Analysis Comp Study Guide

Anlysis Comp Study Guide The Rel nd Complex Number Systems nd Bsic Topology Theorem 1 (Cuchy-Schwrz Inequlity). ( n ) 2 k b k b 2 k. 2 k As ( k tb k ) 2 0, s qudrtic in t it hs t most one root. So the discriminnt must therefore be less thn or equl to 0. As ( k tb k ) 2 = or, fter rerrngement, we get Note: Equlity holds only when k = b k. Theorem 2 (The Archimeden Property). 2 k 2t k b k + t 2 ( n ) 2 4 k b k 4 ( n ) 2 k b k 2 k 2 k n b 2 k 0, b 2 k. If x, y R nd x > 0, then there exists positive integer n such tht nx > y. b 2 k, the discriminnt is Let A be the set of ll nx, where n runs through the positive integers nd ssume towrd contrdiction tht y is n upper bound for A. A hs lest upper bound, so let α = sup A. Since x > 0, α x < α, nd α x is not n upper bound of A. Hence α x < mx for some positive integer m. But then α < (m + 1)x A which is impossible s α is n upper bound of A. So there exists positive integer n such tht nx > y. Theorem 3 (Q is dense in R). If x, y R, x < y, then there exists p Q such tht x < p < y. y x > 0 so, by the Archimeden property, there exists n integer n 1 such tht n(y x) > 1. Agin by the Archimeden property, there exists n integer m such tht m 1 nx < m. So combining these two inequlties, we get nx < m nx + 1 < ny. As n > 0 we my divide nd obtin: x < m n < y, where m n Q. Theorem 4 (Heine-Borel Theorem). If set E R k hs one of the following properties, it hs the other two s well: () E is closed nd bounded. (b) E is compct. (c) Every infinite subset of E hs limit point in E. Our proof requires two lemms in order to proceed. 1

Lemm 1 (Closed subsets of compct sets re compct). 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 O of K. Since K is compct, there is finite subcollection {U n } of O which covers K nd hence F. If F c is member of {U n }, we my remove it from {U n } nd still retin n open cover of F. We hve thus shown tht finite subcollection of {V α } covers F. Lemm 2 (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). Fix member K 1 of {K α } nd put G α = K c α. Assume tht no point of K 1 belongs to every K α. Then the sets G α form n open cover of K 1. Since K 1 is compct there exists subcover K 1 {G 1,..., G n }. However, this mens tht K 1 K2 Kn is empty, contrdicting our hypothesis. Lemm 3 (Every k-cell is compct). Let I be k-cell, consisting of ll points x = (x 1,..., x k ) such tht j x j b j (1 j k), Put 2 k δ = (b j j ) 2. j=1 Then x y < δ if x, y I. Suppose, towrd contrdiction, there exists n open cover {G α } of I which contins no finite subcover of I. Put c j = ( j + b j )/2. The intervls [ j, c j ] nd [c j, b j ] then determine 2 k k-cells Q i whose union is I. At lest one of these, cll it I 1 cnnot be covered by ny finite subcollection of {G α }. We cn next subdivide I 1 nd continue this process. We obtin sequence {I n } with the following properties: (i) I I 1 I 2 ; (ii) I n is not covered by ny finite subcollection of {G α }; (iii) if x I n nd y I n, then x y < 2 n δ. By (i) nd Lemm 2, there is point x which lies in every I n. For some α, x G α. Since G α is open, there exists n r > 0 such tht y x < r implies y G α. There exists n n such tht 2 n δ < r, so by (iii) I n G α, which contrdicts (ii). Proof of Heine-Borel Theorem: (() (b)) As E is bounded, E I, where I is k-cell. By Lemm 3, I is compct. By Lemm 1, s E is closed nd E I, E is compct. ((b) (c)) Let U be n infinite subset of E (if E hs only finitely mny points the result is trivil). If no point of E is limit point of U, then ech q E would hve neighborhood V q which contins t most one point of U, specificlly q if q E. As E is infinite, no finite subcollection of {V q } cn cover U nd the sme is true of E s U E. This contrdicts the compctness of E. So U must hve limit point in E. ((c) ()) If E is not bounded then E contins points x n such tht x n > n, n = 1, 2,.... The set S consisting of these points x n is infinite nd clerly hs no limit point in R k nd hence hs none in E. So (c) implies E is bounded. If E is not closed, then there is point x 0 R k which is limit point of E but not point of E. For n = 1, 2, 3,... there re points x n E such tht x n x 0 < 1/n. Let S be the set of the points x n. Then S is infinite (or else we could find bll round x 0 not contining ny points of E), contins x 0 s limit point nd S hs no other limit points in R k. For if y R k, y x 0, then x n y x 0 y x n x 0 x 0 y 1 n 1 2 x 0 y for ll but finitely mny n. So y is not limit point of S. Thus S hs no limit point in E. By (c) we get then tht E must be closed. 2

Counter Exmple 1 (If K α is collection of closed subsets of metric spce X such tht the intersection of every finite subcollection of {K α } is nonempty, then K α is not necessrily nonempty). Let K n = [n, ). So ech K n is closed nd finite intersection of K i s is nonempty. However, if we ssume there exists some x K n, then, by the Archimeden property, there exists n integer N > x nd so x / K N so x / K n. Counter Exmple 2 (Closed And Bounded Does Not Imply Compct In A Generl Metric Spce). Let X be the discrete metric with infinitely mny points. Then X itself is closed s it is the entire metric spce. Similrly, for ny x X, if we look t B(x, 2) this bounds X s for ny y X, d(x, y) 1. So X is closed nd bounded. However, if we tke the set of {B(x, 1/2) : x X}, then this covers X however no finite subcover (in fct, no proper subcover) cn cover X. So X is not compct. Theorem 5 (Bolzno-Weierstrss). Every bounded infinite subset of R k hs limit point in R k. Being bounded, the set E in question is subset of k-cell I R k. By Lemm 3 bove, I is compct nd so E hs limit point in I by the proof bove, we hve tht E hs limit point in I. Numericl Sequences nd Series Theorem 6 (Cuchy Sequence Condition). () In ny metric spce X, every convergent sequence is Cuchy sequence. (b) If X is compct metric spce nd if {p n } is Cuchy sequence in X, then {p n } converges to some point in X. (c) In R k, every Cuchy sequence converges. Lemm 4 (If E is the closure of set E in metric spce X, then dime =dime.). Since E E, dime dime. Fix ɛ > 0 nd choose p, q E. By the definition of E, there re points p, q E such tht d(p, p ) < ɛ, d(q, q ) < ɛ. Hence d(p, q) d(p, p ) + d(p, q ) + d(q, q) < 2ɛ + dime. It follows tht dime 2ɛ+ dime. As ɛ ws rbitrry, the lemm is proved. () If p n p nd if ɛ > 0, there is n integer N such tht d(p, p n ) < ɛ/2 for ll n N. Hence for n, m N. Thus {p n } is Cuchy sequence. d(p n, p m ) d(p n, p) + d(p, p m ) < ɛ (b) Let {p n } be Cuchy sequence in the compct spce X. For N = 1, 2, 3,... let E N be the set consisting of p N, p N+1,.... Then lim N dime N = 0, by Lemm 4. Being closed subset of the compct spce X, ech E N is compct by Lemm 1. Also, E N E N+1, so E N E N+1. Define E = E n. By Lemm 2, E is nonempty shows now tht there is unique p X which lies n=1 in every E N. Let ɛ > 0 be given. As dime N 0 there is n integer N 0 such tht dime N < ɛ for N 0 N. Since p E N, it follows tht d(p, q) < ɛ for every q E N hence for every q E N. In other words, d(p, p n ) < ɛ if n N 0. This sys exctly tht p n p. 3

