its mx nd min vlues; 3) f() < p < f(b) x : f(x) = p. Proof. 1) f is continuous f([, b]) is compct f([, b]) is closed nd bounded. 2) φ f([, b]) M. Let M 0 = sup f([, b]) M 0 closure(f([, b])) = f([, b]) x 0 [, b] : f(x 0 ) = M 0. 3) [, b] is connected f([, b]) is connected f([, b]) is n intervl. Theorem. f : X Y continuous nd X compct. Then f is uniformly continuous. Proof. 9 Suppose tht f is not uniformly continuous. Then there exist ɛ > 0 nd (setting δ = 1/k in the definition) points x k, z k X 10 such tht x k z k < 1/k while f(x k ) f(z k ) ɛ. Pssing to subsequence, we cn ssume tht x k x X. 11 Since x k z k 0, we lso obtin z k x. Since f is continuous, f(x k ) f(x) nd f(z k ) f(x), so tht f(x k ) f(z k ) f(x k ) f(x) + f(x) f(z k ) 0, contrdiction. 9 Gmelin, Green, p. 26-27; Rudin, p. 91. 10 See the technique of negtion in the beginning of the section. 11 Since {x k } is Cuchy, nd the convergent subsequence cn be constructed, x k x.
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 19 8 Sequences nd Series of Functions 8.1 Pointwise nd Uniform Convergence Suppose {f n }, n = 1, 2, 3,..., is sequence of functions defined on set E, nd suppose tht the sequence of numbers {f n (x)} converges x E. Define function f by f(x) = lim n f n(x), (x E). {f n } converges pointwise to f on E. A sequence of functions {f n } converges uniformly on E to f if ɛ > 0, N, such tht n N, f n (x) f(x) ɛ, x E. Exmple: Consider f(x) = x n on [0, 1] convergent, but not uniformly convergent on [0, 1]. Problem (F 01, #3). If {f n } is sequence of continuous functions on E, nd if f n f uniformly on E, then f is continuous on E. Proof. Fix ɛ > 0. Since f n f uniformly, choose N, s.t. n N f n (x) f(x) < ɛ 3, x E Since f n is continuous t p, choose δ, s.t. x E, x p < δ then f n (x) f n (p) < ɛ 3, n N Thus, f(x) f(p) f(x) f n (x) + f n (x) f n (p) + f n (p) f(p) < ɛ 3 + ɛ 3 + ɛ 3 = ɛ. Hence, given ny ɛ, δ, s.t. x p < δ f(x) f(p) < ɛ, nd f is continuous t p. 8.2 Normed Vector Spces Problem (F 03, #7). C 0 [, b] with the metric d(f, g) sup x [0,1] f(x) g(x) = f g is complete. Proof. Let {ϕ n } 1 be Cuchy sequence of elements of C 0 [, b]. Given ɛ > 0, choose N such tht m, n N ϕ m ϕ n < ɛ 2 (sup norm). Then, in prticulr, ϕ m (x) ϕ n (x) < ɛ/2 for ech x [, b]. Therefore {ϕ n (x)} 1 is Cuchy sequence of rel numbers, nd hence converges to some rel number ϕ(x). It remins to show tht the sequence of functions {ϕ n } converges uniformly to ϕ; if so, it will imply tht ϕ is continuous on [, b], i.e. ϕ C 0 [, b]. Clim: For n N, (N sme s bove, n fixed), ϕ(x) ϕ n (x) < ɛ for ll x [, b]. To see this, choose m N sufficiently lrge (depending on x) s.t. ϕ(x) ϕ m (x) < ɛ/2. ϕ(x) ϕ n (x) ϕ(x) ϕ m (x) + ϕ m (x) ϕ n (x) < ɛ/2+ ϕ m ϕ n < ɛ/2+ɛ/2 = ɛ. Since x [, b] ws rbitrry, it follows tht ϕ n ϕ < ɛ s desired.
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 20 Theorem. Let {f n } C 1 [, b], f n f pointwise, f n g uniformly. Then f n f uniformly, nd f is differentible, with f = g. Proof. By the Fundmentl Theorem of Clculus, we hve f n (x) = f n () + x f n n, x [, b]. From this nd the Uniform Convergence nd Integrtion theorem, we obtin f(x) = lim f n(x) = lim f n() + lim n n f(x) = f() + x g. n Another ppliction of the Fundmentl Theorem yields f = g s desired. To see tht convergence of f n f is uniform, note tht x x x f n (x) f(x) = f n g + f n () f() f n g + f n () f() x f n, (b ) f n g + f n () f(). The uniform convergence of f n therefore follows from tht of f n. Theorem. C 1 [, b], with the C 1 -norm defined by ϕ = mx ϕ(x) + mx x [,b] x [,b] ϕ (x), is complete. Proof. Let {ϕ n } 1 be Cuchy sequence of elements of C1 [, b]. Since mx ϕ m(x) ϕ n (x) ϕ m ϕ n x [,b] (C 1 norm), ϕ n is uniformly Cuchy sequence of continuous functions. Thus ϕ n ϕ C 0 [, b] uniformly. Similrly, since mx x [,b] ϕ m(x) ϕ n(x) ϕ m ϕ n (C 1 norm), ϕ n ψ C 0 [, b] uniformly. By the bove theorem, ϕ is differentible with ϕ = ψ, so ϕ C 1 [, b]. Since mx ϕ n(x) ϕ(x) = ϕ n ϕ (sup norm), nd x [,b] mx x [,b] ϕ n(x) ϕ (x) = ϕ n ϕ (sup norm) the uniform convergence of ϕ n nd ϕ n implies tht ϕ n ϕ with respect to the C 1 -norm of C 1 [, b]. Thus every Cuchy sequence in C 1 [, b] converges.
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 21 8.3 Equicontinuity A fmily F of functions f defined on set E X is equicontinuous on E if: ɛ > 0, δ, such tht x y < δ, x, y E, f F f(x) f(y) < ɛ. 8.3.1 Arzel-Ascoli Theorem Suppose {f n (x)} n=1 is uniformly bounded nd equicontinuous sequence of functions defined on compct subset K of X. Then {f n } is precompct, i.e. the closure of {f n } is compct, i.e. {f n } contins uniformly convergent subsequence, i.e. {f n } contins subsequence {f nk } tht converges uniformly on K to function f X. 9 Connectedness 9.1 Reltive Topology Define the neighborhood of point in R n s N ɛ (x) = {y : x y < ɛ}. Consider the subset of R n, M R n. If ll we re interested in re just points in M, it would be more nturl to define neighborhood of point x M s N M,ɛ = {y M : x y < ɛ}. Thus, the reltive neighborhood is just restriction of the neighborhood in R n to M. Reltive interior points nd reltive boundry points of set, s well s reltive open set nd reltive closed set, cn be defined ccordingly. Alterntive definitions: S M( R n ) is open reltive to M if there is n open set U in R n such tht S = U M. S M( R n ) is closed reltive to M if there is closed set V in R n such tht S = V M. Exmple: A set S = [1, 4) is open reltive to M = [1, 10] R since for the open set U = (0, 4) in R, we hve S = U M. Exmple: A set S = [1, 3] is open reltive to M = [1, 3] [4, 6] R since for the open set U = (0, 4) in R, we hve S = U M. 9.2 Connectedness X is connected if it cnnot be expressed s disjoint 12 union of two nonempty subsets tht re both open nd closed. i.e. M is connected if M = A B, such tht A, B open nd A B = φ, then A or B is empty; or, M is connected if M = A B, A, B open A = φ or B = φ. Fct: X is connected X nd φ re the only subsets which re clopen. X is disconnected if there re closed nd open subsets A nd B of X such tht A B = X, A B = φ, A φ, B φ. Another wy of phrsing: X is disconnected if there is closed nd open U X, such tht U φ nd U X. If there is such U, then the complement V = X \ U of U is lso both closed nd open nd X is the disjoint union of the nonempty sets U nd V. A subset of spce is connected subset if it is connected in the reltive topology. 12 A B = φ, then A nd B re disjoint.
