Lecture Notes on Functional Analysis. Kai-Seng Chou Department of Mathematics The Chinese University of Hong Kong Hong Kong

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1 Lecture Notes on Functionl Anlysis Ki-Seng Chou Deprtment of Mthemtics The Chinese University of Hong Kong Hong Kong My 29, 2014

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3 Contents 1 Normed Spce: Exmples Vector Spces of Functions Zorn s Lemm Existence of Bsis Three Inequlities Normed Vector Spces Normed Spce: Anlyticl Aspects Normed Spce As Metric Spce Seprbility Completeness Sequentil Compctness Arzel-Ascoli Theorem Dul Spce Liner Functionls Concrete Dul Spces Hhn-Bnch Theorem Consequences of Hhn-Bnch Theorem The Dul Spce of Continuous Functions Reflexive Spces Bounded Liner Opertor Bounded Liner Opertors Exmples of Liner Opertors Bire Theorem Uniform Boundedness Principle Open Mpping Theorem

4 4 CONTENTS 4.6 The Spectrum Hilbert Spce Inner Product Inner Product nd Norm Orthogonl Decomposition Complete Orthonorml Sets A Structure Theorem Compct, Self-Adjoint Opertor Adjoint Opertors Compct, Self-Adjoint Opertors An Appliction Wek Compctness Wek Sequentil Compctness Topologies Induced by Functionls Wek nd Wek* Topologies Extreme Points in Convex Sets Nonliner Opertors Fixed-Point Theorems Clculus in Normed Spces Minimiztion Problems

5 ddddddddddddd ddddddddddd ddddddd Chpter 1 Normed Spce: Exmples Generlly speking, in functionl nlysis we study infinite dimensionl vector spces of functions nd the liner opertors between them by nlytic methods. This chpter is of preprtory nture. First, we use Zorn s lemm to prove there is lwys bsis for ny vector spce. It fills up gp in elementry liner lgebr where the proof ws only given for finite dimensionl vector spces. The indequcy of this notion of bsis for infinite dimensionl spces motivtes the introduction of nlysis to the study of function spces. Second, we discuss three bsic inequlities, nmely, Young s, Hölder s, nd Minkowski s inequlities. We estblish Young s inequlity by elementry mens, use it to deduce Hölder s inequlity, nd in term use Hölder s inequlity to prove Minkowski s inequlity. The ltter will be used to introduce norms on some common vector spces. As you will see, these spces form our principl exmples throughout this book. 1.1 Vector Spces of Functions Recll tht vector spce is over field F. Throughout this book it is lwys ssumed this field is either the rel field R or the complex field C. In the following F stnds for R or C. It is true tht mny vector spces cn be viewed s vector spces of functions. To describe this unified point of view, let S be non-empty set nd denote the collection of ll functions from S to F by F (S). It is routine to check tht F (S) forms vector spce over F under the obvious rules of ddition nd sclr multipliction for functions: For f, g F (S) nd α F, (f + g)(p) f(p) + g(p), (αf)(p) αf(p). In fct, these lgebric opertions re inherited from the trget F. First, tke S = {p 1,, p n } set consisting of n mny elements. Every function f F (S) is uniquely determined by its vlues t p 1,, p n, so f cn be identified with the n-triple (f(p 1 ),, f(p n )). It is esy to see tht F ({p 1,, p n }) is linerly isomorphic to F n. More precisely, the mpping f (f(p 1 ),, f(p n )) is liner bijection between F ({p 1,, p n }) nd F n. Second, tke S = {p 1, p 2, }. As bove, ny f F (S) cn be identified with the sequence (f(p 1 ), f(p 2 ), f(p 3 ) ). The vector spce F ({p j } j=1 ) my be clled the spce of sequences over F. Finlly, tking S = [0, 1], F ([0, 1]) consists of ll F-vlued functions. The vector spces we re going to encounter re mostly these spces nd their subspces. 5

6 6 CHAPTER 1. NORMED SPACE: EXAMPLES 1.2 Zorn s Lemm In liner lgebr, it ws pointed out tht every vector spce hs bsis no mtter it is of finite or infinite dimension, but the proof ws only given in the finite dimensionl cse. Here we provide proof of the generl cse. The proof depends criticlly on Zorn s lemm, n ssertion equivlent to the xiom of choice. To formulte Zorn s lemm, we need to consider prtil order on set. A reltion on non-empty set X is clled prtil order on X if it stisfies (PO1) x x, x X; (PO2) x y nd y x implies x = y. (PO3) x y, y z implies x z. The pir (X, ) is clled prtilly ordered set or poset for short. A non-empty subset Y of X is clled chin or totlly ordered set if for ny two y 1, y 2 Y, either y 1 y 2 or y 2 y 1 holds. In other words, every pir of elements in Y re relted. An upper bound of non-empty subset Y of X is n element u, which my or my not be in Y, such tht y u for ll y Y. Finlly, mximl element of (X, ) is n element z in X such tht z x implies z = x. Exmple 1.1. Let S be set nd consider X = P(S), the power set of S. It is cler tht the reltion set inclusion A B is prtil order on P(S). It hs unique mximl element given by S itself. Exmple 1.2. Let X = R 2 nd define x y if nd only if x 1 y 1 nd x 2 y 2. For instnce, ( 1, 5) (0, 8) but ( 2, 3) nd (35, 1) re unrelted. Then (X, ) forms poset without ny mximl element. Zorn s Lemm. Let (X, ) be poset. If every chin in X hs n upper bound, then X hs t lest one mximl element. Although clled lemm by historicl reson, Zorn s lemm, constituent in the Zermelo-Frenkel set theory, is n xiom in nture. It is equivlent to the xiom of choice s well s the Husdorff mximlity principle. You my look up Hewitt-Stromberg s Rel nd Abstrct Anlysis for further informtion. A redble ccount on this lemm cn lso be found in Wikipedi. 1.3 Existence of Bsis As stndrd ppliction of Zorn s lemm, we show there is bsis in ny vector spce. To refresh your memory, let s recll tht subset S in vector spce X is clled linerly independent set if ny finite number of vectors in S re linerly independent. In other words, letting {x 1,, x n } be ny subset of S, if α 1 x α n x n = 0 for some sclrs α i, i = 1,, n, then α i = 0 for ll i. On the other hnd, given ny subset S, denote ll liner combintions of vectors from S by S. It is esy to check tht S forms subspce of X clled the subspce spnned by S. A subset S is clled spnning set of X if S is X, nd it is clled bsis of X if it is lso linerly independent spnning set. When X dmits finite spnning set, it hs bsis consisting of finitely mny vectors. Moreover, ll bses hve the sme number of vectors nd we cll this number the dimension of the spce X. The spce X is of infinite dimension if it does not hve finite spnning set. Theorem 1.1. Every non-zero vector spce hs bsis. This bsis is sometimes clled Hmel bsis. Proof. Let X be the set of ll linerly independent subsets of given vector spce V. Since V is non-zero, X is non-empty set. Clerly the set inclusion mkes it into poset. To pply Zorn s lemm, let s