(c) Let {x n } be sequence in R k. Define E N s in (b), with x i in plce of p i. For some N, dim E N < 1. The rnge of {x n } is the union of E N nd the finite set {x 1,..., x N 1 }. Hence {x n } is bounded. Since every bounded subset of R k hs compct closure in R k (Heine-Borel), (c) follows from (b). Counter Exmple 3 (Q hs Cuchy Sequence tht does not converge). Since Q is dense in R, there exists sequence {p n } of rtionl points in R converging to 2 such tht p n p n+1 < 2 n. Then in Q, {p n } is Cuchy sequence since for ny 2 > ɛ > 0, there exists n N such N tht 2 k > 2 ɛ. So, for m n > N, k=0 N p m p n p m p m 1 + p n+1 p n < 2 k 2 k < ɛ. So our sequence is Cuchy, however 2 is not in Q, so this sequence does not converge in Q. Theorem 7 (Some Specil Sequences). () For p > 0, lim n 1 n p = 0. Fix ɛ > 0. By the Archimeden property, there is n Z + such tht n > (1/ɛ) 1/p. So (b) For p > 0, k=0 1 n p < 1 ((1/ɛ) 1/p ) = ɛ. p lim n p = 1. n If p > 1, let x n = n p 1. So x n > 0, nd 1 + nx n (1 + x n ) n = p, so 0 x n p 1 n. So, by the Squeeze Theorem, x n 0. If p = 1 the result is trivil nd if p < 1 let p = 1/p nd by bove n p 1, so n p 1. k=0 (c) We my ssume n 2, lim n n = 1. n Let x n = n n 1. So x n 0. By the Binomil Theorem, n = (1 + x n ) n 2 0 x n n 1. So gin by the Squeeze Theorem, x n 0. (d) If p > 0, α R, Let k Z + such tht k > α. For n > 2k, ( n (1 + p) n > k So 0 < (e) If x < 1, then ) p k = lim n n α (1 + p) n = 0. n(n 1)... (n k 1) p k > nk p k k! 2 k k!. n α (1 + p) n < 2k k! p k nα k. As α k < 0 by (), n α k 0. lim n xn = 0. As x < 1, x = 1, p > 0. By (d), with α = 0, we get p + 1 lim n ( ) n 1 1 (1 + p) n = lim = lim n 1 + p n xn = 0. n(n 1) x 2 2 n. So 4

Theorem 8 (Cuchy Criterion For Sums). n=1 n m n N. converges if nd only if for every ɛ > 0 there is n integer N such tht m k=n k < ɛ for For proof of this, tret the prtil sums s sequence nd defer to the Cuchy Sequence Conditions. Theorem 9 (Geometric Series). For x 1 x > 0, k=0 x k = 1 xn+1 1 x. Let S n = x k. So xs n = S n 1 + x n+1. Rerrnging, we get k=0 Corollry 1 (Inifinite Geometric Series). S n = 1 xn+1 1 x. For x < 1, If we tke the limit when x < 1. n=0 x n = 1 1 x. lim S n = 1 n 1 x Theorem 10 (Convergence Tests for Series). Given the series n. n=0 () Root Test: Let α = lim sup n n n. If α < 1, the series converges. If α > 1, the series diverges. If α = 1, the result is inconclusive. If α < 1, we cn choose β such tht α < β < 1 nd n integer N such tht n n < β for n N. Tht is for n N, n < β n. As 0 < β < 1, β n converges. Convergence of n follows from the comprison test. Note: This ctully implies bsolute convergence of n. If α > 1, then there is sequence n k such tht n k nk α. 5

Hence n > 1 for infinitely mny vlues of n, so n 0. So n diverges. Consider the series 1 n, 1 n 2. n=1 n=1 For both of these series, α = 1, however the first diverges nd the second converges. (b) Rtio Test: Let α(n) = n+1 n. If lim sup α(n) < 1, the series converges. n If α(n) 1 for n N, for some N Z +, the series diverges. If lim sup n α(n) < 1, we cn find β < 1 nd n integer N such tht n+1 n < β for n N. In prticulr, N+1 < β n,..., N+p < β p n. Tht is, n < N β N β n for ll n N nd our series is convergent by the comprison test since β n converges. If n+1 n for n N, then n 0, so our series diverges. (c) Integrl Test: If N Z nd f is monotone decresing function on [N, ), then N f(x)dx converges to finite vlue. f(n) converges if nd only if Since f is monotone decresing Riemnn-integrble function, we hve f(x) f(n) for x [n, ) nd f(n) f(x) for x [N, n]. So, for every n N, n+1 n f(x)dx n+1 n f(n)dx = f(n) = n n 1 n=n f(n)dx n n 1 f(x)dx. Since the lower estimte is lso vlid for f(n), we get by summtion over ll n from N to some lrger integer M, M+1 M M f(x)dx f(n) f(n) + f(x)dx. N N n=n Letting M tend to infinity, we get the desired result. Theorem 11 (Power Series). Given the series c n z n. Let α = lim sup n cn, R = 1 n α. n=0 Then c n z n converges for z < R nd diverges if z > R. n=0 Proof follows directly from the Rtio Test. Theorem 12 (Summtion by Prts). Given two sequences { n },{b n }, put A n = k=0 k 6

if n 0; put A 1 = 0. Then, if 0 p q, we hve q n b n = n=p Continuity Definition 1. q n b n = n=p q (A n A n 1 )b n = n=p q A n (b n b n+1 ) + A q b q A p 1 b p. n=p q A n b n n=p q 1 n=p 1 A n b n+1 = q A n (b n b n+1 ) + A q b q A p 1 b p. Let X, Y be metric spces E X, p E, f : E Y. f is continuous t p if for every ɛ > 0 there exists δ > 0 such tht d Y (f(x), f(p)) < ɛ for ll points x E for which d X (x, p) < δ. Counter Exmple 4 (A Nowhere Continuous Function). { 1 if x Q Let f(x) = 0 if x Q c Cse 1: x Q. Let 0 < ɛ < 1. For ny δ > 0, s Q c is dense in R, there exists n irrtionl point y such tht x y < δ. However, f(x) f(y) = 1 > ɛ, so f is not continuous t x. Cse 2: x Q c. Let 0 < ɛ < 1. For ny δ > 0, s Q is dense in R, there exists rtionl point y such tht x y < δ. However, f(x) f(y) = 1 > ɛ, so f is not continuous t x. So f is discontinuous t ll points in R. Theorem 13 (Equivlent Definitions of Continuity). () A mpping f of metric spce X into metric spce Y is continuous on X if nd only if f 1 (V ) is open in X for every open set V in Y (b) For the definition of continuity, if we ssume lso tht p is limit point of E, then f is continuous t p if nd only if lim x p f(x) = f(p). () Suppose first tht f is continuous on X nd V is n open set in Y. Suppose p X nd f(p) V. Since V is open, there exists ɛ > 0 such tht y V if d Y (f(p), y) < ɛ; nd since f is continuous t p there exists δ > 0 such tht d Y (f(x), f(p)) < ɛ if d X (x, p) < δ. Thus x f 1 (V ) s soon s d X (x, p) < δ. Conversely, suppose f 1 (V ) is open in X for every open set V in Y. Fix p X nd ɛ > 0, let V be the set of ll y Y such tht d Y (y, f(p)) < ɛ. Then V is open; hence f 1 (V ) is open; hence there exists δ > 0 such tht x f 1 (V ) s soon s d X (p, x) < δ. But if x f 1 (V ), then f(x) V, so tht d Y (f(x), f(p)) < ɛ. So f is continuous. (b) Proof follows directly from the definition of limit nd continuity. Theorem 14 (A Continuous Mpping of Compct Set Is Compct). Suppose f is continuous mpping of compct metric spce X into metric spce Y. Then f(x) is compct. n=p 7