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 22 Problem (S 02, #1). The closed intervl [, b] is connected. Proof. Let [, b] = G H, s.t. G H = φ. Let b H. Then clim: G = φ. If not, let c = sup G. Since G is closed, c G. Since G is open 13, B ɛ (c) G, i.e. [c, c + ɛ) G. Tht contrdicts c = sup G. Thus G = φ. Note: Since [, b) nd (, b) cn be expressed s the union of n incresing sequence of compct intervls, these re lso connected. Theorem. Let S α M, S α connected. Suppose S α φ. Then S α is connected. Proof. Let S = S α = G H, G, H re open in S α. Choose x 0 S α. S α = (S α G) (Sα H). Assume x0 G. Since S α is connected nd x 0 S α G, we get Sα H = φ, α. Therefore, ( Sα ) H = φ. Since H S α H = φ. Therefore, S is connected. Corollry. R is connected. Proof. Let R = n N [ n, n], 0 [ n, n]. Therefore, R is the union of connected subsets. By the theorem bove, R is connected. Theorem. Let f : M N is continuous nd M is connected. Then f(m) is connected. Proof. Sy f(m) = G H, G, H φ. G, H open. Then M = f 1 (G) f 1 (H), where f 1 (G) nd f 1 (H) re both open nd nonempty. Contrdicts connectedness of M. Theorem., b I, nd < c < b, then c I, i.e. I is n intervl I R is connected. Proof. Assume I n intervl. S = [, b]; S = (, b], ; S = [, b), b ; S = (, b),, b. [, b) = n n 0 [, b 1 n ]. [, b 1 n ]. Sy I is not n intervl. < c < b,, b I, c / I. I = ((, c) I) ((c, ) I), i.e. I is not connected. Problem (W 02, #3). The open unit bll in R 2, {(x, y) R 2 : x 2 + y 2 < 1} is connected. Proof. Let f θ (t) = t(cos θ, sin θ), 1 < t < 1. We hve f θ : ( 1, 1) (t cos θ, t sin θ). Since f θ is continuous nd ( 1, 1) is connected, f θ (( 1, 1)) is connected. The unit bll cn be expressed s {(x, y) R 2 : x 2 + y 2 < 1} = 0 θ<π f θ(t). We know tht the origin is contined in the intersection of f θ s. Therefore, 0 θ<π f θ(t) is connected by the theorem bove. 13 If M = G H, G H = φ nd G, H re open, then G is closed nd open, since G = H c. M = [0, 1] [2, 3] is not connected becuse if G = [0, 1], H = [2, 3], M = G H, G, H re clopen in M, G H = φ, nd G, H φ.
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 23 9.3 Pth Connectedness A pth in X from x 0 to x 1 is continuous function γ : [0, 1] X, such tht γ(0) = x 0 nd γ(1) = x 1. X is pth-connected if, for every pir of points x 0 nd x 1 in X, there is pth γ from x 0 to x 1. Theorem. A pth-connected spce is connected. Proof. Fix x 0 X. For ech x X, let γ x : [0, 1] X be pth from x 0 to x. By theorems bove, i.e. (1) X connected nd f : X Y continuous f(x) connected; (2) ny intervl in R is connected; ech γ x ([0, 1]) is connected subset of X. Ech γ x ([0, 1]) contins x 0 nd X = {γx ([0, 1]) : x X}, so tht the theorem bove shows tht X is connected. Theorem. An open subset of R n is connected it is pth-connected. Problem. Any subintervl of R (closed, open, or semiopen) is pth-connected. Proof. If, b belong to n intervl (of ny kind), then γ(t) = (1 t) + tb, 0 t 1, defines pth from to b in the intervl. Problem. If X is pth-connected nd f : X Y is mp, then f(x) is pthconnected. Proof. If p = f(x) nd q = f(y), nd γ is pth in X from p to q, then f γ is pth in f(x) from x to y.
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 24 10 Bire Ctegory Theorem A subset T X is dense in X if T = X, i.e. every point of X is limit point of T, or point of T, or both. A subset Y X is nowhere dense if Y hs no interior points, i.e. int(y ) = φ. Y is nowhere dense X \ Y is dense open subset of X. The interior of E is the lrgest open set in E, i.e. the set of ll interior points of E. Q = R, (R \ Q) = R. [0, 1] Q is not closed; (0, 1) Q is not open in R. Bire s Ctegory Theorem. Let {U n } n=1 be sequence of dense open subsets of complete metric spce X. Then n=1 U n is lso dense in X. (Any countble intersection of dense open sets in complete metric spce is dense.) Corollry. Let {E n } n=1 be sequence of nowhere dense subsets of complete metric spce X. Then n=1 E n hs empty interior. (In complete metric spce, no nonempty open subset cn be expressed s union of countble collection of nowhere dense sets.) Proof. We pply the Bire Ctegory Theorem to the dense open sets U n = X \ E n. Then, n=1 U n is dense in X. U n = (X \ E n ) = X \ n=1 n=1 n=1 E n Therefore, X \ n=1 E n is dense, n=1 E n is nowhere dense n=1 E n hs empty interior. A subset of X is of the first ctegory (i.e. Q) if it is the countble union of nowhere dense subsets. A subset (i.e. I, R) tht is not of the first ctegory is sid to be of the second ctegory. S R is F σ set if S = n=1 F n, F n closed. S R is G δ set if S = n=1 G n, G n open. Problem (W 02, #4). The set of irrtionl numbers I in R is not the countble union of closed sets (not n F σ set). Proof. Suppose I = F n, where ech F n is closed. R = n N F n {q} q Q Thus R cn be expressed s the countble union of closed sets. By (corollry to) the Bire Ctegory Theorem, since R is nonempty open subset, one of these closed sets hs nonempty interior. It cnnot be one of q s, nd since ny nonempty intervl contins rtionl numbers, it cnnot be one of F n s. Contrdiction. Problem (S 02, #2; F 02, #3). The set Q of rtionl numbers is not the countble intersection of open sets of R (not G δ set). Show tht there is subset of R which is not the countble intersection of open subsets. Proof. We tke complements in the preceding theorem. ech G n open. Then, I = R \ Q = R \ ( n N G n ) = n N(R \ G n ), Suppose Q = G n, where nd R \ G n is closed. Thus, I is F σ set, which contrdicts the previous theorem.
