Math829The Arzela-Ascoli Theorem Spring1999
1Introduction
Our tting is a compact metric space X which you can,if you wish,take to be a compact subt of R n,or even of the complex plane(with the Euclidean metric,of cour).Let C(X)denote the space of all continuous functions on X with values in C(equally well,you can take the values to lie in R).In C(X)we always regard the distance between functions f and g in C(X)to be
dist(f,g)=max{|f(x)−g(x)|:x∈X}.
It is easy to check that“dist”is a metric(henceforth:the“max-metric”) on C(X),in which a quence is convergent iffit converges uniformly on X. Similarly,a quence in C(X)is Cauchy iffit is Cauchy uniformly on X. Thus the max-metric,which from now on we always assume to be part of the definition of C(X),makes that space complete.The notes prove the fundamental theorem about compactness in C(X):虾怎么做
1.1The Arzela-Ascoli Theorem If a quence{f n}∞1in C(X)is bounded and equicontinuous then it has a uniformly convergent subquence.
In this statement,
(a)“F⊂C(X)is bounded”means that there exists a positive constant心理健康教育内容
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M<∞such that|f(x)|≤M for each x∈X and each f∈F,and (b)“F⊂C(X)is equicontinuous”means that:for everyε>0there exists
δ>0(which depends only onε)such that for x,y∈X:
d(x,y)<δ⇒|f(x)−f(y)|<ε∀f∈F,
where d is the metric on X.
1.2Exerci.The Arzela-Ascoli Theorem is the key to the following re-sult:A subt F of C(X)is compact if and only if it is clod,bounded,and equicontinuous.
1.3Exerci.You can think of R n as(real-valued)C(X)where X is a t containing n points,and the metric on X is the discrete metric(the distance between any two different points is1).The metric thus induced on R n is equivalent to,but(unless n=1)not the same as,the Euclidean
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Math829The Arzela-Ascoli Theorem Spring1999 one,and a subt of R n is bounded in the usual Euclidean way if and only if it is bounded in this C(X).Show that every bounded subt of this C(X) is equicontinuous,thus establishing the Bolzano-Weierstrass theorem as a generalization of the Arzela-Ascoli Theorem.
2Proof of the Arzela-Ascoli Theorem.
Step I.We show that the compact metric space X is ,has a countable den subt S.
Given a positive integer n and a point x∈X,let
B(x,1/n)={y∈X:d(x,y)<1/n},
the open ball of radius1/n,centered at x.For a given n,the collection of all the balls as x runs through X is an open cover of x,so(becau X is compact)there is afinite subcollection that also covers X.Let S n denote the collection of centers of the balls in thisfinite subcollection.Thus S n is afinite subt of X that is“1/n-den”in the n that every point of X lies within1/n of a point of S n.Clearly the union S of all the ts S n is countable,and den in X.
鬼蝶Step II.Wefind a subquence of{f
n}that converges pointwi on S.
This is a standard diagonal argument.Let’s list the(countably many) elements of S as{x1,x2,...}.Then the numerical quence{f n(x1)}∞n=1 is bounded,so by Bolzano-Weierstrass it has a convergent subquence, which we’ll write using double subscripts:{f1,n(x1)}∞n=1.Now the numer-ical quence{f1,n(x2)}∞n=1is bounded,so it has a convergent subquence {f2,n(x2)}∞n=1.Note that the quence of functions{f2,n}∞n=1,since it is a sub-quence of{f1,n}∞n=1,converges at both x1and x2.Proceeding in this fashion we obtain a countable collection of subquences of our original quence:
惆怅怎么读加盟dq要多少钱f1,1f1,2f1,3···
f2,1f2,2f2,3···
f3,1f3,2f3,3···
......
......
......
我真的生气了where the quence in the n-th row converges at the points x1,...,x n,and each row is a subquence of the one above it.
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Math829The Arzela-Ascoli Theorem Spring1999 Thus the diagonal quence{f n,n}is a subquence of the original -quence{f n}that converges at each point of S.
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Step III.Completion of the proof.
Let{g n}be the diagonal subquence produced in the previous step, convergent at each point of the den t S.Letε>0be given,and choo δ>0by equicontinuity of the original quence,so that d(x,y)<δimplies |g n(x)−g n(y)|<ε/3for each x,y∈x and each positive integer n.Fix M>1/δso that thefinite subt S M⊂S that we produced in Step I is δ-den in X.Since{g n}converges at each point of S M,there exists N>0 such that
(∗)n,m>N⇒|g n(s)−g m(s)|<ε/3∀s∈S M.
Fix x∈X.Then x lies withinδof some s∈S M,so if n,m>M: |g n(x)−g m(x)|≤|g n(x)−g n(s)|+|g n(s)−g m(s)|+|g m(s)−g m(x)| Thefirst and last terms on the right are<ε/3by our choice ofδ(which was possib
le becau of the equicontinuity of the original quence),and the same estimate holds for the middle term by our choice of N in(*).In summary: givenε>0we have produced N so that for each x∈X,
m,n>N⇒|g n(x)−g m(x)|<ε/3+ε/3+ε/3=ε.
Thus on X the subquence{g n}of{f n}is uniformly Cauchy,and there-fore uniformly convergent.This completes the proof of the Arzela-Ascoli Theorem.
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