7 1.4. THREE INEQUALITIES 7 verify tht every chin in it hs n upper bound. Let Y be chin in X, consider the following subset of V, S = C. C Y We clim tht (i) S X, tht s, S is linerly independent set, (ii) C S, C Y, tht s, S is n upper bound of Y. Since (ii) is obvious, it is sufficient to verify (i). To this end, pick v 1,, v n S. By definition, we cn find C 1,, C n in Y such tht v 1 C 1,, v n C n. As Y is chin, C 1,, C n stisfy C i C j or C j C i for ny i, j. After rerrnging the indices, one my ssume C 1 C 2 C n, nd so {v 1,, v n } C n. Since C n is linerly independent set, {v 1,, v n } is linerly independent. This shows tht S is linerly independent set. After showing tht every chin in X hs n upper bound, we ppel to Zorn s lemm to conclude tht X hs mximl element B. We clim tht B is bsis for V. For, first of ll, B belonging to X mens tht B is linerly independent set. To show tht it spns V, we pick v V. Suppose v does not belong to B, so v is independent from ll vectors in B. But then the set B = B {v} is linerly independent set which contins B s its proper subset, contrdicting the mximlity of B. We conclude tht B = V, so B forms bsis of V. The following exmple my help you in understnding the proof of Theorem 1.1. Exmple 1.3. Consider the power set of R 3 which is prtilly ordered by set inclusion. Let X be the subset of ll linerly independent sets in R 3. Then { } Y 1 {(1, 0, 0)}, {(1, 0, 0), (1, 1, 0)}, {(1, 0, 0), (1, 1, 0), (0, 0, 3)} nd re chins but is not chin in X. Y 3 Y 2 { } {(1, 3, 5), (2, 4, 6)}, {(1, 3, 5), (2, 4, 6), (1, 0, 0)} { } {(1, 0, 0)}, {(1, 0, 0), (0, 1, 0)}, {(1, 0, 0), (0, 2, 0), (0, 0, 1)} For finite dimensionl vector spce, it is reltively esy to find n explicit bsis, nd bses re used in mny occsions such s in the determintion of the dimension of the vector spce nd in the representtion of liner opertor s mtrix. However, in contrst, the existence of bsis in infinite dimensionl spce is proved vi non-constructive rgument. It is not esy to write down bsis. For exmple, consider the spce of sequences S {x = (x 1, x 2,, x n ) : x i F}. Letting e j = (0,, 1, ) where 1 ppers in the j-th plce, it is tempting from the formul x = j=1 x je j to ssert tht {e j } 1 forms bsis for S. But, this is not true. Why? It is becuse infinite sums re not liner combintions. Indeed, one cnnot tlk bout infinite sums in vector spce s there is no mens to mesure convergence. According to Theorem 1.1, however, there is rther mysterious bsis. In generl, non-explicit bsis is difficult to work with, nd thus lessens its importnce in the study of infinite dimensionl spces. To proceed further, nlyticl structures will be dded to vector spces. Lter, we will see tht for resonbly nice infinite dimensionl vector spce, ny bsis must consist of uncountbly mny vectors (see Proposition 4.14). Suitble generliztions of this notion re needed. For n infinite dimensionl normed spce, one my introduce the so-clled Schuder bsis s replcement. For complete inner product spces ( Hilbert spce), n even more useful notion, complete orthonorml set, will be much more useful. Mthemtics is deductive science. A limited number of xioms is needed to build up the tower of mthemtics, nd Zorn s lemm is one of them. We will encounter this lemm gin in lter chpters. You my lso google for more of its pplictions. 1.4 Three Inequlities Now we come to Young s, Hölder s nd Minkowski s inequlities.

8 8 CHAPTER 1. NORMED SPACE: EXAMPLES Two positive numbers p nd q re conjugte if 1/p + 1/q = 1. Notice tht they must be greter thn one nd q pproches infinity s p pproches 1. In the following prgrphs q is lwys conjugte to p. Proposition 1.2 (Young s Inequlity). For ny, b > 0 nd p > 1, nd equlity holds if nd only if p = b q. b p p + bq q, Proof. Consider the function ϕ(x) = xp p + 1 q x, x (0, ). From the sign of ϕ (x) = x p 1 1 we see tht ϕ is strictly decresing on (0, 1) nd strictly incresing on (1, ). It follows tht x = 1 is the strict minimum of ϕ on (0, ). So, ϕ(x) ϕ(1) nd equlity holds if nd only if x = 1. In other words, x p p + 1 q x 1 p + 1 q 1, tht is, x p p + 1 q x. Letting x = b/b q, we get the Young s inequlity. Equlity holds if nd only if b/b q = 1, i.e., p = b q. Proposition 1.3 (Hölder s Inequlity). For, b R n, p > 1, n k b k p b q, k=1 where p = ( n k=1 k p ) 1 p nd b q = ( n k=1 b k q ) 1 q. Proof. The inequlity clerly holds when = (0,, 0). We my ssume (0,, 0) in the following proof. By Young s inequlity, for ech ε > 0 nd k, Thus k b k = ε k ε 1 b k εp k p n k b k = 1 b 1 + n b n k=1 εp p n k=1 p k p + ε q q + ε q b k q. q n b k q k=1 = εp p p p + ε q q b q q, (1.1) for ny ε > 0. To hve the best choice of ε, we minimize the right hnd side of this inequlity. Tking derivtive of the right hnd side of (1.1) s function of ε, we obtin tht is, ε p 1 p p ε q 1 b q q = 0, q ε = b p+q q p. p+q is the minimum point. (Clerly this function hs only one criticl point nd does not hve ny mximum.) Plugging this choice of ε into the inequlity yields the Hölder s inequlity fter some mnipultion. p