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 25 Problem (S 03, #3). Find S R such tht both (i) nd (ii) hold for S: (i) S is not the countble union of closed sets (not F σ ); (ii) S is not the countble intersection of open sets (not G δ ). Proof. Let A [0, 1] not F σ, B [2, 3] not G δ A B is neither F σ nor G δ. If A B is F σ, sy A B = F }{{} n closed A = A B [0, 1] = F n [0, 1] = (Fn [0, 1]) } {{ } closed F σ set contrdiction. If A B is G δ, sy A B = G n }{{} open B = A B ( 3 2, 7 2 ) = G n ( 3 2, 7 2 ) = (G n ( 3 2, 7 2 )) } {{ } open G δ set contrdiction. Problem. Q is not open, is not closed, but is the countble union of closed sets (F σ set). Proof. Since ny neighborhood (q ɛ, q + ɛ) of rtionl q contins irrtionls, Q hs no inner points. Q is not open. Since every irrtionl number i is the limit of sequence of rtionls Q is not closed. Since every one-point-set {x} R is closed nd Q is countble, sy (q n ) is sequence of ll rtionl numbers, we find tht Q = n N{q n } is the countble union of closed sets, i.e. Q is F σ. Problem. The set of isolted points of countble complete metric spce X forms dense subset of X. Proof. For ech point x X tht is not n isolted point of X, define U x = X \ {x}. Ech such U x is open nd dense in X, nd the intersection of the U x s consists precisely of the isolted points of X. By the Bire Ctegory Theorem, the intersection of the U x s is dense in X. Problem. Suppose tht F is subset of the first ctegory in metric spce X nd E is subset of F. Prove tht E is of the first ctegory in X. Show by n exmple tht E my not be of the first ctegory in the metric spce F. Proof. If F = F n, where ech F n is nowhere dense, then E = (E F n ), nd ech E F n is nowhere dense. For exmple, note the R is of first ctegory in R 2, but R is not of first ctegory in itself. Problem. Any countble union of sets of the first ctegory in X is gin of the first ctegory in X. Proof. A countble union of countble unions is countble union.
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 26 Problem. ) If, b R stisfy < b, then there exists rtionl number q (, b). b) The set Q of rtionl numbers is dense in R. Problem. The set of irrtionl numbers is dense in R. Proof. If i is ny irrtionl number, nd if q is rtionl, then q + i/n is irrtionl, nd q + i/n q. Problem. Regrd the rtionl numbers Q s subspce of R. Does the metric spce Q hve ny isolted points? Proof. The rtionls hve no isolted points. This does not contrdict the bove, i.e. The set of isolted points of countble complete metric spce X forms dense subset of X becuse the rtionls re not complete. Problem. Every open subset of R is union of disjoint open intervls (finite, semiinfinite, or infinite). Proof. For ech x U, let I x be the union of ll open intervls contining x tht re contined in U. Show tht ech I x is n open intervl (possibly infinite or semi-infinite), ny two I x s either coincide or re disjoint, nd the union of the I x s is U. 11 Integrtion 11.1 Riemnn Integrl Let [, b] be given intervl. A prtition P of [, b] is finite set of points x 0,..., x n : = x 0 < x 1 <... < x n 1 < x n = b, x i = x i x i 1. A Riemnn sum for f correspoding to the given prtition is S(P, f) = n f(x i) x i, x i 1 x i x i. i=1 Definition: f is Riemnn integrble on [, b], if A R such tht: ɛ > 0, δ > 0 such tht whenever S is Riemnn sum for f corresponding to ny prtition of [, b] with mx( x i ) < δ S A < ɛ. In this cse A is clled the Riemnn integrl of f between nd b nd is denoted s f dx. Alterntive Definition: f is Riemnn integrble on [, b] if: ɛ > 0, P such tht U(P, f) L(P, f) < ɛ. If f is bounded, there exist m nd M, such tht m f(x) M, x b. Hence, for every P, m(b ) S(P, f) M(b ), so tht S(P, f) is bounded. bounded function f. This shows tht Riemnn sums re defined for every
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 27 11.2 Existence of Riemnn Integrl Theorem. f is integrble on [, b] ɛ > 0, δ > 0 such tht S 1 (P, f), S 2 (P, f), P with mx( x i ) < δ, then S 1 S 2 < ɛ. Proof. Suppose f is integrble on [, b]. ɛ > 0, δ > 0 such tht S(P, f), P with mx( x i ) < δ, then S f(x)dx < ɛ/2. For such S 1 nd S 2, S 1 S 2 = ( S1 f(x)dx ) ( S 2 f(x)dx ) < ɛ 2 + ɛ 2 = ɛ. ɛ > 0, δ > 0 such tht S 1 (P, f), S 2 (P, F ), P with mx( x i ) < δ S 1 S 2 < ɛ. For n = 1, 2,..., choose S (n) (P, f), P with mx( x i ) < 1/n. Then, ɛ > 0, N > 0 (δ = 1/N), such tht S (n) S (m) < ɛ, n, m N S (n) is Cuchy sequence of rel numbers S (n) converges to some A R S (N) A < ɛ, 1/N < δ. Thus for ny S(P, f), P with mx( x i ) < δ, we hve S A S S (N) + S (N) A < 2ɛ. Theorem. If f is continuous on [, b] then f is integrble on [, b]. Proof. Since f is uniformly continuous on [, b], ɛ > 0, δ > 0 such tht x, z [, b], x z < δ f(x) f(z) < ɛ. If P is ny prtition of [, b] with mx( x i ) < δ, then implies tht M i m i ɛ, i = 1,..., n, nd therefore n n U(P, f) L(P, f) = (M i m i ) x i ɛ x i = ɛ(b ). i=1 Thus, ɛ > 0, P such tht U(P, f) L(P, f) < Cɛ f is integrble. 11.3 Fundmentl Theorem of Clculus Theorem. If f : [, b] R is continuous, nd F : [, b] R is defined by F (x) = x f(t)dt, then F is differentible nd F = f. Proof. Since f is continuous, F (x) = x f(t)dt is defined for ll x [, b]. We hve to show tht for ny fixed x 0 [, b] F (x) F (x 0 ) lim = f(x 0 ). x x 0 x x 0 For ny x [, b], x x 0, we hve F (x) F (x 0 ) x f(x 0 ) x x 0 = f(t)dt x0 f(t)dt x f(x 0 ) x x 0 = x 0 f(t)dt f(x 0 ) x x 0 x x x x = x 0 f(t)dt x 0 f(x 0 )dt x x 0 x x 0 = x 0 (f(t) f(x 0 ))dt x x 0 x 0 f(t) f(x 0 ) dt x x 0 Since f is continuous t x 0, given ɛ > 0, δ, such tht x [, b], x x 0 < δ f(x) f(x 0 ) < ɛ. Then for ny t between x nd x 0, we hve f(t) f(x 0 ) < ɛ. x x < 0 ɛ dt x x 0 = ɛ. Thus F (x 0 ) = f(x 0 ). i=1
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 28 Corollry. If f : [, b] R is continuous nd F = f on [, b], then f(t)dt = F (b) F (). Proof. Since d dx ( x f(t)dt F (x)) = f(x) f(x) = 0, x f(t)dt F (x) is constnt. Thus x f(t)dt = F (x) + c, for some c R. In prticulr, 0 = f(t)dt = F () + c, so tht c = F (). Therefore, x f(t)dt = F (x) F (). Hence, f(t)dt = F (b) F (). Integrtion by Prts. Suppose f nd g re differentible functions on [, b], f, g R. Then f(x)g (x)dx = f(b)g(b) f()g() f (x)g(x)dx. Proof. Let h(x) = f(x)g(x) nd pply the Fundmentl Theorem of Clculus to h nd its derivtive. Note tht h R. h (x)dx = h(b) h() (f (x)g(x) + f(x)g (x))dx = f(b)g(b) f()g(). Men Vlue Theorem for Integrls. If f : [, b] R is continuous, then f(x)dx = f(c)(b ) for some c [, b]. Proof. Since f is continuous, by the Fundmentl Theorem of Clculus, there is function F : [, b] R such tht F (x) = f(x) for x (, b), nd f(x)dx = F (b) F (). By the Men Vlue Theorem for Differentition, c (, b) such tht F (b) F () = F (c)(b ). Thus, f(x)dx = }{{} F T C F (b) F () = }{{} MV T F (c)(b ) = Thus, c (, b) such tht f(x)dx = f(c)(b ). }{{} F T C f(c)(b ). Generlized Men Vlue Theorem for Integrls. If f, g : [, b] R re continuous nd g(x) > 0 for ll x [, b], then there exists c [, b] such tht f(x)g(x)dx = f(c) g(x)dx. Proof. Since g(x) > 0 for ll x [, b] nd since g is continuous, g(x) > 0. Suppose f(x) ttins its mximum M t x 2 nd minimum m t x 1. Then m = f(x 1 ) f(x) f(x 2 ) = M for x [, b], nd m nd hence m g(x)dx f(x)g(x)dx g(x)dx M. f(x)g(x)dx M g(x)dx, Since f is continuous on compct [, b], c such tht f(c) = f(x)g(x)dx g(x)dx.
