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Equicontinuity and uniform convergence

  • Thread starter radou
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  • #1
radou
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Homework Statement



This is a nice one, if it's correct.

Show that if (fn) is a sequence of elements of C(X, Y) (where Y is a metric space) which converges uniformly, then the collection {fn} is equicontinuous.

The Attempt at a Solution



Let ε > 0 be given and let x0 be a point of X. The proof is based on the following:

d(fn(x0), fn(x)) <= d(fn(x0), f(x0)) + d(f(x0), f(x)) + d(f(x), fn(x)).

Since fn converges uniformly, we can find a positive integer N such that, for all x in X, and for n >= N, we have d(f(x), fn(x)) < ε/3. Since f is continuous, for ε/3 choose a neighborhood U of x0 such that whenever x is in U, we have d(f(x0), f(x)) < ε/3. Now the upper inequality becomes d(fn(x0), fn(x)) < ε, for all x in X (and hence for all x in U), and for all n >= N, i.e. for all but finitely members of the collection {fn}. Now, for every of the remaining members of {fn], use continuity of fn at x0 to find neighborhoods of x0 such that d(fn(x0), fn(x)) < ε. Now, intersect this finite number of neighborhoods of x0 with the neighborhood U, call this intersection U'. Clearly, for any x in U, we have d(fn(x0), fn(x)) < ε, for all elements of {fn}.
 

Answers and Replies

  • #2
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Is X compact here? Because if it is not, i don't think the statement is even true. The problem is that if X is not compact than f may or may not be uniformly continuous, which is crucial in this case. For, if f is not uniformly continuous you won't be able to find an epsilon that will work for all N and all x.

Other than this, your proof seems correct.
 
  • #3
radou
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No, Y is not assumed to be compact.
 
  • #4
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The proof is correct. I don't think you need compactness or uniform continuity here...
 
  • #5
radou
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The proof is correct. I don't think you need compactness or uniform continuity here...
Thanks!

Btw, it seems I found the first error in Munkres, after 278 pages!

Definition. If (Y, d) is a metric space, a subset F of C(X, Y) is said to be pointwise bounded under d if for each x in X, the subset Fa = {f(a) : f is in F} of Y is bounded under d.

I assume he meant Fx = {f(x) : f is in F}, right? I mean, it's not a big mistake (if!) but I want to make sure, since I need this definition in another proof...
 
  • #6
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Haha! You're correct, Munkres indeed made a mistake here :smile:
 
  • #7
radou
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Haha! You're correct, Munkres indeed made a mistake here :smile:
OK! Now off to proving Arzela's theorem... :biggrin:
 

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