9 1.4. THREE INEQUALITIES 9 Proposition 1.4 (Minkowski s Inequlity). For, b F n nd p 1, + b p p + b p. Proof. The inequlity clerly holds when p = 1 or + b = 0. In the following proof we my ssume p > 1 nd + b > 0. For ech k, k + b k p = k + b k k + b k p 1 k k + b k p 1 + b k k + b k p 1. (1.2) Applying Hölder s inequlity to the two terms on right hnd side of (1.2) seprtely (more precisely, to the pirs of rel vectors ( 1,, n ) nd ( 1 + b 1 p 1,, n + b n p 1 ), nd ( b 1,, b n ) nd ( 1 + b 1 p 1,, n + b n p 1 )), we hve ( n n ) 1 q k + b k p p k + b k (p 1)q k=1 nd Minkowski s inequlity follows. k=1 = ( p + b p ) ( n k=1 k + b k p ) 1 q, + b p ( n k=1 k + b k (p 1)q ) 1 q Look up Wikipedi for the gret mthemticin Hermnn Minkowski ( ), the best friend of Dvid Hilbert nd techer of Albert Einstein, who died unexpectedly t forty-five. The biogrphy Hilbert by C. Reid contins n interesting ccount on Minkowski nd Hilbert. The lst two inequlities llow the following generliztion. Hölder s Inequlity for Sequences. For ny two sequences nd b in F, nd p > 1, k b k p b q, where now the summtion in the sums on the right runs from 1 to. k=1 Since the norms p nd b q re llowed to be zero or infinity, we dopt the convention 0 = 0 in the bove inequlity. Minkowski s Inequlity for Sequences. For ny two sequences nd b in F nd p 1, + b p p + b p, where now the summtion in the sums runs from 1 to. Hölder s Inequlity for Functions. For p > 1 nd Riemnn integrble functions f nd g on [, b], we hve b fg ( b f p ) 1 p ( b g q ) 1 q. Minkowski s Inequlity for Functions. For p 1 nd Riemnn integrble functions f nd g on [, b], we hve

10 10 CHAPTER 1. NORMED SPACE: EXAMPLES ( b ) 1 ( p ) 1 ( b p ) 1 b p f + g p f p + g p, We leve the proofs of these generliztions s exercises. 1.5 Normed Vector Spces Let (X, +, ) be vector spce over F. A norm on X is function from X to [0, ) stisfying the following three properties: For ll x, y X nd α F, (N1) x 0 nd = holds if nd only if x = 0, (N2) x + y x + y, (N3) αx = α x. The vector spce with norm, (X, +,, ), or (X, ), or even stripped to single X when the context is cler, is clled normed vector spce or simply normed spce. Here re some normed vector spces. Exmple 1.4. (F n, p ), 1 p <, where x p = ( n k=1 x k p ) 1 p. Clerly, (N1) nd (N3) hold. According to the Minkowski s inequlity (N2) holds too. When p = 2 nd F n = R n or C n, the norm is clled the Eucliden norm or the unitry norm. Exmple 1.5. is clled the sup-norm. (F n, ) where x = mx x k. k=1,,n Exmple 1.6. Let l p, 1 p <, be the collection of ll F-vlued sequences x = (x 1, x 2, ) stisfying x k p <. k=1 First of ll, from the Minkowski s inequlity for sequences the sum of two sequences in l p belongs to l p. With the other esily checked properties, l p forms vector spce. The function p, i.e., x p = ( k=1 x k p ) 1 p clerly stisfies (N1) nd (N3). Moreover, (N2) lso holds by Minkowski s inequlity for sequences. Hence it defines norm on l p. Exmple 1.7. Let l be the collection of ll F-vlued bounded sequences. Define the sup-norm Clerly l forms normed vector spce over F. x = sup x k. k

11 1.5. NORMED VECTOR SPACES 11 Exmple 1.8. Let C[, b] be the vector spce of ll continuous functions on the intervl [, b]. For 1 p <, define ( ) 1 b p f p = f(x) p dx. By the Minkowski s inequlity for functions, one sees tht (C[, b], p ) forms normed spce under this norm. Exmple 1.9. defines norm on B([, b]). Let B([, b]) be the vector spce of ll bounded functions on [, b]. The sup-norm f = sup f(x) x [,b] Exmple One my hve lredy observed tht the normed spces in Exmples 1.5, 1.7 nd 1.9 re of the sme nture. In fct, let F b (S) be the vector subspce of F (S) consisting of ll bounded functions from S to F. The sub-norm cn be defined on F b (S) nd these exmples re specil cses obtined by tking different sets S. Exmple Any vector subspce of normed vector spce forms normed vector spce under the sme norm. In this wy we obtin mny mny normed vector spces. Here re some exmples: The spce of ll convergent sequences, C, the spce of ll sequences which converges to 0, C 0, nd the spce of ll sequences which hve finitely mny non-zero terms, C 00, re normed subspces of l under the sup-norm. The spce of ll continuous functions on [, b], C[, b], is n importnt normed subspce of B([, b]). The spces {f : f() = 0, f C[, b]}, {f : f is differentible, f C[, b]} nd {f : f is the restriction of polynomil on [, b]} re normed subspces of C[, b] under the sup-norm. But the set {f : f() = 1, f C[, b]} is not normed spce becuse it is not subspce. To ccommodte more pplictions, one needs to replce [, b] by more generl sets in the exmples bove. For ny closed nd bounded subset K in R n, one my define C(K) to be the collection of ll continuous functions in K. As ny continuous function in closed nd bounded set must be bounded (with its mximum ttined t some point), its sup-norm is well-defined. Thus (C(K), ) forms normed spce. On the other hnd, let R be ny rectngulr box in R n. We know tht Riemnn integrtion mkes sense for bounded, continuous functions in R. Consequently, we my introduce the normed p = ( R f p ) 1/p to mke ll bounded, continuous functions in R normed spce. However, this p-norm does not form norm on the spce of Riemnn integrble functions. Which xiom of the norm is not stisfied? In ddition to Exmple 10 where new normed spces re found by restricting to subspces, there re two more generl wys to obtin them. For ny two given normed spces (X, 1 ) nd (Y, 2 ) the function (x, y) = x 1 + y 2 defines norm on the product spce X Y nd thus mkes X Y the product normed spce. On the other hnd, to ech subspce of normed spce one my form corresponding quotient spce nd endow it the quotient norm. We will do this in the next chpter. These exmples of normed spces will be used throughout this book. For simplicity the norm of the spce will usully be suppressed. For instnce, F n lwys stnds for the normed spce under the Eucliden or the unitry norm, l p nd l re lwys under the p-norms nd sup-norm respectively nd single C(K) refers to the spce of continuous functions on the closed, bounded set K under the sup-norm. Exercise 1 1. Find reltion which stisfies (PO1) nd (PO2) but not (PO3), nd one which stisfies (PO1) nd (PO3) but not (PO2). 2. Let V be vector spce. Two subspces U nd W form direct sum of V if for every v V, there exist unique u U nd w W such tht v = u + w. Show tht for every subspce U, there exists subspce W so tht U nd W forms direct sum of V. Suggestion: Try Zorn s lemm.