Let {V α } be n open cover of f(x). Since f is continuous, ech set f 1 (V α ) is open. Since X is compct, there re finitely mny indices, α 1,..., α n such tht Since f(f 1 (E)) E for every E Y, we get Corollry 2 (Extreme Vlue Theorem). X f 1 (V α1 ) f 1 (V αn ). f(x) V α1 Vαn. Suppose f is continuous rel function on compct metric spce X, nd M = sup f(p), m = inf f(p). p X p X Then there exist points p, q X such tht f(p) = M nd f(q) = m. By the Theorem bove, s f is compct on X, f(x) is compct nd thus closed nd bounded. So f(x) contins sup f(x) nd inf f(x) s these exist nd re limit points of f(x). Theorem 15 (Continuity Preserves Connectedness). If f is continuous mpping of metric spce X into metric spce Y, nd if E is connected subset of X, then f(e) is connected. Assume, to the contrry, tht f(e) = A B, where A nd B re nonempty seprted subsets of Y. Put G = E f 1 (A), H = E f 1 (B). Then E = G H nd neither G nor H is empty. Since A A, we hve G f 1 (A); the ltter set is closed, since f is continuous; hence G f 1 (A). It follows tht f(g) A. Since f(h) = B nd A B is empty, we conclude tht G H is empty. Similrly, G H is empty. Thus we hve reched contrdiction. Corollry 3 (Intermedite Vlue Theorem). Let f be continuous rel function on the intervl [, b]. If f() < f(b) nd if c is number such tht f() < c < f(b), then there exists point x (, b) such tht f(x) = c. [, b] is connected; hence, by the Theorem bove f([, b]) is connected subset of R nd the ssertion follows by the definition of connectedness in R. Definition 2 (Uniform Continuity). Let f be mpping of metric spce X into metric spce Y. We sy tht f is uniformly continuous on X if for every ɛ > 0 there exists δ > 0 such tht d Y (f(x), f(y)) < ɛ for ll x, y X such tht d X (x, y) < δ. Counter Exmple 5 (A Not Uniformly Continuous Function). Let f(x) = x 2. As f is polynomil, it is continuous on R. Let ɛ = 1 nd let δ > 0. By the Archimeden property, there exists n integer n such tht nδ > 1. Let x = n, y = n + δ/2. So x y < δ. f(x) f(y) = (n + δ/2 n)(n + δ/2 + n) = nδ + δ 2 /4 > 1 = ɛ. So f cnnot be uniformly continuous on R. Theorem 16 (A Continuous Function Is Uniformly Continuous on Compct Set). Let f be continuous mpping on compct metric spce X into metric spce Y. Then f is uniformly continuous on X. 8

Let ɛ > 0 be given. Since f is continuous, we cn ssocite to ech point p X positive number φ(p) such tht q X, d X (p, q) < φ(p) implies d Y (f(p), f(q)) < ɛ 2. Let J(p) be the set of ll q X for which d X (p, q) < 1 2 φ(p). Since p J(p), the collection of ll sets J(p) is n open cover of X; nd since X is compct, there is finite set of points p 1,..., p n in X, such tht X J(p 1 ) J(p n ). Let δ = 1 2 min[φ(p 1),..., φ(p n )]. So δ > 0. Now let q nd p be points of X, such tht d X (p, q) < δ. There is n integer m, 1 m n such tht p J(p m ); hence d X (p, p m ) < 1 2 φ(p m), nd we lso hve Finlly, we hve d x (q, p m ) d x (p, q) + d x (p, p m ) < δ + 1 2 φ(p m) φ(p m ). d Y (f(p), f(q)) d Y (f(p), f(p m )) + d Y (f(q), f(p m )) < ɛ. Theorem 17 (A Convex Function on n Open Set Is Continuous). If f is convex function on n open set U R, then f is continuous on U. Note: A function is convex on [, b] if for ny x, y [, b] nd ny t [0, 1], f(tx + (1 t)y) tf(x) + (1 t)f(y). Let b U. As U is open there exist, c U such tht < b < c. Counter Exmple 6 (A Convex Function on Closed Set Need Not Be Continuous). If we look t the closed set [0, 1] nd let f(x) = 0 for x [0, 1) nd f(1) = 1, then f is convex, but f is not continuous. Differentition Definition 3. Let f be defined on [, b]. For ny x [, b] form the quotient nd define φ(t) = f(t) f(x) t x f (x) = lim t x φ(t). Then f is defined to be the derivtive of f t x provided the limit exists. Theorem 18 (Differentibility Implies Continuity). Let f be defined on [, b]. If f is differentible t point x [, b], then f is continuous t x. As t x, we hve, f(t) f(x) = f(t) f(x) t x (t x) f (x)0 = 0. Counter Exmple 7 (A Continuous Function Need Not Be Differentible). 9

Let f(x) = x on contining neighborhood of 0. This function is continuous s x y < x y for ll x, y R. However, we hve t lim t 0+ t = lim t t 0 + t = 1 nd t lim t 0 t = lim t = 1 t 0 + t so our two sided limits do not gree, so our function is not differentible t 0. Counter Exmple 8 (The Derivtive of Function Need Not Be Continuous). { x Let f(x) = 2 sin(1/x) if x 0;, f is continuous for ll positive x nd, by the Squeeze Theorem, 0 if x > 0 f is continuous t 0 s well. f (x) is defined on ll nonnegtive rel numbers s, for x 0 f (x) = 2x sin(1/x) cos(1/x) nd for x = 0, f f(x) 0 (0) = lim = lim x sin(1/x) = 0 x 0 + x 0 x 0 + by the Squeeze Theorem. So f (x) is differentible, but is discontinuous t x = 0. Counter Exmple 9 (A Nowhere Differentible Continuous Function). Let φ(x) = x for 1 < x 1 nd extend this definition to ll rel numbers by letting φ(x + 2) = φ(x). Then, for ll s nd t, φ(s) φ(t) s t since if we let s = s + 2k where s ( 1, 1] nd k is n integer. Similrly define t = t + 2l. So phi(s) φ(t) = s t < 2, nd s t = (s t ) + 2(k l). So we hve in prticulr tht φ is continuous on R. Define ( ) n 3 f(x) φ(4 n x). 4 n=0 Since 0 φ 1, by the Weierstrss M-test, this series converges uniformly on R. Thus, since prtil sum is continuous, f is continuous on R. Fix rel number x nd positive integer m. Put δ m = ± 1 2 4 m where the sign is so chosen tht no integer lies between 4 m x nd 4 m (x + δ m ). This cn be done since 4 m d m = 1 2. Define γn = φ(4n (x + δ m )) φ(4 n x) δ m. When n > m, 4 n δ m is n even integer so tht γ n = 0. When 0 n m, we hve tht γ n 4 n, by the fct tht for ll s, t, φ(s) φ(t) s t. Since γ m = 4 m, we conclude tht f(x + δ m ) f(x) δ m = m ( ) n 3 γ n 4 3m n = 0 m 1 3 n = 1 2 (3m + 1). n=0 As m, δ m 0, it follows tht f is not differentible t x. Theorem 19 (Fermt s Theorem). Let f be defined on [, b]; if f hs locl mximum t point x (, b) nd if f (x) exists, then f (x) = 0. 10

Choose δ such tht for ll points y [, b] such tht x y < δ implies f(y) f(x). So < x δ < x < x + δ < b. If x δ < t < x, then f(t) f(x) 0. t x Letting t x, we see tht f (x) 0. Similrly, for x < t < x + δ, f(t) f(x) t x Once gin letting t x, we get f (x) 0. So f (x) = 0. Theorem 20 (Generlized Men Vlue Theorem). If f nd g re continuous rel functions on [, b] which re differentible in (, b), then there is point x (, b) t which [f(b) f()]g (x) = [g(b) g()]f (x). 0. Put h(t) = [f(b) f()]g(t) [g(b) g()]f(t) for t b. Then h is continuous on [, b], h is differentible in (, b) nd h() = f(b)g() f()g(b) = h(b). It suffices to show tht h (t) = 0 somewhere in (, b). If h is constnt, this holds for ll x (, b). If h(t) > h() for some t (, b), let x be point on [, b] t which h ttins its mximum, which exists by the Extreme Vlue Theorem. So x (, b) nd Fermt s Theorem shows tht h (x) = 0. If h(t) < h(), we my look t h(t) nd show h (x) = 0 for some x (, b). Corollry 4 (Men Vlue Theorem). If f is continuous on [, b], differentible on (, b), then there exists c (, b) such tht f (c)(b ) = f(b) f(). If we set g(x) = x in the bove theorem we get the desired result. Theorem 21 (L Hôpitl s Rule). Suppose f nd g re rel nd differentible in (, b) nd g (x) 0 for ll x (, b), where < b. Suppose f (x) g A s x. (x) If f(x) 0 nd g(x) 0 s x or if g(x) s x, then f(x) A s x. g(x) We first consider the cse in which A < +. Choose rel number q such tht A < q, nd then choose r such tht A < r < q. There is point c (, b) such tht < x < c implies f (x) g (x) < r. If < x < y < c, then the Men Vlue Theorem furnishes point r (x, y) such tht f(x) f(y) g(x) g(y) = f (t) g (t) < r. 11

