10 Compactness in function spaces: Ascoli-Arzelá theorem

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1 10 Compactness in function spaces: Ascoli-Arzelá theorem Recall that if (X, d) isacompactmetricspacethenthespacec(x) isavector space consisting of all continuous function f : X R. The space C(X) is equipped with the norm f := max{ f(x) x }. Thenorminducesthemetric σ(f,g) = f g =max{ f(x) g(x) x X}. Definition The family F C(X) is called equicontinuous if for every ε>0thereisδ>0suchthat f(x) f(y) <εfor all x, y X satisfying d(x, y) <δand all f F. The family F is called equibounded if there is a constant M such that for all f F and all x X. f(x) M Example Let (X, d) beametricspaceandletm>0. By F denote the family of functions f : X R satisfying f(x) f(y) Md(x, y) for all x, y X. Then the family F is equicontinuous. To see this, take ε>0andsetδ = ε/m. Then for every y B δ (x) andeveryf F,wehave f(x) f(y) Md(x, y) <M ε/m = ε. Let M > 0andletF be the set all differentiable functions f : [a, b] R satisfying f (x) M for all x (a, b). In view of the mean value theorem, f(x) f(y) M x y for all x, y (a, b). Using the previous example, F is a equicontinuous family. If (X, d) is a metric space,we call a subset F X precompact (or relatively compact) if F is compact in X. Theorem 10.3 (Ascoli Arzelá Theorem). Let (X, d) be a compact space. AsubsetF of C(X) is relatively compact if and only if F is equibounded and equicontinuous. Proof. Assume that F is relatively compact. This is means that F is compact. We claim that F is equibounded and equicontinuous. Since F compact, it is totally bounded. In particular, F is totally bounded. This implies that F is equibounded. To see that F is equicontinuous, take ε>0. Then, there are f 1,...,f N C(X) suchthat F B ε/3 (f 1 )... B ε/3 (f N ). (1) 60

2 Each f i is uniformly continuous since (X, d) iscompact. Hencethereexists δ>0suchthat f i (x) f i (y) <ε/3 for all x, y such that d(x, y) <δand all 1 i N. Given f F,inviewof(10)thereis1 j N such that σ(f,f j ) <ε/3. Now if x, y X satisfy d(x, y) <δ,then f(x) f(y) f(x) f j (x) + f j (x) f j (y) + f j (y) f(y) σ(f,f j ) f j (x) f j (y) + σ(f,j j ) < 3ε/3 =ε showing that F is equicontinuous. Conversely, assume that F is equibounded and equicontinuous. It suffices to show that F is totally bounded. Indeed, if F is totally bounded, then F is totally bounded and since C(X) is complete, the set F is also complete. Hence F is compact. Take ε>0. Since F is equicontinuous, for every x X there exists δ x > 0suchthat f(x) f(y) <ε/4 for all y such that d(x, y) <δ x and all f F. The collection {B δx (x)} x X is an open cover of a compact metric space X. Hence there are x 1,...,x N such that In particular, f(x) f(x i ) <ε/4 X = B δx1... B δxn. for all x B δi (x i )andallf F, where we have abbreviated δ i = δ xi. Since F is equibounded, the set F := {f(x i ) 1 i N, f F}is bounded. Since a bounded set in R (with the standard metric) is totally bounded, there are points y 1,...,y K in R such that F B ε/4 (y i ). 1 i K For any map ϕ : {1,...,N} {1,...,K}, define F ϕ := {f F f(x i ) B ε/4 (y ϕ(i) ),i=1,...,n}. Note that there are only finitely many sets F ϕ and that every f Fbelongs to one of the sets F ϕ.weclaimthatthediameteroff ϕ is finite. Indeed, take f,g F ϕ and x X. Thenx B δi (x i )forsomei and f(x) g(x) f(x) f(x i ) + f(xi ) y ϕ(i) + yϕ(i) g(x i ) + g(xi ) g(x) 4 ε/4 =ε. Hence σ(f,g) <εshowing that diam F ϕ ε. Consequently,F can be covered by finitely many sets of diameter less thanε. HenceF is totally bounded and the proof is complete. 61