12 12 CHAPTER 1. NORMED SPACE: EXAMPLES 3. Prove or disprove whether B is bsis for the vector spce V in the followings: () V = l 1, nd B = {e j } j=1. (e j is the j th cnonicl vector.) (b) V = C 00, nd B = {e j } j=1. (C 00 consists of ll sequences with finitely mny non-zero terms.) (c) V = { ll continuous functions on [0, 1]}, nd B = {x k } k=0. (d) V = { ll smooth, 2π-periodic functions}, nd B = {1, cos nx, sin nx} Determine when equlity in Hölder s inequlity (Proposition 1.3) holds. Hint: Keep trcking the equlity sign in the proof of the proposition. 5. Let ϕ be strictly incresing function on (0, ) stisfying ϕ(0) = 0. Denote its inverse function by ψ. Estblish the following generl form of Young s inequlity: For, b > 0, b 0 ϕ(x)dx + b 0 ψ(x)dx. 6. Prove Hölder s nd Minkowski s inequlities for sequences stted in 1.4, Chpter Recll tht f is Riemnn integrble if ε > 0 there exists δ > 0 such tht b f R(f, P ) < ε, P < δ, where R(f, P ) is the Riemnn sum. (Nottions s in MATH2060.) Use this fct nd Propositions 1.3 nd 1.4 to prove Hölder s inequlity nd Minkowski s inqulity for functions stted in 1.4, Chpter Apply Hölder s inequlity to estblish the following interpoltion inequlity: R n, p, q 1, r = (1 λ)p + λq, λ [0, 1], r 1 λ p λ q. Then extend this interpoltion inequlity to functions in C[, b]. 9. Is p norm on the spce of ll Riemnn integrble functions on [, b]? If not, discuss how to mke ll Riemnn integrble functions normed spce under this norm. This problem involves the concept of sets of mesure zero. Skip it if you hve not lernt it. 10. Let f be continuously differentible function on [, b]. For p 1, define f 1,p f p + f p, where f is the derivtive of f. Show tht 1,p forms norm on the spce C 1 [, b] {f : f nd f re continuous on [, b]}. 11. Let X Y be the product spce of two normed spces X nd Y. Show tht it is lso normed spce under the product norm (x, y) = x X + y Y. 12. Give n exmple to show tht p is not norm on F n when n 2 nd p (0, 1). Note: In fct, there re reverse Hölder s nd Minkowski s inequlities when p (0, 1). Google for them.

13 dddddddddddddddd...dddddddddddddddd dddddddd Chpter 2 Normed Spce: Anlyticl Aspects When vector spce is endowed with norm, one cn tlk bout the distnce between two vectors nd consequently it mkes sense to tlk bout limit, convergence nd continuity. The underlying structure is tht of metric spce. We give brief introduction to metric spce in the first section nd use it to discuss three nlyticl properties of normed vector spce, nmely, seprbility, completeness nd Bolzno-Weierstrss property, in lter sections. Emphsis is on how these properties re preserved, modified or lost when one psses from finite to infinite dimensions. Our discussion on metric spces is miniml in order to void possible overlp with course on point set topology. Chpters 2-4 in Rudin s Principles in Mthemticl Anlysis on metric spces contin more thn enough mterils for us. 2.1 Normed Spce As Metric Spce Let M be non-empty set. A function d : M M [0, ) is clled metric on M if p, q, r M (D1) d(p, q) 0, nd = holds if nd only if p = q. (D2) d(p, q) = d(q, p). (D3) d(p, q) d(p, r) + d(r, q). The pir (M, d) is clled metric spce. Notions such s convergence of sequences, Cuchy sequences, continuity of functions,... which we discussed in R or R n in Elementry Anlysis nd Advnced Clculus mke sense nturlly in metric spce. To be precise, we hve Let {p n } be sequence in (M, d). We cll p M the limit of {p n } if for ny ε > 0, there exists n 0 such tht d(p n, p) < ε for ll n n 0. Write p = lim n p n or simply p n p.. The sequence {p n } is clled Cuchy sequence if for ny ε > 0, there exists n 0 d(p n, p m ) < ε, for ll n, m n 0. such tht Let f : (M, d) (N, ρ) where (N, ρ) is nother metric spce be function nd p 0 M. f is continuous t p 0 if f(p 0 ) = lim n f(p n ) whenever lim n p n = p. Alterntively, for ny ε > 0, it is required tht there exists δ > 0 such tht ρ(f(p), f(p 0 )) < ε whenever d(p, p n ) < δ. f is clled continuous function on M if it is continuous t every point. Very often it is more convenient to use the lnguge of topology (open nd closed sets) to describe these concepts. To introduce it let s denote the metric bll centered t p, {q M : d(q, p) < r}, by B r (p). A non-empty subset G of M is clled n open set if p G, there exists positive r (depending 13

14 14 CHAPTER 2. NORMED SPACE: ANALYTICAL ASPECTS on p) such tht B r (p) G. We define the empty set to be n open set. Also the whole M is open becuse it contins every bll. It is esy to see tht ny metric bll B R (p 0 ) is n open set. For, let p B R (p 0 ), we clim tht B r (p), r = R d(p, p 0 ), is contined inside B R (p 0 ). This is consequence of the tringle inequlity (D3): Let q B r (p), then d(q, p 0 ) d(q, p) + d(p, p 0 ) < r + d(p, p 0 ) = R, so q B R (p 0 ). Roughly speking, n open set is set without boundry. A subset E is clled closed set if its complement M \ E is n open set. The empty set is closed set s its complement is the whole spce. By the sme reson M is lso closed. So the empty set nd the whole spce re both open nd closed. Proposition 2.1. Let (M, d) be metric spce. Then the union of open sets nd the intersection of finitely mny open sets re open. The intersection of closed sets nd the union of finitely mny closed sets re closed. Proof. Tht ny countble or uncountble open sets still form n open set comes from definition. As for finite intersections, let G = n k=1 G k where G k is open. For x G G k, we cn find metric bll B rk (x) G k for ech k since G k is open. It follows tht the bll B r (x), r = min{r 1, r 2,, r n } is contined in G, so G is open. The ssertions on closed sets come from tking complements of the ssertions on open sets. Notice tht infinite intersection of open sets my not be open. Let us consider the open intervls I n = ( 1/n, 1 + 1/n), n N in R under the Eucliden metric. Then I n = [0, 1] which is not open. Similrly, let { 1, 2, 3,, } be n enumertion of ll rtionl numbers nd set F n = { 1, 2,, n }. Then ech F n is closed, but F n is the set of ll rtionl numbers which is clerly not closed in R. To hve better picture bout the closed set we introduce the notion of the limit point of set. We cll point p M limit point of set E if for ll r > 0, B r (x) \ {x} E. The limit point is relted to set, but the limit, lthough it is lso point, is relted to sequence. They re not the sme. For exmple, consider the sequence {1, 1 2, 1 3, 1 4, }, its limit is clerly 0. If we regrd {1, 1 2, 1 3, 1 4, } s set, 0 is its unique limit point. However, for the sequence {0, 2, 1, 1, 1, }, the limit is 1 but s set it hs no limit point. Proposition 2.2. A non-empty set E is closed if nd only if it contins ll its limit points. Proof. Let E be closed set. By definition M \ E is open. If p M \ E, there exists r such tht B r (p) M \ E, ie, B r (p) E =. It follows tht p cnnot be limit point of E. This shows tht ny limit point of E must belong to E. Conversely, we need to show M \ E is open. Since E lredy contins ll limit points, ny point p M \ E cnnot be limit point of E. Therefore, there is n r such tht B r (p) E =, but tht mens B r (p) M \ E, so M \ E is open. The closure of E, denoted by E, is defined to be the union of E nd its limit points. By Proposition 2.1 it is esily shown tht E is the smllest closed set contining E, tht is, E F whenever F is closed set contining E. In terms of the lnguge of open-closed sets (or topology), sequence {x n } x cn be expressed s, for ech open set G contining x, there exists n n 0 such tht x n G for ll n n 0. For f : (M, d) (N, ρ) where (N, ρ) is nother metric spce. In terms of topology, we hve the following chrcteriztion of continuity: Proposition 2.3. f : (M, d) (N, ρ) is continuous if nd only if f 1 (G) is open for ny open G in N. Proof. Assume on the contrry tht there is n open set G in N whose preimge is not open. We cn find some p 0 f 1 (G) nd p n M \ f 1 (G) with {p n } p 0. By continuity, {f(p n )} f(p 0 ). As G is

