# Pointwise convergence of continuous functions implies uniform convergence on some open subset

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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Can you prove this for $A=\mathbb{R}$? –  Chris Eagle Sep 30 '12 at 10:00
No, i'm not sure how to do this either. –  davidg Sep 30 '12 at 10:06
Are you sure it holds? It might assume enough compact sets, and I'm not sure that every Banach space has such.. Also, is $g$ known to be continuous, as well? –  Berci Sep 30 '12 at 11:28
What's wrong with the following?: Fix any $a \in A$. Since $g_n$ are continuous there is $\delta$ such that $|g_n(a) - g_n (x)| < \varepsilon / 3$ for $\|a-x\| < \delta$. –  Rudy the Reindeer Sep 30 '12 at 11:31
Since $g_n \to g$ pointwise there is $N$ such that $|g_n(a) -g(a)| < \varepsilon / 3$ for $n > N$. –  Rudy the Reindeer Sep 30 '12 at 11:32

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.