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 29 Uniform Convergence nd Integrtion. Let {f n } be sequence of continuous functions on [, b] nd f n f uniformly on [, b]. Then lim f n(x) dx = lim n n } {{ } f(x) f n (x)dx. Proof. Let f = lim n f n. Since ech f n is continuous nd f n f uniformly f is continuous. In prticulr, f is integrble on [, b]. By the definition of uniform convergence, ɛ > 0, N > 0 such tht n > N, f(x) f n (x) < ɛ/(b ), x [, b]. Thus, ɛ b f(x) f n(x) or ɛ (f(x) f n (x))dx ɛ f(x)dx ɛ, x [, b] b f n (x)dx ɛ. The lst inequlity holds for ll n > N, nd therefore, lim n f n(x)dx = f(x)dx. Uniform Convergence nd Differentition. Let {f n } be sequence of differentible functions on [, b] such tht {f n (x 0 )} converges for some x 0 [, b]. If f n f uniformly, then f n f uniformly on [, b], nd ( ) lim f n = lim n n f n Proof. See the section on Sequences nd Series of Functions: Normed Vector Spces where the weker sttement is proved, i.e. {f n } C 1, f n f pointwise on [, b].
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 30 12 Differentition 12.1 R R 12.1.1 The Derivtive of Rel Function Let f : (, b) R, x 0 (, b). f is differentible t x 0 if f (x 0 ) = lim x x 0 f(x) f(x 0 ) x x 0 exists. f is the derivtive of f. Theorem. f : (, b) R. f is differentible t x 0 f is continuous t x 0. Proof. f(x) f(x 0 ) lim (f(x) f(x 0 )) = lim lim (x x 0 ) = f (x 0 ) 0 = 0. x x 0 x x 0 x x 0 x x 0 Since lim x x0 f(x) = f(x 0 ), f is continuous t x 0. Lemm. If f (c) > 0, then f is loclly strictly incresing t c, i.e., δ > 0 such tht: c δ < x < c f(x) < f(c), c < x < c + δ f(c) < f(x). Proof. lim x c f(x) f(c) x c > 0 δ > 0: f(x) f(c) x c > 0 whenever 0 < x c < δ. Thus since f(x) f(c) = f(x) f(c) x c (x c), (x c), we hve c δ < x < c f(x) f(c) < 0, c < x < c + δ f(x) f(c) > 0. Corollry. If f hs mx (or min) t c (, b), i.e. f(x) f(c) (or f(x) f(c)) for ll x, then f (c) = 0. Proof. Sy f (c) 0. Sy f (c) > 0. Then x > c f(x) f(c) (since f hs mx t c), contrdicting the lemm bove. Proofs of other conditions re similr. 12.1.2 Rolle s Theorem Theorem. Let f be continuous on [, b], differentible on (, b), nd f() = f(b). Then c (, b) : f (c) = 0. Proof. Let M = sup{f(x) : x [, b]}, m = inf{f(x) : x [, b]}. Then m f() = f(b) M. If M = m f(x) = f(), x f (c) = 0, c (, b). Sy f() = f(b) < M. Then choose c [, b] : f(c) = M. From, c (, b). We hve from corollry, f (c) = 0. Similrly, sy m < f() = f(b). Then choose c [, b] : f(c) = m. From, c (, b). We hve from corollry, f (c) = 0. 12.1.3 Men Vlue Theorem Theorem. Let f be continuous on [, b], differentible on (, b). Then c (, b): f (c) = f(b) f(). b
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 31 Proof. Let g(x) = f(x) f(b) f() b (x ). Then g() = g(b) = f(). By Rolle s Theorem, c (, b), such tht 0 = g (c) = f (c) f(b) f() b. Corollry. Let f be continuous on [, b], differentible on (, b). ) f (x) = 0, x (, b) f is constnt. b) f (x) > 0, x (, b) f is strictly incresing. Proof. b) x 1 < x 2 b. By men vlue theorem, f(x 2 ) f(x 1 ) = f (c)(x 2 x 1 ) for some c (x 1, x 2 ). Therefore, f(x 2 ) f(x 1 ) > 0 for ll such x 1, x 2. Proof of ) is similr. 12.2 R R m f : R R m is differentible t c R if there exists liner mp L : R R m such tht f(c + h) f(c) L(h) lim = 0. h 0 h f 1 (c) in which cse L is defined by L = df c = f (c) =. f m(c) The liner mpping df c : R R m is clled the differentil of f t c. The mtrix of the liner mpping f (c) is the derivtive. The differentil is the liner mpping whose mtrix is the derivtive. 12.3 R n R m Let U R n open, c U. f : U R m is differentible t c if there exists liner mp L : R n R m such tht f(c + h) f(c) L(h) lim = 0. h 0 h in which cse L is defined by L = df c = f (c) = f 1 f 1 x n x 1 f m x 1 f m x n Let v R n, c G R n. The directionl derivtive with respect to v of f t c is D v f(c) = lim t 0 f(c + tv) f(c) t provided tht the limit exists. In prticulr, the prtil derivtives of f t c re x=c. f f(c + te j ) f(c) f(c 1,..., c j + t,..., c n ) f(c 1,..., c j,..., c n ) (c) = D ej f(c) = lim = lim x j t 0 t t 0 t
