Note that when f,g:X→Y are continuous, then since d:Y×Y→ℝ is continuous, we observe d∘(f×g):X→ℝ is continuous and due to compactness of X indeed attains a maximum value, so that the function h:F×F→ℝ 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 x∈X, i.e. f=g. As well, h(f,g)=maxx∈X d(f(x),g(x))=maxx∈X d(g(x),f(x))=h(g,f) Finally, let a,b,c∈F. Then h(a,c)=maxx∈X d(a(x),c(x))≤maxx∈X [d(a(x),b(x))+d(b(x),c(x))]≤maxx∈X [d(a(x),b(x))]+maxx∈X [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 x∈X, 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))≤maxx∈X d(fn(x),fm(x))=h(fn,fm)<ε. Thus since Y is complete write fn(x)→f(x) for each x∈X. It now suffices to show f:X→Y is continuous and fn→f. 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 x∈X. 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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