15 2.1. NORMED SPACE AS METRIC SPACE 15 open, there exists some n 0 such tht f(p n ) G for ll n n 0. But this mens tht f 1 (G) contins p n for ll n n 0, contrdiction holds. We conclude tht f 1 (G) must be open when G is open. On the other hnd, suppose f is not continuous, then there exists {p n } p 0 in M but {f(p n )} does not converge to f(p 0 ). Then there exists ρ > 0 nd subsequence {f(p nj )}, f(p nj ) / B ρ (f(p 0 )), n j. As B ρ (f(p 0 )) is open, f 1 (B ρ (f(p 0 ))) is open in M, so there exists n 0 such tht p n f 1 (B ρ (f(p 0 ))). But then f(p n ) B ρ (f(p o )) for ll n n 0, contrdiction holds. Let E be ny nonempty subset of (M, d). Then it is cler tht (E, d) forms metric spce. It is clled metric subspce or simply subspce. As we will see, the subspces formed by closed subsets re prticulrly importnt since they inherit mny properties of M. Now, let us return to normed spces. Let (X, ) be normed spce. Define d(x, y) = x y. Using (N1)-(N3), it is esy to verify (D1)-(D3) hold for d, so (X, d) becomes metric spce. This metric is clled the induced metric of the norm. Of course, there re mny metrics which re not induced by norms. But in functionl nlysis most metrics re induced in this wy. From now on whenever we hve normed spce, we cn tlk bout convergence nd continuity implicitly referring to this metric. The following sttements show tht the norm nd the lgebric opertions on the vector spces interct nicely with the metric. Proposition 2.4. Let (X, ) be normed spce. Then () The norm is continuous function from X to [0, ); (b) Addition, s considered s mp X X X, nd sclr multipliction, mp F X X, re continuous in X X nd F X respectively. Proof. () p n p mens d(p n, p) 0. But then p n p = d(p n, p) 0, p n p p n p 0. (b) We need to show p n p nd q n q implies p n + q n p + q. This is cler from d(p n + q n, p + q) = (p n + q n ) (p + q) p n p + q n q = d(p n, p) + d(q n, q). For sclr multipliction, need to show α n α nd p n p implies α n p n αp. By (), we hve p n p. Hence for ε = 1, there exists some n 0 such tht p n p < 1, or p n 1 + p, for ll n n 0. s n. d(α n p n, αp) = α n p n αp = α n p n αp n + α(p n p) α n α p n + α p n p α n α ( p + 1) + α p n p = α n α ( p + 1) + α d(p n, p) 0 As n interesting ppliction of the continuity of norm, we study the equivlence problem for norms. Consider two norms defined on the sme spce (X, 1 ) nd (X, 2 ). We cll 2 is stronger thn 1 if there exists C > 0 such tht x 1 C x 2, x X. In prticulr, it mens x n x in 2 implies x n x in 1. Two norms re equivlent if 1 is stronger thn 2 nd 2 is stronger thn 1. In other words, there exists C 1, C 2 > 0 such tht C 1 x 2 x 1 C 2 x 2, x X. Exmple 2.1. On F n consider the p-metric d p (x, y) = x y p induced from the p-norm (1 p ). In previous exercise we were sked to show ll these metrics re equivlent. In fct, this is generl fct s estblished by the following result.

16 16 CHAPTER 2. NORMED SPACE: ANALYTICAL ASPECTS Theorem 2.5. Any two norms on finite dimensionl spce re equivlent. Proof. In the following proof we ssume the spce is over R. The sme rguments work for spces over C. Step 1: Tke X = R n first. It suffices to show tht ny norm on R n is equivlent to the Eucliden norm. Let be norm on R n. For x = α j e j, reclling tht x 2 = α j 2, we hve x α j e j αj 2 ej 2 = C x 2, where C = ( j e j 2 ) 1/2. This shows tht 2 is stronger thn. To estblish the other inequlity, letting ϕ(x) = x, from the tringle inequlity ϕ(x) ϕ(y) x y C x y 2 ϕ is continuous function with respect to the Eucliden norm. Consider α inf{ϕ(x) : x R n, x 2 = 1}. As the function ϕ is positive on the unit sphere of 2, α is nonnegtive number. The second inequlity will come out esily if α is positive. To see this we observe tht for every nonzero x R n, x 0 < α ϕ( ) = x, x 2 x 2 i.e., α x 2 x, x. To show tht α is positive, we use the fct tht every continuous function on closed nd bounded subset of R n must ttin its minimum. Applying it to ϕ nd the unit sphere { x 2 = 1}, the infimum α is ttined t some point x 0 nd so in prticulr α = ϕ(x 0 ) > 0. Step 2: For ny n dimensionl spce X, fix bsis {x 1, x 2,, x n }. For ny x X, we hve unique representtion x = n k=1 α kx k. The mp x (α 1,, α n ) is liner isomorphism from X to R n. Any norm on X induces norm on R n by (α 1,, α n ) = α k x k. Let 1 nd 2 be two norms on X nd let 1 nd 2 be the corresponding norms on R n. From Step 1, there exist C 1, C 2 > 0 such tht C 1 x 2 = C 1 α 2 α 1 = x 1 C 2 α 2 = C 2 x 2. Exmple 2.2. Consider the norms nd 1 on C[, b]. On one hnd, from the obvious estimte f g 1 = b f g (x)dx (b ) f g we see tht is stronger thn 1. But they re not equivlent. It is esy to find sequence of functions {f n } in C[, b] which stisfies f n = 1 but f n 1 0. Consequently, it is impossible to find constnt C such tht f C f 1 for ll f. In other words, 1 cnnot be stronger thn.

