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Thursday, October 2, 2014

Function Spaces from Compact Sets into Complete Metric Spaces

MathJax TeX Test Page Let X be a compact space, and let (Y,d) be a complete metric space. We show that the set F of continuous functions from X to Y under the metric h(f,g)=maxxX d(f(x),g(x))     for f,gF induces a complete metric space.

Note that when f,g:XY are continuous, then since d:Y×Y is continuous, we observe d(f×g):X is continuous and due to compactness of X indeed attains a maximum value, so that the function h:F×F is well defined. To verify that it is a metric, we see h(f,g)=0 iff d(f(x),g(x))=0 iff f(x)=g(x) for all xX, i.e. f=g. As well, h(f,g)=maxxX d(f(x),g(x))=maxxX d(g(x),f(x))=h(g,f) Finally, let a,b,cF. Then h(a,c)=maxxX d(a(x),c(x))maxxX [d(a(x),b(x))+d(b(x),c(x))]maxxX [d(a(x),b(x))]+maxxX [d(b(x),c(x))]=h(a,b)+h(b,c) so that h(a,c)h(a,b)+h(b,c) and (F,h) is a metric space.

Now we show completeness. Let (fn) be a Cauchy sequence in F. We show that for each xX, the sequence (fn(x)) is Cauchy in Y. To wit, given ε>0, choose N such that n,m>N implies h(fn,fm)<ε. Then d(fn(x),fm(x))maxxX d(fn(x),fm(x))=h(fn,fm)<ε. Thus since Y is complete write fn(x)f(x) for each xX. It now suffices to show f:XY is continuous and fnf. We shall prove the latter convergence is uniform and the former will follow by the uniform limit theorem.

Let ε>0 be given. Since (fn) is Cauchy, let N be such that n,m>N imply h(fn,fm)<ε/2. Then d(fn(x),fm(x))<ε/2 for all xX. Now let n>N be given. Since fn(x)f(x) choose m>N such that d(fm(x),f(x))<ε/2. Then d(fn(x),f(x))d(fn(x),fm(x))+d(fm(x),f(x))<ε so the convergence is uniform.

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