If we ssume tht f(x), g(x) 0, then letting x in the bove eqution yields f(y) g(y) r < q If we ssume g(x), keeping y fixed in the bove sitution, we cn choose pint c 1 (, y) such tht g(x) > g(y) nd g(x) > 0 if < x < c 1. Multiplying the equting by [g(x) g(y)]/g(x), we obtin f(x) g(x) < r r g(y) g(x) + f(y) g(x). If we let x, then s g(x), there is point c 2 (, c 1 ) such tht Summing two of the equtions bove shows, we get f(x) g(x) < q. f(y) g(y) + f(x) g(x) < 2q. So, for ny q subject only to A < q there is point c 2 such tht f(x)/g(x) < q if < x < c 2. Similrly, if < A we cn obtin the sme result. The result holds lso if g(x) by replcing g(x) with g(x). Theorem 22 (Tylor s Theorem). Suppose f is rel function on [, b], n is positive integer, f (n 1) is continuous on [, b], f (n) (t) exists for every t (, b). Let α, β be distinct points of [, b], nd define P (t) = n 1 k=0 Then there exists point x between α nd β such tht f (k) (α) (t α) k. k! f(β) = P (β) + f (n) (x) (β α) n. n! Let M be the number defined by nd put f(β) = P (β) + M(β α) n g(t) = f(t) P (t) M(t α) n for t b. We hve to show tht n!m = f (n) (x) for some x between α nd β. By the definition of P (t) nd g(t), we hve g (n) (t) = f (n) (t) n!m for < t < b. Hence the proof will be complete if we cn show tht g (n) (x) = 0 for some x between α nd β. Since P (k) (α) = f (k) (α) for k = 0, 1,..., n 1 we hve g(α) = g (α) = = g (n 1) = 0 Our choice of M shows tht g(β) = 0, so tht g (x 1 ) = 0 for some x 1 between α nd β, by the men vlue theorem. Since g (α) = 0, we conclude similrly tht g (x 2 ) = 0 for some x 2 between α nd x 1. After n steps we rrive t the conclusion tht g (n) (x n ) = 0 for some x n between α nd x n 1, tht is, between α nd β. Theorem 23 (Convexity nd f (x) 0 Are Equivlent). 12

A twice-differentible function f is convex on [, b] if nd only if f (x) 0 for ll x (, b). Assume first tht f (x) 0. For ny x 1, x 2 (, b), x 1 < x 2 nd λ (0, 1). f(λx 1 +(1 λ)x 2 )λf(x 1 ) (1 λ)f(x 2 ) = λ[f(λx 1 +(1 λ)x 2 ) f(x 1 )]+(1 λ)[f(λx 1 +(1 λ)x 2 ) f(x 2 )]. By the Men Vlue Theorem, where Rerrnging the bove eqution, we get = λf (ξ 1 )((λ 1)x 1 + (1 λ)x 2 ) + (1 λ)f (ξ 2 )(λx 1 λx 2 ) ξ 1 (x 1, λx 1 + (1 λ)x 2 ), ξ 2 (λx 1 + (1 λ)x 2, x 2 ). = λ(1 λ)f (ξ 1 )(x 2 x 1 ) + λ(1 λ)f (ξ 2 )(x 2 x 1 ) = λ(1 λ)(x 2 x 1 )[f (ξ 1 ) f (ξ 2 )]. Utilizing the Men Vlue Theorem gin, we get tht this is equl to λ(1 λ)(x 2 x 1 )f (ξ)(ξ 1 ξ 2 ) 0 for ξ (ξ 1, ξ 2 ) since ξ 1 < ξ 2 nd we hve ssumed tht f (ξ) 0. So f is convex. Conversely, ssume f is convex. For ny x (, b), there exists n h 0 R smll enough so tht x ± h (, b). Since f is convex, f(x) = f( 1 2 (x + h) + 1 2 (x h)) 1 (f(x + h) + f(x h)). 2 So f(x + h) + f(x h) 2f(x) h 2 0 This result is independent of our choice of h, so we my mke h s smll s we like. Thus we cn tke the limit s h 0 nd the result will still hold f(x + h) + f(x h) 2f(x) lim h 0 h 2 is n indeterminte form s f is continuous, so L Hôpitl s Rule pplies. So Adding nd subtrcting f (x), we get h So f (x) 0 for ll x (, b). Integrtion Definition 4. f(x + h) + f(x h) 2f(x) f (x + h) f (x h) 0 lim h 0 h 2 = lim. h 0 2h 1 0 lim h 0 2 ( f (x + h) f (x) h = 1 2 (f (x) + f (x)) = f (x) + f (x) f ) (x h) h 13

Let [, b] be given intervl. Suppose f is bounded rel function defined on [, b]. Corresponding to ech prtition P of [, b], we put M i = sup f(x), (x i 1 x x i ), m i = inf f(x), (x i 1 x x i ), U(P, f) = M i x i, L(P, f) = i=1 m i x i, nd finlly i=1 fdx = inf U(P, f), fdx = sup U(P, f), where the inf nd sup re tken over ll prtitions P of [, b]. If the upper nd lower integrls re equl, we sy tht f is Riemnn-integrble on [,b] nd denote the common vlue by Theorem 24 (Equivlent Definition of Integrbility). f(x)dx. f is Riemnn-integrble on [, b] if nd only if for every ɛ > 0 there eixsts prtition P such tht U(P, f) L(P, f) < ɛ. Lemm 5 (If P is refinement of P, then L(P, f) L(P, f), U(P, f) U(P, f)). It suffices to prove only the first clim s the proof for the second is identicl. Suppose first tht P contins just one more point thn P. Let this extr point be x nd suppose x i 1 < x < x i. Put w 1 = inf f(x), x [x i 1,x ] w 2 = inf f(x). x [x,x i] It is cler tht w 1 m i nd w 2 m i where m i is defined s bove. Hence, L(P, f) L(P, f) = w 1 [x x i 1 ] + w 2 [x i x ] m i [x i x i 1 ] = (w 1 m i )(x x i 1 ) + (w 2 m i )(x i x ) 0. We my repet this resoning for ll points in P tht re not in P. Lemm 6 (L(P, f) fdx fdx U(P, f)). It suffices to show tht fdx fdx s the outer inequlities hold by definition of sup nd inf. Let P be the common refinement of two prtitions P 1 nd P 2. By Lemm 5, L(P 1, f) L(P, f) U(P, f) U(P 2, f). Hence L(P 1, f) U(P 2, f). If P 2 is fixed nd the sup is tken over ll P 1, we obtin fdx U(P 2, f). If we now tke the inf over ll P 2 we get the desired result. For every P, by Lemm 6 we hve L(P, f) fdx fdx U(P, f). 14