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 32 Theorem. Let f be differentible t c (i.e., df c : R n R m is defined). Then: ) 14 The directionl derivtive D v f(c) exists v R n, nd D v f(c) = df c (v) ( = f (c)v). b) 15 The mtrix f (c) of df c is df c = f (c) = [ f i x j (c)]. Proof. ) Since f is differentible, nd letting h = tv, f(c + tv) f(c) df c (tv) lim = 0, t 0 tv 1 [ v lim t 0 f(c + tv) f(c) t f(c + tv) f(c) lim = df c (v). t 0 t ] df c (v) = 0, i.e. D v f(c) exists, nd equls to df c (v). Proof. b) df c = df c (e 1 ) df c (e n ) = D e1 f(c) D en f(c) = f x 1 (c) [ ] fi x j (c). m n f : R n R m is sid to be continuously differentible t c if the prtil derivtives f i x j exist on B ɛ (c), nd re continuous t c. 14 Edwrds, Theorem 2.1 15 Edwrds, Theorem 2.4 f x n (c) =
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 33 Theorem 16. If f i x j exist nd re continuous on G, then t ech c G, df c = f (c) = [ fi ] (c). (i.e. f i x j x j continuous f differentible). Proof. Since f : R n R m is differentible t c iff ech of its component functions f 1,..., f m is, we my ssume m = 1; f : U R. Given h = (h 1,..., h n ), let h 0 = (0,..., 0), h j = (h 1,..., h j, 0,..., 0), j = 1,..., n. We hve f(c + h) f(c) = n [f(c + h j ) f(c + h j 1 )]. j=1 f(c + h j ) f(c + h j 1 ) = f(c 1 + h 1,..., c j 1 + h j 1, c j + h j, c j+1,..., c n ) for some 0 t h j, by men-vlue theorem. Thus f(c 1 + h 1,..., c j 1 + h j 1, c j, c j+1,..., c n ) = f x j (c 1 + h 1,..., c j 1 + h j 1, c j + t, c j+1,..., c n ) h j f(c + h j ) f(c + h j 1 ) = f x j (d j ) h j, for some d j, d j c h. f(c + h) f(c) = Also considering f (c) h = n j=1 f x j (d j ) h j [ f x 1 (c) f x n (c) f(c + h) f(c) f n (c) h) j=1 lim = lim h 0 h h 0 n lim h 0 j=1 ] h 1. h n since ech d j c s h 0, nd ech f x j is continuous t c., we hve [ f x j (d j ) f x j (c) ] h j h f (d j ) f (c) = 0, x j x j 16 Edwrds, Theorem 2.5
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 34 12.3.1 Chin Rule Theorem. U R n, V R m. If the mppings F : U R m nd G : V R k re differentible t U nd F() V respectively, then their composition H = G F is differentible t, nd dh = dg F () df (composition of liner mppings) In terms of derivtives H () = G (F ()) F () (mtrix multipliction) The differentil of the composition is the composition of the differentils; the derivtive of the composition is the product of the derivtives. Proof. We must show tht H( + h) H() dg F () df (h) lim = 0. Define h 0 h ϕ(h) = F ( + h) F () df (h) h F ( + h) F () = df (h)+ h ϕ(h) nd (12.1) ψ(k) = G(F () + k) G(F ()) dg F ()(k) k G(F ()+k) G(F ()) = dg F () (k)+ k ψ(k) (12.2) The fct tht F nd G re differentible t nd F (), respectively, implies tht lim ϕ(h) = lim ψ(k) = 0. Then h 0 k 0 H( + h) H() = G(F ( + h)) G(F ()) = G(F () + (F ( + h) F ())) G(F ()) = [k=f (+h) F () in (12.2)] = dg F () (F ( + h) F ()) + F ( + h) F () ψ(f ( + h) F ()) = (12.1) = dg F () (df (h) + h ϕ(h)) + df (h) + h ϕ(h) ψ(f ( + h) F ()) = dg F () df (h) + h dg F () (ϕ(h)) + h df ( h h ) + ϕ(h) ψ(f ( + h) F ()) H( + h) H() dg F () df (h) = dg h F () (ϕ(h)) + df ( h h ) + ϕ(h) ψ(f ( + h) F ()) dg F () is liner continuous, nd lim h 0 ϕ(h) = 0 lim h 0 dg F () (ϕ(h)) = 0. Since F is continuous t nd lim k 0 ψ(k) = 0 lim h 0 ψ(f ( + h) F ()) = 0. df is liner continuous bounded M, df (x) M x. Therefore, the limit of the entire expression bove 0 dh = dg F () df.
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 35 Theorem. Let U R n be open nd connected. F : U R m. F (x) = 0 for ll x U F is constnt. Proof. Since F is constnt ech of its component functions is constnt, nd the mtrix F (x) = 0 ech of its rows is 0, we my ssume F = f : U R. Suppose f (x) = f(x) = 0, x U. Given nd b U, let γ : R U be differentible mpping with γ(0) =, γ(1) = b. R γ R n f R If g = f γ : R R, then g (t) = f(γ(t)) γ (t) = 0 t R g is constnt on [0, 1], so f() = f(γ(0)) = g(0) = g(1) = f(γ(1)) = f(b). Exmple 1. w = w(x, y), x = x(r, θ), y = y(r, θ) [ w r w θ ] = [ w x w y ] [ x r y r x θ y θ ] [ w x w y ] = [ w r w θ ] [ x r y r x θ y θ ] 1 f z Exmple 2. f(x, y, z) = 0, 0 z = h(x, y), f(x, y, h(x, y)) 0. For exmple, cn solve for z x : 0 = f x xf(x, y, h(x, y)) = x x + f y f x + f z z x. y x + f z z x = f x / f z. Similrly, cn solve for y x z nd y, from z f(x, y(x, z), z) nd y f(x(y, z), y, z), respectively, nd show x y z y z x = 1. 12.3.2 Men Vlue Theorem z x = Theorem. Let f : U R n R be differentible, nd nd b U, such tht [, b] U. Then c (, b), such tht f(b) f() = f(c) (b ). Proof. Define γ : [0, 1] [, b] s γ(t) = + t(b ) = (1 t) + tb, t [0, 1]. Then γ (t) = b. Let g = f γ. R γ R n f R Since g : [0, 1] R, then by single-vrible MVT, ξ [0, 1], such tht g(1) g(0) = g (ξ). If c = γ(ξ), then f(b) f() = f(γ(1)) f(γ(0)) = g(1) g(0) = g (ξ) = = Chin Rule = f(γ(ξ)) γ (ξ) = f(c) (b ).