3 Asimpleconsequencewehavethefollowingcorollary. Corollary Let X be a compact metric space and let (f n ) C(X) be asequencewhichisequiboundedandequicontinuousinc(x). Then every the sequence (f n ) has a uniformly convergent subsequence. 11 Structure of complete metric spaces-baire s theorem Let (X, d) beametricspace. AsubsetU of X is called dense if U = X. If U and V are open and dense, then U V is also open and dense. To see that U V is dense, we have to show that O U V is non-empty for any open set O. Since U is dense, there is u O U, andsinceo U is open, B(u, r) O U for some r>0. Since V is dense, B(u, r) V so that, = B(u, r) V O U V.IfU and V are assumed to be dense but not necessarily open, then the intersection U V does not have to be dense. For example, let U be the set of rational numbers and V the set of irrational numbers Q c.thenbothsetsaredenseinr with the usual metric, however, U V =. Consider,nowasequenceofdense and open sets U n.ingeneral,theintersection n 1 U n may be empty. Example Consider (Q,d)withtheusualmetricd. Let {q n n N} be an enumeration of rational numbers, and let U n = Q \{q n }. Then each U n is open since it is a complement of a closed set {q n },andisdense. However, n 1 U n = n 1[ Q \{qn } ] = Q \ n 1 {q n} =. AsubsetF of X is called nowhere dense if (F ) =. Theorem 11.2 (Baire). Let (X, d) be a complete metric space. Then: (a) If {Un} is a sequence of open and dense subsets of X, then n 1 U n is dense. (b) If {F n } is a sequence of nowhere dense subsets of X, then F n has empty interior. Proof. (a) It suffices to show that B(x, r) containsapointbelongingto n 1 U n for any open ball B(x, r). Since U 1 is open and dense, B(x, r) U 1 is nonempty and open. So, there exists an open ball B(x 1,R)withR<1suchthat B(x 1,R) B(x, r) andb(x 1,R) U 1.Takingr 1 <R,wegetthatB(x 1,r 1 ) B(x, r) andb(x 1,r 1 ) U 1.Similarly,sinceU 2 is open and dense, there exists x 2 and r 2 < 1/2 suchthatb(x 2,r 2 ) B(x 1,r 1 ) U 2. Continuing in this way we find a sequence of balls B(x n,r n )withr n < 1/n and B(x n+1,r n+1 ) B(x n,r n ) U n.weclaimthat{x n } is Cauchy. By construction, B n (x n,r n ) B(x k,r k )foralln k. Givenε>0choosek N so that 1/k < ε/2. Then, if n, m k, d(x n,x m ) d(x n,x k )+d(x k,x m ) < 1/k +1/k < ε. 62

4 Because (X, d) iscomplete,{x n } converges, say to y. The point y lies in all balls B(x k,r k )sincex n B(x k,r k )foralln k and B(x k,r k )isclosedforall k, sothataftertakingalimitasn, y B(x k,r k )forallk. Inparticular, y B(x 1,r 1 ) B(x, r) andy B(x n+1,r n+1 ) U n for all n. Consequently, y B(x, r) n 1 U n,andtheproofisfinished. (b) Arguing by contradiction assume that F n has non-empty interior. So B(x, r) F n for some x and r>0. Define U n = X \ F n. Clearly, U n is open and we claim that it is dense. Indeed, if for some open set V we have V U n =, thenv X \ U n = F n contradicting that F n has empty interior. By (a), n 1 U n is dense. So B(x, r) n 1 U n. On the other hand, B(x, r) F n F n so that = B(x, r) [ X \ n 1 F n] = B(x, r) [ ] X \ Fn = B(x, r) n 1 U n,acontradiction. n 1 Example The metric space R with the standard metric space cannot be written as a countable union of nowhere dense sets since it is complete. By contrast, Q with the standard metric can be written as the union of one point sets {q n },where{q n n N} is an enumeration of Q. Every one point set {q n } is closed in Q and its interior is empty, so nowhere dense. This does not contradict Baire s theorem since Q with the standard metric is not complete. Application of the Baire s theorem Theorem There exists a continuous function f :[0, 1] R which is not differentiable at any point x [0, 1). Proof. Recall that f has a right-hand derivative at x if lim [(f(x + h) f(x))/h] exists. h 0 + We denote this limit by f + (x). In particular, if f is differentiable at x [0, 1) then f +(x) existsandisequaltof (x). Let D = {f C([0, 1], R) there exists x [0, 1) such that f + (x) exists} and let D n,m be the set of all f C([0, 1], R) forwhichthereexistssome x [0, 1 1/m] suchthat f(x + h) f(x) n h for all h [0, 1/m]. We shall show that D n,m D n,m and that each D n,m is nowhere dense. By Theorem 11.2 (b), n,m D n,m has empty interior. Consequently, the set D has empty interior showing that D C[0, 1]. Claim 1: D n,m D n,m. Toseethis,letf D. Thenthereexistsx [0, 1) such that f + (x) exists.inparticular, lim f(x + h) f(x) h 0 + h = f + (x). (1) 63

5 Take an integer n such that f + (x) <n.thenthereexistsδ>0suchthat f(x + h) f(x) n h for all 0 h δ. Now choose m N large so that x 1 1/m and 1/m < δ. Thenf D n,m as claimed. Claim 2: D n,m is closed. Take a sequence (f k ) D n,m such that σ(f k,f) 0 for some f C[0, 1]. To prove the claim we have to show that f D n,m.since f k D n,m,thereisasequence(x k )satisfying0 x k 1 1/m and f k (x k + h) f k (x k ) nh for all 0 h 1 1/m. (2) Without loss of generality we may assume that x k x [0, 1 1/m]. Then, using the triangle inequality, f(x + h) f(x ) f(x + h) f(x k + h) + f(x k + h) f k (x k + h) + f k (x k + h) f k (x k ) + f k (x k ) f k (x ) + f k (x ) f(x ) f(x + h) f(x k + h) + f k (x k ) f k (x ) +2d(f k,f)+n h for all 0 h 1/m. Since d(f k,f) 0, f(x + h) f(x k + h) 0, and f k (x k ) f(x k ) 0, we conclude that f(x + h) f(x) n h for all 0 h 1/m. Consequently,f D n,m showing that D n,m is closed. Claim 3: Dn,m =. ThistogetherwithClaim2impliesthatM m is nowhere dense. To prove the claim it suffices to show that if f D n,m,thenanyopenball B ε contains g which doesn t belong to D n,m. Take a piecewise linear function g :[0, 1] R such that σ(f,g) =sup{ f(x) g(x) 0 x 1} <εand g + (x) >nfor all x [0, 1). Then g B ε (f) andg D n,m.so,dn,m =, as claimed. 64