Thus, if we ssume, for ny ɛ > 0, there is prtition P such tht U(P, f) L(P, f) < ɛ, then we hve fdx fdx < ɛ. As this holds for ny ɛ, we hve fdx = fdx. Conversely, suppose f is Riemnn-integrble nd let ɛ > 0 be given. Then there exist prtitions P 1 nd P 2 such tht U(P 2, f) fdx < ɛ 2, fdx L(P, f) < ɛ 2. Choosing P to be the common refinement of P 1 nd P 2, we get, by Lemm 5, U(P, f) U(P 2, f) < fdx + ɛ 2 < L(P 1, f) + ɛ L(P, f) + ɛ. After rerrngement, we get Theorem 25 (Conditions For Integrbility). U(P, f) L(P, f) < ɛ. () If f is continuous on [, b] then f is Riemnn-integrble on [, b]. (b) If f is monotonic on [, b], then f is Riemnn-integrble. (c) Suppose f is Riemnn-integrble on [, b], m f(x) M, φ is continuous on [m, M], nd h(x) = φ(f(x)) on [, b]. Then h is Riemnn-integrble on [, b]. (z) If f hs countbly mny discontinuities, then f is Riemnn-integrble. () Let ɛ > 0 be given. Choose η > 0 so tht η < exists δ > 0 such tht ɛ. Since f is uniformly continuous on [, b], there b f(x) f(y) < η for x, y [, b], x y < δ. If P is ny prtition of [, b] such tht x i < δ for ll i, then the bove implies tht M i m i η nd therefore U(P, f) L(P, f) = (M i m i ) x i η i=1 x i = η(b ) < ɛ. i=1 So, by the Theorem bove, f is Riemnn-integrble. (b )(f(b) f()) (b) Let ɛ > 0 be given. Choose n integer n such tht n >. Choose prtition such ɛ tht x i = b n. We suppose tht f is monotoniclly incresing (the proof is nlogous in the other cse). Then M i = f(x i ), m i = f(x i 1 ) for i = 1,..., n, so tht U(P, f) L(P, f) = b n 15 [f(x i ) f(x i 1 )] i=1

= b (f(b) f()) < ɛ n by our choice of n. So f is Riemnn-integrble. (c) Fix ɛ > 0. Since φ is uniformly continuous on [m, M], there exists δ > 0 such tht δ < ɛ nd φ(s) φ(t) < ɛ if s t δ nd s, t [m, M]. Since f is Riemnn-integrble, there is prtition P = {x 0, x 1,..., x n } of [, b] such tht U(P, f) L(P, f) < δ 2. Let M i, m i hve the sme mening s in the definition of integrl nd let Mi, m i be the nlogous numbers for h. Divide the numbers 1,..., n into two clsses: i A if M i m i < δ, i B if M i m i δ. For i A, our choice of δ show tht Mi m i ɛ. For i B, Mi m i 2K where K = sup φ(t), m t M. By our choice of prtition, we hve δ x i (M i m i ) x i < δ 2 i B i B so tht i B x i < δ. It follows tht U(P, h) L(P, h) = i A (M i m i ) x i + i B(M i m i ) x i ɛ[b ] + 2Kδ < ɛ[b + 2K]. Since ɛ ws rbitrry, we hve tht h is Riemnn-integrble. Theorem 26 (Fundmentl Theorem of Clculus, Prt I). Let f be Riemnn-integrble on [, b]. For x b, put F (x) = x f(t)dt. Then F is continuous on [, b]; furthermore, if f is continuous t point x 0 of [, b], then F is differentible t x 0, nd F (x 0 ) = f(x 0 ). Since f is Riemnn-integrble, f is bounded. Suppose f(t) M for t b. If x < y b, then F (y) F (x) = y Given ɛ > 0, we see tht F (y) F (x) < ɛ, when y x < ɛ M. x f(t)dt M(y x). Now suppose f is continuous t x 0. Given ɛ > 0 choose δ > 0 such tht f(t) f(x 0 ) < ɛ if t x 0 < δ, nd t b. Hence, if x 0 δ < s x 0 t < x 0 + δ nd we hve tht It follows tht F (x 0 ) = f(x 0 ). s < t b, F (t) F (s) f(x 0 ) t s = 1 t s Theorem 27 (Fundmentl Theorem of Clculus, Prt II). t s f(u) f(x 0 )du < ɛ. 16

If f is Riemnn-integrble on [, b] nd if there is differentible function F on [, b] such tht F = f, then f(x)dx = F (b) F (). Let ɛ > 0 be given. Choose prtition P = {x 0,..., x n } of [, b] so tht U(P, f) L(P, f) < ɛ. The Men Vlue Theorem furnishes points t i [x i 1, x i ] such tht for i = 1,..., n. Thus It now follows tht F (x i ) F (x i 1 ) = f(t i ) x i f(t i ) x i = F (b) F (). i=1 F (b) F () Since this holds for every ɛ > 0, the proof is complete. Corollry 5 (Integrtion by Prts). f(x)dx < ɛ. Suppose F nd G re differentible functions on [, b], F = f G = g, both of which re Riemnnintegrble, then F (x)g(x)dx = F (b)g(b) F ()G() f(x)g(x)dx. Put H(x) = F (x)g(x) nd pply the Fundmentl Theorem of Clculus Prt II to H nd its derivtive. Theorem 28 (Arc Length). If γ is continuous on [, b], then the rc length of γ is given by Theorem 29 (Hölder s Inequlity). Λ(γ) = γ (t) dt. Let f(x) nd g(x) be two integrble functions on [, b], both bounded. Then where 1 p, q < nd 1/p + 1/q = 1. Lemm 7 (Young s Inequlity). ( ) 1/p ( b f(x)g(x)dx f(x) p dx f(x) q dx If u, v 0, p, q s bove, then uv up p + vq q. If u or v re 0, then the inequlity hold trivilly. If u, v > 0, there exist x, y such tht u = e x/p nd v = e y/q. As the second derivtive of e t is e t nd e t > 0, for ll t, e t is convex. So if we let t = 1/p nd 1 t = 1/q, then we get uv = e x/p e y/q = e tx e (1 t)y = e tx+(1 t)y te x + (1 t)e y = up p + vq q. Lemm 8. ) 1/q 17

If f nd g re integrble p nd q s bove, both re nonnegtive f(x)g(x)dx 1. f p dx = 1 = g q dx, then f p nd g q re integrble by composition of continuous nd integrble functions nd fg is integrble. Using the inequlity found in the lemm bove, we get tht fg f p /p + g q /q. Integrting both sides (which is llowed s the inequlity holds for ll vlues), we get f(x)g(x)dx 1/p f(x) p dx + 1/q = 1/p + 1/q = 1. g(x) q dx Let f nd g be two integrble functions on [, b]. By the Cuchy-Schwrz Inequlity, f(x)g(x)dx f(x) g(x) dx. Let F = f p dx, G = g q dx. We cn use the lemm bove on the integrble functions f /(F 1/p ) nd ( b ) b 1/p ( ) b 1/q g /(G 1/q ). Thus we get f(x)g(x)dx F 1/p G 1/q = f(x) p dx g(x) q dx. Note if F or G re equl to zero then the result holds trivilly. Multivrible Differentition Definition 5. Suppose E is n open set in R n, f mps E into R m, nd x E. If there exists liner trnsformtion A of R n into R m such tht f(x + h) f(x) Ah lim = 0, h 0 h then we sy tht f is differentible t x nd we write Theorem 30 (Multivrible Derivtive Is Unique). f (x) = A. Suppose E nd f re s in the definition bove nd holds for A = A 1 nd A = A 2. Then A 1 = A 2. If B = A 1 A 2, then the inequlity f(x + h) f(x) Ah lim = 0 h 0 h Bh f(x + h) f(x) A 1 h + f(x + h) f(x) A 2 h shows tht Bh 0 s h 0. For fixed h 0, it follows tht B(th) 0 s t 0. The linerity of B h th shows tht this is independent of t, so Bh = 0 for ll h. So B = 0. 18