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 36 12.3.3 x ( f y ) = y ( f x ) Theorem. Let f : U R n R. If f, f 2 f x y, 2 f y x x x, f y exist on U nd re continuous t, then ( f ) = y y ( f ) x on U. exist nd re continuous on U nd Proof. Consider S h (x, y) = f(x + h, y + h) f(x + h, y) f(x, y + h) + f(x, y). Let g(x, y) = f(x + h, y) f(x, y). Then, S h (x, y) = f(x + h, y + h) f(x + h, y) f(x, y + h) + f(x, y) = g(x, y + h) g(x, y) = = MV T = h g y (x, y + βh) = h[ f f (x + h, y + βh) y y (x, y + βh)] = = MV T = h 2 ( f ) (x + αh, y + βh) x y where 0 < α, β < 1. Let r(x, y) = f(x, y + h) f(x, y). Then, S h (x, y) = f(x + h, y + h) f(x + h, y) f(x, y + h) + f(x, y) = r(x + h, y) r(x, y) = = MV T = h r x (x + α h, y) = h [ f x (x + α h, y + h) f x (x + α h, y) ] = = MV T = h 2 ( f ) (x + α h, y + β h) y x where 0 < α, β < 1. ( For ech smll enough h > 0, f ) ( x y (x + αh, y + βh) = f y x) (x + α h, y + β h). Since the mixed prtil derivtives re continuous t = (x, y), let h 0 ( f ) ( x y (x, y) = f y x) (x, y).
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 37 12.4 Tylor s Theorem n-th degree Tylor polynomil of f t ; (h = x ) P n (h) = f() + f ()h + + f (n) () h n n! Men Vlue Theorem (R R, Revisited). f : [, b] R. Suppose tht f exists on [, b]. h = b. Then ξ between nd b such tht R 0 (h) = f (ξ)h f( + h) = f() + f (ξ)h = P 0 (h) + R 0 (h) Proof. Need to show: R 0 (h) = f (ξ)h. For t [0, h], define R 0 (t) = f( + t) P 0 (t) = f( + t) f(). So, R 0 (t) = f ( + t). Define ϕ : [0, h] R by ϕ(t) = R 0 (t) R 0(h) t ϕ(0) = ϕ(h) = 0 h By Rolle s theorem, c (0, h) such tht 0 = ϕ (c) = R 0(c) R 0(h) h For ξ = + c, R 0 (h) = f (ξ)h. = f ( + c) R 0(h) h Tylor s Theorem (R R). f : [, b] R. Suppose tht f (n+1) exists on [, b]. h = b. Then ξ between nd b such tht R n (h) = f (n+1) (ξ) (n + 1)! hn+1. f( + h) = f() + f ()h + + f (n) () h n + f (n+1) (ξ) n! (n + 1)! hn+1 = P n (h) + R n (h). Proof. Need to show: R n (h) = f (n+1) (ξ) (n+1)! h n+1. For t [0, h], define R n (t) = f( + t) P n (t), nd note tht R n (0) = R n(0) = = R (n) n (0) = 0 (12.3) since the first n derivtives of P n (t) t 0 gree with those of f t. Also, R (n+1) n (t) = f (n+1) ( + t) (12.4) since P (n+1) n (t) 0 becuse P n (t) is polynomil of degree n. Define ϕ : [0, h] R by ϕ(t) = R n (t) R n(h) h n+1 tn+1 ϕ(0) = ϕ(h) = 0 By Rolle s theorem, c 1 (0, h) such tht ϕ (c 1 ) = 0. It follows from (12.3) nd (12.4) tht ϕ(0) = ϕ (0) = = ϕ (n) (0) = 0 (12.5)
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 38 ϕ (n+1) (t) = f (n+1) ( + t) R n(h) (n + 1)! (12.6) hn+1 Therefore, we cn pply Rolle s theorem to ϕ on [0, c 1 ] to obtin c 2 (0, c 1 ) such tht ϕ (c 2 ) = 0. By (12.5), ϕ stisfies the hypothesis of Rolle s theorem on [0, c 2 ], so we cn continue in this wy. After n + 1 pplictions of Rolle s theorem, we obtin c n+1 (0, h) such tht ϕ (n+1) (c n+1 ) = 0. From (12.6) we obtin R n (h) = f (n+1) (ξ) (n+1)! h n+1 where ξ = + c n+1. Problem (F 03, #5). Assume f : R 2 R is function such tht ll prtil derivtives of order 3 exist nd re continuous. Write down (explicitly in terms of prtil derivtives of f) qudrtic polynomil P (x, y) in x nd y such tht f(x, y) P (x, y) C(x 2 + y 2 ) 3 2 for ll (x, y) in some smll neighborhood of (0, 0), where C is number tht my depend on f but not on x nd y. Then prove the bove estimte. Proof. Tylor expnd f(x, y) round (0, 0): f(x, y) = f(0, 0) + xf x (0, 0) + yf y (0, 0) + 1 2! [ x 2 f xx (0, 0) + 2xyf xy (0, 0) + y 2 f yy (0, 0) ] + 1 3![ x 3 f xxx (0, 0) + 3x 2 yf xxy (0, 0) + 3xy 2 f xyy (0, 0) + y 3 f yyy (0, 0) ] + O(x 4 ) f(x, y) P 2 (x, y) = 1 3![ x 3 f xxx (ξ, η) + 3x 2 yf xxy (ξ, η) + 3xy 2 f xyy (ξ, η) + y 3 f yyy (ξ, η) ] Note tht x 3, x 2 y, xy 2, y 3 (x 2 + y 2 ) 3 2. Also, since 3 rd order prtil derivtives re continuous, C 1 R s.t. mx{f xxx, 3f xxy, 3f xyy, f yyy } < C 1 4 in some nbd of (0, 0). Thus, f(x, y) P 2 (x, y) 1 3! (x2 + y 2 ) 3 2 C 1 C(x 2 + y 2 ) 3 2.
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 39 Problem (F 02, #5). Suppose f : R 2 R hs prtil derivtives t every point bounded by A > 0. () Show tht there is n M such tht f((x, y)) f((x 0, y 0 )) M((x x 0 ) 2 + (y y 0 ) 2 ) 1 2 (b) Wht is the smllest vlue of M (in terms of A) for which this lwys works? (c) Give n exmple where tht vlue of M mkes the inequlity n equlity. Proof. () Since f f x A, y A, by the Men Vlue Theorem, f(x, y) f(x 0, y) A x x 0 f(x, y) f(x, y 0 ) A y y 0 f(x, y) f(x 0, y 0 ) f(x, y) f(x 0, y) + f(x 0, y) f(x 0, y 0 ) A( x x 0 + y y 0 ) A 2( x x 0 2 + y y 0 2 ) 1 2 (b) This lwys works for M = A 2. (c) If x x 0 = y y 0, then we hve n equlity in, since then A( x x 0 + y y 0 ) A 2( x x 0 2 + y y 0 2 ) 1 2 2A x x 0 A 2(2 x x 0 2 ) 1 2 2A x x 0 2A x x 0 Problem (F 03, #2). Let f : R R be infinitely differentible function. Assume tht x [0, 1], m > 0, such tht f (m) (x) 0. Prove tht M such tht the following stronger sttement holds: x [0, 1], m > 0, m M such tht f (m) (x) 0. Proof. There re uncountbly mny x α s in [0, 1], nd for ech x α [0, 1], m α > 0 such tht f (m α) 0 for [x α ɛ α, x α + ɛ α ], for some ɛ α > 0 (since f (m α) is continuous). Let ɛ = min α (ɛ α ). Prtition [0, 1] into n = 1/ɛ subintervls (since ɛ > 0, n < ): 0 < ɛ = x 0 < x 1 <... < x n = 1 ɛ < 1, such tht x 0 = ɛ, x i = x 0 + iɛ, i = 1,..., n. Thus [0, 1] is covered by finitely mny overlpping intervls [x i ɛ, x i + ɛ]. For ech x i, i = 1,..., n, m i > 0 such tht f (mi) (x i ) 0 on [x i ɛ, x i + ɛ]. Tke M = mx 0 i n (m i ). Thus, x [0, 1], m > 0, m M such tht f (m) (x) 0.
