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Suppose $H_n,\ n\in\mathbb{N}$, and $H_0$ are subsets of a semimetric space $H$ such that

  1. every $h\in H_0$ is a limit of a sequence $h_n\in H_n$,
  2. if a subsequence $h_{n_j}$ converges to a limit $h$, then $h\in H_0$.

Suppose that $x$, $x_n,\ n\in\mathbb{N}$, are real bounded functions defined on $H$, such that $x_n$ converges to $x$ uniformly and $x$ is uniformly continuous. I need to show that $$S_n:=\sup_{h\in H_n}x_n(h)\to\sup_{h\in H_0}x(h)=:S_0,\ n\to\infty.$$

I succeeded in showing that for an arbitrary $\varepsilon$ and $n$ large enough $S_0<S_n+\varepsilon$. An opposite inequality is where I have trouble. By definition there exist $h_n\in H_n$ such that $x_n(h_n)\leq S_n<x_n(h_n)+\varepsilon$. Due to the uniform convergence for $n$ large enough $S_n<x(h_n)+\varepsilon$. If I knew that $h_n$ was convergent, then its limit would belong to $H_0$ and that would prove the statement due to the uniform continuity of $x$. 2. suggests that I should look for a convergent subsequence of $h_n$, but as far as I can see that would result in a particular subsequence of $S_n$ converging to $S_0$. And I don't see how to find such a subsequence.

P.S. $H$ can be assumed to be totally bounded.

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1 Answer 1

The following rather trivial fact might help: $S_n\to S_0$ if and only if every subsequence $n_k$ has another subsequence with $S_{n_{k_j}} \to S_0$.

If then $S_{n_k} \cong x_{n_k}(h_{n_k})$ and $H$ would not only be totally bounded but actually (sequentially) compact you could choose a convergent subsequence of $h_{n_k}$ whose limit would be in $H_0$ because of your second assumption.

If $H$ is only totally bounded one could try to extend the situation to the completion of $H$ which is compact (this requires some care if $H$ is only semimetric).

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Thanks, subsequences would work indeed. This is a problem from a textbook, so I thought it was possible to avoid assuming compactness. Regarding the extension: how would one be sure that the limit of the subsequence belongs to $H$? –  nokiddn Aug 29 '12 at 12:30

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