17 2.2. SEPARABILITY Seprbility There re some importnt nd bsic properties of the spce of ll rel numbers which we would like to study in generl normed spce. They re Seprbility Completeness Bolzno-Weierstrss property. We study the first item in this section. As we ll know, the rtionl numbers re dense in the spce of ll rel numbers. The notion of dense set mkes perfect sense in metric spce. A subset E of (M, d) is dense set if its closure is the whole M, or equivlently, for every p M, there exists {p n } in E, p n p. A metric spce is clled seprble if it hs countble dense subset. Thus R is seprble becuse it contins the countble dense subset Q. The following two results show tht there re mny seprble normed spces. Proposition 2.6. The following normed spces re seprble: () (F n, p ) (1 p ), (b) l p (1 p < ), (c) (C[, b], p ) (1 p ). Proof. We only prove (c) nd leve () nd (b) to you. For ny continuous, rel-vlued f, given ny ε > 0, by Weierstrss pproximtion theorem there exists polynomil p such tht f p < ε. The coefficients of p re rel numbers in generl, but we cn pproximte them by rtionl numbers, so without loss of generlity we my ssume its coefficients re rtionl. The set E = {p C[, b] : p is polynomil with rtionl coefficients} forms countble, dense subset of (C[, b], ). For ny finite p, we observe tht f p = ( b f(x) p dx) 1 p (b ) 1 p f. As for every f, there exists p n E, p n f 0, we lso hve p n f p 0, so E is lso dense in (C[, b], p ). When the function is complex-vlued, simply pply the bove result to its rel nd imginry prts. Proposition 2.7. Any subset of seprble metric spce is gin seprble. Proof. Let Y X nd E countble, dense subset of X. Write E = {p k } k=1. For ech m, B 1 (p k) Y m my or my not be empty. Pick point p m,k if it is not empty. The collection of ll these p m,k points forms countble subset S of Y. We clim tht it is dense in Y. For, ny p Y, nd m > 0, there exists p k B 1 (p), p k E by ssumption. But then p B 1 (p k) which mens B 1 (p k) Y. Then we hve m m m p m,k B 1 (p k) nd so d(p, p m m,k ) d(p, p k ) + d(p k, p m,k ) < 2/m. Now we give n exmple of non-seprble spce. Proposition 2.8. l is not seprble.

18 18 CHAPTER 2. NORMED SPACE: ANALYTICAL ASPECTS Proof. Consider the subset F of l consisting of ll sequences of the form ( 1, 2, 3, ) where k = 1 or 0. In view of Proposition 2.7, it suffices to show tht F is not seprble. First of ll, it is n uncountble set s esily seen from the correspondence ( 1, 2, 3, ) in binry representtion of rel number which mps F onto [0, 1]. For ech x, y F, we hve d(x, y) = x y = sup k x k y k = 1. It follows tht the blls B 1/2 (x), x F, re mutully disjoint. Let E be dense set in F. By definition there exists some p x B 1/2 (x) E. Since these blls re disjoint, ll p x re distinct, so {p x } forms n uncountble subset of E. Thus E is lso uncountble. We hve shown tht there re no countble dense subsets in F, tht is, F is not seprble. Cn you find more non-seprble normed spces? 2.3 Completeness A metric spce (M, d) is complete if every Cuchy sequence converges. As we ll know, R is complete metric spce. Proposition 2.9. The following spces re complete: () (F n, p ) (1 p ), (b) l p (1 p ), (c) (C[, b], ). Proof. () Let {p k } be Cuchy sequence in F n. For p k = (p k 1,, p k n), from p k j p l j p k p l p we see tht {p k j } is Cuchy sequence in F for ech j = 1, 2,, n. By the completeness of F there exists p j such tht p k j p j s k for ech j. Given ε > 0, there exists k 0 such tht p k j p j ε, k k 0. Summing up over j, p k p p < n 1/p mx j p k j p j < n 1/p ε, k k 0, which shows tht p k p (p 1,, p n ). We leve the proofs of (b) nd (c) to the reder. Note tht (c) ws theorem on uniform convergence in elementry nlysis. nd But C[, b] is not complete in the L p -norm (1 p < ). To find divergent Cuchy sequence we let 1, x [ 1, 0] ϕ n (x) = nx + 1, 0, x [0, 1 n ] x [ 1 n, 1]. ϕ(x) = It is esy to see tht ϕ n ϕ p 0. Therefore, { 1, x [ 1, 0] 0, x (0, 1]. f n f m p f n f p + f m f p 0,

19 2.3. COMPLETENESS 19 s n, m, tht is, {f n } is Cuchy sequence in p-norm. To show tht it does not converge to continuous function let s ssume on the contrry it converges to some continuous f. From ( 0 f ϕ p) 1/p ( 0 f ϕ n p) 1/p ( 1 + ϕ n ϕ p) 1/p 0, 1 1 s n, we see tht f is identicl to ϕ on [ 1, 0] since both functions re continuous on [ 1, 0]. In prticulr, f(0) = 1, so by continuity f > 0 on [0, δ] for some positive δ. However, since f nd ϕ re continuous on [δ, 1] by similr rgument s bove f is identicl to ϕ on [δ, 1], but then g(δ) = ϕ(δ) = 0, contrdiction holds. Fortuntely, one cn mke ny metric spce complete by putting in idel points. In generl, mp f : (M, d) (N, ρ) is clled metric preserving mp if ρ(f(x), f(y)) = d(x, y) for ll x, y in M. Note tht metric preserving mp is necessrily injective. In some texts the nme n isometry is used insted of metric preserving mp. We prefer to use the former nd reserve the ltter for metric preserving nd surjective mp. A complete metric spce ( M, d) is clled the completion of metric spce (M, d) if there exists n metric preserving mp Φ of M into M such tht Φ(M) is dense in M. Theorem Every metric spce hs completion. Proof. Let C be the collection of ll Cuchy sequences in (M, d). We introduce reltion on C by x y if nd only if d(x n, y n ) 0 s n. It is routine to verify tht is n equivlence reltion on C. Let M = C / nd define mp: M M [0, ) by d( x, ỹ) = lim n d(x n, y n ) where x = (x 1, x 2, x 3, ) nd y = (y 1, y 2, y 3, ) re respective representtives of x nd ỹ. We note tht the limit in the definition lwys exists: For nd, fter switching m nd n, 1 d(x n, y n ) d(x n, x m ) + d(x m, y m ) + d(y m, y n ) d(x n, y n ) d(x m, y m ) d(x n, x m ) + d(y m, y n ). As x nd y re Cuchy sequences, d(x n, x m ) nd d(y m, y n ) 0 s n, m, so {d(x n, y n )} is Cuchy sequence of rel numbers. Step 1. Well-definedness of d. To show tht d( x, ỹ) is independent of their representtives let x x nd y y. We hve d(x n, y n ) d(x n, x n) + d(x n, y n) + d(y n, y n ). After switching x nd x, nd y nd y, d(x n, y n ) d(x n, y n) d(x n, x n) + d(y n, y n). As x x nd y y, the right hnd side of this inequlity tends to 0 s n. Hence lim n d(x n, y n ) = lim n d(x n, y n). Step 2. d is metric. This is strightforwrd nd is left s n exercise. Step 3. Φ is metric preserving nd hs dense imge in M. More precisely, we need to show tht there is mp Φ : M M so tht d(φ(x), Φ(y)) = d(x, y) nd Φ(M) is dense in M. Given ny x in M, the constnt sequence (x, x, x, ) is clerly Cuchy sequence. Let x be its equivlence clss in C. Then Φx = x defines mp from M to M. Clerly d(φ(x), Φ(y)) = lim n d(x n, y n ) = d(x, y)

