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Similar to the Egoroff theorem, we can get the following theorem:

Theorem: Let $X$ be a locally compact Hausdorff space(non-empty),and $\{f_n\}$ be a pointwise bounded sequence of continuous functions defined on $X$,then $\{f_n\}$ are bounded uniformly on an open subset of $X$

From this, I want to judge whether the theorem holds for pointwise convergent sequence:

Let $X$ be a locally compact Hausdorff space(non-empty),and $\{f_n\}$ be a pointwise convergent sequence of continuous functions defined on $X$,then $\{f_n\}$ are convergent uniformly on an open subset of $X$.

Since a bounded sequence can contain a convergent subsequence, I guess the above theorem is also true, but I don't know where to start.

LudvigH
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mathon
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  • You should add that the functions $(f_n)$ are continuous; otherwise the first theorem does not hold. – Etienne Sep 21 '13 at 15:00
  • There are examples of a sequence of continuous functions on $[0,1]$ that converge pointwise to the zero function, but converge uniformly on no open interval in $[0,1]$. – David Mitra Sep 21 '13 at 15:02
  • Interesting,show me a counterexample,please. – mathon Sep 21 '13 at 15:04
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    From Hewitt and Stromberg Real and Abstract Analysis, Exercise 11.43: For each $n$, let $F_n$ be the set of all numbers in $(0,1]$ having the form $k/2^m$ for an integer $k$ and an integer $m\in{0,1,\ldots,n}$. Let $f_n$ be $0$ on $F_n$. For $k/2^m\in F_n$, where $k$ is odd, let $f_n({k\over2^m}-{1\over2^{n+1}})={1\over2^m}$. Let $f_n$ be linear in all subintervals of $[0,1]$ where it is not yet defined. (Drawing the graphs of these should convince you the sequence $(f_n)$ is as proclaimed.) – David Mitra Sep 21 '13 at 15:11
  • How can the first theorem be proved? (Even a vague sketch of a proof would be interesting for me - thanks) – Giuseppe Negro Apr 13 '14 at 13:43

1 Answers1

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Here is another example (perhaps in fact the same as David's). It is a bit too long for a comment.

Enumerate all non trivial closed sub-intervals of $[0,1]$ with rational endpoints as a sequence $(I_k)_{k\in\mathbb N}$. For each $k$, choose your favourite sequence of continuous functions converging pointwise to $0$ but not uniformly on $I_k$. More precisely, choose a sequence of continuous functions $(f_{n,k})_{n\in\mathbb N}$ on $[0,1]$ with $f_{n,k}\equiv 0$ outside $I_k$ and $0\leq f_{n,k}\leq 1$, such that $f_{n,k}\to 0$ pointwise as $n\to\infty$ but $\sup_{x\in I_k} f_{n,k}(x)=1$ for any $n,k$.

Then define $f_n(x)=\sum_{k=1}^\infty 2^{-k}f_{n,k}(x)$. The $f_n$ are continuous (uniform convergence) with $0\leq f_n\leq 1$. It is not hard to check that $f_n\to 0$ pointwise (you have to interchange a limit and a $\Sigma$). On the other hand, for any $k, n$ you have $f_n(x)\geq 2^{-k} f_{k,n}(x)$; so $\sup_{x\in I_k} f_n(x)\geq 2^{-k}$ for each $k$ and all $n\in\mathbb N$, and hence no subsequence of $(f_n)$ can converge uniformly to $0$ on any $I_k$. Since any open subset of $[0,1]$ contains an $I_k$, it follows that no subsequence of $(f_n)$ converges uniformly on any open set.

Etienne
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