Theorem 31 (f Is Differentible Implies Prtil Derivtives Exist). Suppose f mps n open set E R n into R m, nd f is differentible t point x E. Then the prtil derivtives (D j f i )(x) exist, nd m f (x)e j = (D j f i )(x)u i for 1 j m. Fix j. Since f is differentible t x, i=1 f(x + te j ) f(x) = f (x)(te j ) + r(te j ) where r(te j ) /t 0 s t 0. The linerity of f (x) shows therefore tht f(x + te j ) f(x) lim = f (x)e j. t 0 t If we now represent f in terms of components, then the bove becomes lim m t 0 i=1 f i (x + te j ) f i (x) u i = f (x)e j. t It follows tht ech quotient in this sum hs limit, s t 0, so ech (D j f i )(x) exists. Counter Exmple 10 (Existence of Prtil Derivtives Do Not Imply Differentibility). Let f(x, y) = { xy x 2 +y 2, (x, y) (0, 0) 0, (x, y) = (0, 0) f x (x, y) = y(x2 y 2 ) (x 2 + y 2 ) 2 for (x, y) (0, 0) nd if we fix y = 0 we get f(x, 0) = 0, so f x(x, y) is defined for ll vlues. We get similr vlues for f y. However, f is not differentible since... Theorem 32 (Men Vlue Estimte). Suppose f mps convex open set E R n into R m, f is differentible in E nd there is rel number M such tht f (x) M for every x E. Then for ll, b E. f(b) f() M b Lemm 9 (Men Vlue Theorem for Vector Vlued Functions). Suppose f is continuous mpping of [, b] into R k nd f is differentible in (, b). Then there exists x (, b) such tht f(b) f() (b ) f (x). Put z = f(b) f() nd define φ(t) = z f(t). Then φ is rel-vlued continuous function on [, b] which is differentible in (, b). So by the Men Vlue Theorem, φ(b) φ() = (b )φ (x) = (b )z f (x) for some x (, b). On the other hnd, φ(b) φ() = z f(b) z f() = z z = z 2. 19

The Cuchy-Schwrz Inequlity yields Hence z (b ) f (x). Fix, b E. Define z 2 = (b ) z f (x) (b ) z f (x). γ(t) = (1 t) + tb for ll t R such tht γ(t) E. Since E is convex γ(t) E if 0 t 1. Put Then, by the chin rule, we hve so tht for ll t [0, 1]. Thus, by the bove Lemm, But g(0) = f() nd g(1) = f(b). g(t) = f(γ(t)). g (t) = f (γ(t))γ (t) = f (γ(t))(b ), g (t) f (γ(t)) b M b g(1) g(0) M b. Theorem 33 (Equivlence of Continuous Differentibility nd Continuity of Prtil Derivtives). Suppose f mps n open set E R n into R m. Then f is continuously differentible on E if nd only if the prtil derivtives D j f i exist nd re continuous on E for 1 i m, 1 j n. Assume first tht f is continuously differentible on E. By bove, we get for ll i, j nd ll x E. Hence nd since u i = e j = 1, it follows tht (D j f i )(x) = (f (x)e j ) u i (D j f i )(y) (D j f i )(x) = {[f (y) f (x)]e j } u i (D j f i )(y) (D j f i )(x) [f (y) f (x)]e j f (y) f (x). Hence D j f i is continuous s f is continuous. For the converse, it suffices to consider the cse where m = 1. Fix x E nd ɛ > 0. Since E is open, there is n open bll S E with center t x nd rdius r, nd the continuity of the functions D j f shows tht r cn be chosen so tht (D j f)(y) (D j f)(x) < ɛ n for y S nd 1 j n. Suppose h = h j e j, h < r, put v 0 = 0 nd v k = h 1 e 1 + + h k e k for 1 n. Then j=1 f(x + h) f(x) = [f(x + v j ) f(x + v j 1 )]. j=1 Since v k < r for 1 k n nd since S is convex, the segments with end points x + v j 1 nd x + v j lie in S. Since v j = v j 1 + h j e j, the Men Vlue Theorem shows tht the jth summnd in the bove equlity is equl to h j (D j f)(x + v j 1 + θ j h j e j ) 20

for some θ j (0, 1), nd this differs from h j (D j f)(x) by less thn h j ɛ/n by our choice of r. By the bove equlity, it follows tht f(x + h) f(x) h j (D j f)(x) 1 h j ɛ h ɛ n for ll h such tht h < r. Dividing by h gives the desired result. Theorem 34 (f Is Convex If nd Only If Every Criticl Point Is A Globl Minimizer). j=1 If f is twice differentible nd convex on convex domin D, then every criticl point of f in D must be globl minimizer. Lemm 10 (f Is Convex If nd Only If H(f) Is Positive Semidefinite). f is convex in D (D s bove) if nd only if H(f) is positive semidefinite, where H(f) is the Hessin of f nd mtrix is positive semidefinite if ll eigenvlues re greter thn or equl to 0. For ny fixed x, y D, let g xy (t) = f(x + t(y x)), where t is defined such tht x + t(x y) D. So g xy (t) is convex since f is. So, by n bove theorem, g xy(t) 0. So So f is convex if nd only if 0 (g xy(t)) = = l=1 i,j=1 ( n j=1 ) f (x + t(y x))(y k x k ) x k 2 f x l x k (x + t(y x))(y k x k )(y l x l ). 2 f x i x j (x)w i w j 0, for w = (w 1,..., w n ) R n, x D. Let x 0 be criticl point of f. So f x j (x 0 ) = 0 for j = 1, 2,..., n. For ny x D, by Tylor s Theorem, f(x) = f(x 0 ) = j=1 f x j (x 0 )(x j x 0j ) + i,j=1 2 f x i x j (ζ)(x i x 0i )(x j x 0j ) for ζ picked ppropritely. However, since x 0 is criticl point nd by the lemm bove, we hve So x 0 is globl minimizer. Definition 6 (Contrction). f(x) f(x 0 ) + 0 + 0 = f(x ). Let X be metric spce, with metric d. If φ mps X into X nd if there is number c < 1 such tht d(φ(x), φ(y)) cd(x, y) for ll x, y X, then φ is sid to be contrction of X into X. Theorem 35 (The Contrction Principle). If X is complete metric spce, nd if φ is contrction of X into X, then there exists one nd only one x X such φ(x) = x. Pick x 0 X rbitrrily, nd define {x n } recursively, by setting x n+1 = φ(x n ). 21

As φ is contrction, there is c < 1 stisfying the definition. For n 1 we then hve By induction, we get If n < m, it follows tht d(x n+1, x n ) = d(φ(x n ), φ(x n 1 )) cd(x n, x n 1 ). d(x n, x m ) m i=n+1 d(x n+1, x n ) c n d(x 1, x 0 ). d(x i, x i 1 ) (c n + + c m 1 )d(x 1, x 0 ) [(1 c) 1 d(x 1, x 0 )]c n s this is geometric series. Since c < 1, s n, this goes to 0, so {x n } is Cuchy Sequence. Since X is complete, lim n x n = x for some x X. Since φ is contrction, φ is (uniformly) continuous on X. Hence φ(x) = lim n φ(x n) = lim n x n+1 = x. Counter Exmple 11 (If contrction hs c = 1, the bove conclusion need not hold). Let f(x) = x + 1. So f(x) f(y) = 1 + ex x y + e y e x (1 + e x )(1 + e y ) < x y. However, f(x) x s e x > 0 for ll x R. Theorem 36 (Inverse Function Theorem). Suppose f is continuously differentible mpping on E of n open set E R n into R n, f () is invertible for some E, nd b = f(). Then () there exist open sets U nd V in R n such tht U, b V, f is one-to-to on U, nd f(u) = V ; (b) if g is the inverse of f, which exists by (), defined in V by g(f(x)) = x, (x U), then g is continuously differentible on V nd g (y) = {f (g(y))} 1 for y V. () Put f () = A, nd choose λ so tht 2λ A 1 = 1. Since f is continuous t, there is n open bll U E, with center t, such tht We ssocite to ech y R n function φ, defined by f (x) A < λ, (x U). φ(x) = x + A 1 (y f(x)), (x E). (Note tht f(x) = y if nd only if x is fixed point of φ). Since φ (x) = I A 1 f (x) = A 1 (A f (x)), by the bove, we hve φ (x) < 1, (x U). 2 Hence φ(x 1 ) φ(x 2 ) 1 2 x 1 x 2, (x 1, x 2 U), by the Men Vlue Estimte. It follows tht φ hs t most one fixed point in U. So f is one-to-one in U. Next, put V = f(u) nd pick y 0 V. Then y 0 = f(x 0 ) for some x 0 U. Let B be n open bll with center 22