20 20 CHAPTER 2. NORMED SPACE: ANALYTICAL ASPECTS since x n = x nd y n = y for ll n, so Φ is metric preserving nd it is injective in prticulr. To show tht Φ(M) is dense in M we observe tht ny x in M is represented by Cuchy sequence x = (x 1, x 2, x 3, ). Consider the constnt sequence x n = (x n, x n, x n, ) Φ(M). We hve d( x, x n ) = lim m d(x m, x n ). Given ε > 0, there exists n 0 such tht d(x m, x n ) < ε/2 for ll m, n n 0. Hence d( x, x n ) = lim m d(x m, x n ) < ε for n n 0. Tht is x n x s n, so Φ(M) is dense in M. Step 4. d is complete metric on M. Let { x n } be Cuchy sequence in M. As Φ(M) is dense in M, for ech n we cn find ỹ n in Φ(M) such tht d( x n, ỹ n ) < 1 n. So {ỹ n } is Cuchy in d. Let y n be the point in M so tht y n = (y n, y n, y n, ) represents ỹ n. Since Φ is metric preserving nd {ỹ n } is Cuchy in d, {y n } is Cuchy sequence in M. Let (y 1, y 2, y 3, ) ỹ in M. We clim tht ỹ = lim n x n in M. For, we hve d( x n, ỹ) d( x n, ỹ n ) + d(ỹ n, ỹ) s n. 1 n + lim m d(y n, y m ) 0 The ide of this proof is due to Cntor, who used equivlence clsses of Cuchy sequences of rtionl numbers to construct rel numbers. Another populr pproch for the rel number system is by Dedekind cut. See the old book by E. Lndu Foundtions of Anlysis for more. The uniqueness of the completion could be formulted s follows. Let Ψ i : (M, d) (M i, d i ), i = 1, 2, be two metric preserving mps with dense imges. Then the mp Ψ 2 Ψ 1 1 : Ψ 1 (M 1 ) M 2 cn be extended to be n isometry between M 1 nd M 2. I leve the proof of this fct to you. When given metric spce is induced from normed spce, it is nturlly to sk it is possible to mke the completion into normed spce so tht the complete metric is induced by the norm on the completion. To this question we hve stisfctory nswer s stted in the theorem below. You my lso formulte uniqueness result for the completion of normed spce. Theorem Let (X, ) be normed spce nd X its completion under the induced metric of X. There is unique normed spce structure on X so tht the quotient mp x x becomes liner nd norm-preserving. Moreover, the metric induced by this norm X is identicl to the completion metric. Proof. We only give the outline of the proof nd leve the ptient reder to provide ll detils. Step 1. Let Φ : X X be the quotient mp. As Φ(X) is dense in X, for ny x, ỹ in X, we cn find sequences { x n }, {ỹ n } converging to x, ỹ respectively. We define n ddition nd sclr multipliction on X by x + ỹ lim n Φ(x n + y n ), nd α x lim n Φ(αx n), where x n nd y n re representtives of x n nd ỹ n respectively. You need to estblish three things. First, these opertions re well-defined, tht s, they re independent of the representtives. Second, they mke

21 2.4. SEQUENTIAL COMPACTNESS 21 X into vector spce. Third, the mp Φ is liner from X to X. (In fct, this follows immeditely from the definitions.) Step 2. Introduce mp on X by x d( x, 0). Then verify the following three fcts: First, trnsltionl invrince: d( x + ỹ, ỹ) = d( x, 0) for ll x nd ỹ. Second, use trnsltionl invrince to show tht this mp relly defines norm on X. Third, show tht the metric induced by this norm coincides with the completion metric. This in fct follows from the definition of the norm. Step 3. Show tht if there is nother normed spce structure on X so tht the quotient mp Φ becomes liner nd norm-preserving, then this normed spce structure is identicl to the one given by Steps 1 nd 2. Essentilly this follows from the fct tht Φ(X) is dense in X. As n immedite ppliction of these results, we let L p (, b) be the completion of C[, b] under the L p -norm. We shll cll n element in L p (, b) n L p -function, lthough it mkes sense only when the element is relly in C[, b]. Such terminology is bsed on nother construction in rel nlysis where we relly identify L p (, b) s the function spce consisting of L p -integrble functions. We do not need this fct in this course. A complete normed spce is clled Bnch spce. Bnch spce is one of the fundmentl concepts in functionl nlysis. Now we know tht even spce is not complete, we cn mke it into Bnch spce. The following nice properties of Bnch spces hold: Any closed subspce of Bnch spce is Bnch spce. The product spce of two Bnch spces is Bnch spce under the product norm. For ny closed subspce Z of Bnch spce X, the quotient spce X/Z is Bnch spce under the quotient norm. We hve shown tht the spces F n, l p (1 p ), C[, b] nd L p [, b], p [1, ), re Bnch spces. In fct, for ny metric spce X, the spce C b (X) = {f : f is bounded nd continuous in X} forms Bnch spce under the sup-norm. For ny mesure spce (X, µ), the spce L p (X, µ) = {f : f is L p -integrble} forms Bnch spce under the L p -norm. Finlly, without requiring ny topology or integrbility on the functions, the spce L (X) consisting of ll bounded functions in nonempty set X is Bnch spce under the sup-norm. It is n exercise to show tht every subspce in finite dimensionl normed spce is closed. Cn you find non-closed subspce in C[, b]? So fr we hve encountered three types of mthemticl structure, nmely, those of vector spce, metric spce nd normed spce. How do we identify two spces from the sme structure? Well, first of ll, we view two vector spces the sme if there exists bijective liner mp between them. A bijective liner mp is lso clled liner isomorphism. Next, two metric spces re the sme if there exists metric preserving bijective mp, tht is, n isometry, from one to the other. Finlly, two normed spces re the sme if there exists norm-preserving liner isomorphism from one to the other. 2.4 Sequentil Compctness In the spce of rel numbers, ny bounded sequence hs convergent subsequence. This property is clled the Bolzno-Weierstrss property. In generl setting, it is more convenient to put this concept in nother wy. Let E be subset of the metric spce (M, d). E is clled sequentilly compct if every sequence in E enjoys the Bolzno-Weierstrss property, tht is, it contins convergent subsequence, in E. Any sequentilly compct set is necessrily closed set. It is cler tht the Bolzno-Weierstrss