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 40 Problem (S 03, #4). Consider the following eqution for function F (x, y) on R 2 : 2 F x 2 = 2 F y 2 () Show tht if function F hs the form F (x, y) = f(x+y)+g(x y) where f : R R nd g : R R re twice differentible, then F stisfies the eqution. (b) Show tht if F (x, y) = x 2 + bxy + cy 2,, b, c R, stisfies then F (x, y) = f(x + y) + g(x y) for some polynomils f nd g in one vrible. Proof. () Let ξ(x, y) = x + y, η(x, y) = x y, so F (x, y) = f(ξ(x, y)) + g(η(x, y)). By Chin Rule, nd similrly F x = df dξ ξ x + dg dη η x = df dξ 1 + dg dη 1 = df dξ + dg dη, 2 F x 2 = ( df x dξ + dg ) = d2 f dη dξ 2 ξ x + d2 g dη 2 η x = d2 f dξ 2 + d2 g dη 2, F y = df dξ dg dη, nd 2 F x 2 = d2 f dξ 2 + d2 g dη 2, nd thus, 2 F x 2 = 2 F y 2. (b) Suppose F (x, y) = x 2 + bxy + cy 2,, b, c R stisfies, then F x = 2x + by 2 F x 2 = 2, F y = 2cy + bx 2 F = 2c = c. y2 F (x, y) = x 2 + bxy + y 2 = (x 2 + y 2 ) + bxy = (x + y)2 + (x y) 2 + b (x + y)2 (x y) 2. 2 4 12.5 Lgrnge Multipliers Theorem. Let f nd g be C 1 on R 2. Suppose tht f ttins its mximum or minimum vlue on the zero set S of g t the point p where g(p) 0. Then f(p) = λ g(p) for some number λ. Problem (S 03, #5). Consider the function F (x, y) = x 2 + 2bxy + cy 2 on the set A = {(x, y) : x 2 + y 2 = 1}. () Show tht F hs mximum nd minimum on A. (b) Use Lgrnge multipliers to show tht if the mximum of F ( on A occurs ) t point b (x 0, y 0 ), then the vector (x 0, y 0 ) is n eigenvector of the mtrix. b c Proof. () Since F is continuous nd the circle is closed nd bounded, F ttins both mximum nd minimum vlues on the unit circle g(x, y) = x 2 + y 2 1 = 0. (b) Applying the bove theorem, we obtin the three equtions (for x, y, λ) 2x + 2by = 2λx, 2bx + 2cy = 2λy, x 2 + y 2 = 1. ( ) ( ) ( ) ( ) b x x x0 = λ, nd is n eigenvector of this mtrix. b c y y y 0
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 41 13 Successive Approximtions nd Implicit Functions 13.1 Contrction Mppings C R n. The mpping ϕ : C C is contrction mpping with contrction constnt k < 1 if ϕ(x) ϕ(y) k x y x, y C. Contrction Lemm. If M is complete nd ϕ : M M is contrction mpping with k < 1, then ϕ hs unique fixed point x. Proof. x n+1 x n = ϕ(x n ) ϕ(x n 1 ) k x n x n 1 k n x 1 x 0. If m > n > 0, then x m x n x m x m 1 + + x n+1 x n (k m 1 + + k n ) x 1 x 0 k n x 1 x 0 (1 + k + k 2 + ) kn 1 k x 1 x 0. Thus the sequence {x n } is Cuchy sequence, nd therefore converges to point x. Letting m, we get x x n kn 1 k x 1 x 0. Since ϕ is contrction, it is continuous. Therefore, ϕ(x ) = lim n ϕ(x n) = lim n x n+1 = x. If x were nother fixed point of ϕ, we would hve x x = ϕ(x ) ϕ(x ) k x x. Since k < 1, it follows tht x = x, so x is the unique fixed point of ϕ. 13.2 Inverse Function Theorem Lemm. 17 Suppose f : R n R n is C 1, 0 W, f(0) = 0, f (0) = I. f (x) I ɛ, x B r. Then Suppose B (1 ɛ)r f(b r ) B (1+ɛ)r. (13.1) If U = B r f 1 (B (1 ɛ)r ), then f : U B (1 ɛ)r is bijection, nd the inverse mpping g : V U is differentible t 0. Proof. f (x) I ɛ < 1, x B r. Apply MVT 18 to g(x) = f(x) x., b B r, then (f(b) b) (f() ) = g(b) g() g (ξ) b = f (ξ) I b ɛ b (13.2) (1 ɛ) b f(b) f() (1 + ɛ) b (13.3) The left-hnd inequlity shows tht f is 1-1 on B r. The right-hnd inequlity (with = 0) shows f(b r ) B (1+ɛ)r. 17 Problem F 01, # 6. 18 f : U R n R m is C 1 f(b) f() b mx x L f (x).
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 42 To show B (1 ɛ)r f(b r ), we use contrction mpping theorem. Given y B (1 ɛ)r, define ϕ : R n R n by ϕ(x) = x f(x) + y. We wnt to show tht ϕ is contrction mpping of B r ; its unique fixed point will then be the desired point x B r such tht f(x) = y. To see tht ϕ mps B r into itself: ϕ(x) f(x) x + y (by (13.2) with = 0) ɛ x +(1 ɛ)r ɛr+(1 ɛ)r = r, so if x B r, then ϕ(x) B r. Thus, ϕ(b r ) B r. To see ϕ : B r B r is contrction mpping, note tht ϕ(b) ϕ() = f(b) f() (b ) ɛ b Thus, ϕ hs unique fixed point x, ϕ(x ) = x, such tht f(x ) = y. From the sttement of theorem, U = B r f 1 (B (1 ɛ)r ). f is bijection of U onto B (1 ɛ)r. It remins to show tht g : V U is differentible t 0, where g(0) = 0. Need to show g(h) g(0) h g(h) h lim = lim = 0. h 0 h h 0 h This will prove tht g (0) = I. Applying (13.2) with = 0, b = g(h), h = f(b), we get g(h) h ɛ b by (13.3) Therefore, lim h 0 g(h) h h = 0, with g (0) = I. ɛ 1 ɛ f(b) = ɛ 1 ɛ h Theorem. 19 Suppose f : W R n R n is C 1, W, b = f(), nd the mtrix f () is nonsingulr. 20 Then there exist 21 open sets U W of nd V of b, such tht f mps U bijectively onto V. ( 1-1 C 1 mpping g : V W such tht g(f(x)) = x for x U, f(g(y)) = y for y V. ) Also, for ll y V (y = f(x)), g = f 1 stisfies g (y) = g (f(x)) = f (x) 1. Proof. Fix U nd let b = f(). Put T = f (), mtrix / liner mp. Define f(h) = T 1 (f( + h) b) = T 1 (f( + h) f()) Note f(0) = T 1 (0) = 0; f (0) = T 1 f () = T 1 T = I. Thus, by previous Lemm, U 0 open, 0 U 0, such tht f : U 0 V 0 is bijection, 0 V 0. f mps U 0 bijectively onto V 0 contining 0. Lets express f in terms of f. f( + h) f() = T f(h). Let x = + h: f(x) f() = T f(x ), 19 f is loclly one-to-one in E ech point x E hs neighborhood in which f is 1-1. 20 Jcobin of f = det f () = f i x j () = 0. 21 i.e. A C 1 mp f : W V is loclly invertible t there exist open sets U W of nd V of b = f(), nd C 1 mp g : V U such tht f nd g re inverse to ech other.