x 0 nd rdius r > 0, so smll tht its closure B lies in U. Fix y, y y 0 < λr. With φ s bove, If x B, it follows by the contrction tht φ(x 0 ) x 0 = A 1 (y y 0 ) < A 1 λr = r 2. φ(x) x 0 φ(x) φ(x 0 ) + φ(x 0 ) x 0 < 1 2 x x 0 + r 2 r; hence φ(x) B. Thus φ is contrction of B into B. Being closed, bounded subset of R n, B is complete. The Contrction Principle implies tht φ hs fixed point x B. For this x, f(x) = y. Thus y f(b) f(u) = V. So V is open in R n. Corollry 6 (Open Mpping Theorem). If f is continuously differentible mpping on n open set E R n into R n nd if f (x) is invertible for every x E, then f(w ) is n open subset of R n for every open set W E. Let W be n open set in R n. U() W such tht, for x, y U(), f (y) f (x) < For every point W, since f is invertible t, there is n open bll 1 2 (f ()) 1. Note tht W = U(x). Thus, by x W the Inverse Function Theorem, V (x) = f(u(x)) is open. As ny union of open sets is open, we hve tht f(w ) = f( U(x)) = V (x) is open in R n. Theorem 37 (Implicit Function Theorem). Let f be continuously differentible mpping of n open set E R n+m into R n, such tht f(, b) = 0 for some point (, b) E. Put A = f (, b) nd ssume tht A x is invertible. Then there exist open sets U R n+m nd W R m, with (, b) U nd b W, hving the following property: To every y W corresponds unique x such tht (x, y) U nd f(x, y) = 0. If this x is defined to be g(y), then g is continuously differentible mpping of W into R n, g(b) =, f(g(y), y) = 0, y W, nd g (b) = (A x ) 1 A y. Theorem 38 (Condition For Chnging Order of Derivtives). Suppose f is defined in n open set E R 2, suppose tht D 1 f, D 21 f nd D 2 f exist t every point of E, nd D 21 f is continuous t some point (, b) E. Then D 12 f exists t (, b) nd (D 12 f)(, b) = (D 21 f)(, b). Counter Exmple 12 (A Function Whose Derivtives Are Not Interchngble). So Let f x (x, y) = f(x, y) = { xy(y 2 x 2 ) x 2 +y 2 if(x, y) (0, 0) 0 if(x, y) = (0, 0) { (3x 2 y 3 )(x 2 +y 2 ) 2x(x 3 y xy 3 ) (x 2 +y 2 ) 2 if(x, y) (0, 0) 0 if(x, y) = (0, 0) 23

So f xy (0, 0) f yx (0, 0). f y (x, y) = { ( 3xy 2 +x 3 )(x 2 +y 2 ) 2y(x 3 y xy 3 ) (x 2 +y 2 ) 2 if(x, y) (0, 0) 0 if(x, y) = (0, 0) f xy (0, 0) = lim = f x(0, h) f x (0, 0) h 0 = lim = 1, h 0 h h 0 h Integrtion of Differentil Forms Theorem 39 (Stokes Theorem). f yx (0, 0) = lim = f y(h, 0) f y (0, 0) h 0 = lim = 1. h 0 h h 0 h Let D be bounded domin in R n with piecewise-c 1 boundry. Let f be differentible (n 1)-form on D. Then f = df, D D f = f i dx (j), dx (j) = dx 1 dx j 1 dx j+1 dx n. j=1 Theorem 40 (Green s Theorem). Let E be n open subset of R 2, α, β C 1 (E) nd D be closed subset of E with positively oriented boundry D. Then β αdx + βdy = x α y da. Let λ = αdx + βdy. Then So, by Stokes Theorem D dλ = d(αdx + βdy) = α y D Theorem 41 (Divergence Theorem). αdx + βdy = D ( β β dy dx + dx dy = x x α ) da. y D λ = D dλ = D β x α y da. Let D be bounded domin in R n with piecewise C 1 boundry. If f = (f 1,..., f n ) : D R n wiht f j C 1 (D) C(D) for 1 j n. Then fdx = f νdσ(x), where ν is the unit outer norml vector of D t x, nd ν is given by Sequences nd Series of Functions Definition 7. D D f f f. A sequence {f n } n=1 of functions on E is pointwise convergent if, for ny fixed x E, f n (x) f(x) s n. A sequence {f n } of functions on E is uniformly convergent to function f(x) if for every ɛ > 0 there is n integer N such tht n N implies f n (x) f(x) < ɛ for ll x E. 24

Counter Exmple 13 (Pointwise Convergence Does Not Imply Uniform Convergent). Let ɛ > 0 be given nd let f n (x) = x. For ny fixed x R, if we choose n integer n such tht x /ɛ < n, n then f n (x) 0 < ɛ, so f n converges pointwise to f(x) = 0. Assumet towrd contrdiction tht f n f uniformly nd let ɛ < 1. So there exists n integer N such tht for ll n N, f n (x) f(x) < ɛ for ll x R. Fix n = N nd choose x = N. We hve f N (N) 0 = N N = 1 ɛ. So f n does not converge to f uniformly. Note: Uniform Convergence does imply pointwise convergence. Theorem 42 (Cuchy Criterion). The sequence of functions {f n } on E converges uniformly if nd only if for every ɛ > 0 there exists n integer N such tht for m, n N, x E imply f n (x) f m (x) < ɛ. Suppose {f n } converges uniformly on E nd let f be the limit function. Then there is n integer N such tht n N, x E implies f n (x) f(x) < ɛ 2, so tht f n (x) f m (x) f n (x) f(x) + f(x) f m (x) < ɛ if n, m N, x E. Conversely, suppose the Cuchy criterion holds. By the Cuchy Sequence Condition, the sequence {f n (x)} converges, for every x, to limit which me wy cll f(x). Thus the sequence {f n (x)} converges on E, to f. Let ɛ > 0 be given, nd choose N such tht f n (x) f m (x) < ɛ for n, m N, x E. Fix n, nd let m. Since f m (x) f(x) s m, this gives for every n N nd every x E. Theorem 43 (Weierstrss M-Test). Then f n (x) f(x) < ɛ Suppose {f n (x)} is sequence of functions defined on E nd suppose f n (x) converges uniformly on E if n=1 If M n converges, then, for rbitrry ɛ > 0, n=1 f n (x) M n, (x E, n = 1, 2, 3,... ). M n converges. n=1 m m f i (x) M i < ɛ, (x E), i=n provided m nd n re lrge enough. Uniform convergence follows by the Cuchy Criterion for sums. i=n 25