22 22 CHAPTER 2. NORMED SPACE: ANALYTICAL ASPECTS property essentilly refers to the fct tht the intervl [, b] is sequentilly compct in R. The sme s in the cse of R, one cn show tht every closed nd bounded set in R n is sequentilly compct. Surprisingly, this property is chrcteriztion of finite dimensionlity. Theorem Any closed bll in normed spce is sequentilly compct if nd only if the spce is of finite dimension. Lemm Let Y be ny proper finite dimensionl subspce of the normed spce (X, ). Then for ny x X \ Y, there exists y 0 Y such tht is relized t y 0. d dist(x, Y ) inf x y > 0. y Y The distnce d is positive becuse Y is closed due to finite dimensionlity nd x stys outside Y. Proof. Let {y k } be minimizing sequence of the distnce, tht is, d = lim k x y k. We my ssume x y k d + 1, for ll k. Then y k x + y k x x + d + 1, which mens tht {y k } is bounded sequence in Y. Since Y is finite dimensionl, it is closed nd Bolzno-Weierstrss property holds in it, there exists subsequence {y nj } converging to some y 0 in Y. We hve d = lim nj x y nj = x y 0, hence y 0 relizes the distnce between x nd Y. Proof of Theorem It suffices to show tht the closed unit bll {x X : x 1} is not sequentilly compct when X is of infinite dimension. Let {x 1, x 2, x 3, } be sequence of linerly independent vectors in X. We re going to construct sequence {z n }, z n x 1, x 2,, x n, z n = 1 stisfying tht z n x 1, for ll x x 1, x 2,, x n 1, n 2. Set z 1 = x 1 / x 1. For x n / x 1, x 2,, x n 1, n 2, let y n 1 be the point in x 1, x 2,, x n 1 relizing dist(x n, x 1,, x n 1 ). Let z n = x n y n 1 x n y n 1. Then z n = 1 nd, for ll y x 1,, x n 1, z n y = x n y n 1 x n y n 1 y x n y = x n y n 1 1, where y = y n 1 + x n y n 1 y x 1,, x n 1, since x n y n 1 x n y. We clim tht the bounded sequence {z n } does not hve convergent subsequence. For, if it hs, this subsequence is Cuchy sequence. Tking ε = 1, we hve z nk z nj < 1, k, j sufficiently lrge. Tking n k > n j, s z nk x 1, for ll x x 1,, x nk 1 nd z nj x 1,, x nk 1, we hve z nk z nj 1, contrdiction holds. We conclude tht the closed unit bll is not sequentilly compct in n infinite dimensionl normed spce. Digressing bit, let x be point lying outside Y, proper subspce of the normed spce X. A point in Y relizing the distnce from x to Y is clled the best pproximtion from x to Y. It lwys exists

23 2.5. ARZELA-ASCOLI THEOREM 23 when Y is finite dimensionl subspce. However, things chnge drmticlly when the subspce hs infinite dimension. For instnce, let Y be the closed subspce of C[ 1, 1] given by 0 1 nd h continuous function stisfying 0 f(x)dx = 0, h(x)dx = 1, f(x)dx = 0, f Y, 1 0 h(x)dx = 1. One cn show tht the distnce from h to Y is equl to 1, but it is not relized t ny point on Y. You my try to prove this fct or consult chpter 5 of [L]. Lter we will see tht the best pproximtion problem hs lwys solution when the spce X is reflexive. 2.5 Arzel-Ascoli Theorem From the lst section, we know tht not ll bounded sequences in n infinite dimensionl normed spce hve convergent subsequences. It is nturl to sk wht dditionl conditions re needed to ensure this property. For the spce C[, b], complete nswer is provided by the Arzel-Ascoli theorem. This theorem gives necessry nd sufficient condition when closed nd bounded set in C[, b] is sequentilly compct. In order to hve wider pplictions, we shll work on more generl spce C(K), where K is closed, bounded subset of R n, insted of C[, b]. As every continuous function in K ttins its mximum nd minimum, its sup-norm is lwys finite. It cn be shown tht C(K) is seprble Bnch spce under the sup-norm. The crux for sequentil compctness for continuous functions lies on the notion of equicontinuity. Let E be subset of R n. A subset F of C(E) is equicontinuous if for every ε > 0, there exists some δ such tht f(x) f(y) < ε, for ll f F, nd x y < δ, x, y E. Recll tht function is uniformly continuous in E if for ech ε > 0, there exists some δ such tht f(x) f(y) < ε whenever x y < δ, x, y E. So, equicontinuity mens tht δ cn further be chosen independent of individul functions in F. There re vrious wys to show tht fmily of functions is equicontinuous. A function f defined in subset E of R n is clled Hölder continuous if there exists some α (0, 1) such tht f(x) f(y) L x y α, for ll x, y E, (2.1) for some constnt L. The number α is clled the Hölder exponent. The function is clled Lipschitz continuous if (2.1) holds for α equls to 1. A fmily of functions F in C(E) is sid to stisfy uniform Hölder or Lipschitz condition if ll members in F re Hölder continuous with the sme α or Lipschitz continuous nd (2.1) holds for the sme constnt L. Clerly, such F is equicontinuous. The following sitution is commonly encountered in the study of differentil equtions. The philosophy is tht equicontinuity cn be obtined if there is good, uniform control on the derivtives of functions in F. Proposition Let F be subset of C(G) where G is convex set in R n. Suppose tht ech member in F is differentible nd there is uniform bound on the prtil derivtives of the functions in F. Then F is equicontinuous. Proof. For, x nd y in G, (1 t)x + ty, t [0, 1], belongs to G by convexity. Let ψ(t) f((1 t)x + ty). From the men-vlue theorem ψ(1) ψ(0) = ψ (t )(1 0), t [0, 1],