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 43 f(x) = T f(x ) + f(), f : U 0 + T } {{ } f(v 0 ) + f(), bijection. U Let s compute f 1 : Let y = T f(x ) + f() T f(x ) = y f() f(x ) = T 1 (y f()) x = f 1 (T 1 (y f())) = g(t 1 (y f())) f 1 (y) = x = g(t 1 (y f())) + (f 1 ) (y) = ( g) (T 1 (y f())) T 1 (f 1 ) (b) = ( g) (0) T 1 = T 1 = f () 1.
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 44 Problem (S 02, #7; W 02, #7; F 03, #6). Suppose F : R 2 R 2 is C 1 nd tht the Jcobin mtrix of F is everywhere nonsingulr. Suppose tht F (0) = 0 nd tht F (x, y) 1 for ll (x, y) with (x, y) = 1. Denote U = {(x, y) : (x, y) < 1}. Prove tht F (U) U. Hint: Show tht F (U) U is both open nd closed in U. Proof. Since U is connected, clopenness of F (U) U in U implies tht either F (U) U = U or F (U) U = φ. Since there exists point, nmely 0 such tht it is inside both U nd F (U), F (U) U cnnot be empty, nd thus clopenness of F (U) U in U would imply tht F (U) U = U (which would men U F (U)). 1) Show F (U) U is open in U. F (U) is open in R 2. Sy y 0 F (U), y 0 = F (x 0 ), F (x 0 ) invertible. By inverse function thm, F mps open set U 0 onto open set V 0 ; x 0 U 0 y 0 = F (x 0 ) V 0. y 0 V 0 F (U) F (U) U is open in U. 2) Show F (U) U is closed in U. Sy x n F (U) U, x n x U. x n = F (y n ), y n U U. There is subsequence y nk y U. Since F is continuous, F (y nk ) F (y) = x. y = 1 F (y) 1 F (y) = x / U. Contrdiction. 13.3 Implicit Function Theorem (x, y) = (x 1,..., x m, y 1,..., y n ) Theorem. Suppose G : R m+n R n is C 1. G(, b) = 0 for some point (, b). Prtil derivtive mtrix G y (, b) is invertible. Then there exist open sets U of in R m nd W of (, b) in R m+n nd C 1 mpping h : U R n, such tht y = h(x) solves the eqution G(x, y) = 0 in W. Exmple. f(x, y, z) = 0, f z 0 cn solve for z = h(x, y), f(x, y, h(x, y)) 0. Exmple. m = n = 1. G(x, y) = 0, y = h(x) G(x, h(x)) = 0 d dxg(x, h(x)) = 0 G x + G h h (x) = 0 h (x) = G x / G h, G h 0. Sy G(x, y) = x 2 + y 2 1, x 2 + y 2 1 = 0 y = ± 1 x 2, problem t (1, 0). G h cnnot be equl to 0. Problem (S 02, #6). Suppose f : R 3 R is C 1 with f 0 t 0. Show tht there re two other C 1 functions g, h : R 3 R, such tht the function (x, y, z) (f(x, y, z), g(x, y, z), h(x, y, z)) from R 3 to R 3 is one-to-one on some neighborhood of 0. Proof. f 0 one of f F = f x g x h x f y g y h y f z g z h z x, f y, f z is not 0. F = (f, g, h). Need to produce functions g, h such tht the mtrix bove is invertible. If f x (0) 0, let g(x, y, z) = z, h(x, y, z) = y. Then f x 0 0 0 0 1 nd F = F (f(x, y, z), z, y). 0 1 0
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 45 Similrly, we cn find set of functions g, h by choosing mtrix in ech of the other two cses, i.e. when f f y (0) 0 nd z (0) 0. Problem (F 02, #6). Suppose F : R 3 R 2 is C 1. Suppose for some v 0 R 3 nd x 0 R 2 tht F (v 0 ) = x 0 nd F (v 0 ) : R 3 R 2 is onto. Show tht there is C 1 function γ : ( ɛ, ɛ) R 3 for some ɛ > 0, such tht (i) γ (0) 0 R 3, nd (ii) F (γ(t)) = x 0 for ll t ( ɛ, ɛ). Proof. Since F (v 0 ) is onto, the mtrix F (v 0 ) hs rnk 2. So, 2 of the 3 columns of F (v 0 ) re linerly independent. ] F (v 0 ) = [ f1 f 1 f 1 x 1 f 2 x 2 f 2 x 3 f 2 x 1 x 2 x 3 v 0. Assume the lst two columns re linerly independent. Consider the function G(x 1, x 2, x 3 ) = F (x 1, x 2, x 3 ) x 0. Write (x 1, x 2, x 3 ) = (s 1, s 2 ) where s 1 = x 1, s 2 = (x 2, x 3 ). Write v 0 = (u 1, u 2 ). Then G(u 1, u 2 ) = 0 nd G s 2 (v 0 ) is invertible. By Implicit Function Theorem, ɛ > 0 nd h C 1, such tht h : (u 1 ɛ, u 1 + ɛ) R 2 nd G(s 1, h(s 1 )) = 0, s 1 (u 1 ɛ, u 1 + ɛ). F (s 1, h(s 1 )) = x 0, s 1 (u 1 ɛ, u 1 + ɛ). Define γ(t) : ( ɛ, ɛ) R 3 by γ(t) = (u 1 + t, h(u 1 + t)). Then γ(t) is differentible curve stisfying (i) γ (t) = (1, h (u 1 + t)) 0, (ii) F (γ(t)) = x 0 for ll t ( ɛ, ɛ).
Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 46 13.4 Differentition Under Integrl Sign Problem (W 02, #1). f : [, b] [c, d] R ((x, y) f(x, y)). Suppose f y [, b] (c, d) nd extends to continuous function on [, b] [c, d]. Let exists on F (y) = f(x, y)dx. Then F is differentible in [, b] nd d b dy F (y) = f (x, y)dx. y d dy Proof. F (y + h) F (y) h b f(x, y)dx = f y By MVT, c, 0 < c < 1, such tht f (x, y)dx. y b [ (x, y)dx f(x, y + h) f(x, y) = f ] h y (x, y) dx f(x, y + h) f(x, y) f (x, y) h y dx f(x, y + h) f(x, y) h = f (x, y + ch). y (x, y + ch) (x, y) = c h h. Since f f y is continuous on [, b] [c, d] y Choose δ such tht (x, y) (x, y ) δ f y (x, y ) f y = f f (x, y + ch) (x, y) y y is uniformly continuous on [, b] [c, d]. dx (x, y) ɛ b ɛ ɛ dx = (b ) b b = ɛ.