Theorem 44 (Uniform Convergence nd Continuity). Suppose {f n } is sequence of continuous functions nd f n f uniformly. Then f is continuous. Let ɛ > 0 be fixed. Let x 0 E. If x 0 is n isolted point of E, then f is trivilly continuous t x 0. We my suppose then tht x 0 is limit point of E. Since we hve uniform convergence, there is n integer N such tht for ll n N, f n (x) f(x) < ɛ 3 for ll x E. Since f N is continuous t x 0, there is neighborhood B(x 0 ) such tht for ll x B(x 0 ) E, So, for ny x B(x 0 ) E, f N (x 0 ) f N (x) < ɛ 3. f(x 0 ) f(x) f(x 0 ) f N (x 0 ) + f N (x 0 ) f N (x) + f N (x) f(x) < ɛ. Counter Exmple 14 (Sequence of Continuous Functions tht Converge to Discontinuous Function). Let f(x) = 0 if x < 1 n nx + 1 if 1 n x < 0 1 if x = 0 nx + 1 if 0 < x 1 n 0 if 1 n < x Ech f n (x) is continuous for ll n. For x 0, f n (x) 0 s there exists some N > 1. For x = 0, x f n (x) = 1 1. So f n (x) f(x) where f(x) is not continuous t 0. Theorem 45 (Prtil Converse to Uniform Convergence nd Continuity). Suppose K is compct nd () {f n } is sequence of continuous functions on K; (b) {f n } converges pointwise to continuous function on K; (c) f n (x) f n+1 (x) for ll x K, n = 1, 2, 3,.... Then f n f uniformly on K. Put g n = f n f. Then g n is continuous, g n 0 pointwise, nd g n g n+1. It suffices to show tht g n 0 uniformly on K. Let ɛ > 0 be given. Let K n be the set of ll x K with g n (x) ɛ. Since g n is continuous, K n is closed, hence compct by Heine-Borel becuse K n is bounded by the fct tht g n (K) is compct. Since g n g n+1, we hve K n K n+1. Fix x K. Since g n (x) 0, we see tht x / K n if n is sufficiently lrge. This x / K n. In other words, K n is empty. Hence K N is empty for some N. It follows tht 0 g n (x) < ɛ for ll x K nd for ll n N. Counter Exmple 15. Note tht ll of the bove conditions re necessry for this result to hold. For instnce, if we omit 1 compctness nd look t f n (x) = nx + 1 on (0, 1), then f n is continuous, converges pointwise to 0 nd is monotonic. However, f n does not converge uniformly since for ny n Z +, 1/n (0, 1) nd f n (1/n) = 1/2 0. If we insted omit the monotonicity of f nd look t f n (x) = nx(1 x 2 ) n on [0, 1], we hve K is compct, f n is continuous, f n 0 pointwise by L Hôpitl s Rule, but f n does not converge uniformly to 0, since if we pick ɛ < e 1, nd let x = 1/n, f n (x) = (1 + 1 n )n (1 1 n )n > e 1 s n. So f n cnnot converge uniformly. 26

Theorem 46 (Integrbility nd Uniform Convergence). Suppose f n is Riemnn-integrble on [, b], for n = 1, 2, 3,.... nd suppose f n f uniformly on [, b], then f is Riemnn-integrble nd lim n f n (x)dx = f(x)dx. Let ɛ n = sup f n (x) f(x), the supremum being tken over x b. Then f n ɛ n f f n + ɛ n, so tht the upper nd lower integrls of f stisfy f n ɛ n dx fdx fdx f n + ɛ n dx. Hence 0 fdx fdx 2ɛ n [b ]. Since ɛ n 0 s n, the upper nd lower integrls re equl. Thus f is Riemnn-integrble. Counter Exmple 16. If f n f pointwise on [, b], f n (x)dx does not necessrily converge to f(x)dx. Let f n (x) = nx(1 x 2 ) n on [0, 1]. As hs been shown, f n 0. We hve n x(1 x 2 ) n dx = n 2n + 2. So 1 0 nx(1 x 2 ) n dx 1 2 Counter Exmple 17. If f n (x)dx Let f n (x) = sin(x), f(x) = 0. 1 However, f n (x) = sin(x) f(x) = 0. 0 f(x)dx = 0. f(x)dx, it does not gurntee tht f n f t ll (let lone uniformly). 2π 0 sin(x)dx = 0 2π Theorem 47 (Differentibility nd Uniform Convergence). 0 f(x)dx. Suppose {f n } is sequence of functions, differentible on [, b] nd such tht {f n (x 0 )} converges for some point x 0 on [, b]. If {f n} converges uniformly on [, b], then {f n } converges uniformly on [, b], to function f, nd f (x) = lim n f n(x), ( x b). Let ɛ > 0 be given. Choose N such tht n, m N implies f n (x 0 ) f m (x 0 ) < ɛ 2 nd f n(t) f m(t) < ɛ, ( t b). 2(b ) 27

If we pply the Men Vlue Theorem to the function f n f m, the bove inequlity furnishes (f n (x) f m (x)) f n (t) + f m (t) for ny x nd t on [, b], if n, m N. The inequlity x t ɛ 2(b ) < ɛ 2 f n (x) f m (x) f n (x) f m (x) f n (x 0 ) + f m (x 0 ) + f n (x 0 ) f m (x 0 ) implies, by bove inequlities, tht so tht {f n } converges uniformly on [, b]. Let Let us now fix point x on [, b] nd define f n (x) f m (x) < ɛ, ( x b, n, m N), f(x) = lim n f n(x), ( x b). φ n (t) = f n(t) f n (x) f(t) f(x), φ(t) = t x t x for t b, t x. Then lim φ n(t) = f n(x). t x From wht we obtined by using the Men Vlue Theorem, we get φ n (t) φ m (t) < ɛ, n, m N, 2(b ) so tht {φ n } converges uniformly, for t x. Since {f n } converges to f, we conclude by the definition of φ n nd φ tht lim φ n(t) = φ(t) n uniformly for t b, t x. Since {φ n } is set of continuous functions tht converge uniformly to φ, we hve lim φ(t) = lim f t x n(x). So, n specificlly, f (x) = lim f n(x) for x b. n Definition 8. A fmily F of functions f defined on E in metric spce (X, d) is sid to be equicontinuous on E if for every ɛ > 0 there exists δ > 0 such tht whenever d(x, y) < δ, x, y, E nd f F. f(x) f(y) < ɛ Theorem 48 (Convergent Subsequence on Countble Set). If {f n } is pointwise bounded sequence of complex functions on countble set E, then {f n } hs subsequence {f nk } such tht {f nk (x)} converges for every x E. Let {x i } i = 1, 2, 3,..., be the points of E rrnged in sequence. Since {f n (x 1 )} is bounded, by Bolzno Weierstrss there exists subsequence, which we shll denote by {f 1,k }, such tht {f 1,k (x 1 )} converges s k. Let us now consider sequences S 1, S 2,, which we represent by the rry: S 1 : f 1,1, f 1,2, f 1,3, nd which hve the following properties: S 2 : f 2,1, f 2,2, f 2,3, S 3 : f 3,1, f 3,2, f 3,3, 28

() S n is subsequence of S n 1 for n = 2, 3, 4,... ; (b) {f n,k (x n )} converges s k (the boundedness of {f n (x k )} mkes it possible to choose S n in this wy); (c) The order in which the functions pper is the sme in ech sequence, i.e., if one function precedes nother in S 1, they re in the sme reltion in every S n, until one or the other is deleted. Hence, when going from one row in the bove rry to the next below, functions my move to the left but never to the right. We now consider the sequence S : f 1,1, f 2,2, f 3,3,.... By (c), the sequence S (except possibly the first n 1 terms) is subsequence of S n for n = 1, 2, 3,.... Hence (b) implies tht {f n,n (x i )} converges s n for every x i E. Theorem 49 (Arzelà-Ascoli). Let K be compct set. If F is fmily of continuous functions on K, the following re equivlent: () {f n } is pointwise bounded nd equicontinuous. (b) {f n } hs uniformly convergent subsequence. ((b) ()) Let ɛ > 0 be given nd let E be countble dense subset of K. By the bove Theorem, {f n } hs subsequence {f nk } such tht {f nk (x)} converges for every x E. Put f nk = g k nd pick δ > 0 such tht f n (x) f n (y) < ɛ/3 for ll n provided tht d(x, y) < δ. Let V (x, δ) be the set of ll y K with d(x, y) < δ. Since E is dense in K, nd K is compct, there re finitely mny points x 1,..., x m in E such tht K V (x 1, δ) V (x, δ). Since {g k (x)} converges for every x E, there is n integer N such tht g k (x s ) g l (x s ) < ɛ/3 whenever k, l N, 1 s m. If x K, then by bove x V (x s, δ) for some s, so tht for every k. If k, l N, it follows tht g k (x) g k (x s ) < ɛ/3 g k (x) g l (x) g k (x) g k (x s ) + g k (x s ) g l (x s ) + g l (x s ) g l (x) < ɛ. Theorem 50 (Stone Weierstrss Theorem). If f is continuous complex function on [, b], there exists sequence of polynomils P n such tht uniformly on [, b]. If f is rel, P n my be tken rel. Some Specil Functions Theorem 51 (Uniqueness of the Gmm Function). If f is positive function on (0, ) such tht () f(x + 1) = xf(x), (b) f(1) = 1, (c) log f is convex, lim P n(x) = f(x) n 29