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For some Banachspace $A$ we have a sequence of continuous functions $g_n:A\rightarrow \mathbb{R}$ pointwise converging to some $g:A\rightarrow\mathbb{R}$. Prove that for any $\epsilon>0$ there exist $\emptyset\not=U\subset A$ open and $N\in\mathbb{N}$ such that for all $n>N$ we have $\sup_{x\in U}\left|g_n(x)-g(x)\right|<\epsilon$.

I'm not sure how to approach this problem. Is it a good idea to prove something like local boundedness first?

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    @JoelCohen Oh right, thank you!2012-09-30

1 Answers 1

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Fix $\epsilon > 0$. For a given $N$ define $B_N = \{x \in A : \forall m,n > N\,\,|g_m(x) - g_n(x)| \leq 2\epsilon\}$ $ = \bigcap_{m,n > N} \{x \in A : |g_m(x) - g_n(x)| \leq 2\epsilon\}$ Each set $\{x \in A : |g_m(x) - g_n(x)| \leq 2\epsilon\}$ is closed since $g_m$ and $g_n$ are continuous. Therefore the intersection $B_N$ is also closed. Because for each $x$ the sequence $\{g_n(x)\}$ is convergent, for each $x$ it is also a Cauchy sequence, so $\bigcup_N B_N$ is all of $A$. By the Baire Category theorem, some $B_N$ contains an open set $U$. For all $x \in U$ and all $m,n > N$ one has $|g_m(x) - g_n(x)| \leq 2\epsilon$. Taking limits as $m$ goes to infinity, one has $|g(x) - g_n(x)| \leq 2\epsilon$ for $n > N$, so in particular $|g(x) - g_n(x)| < \epsilon$ for $n > N$ as needed.