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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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    Can you prove this for $A=\mathbb{R}$?2012-09-30
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    No, i'm not sure how to do this either.2012-09-30
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    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?2012-09-30
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    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 $\|a-x\| < \delta$.2012-09-30
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    Since $g_n \to g$ pointwise there is $N$ such that $|g_n(a) -g(a)| < \varepsilon / 3$ for $n > N$.2012-09-30
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    Let $U = B_\delta(a)$ and $n >N$. Then $\sup_{x \in U} |g_n (x) - g(x)| = \sup_{x \in U} |g_n (x) - g_n(a) + g(a) -g(a) + g(a) - g(x)| \leq \sup_{x \in U} |g_n (x) - g_n(a)| + |g_n(a) -g(a)| + \sup_{x \in U} |g(a) - g(x)| \leq \varepsilon$2012-09-30
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    @Matt : The problem I see is that $\delta$ depends a priori on $n$. You'd have to show that you can find a $\delta$ that works for every $n > N$.2012-09-30
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    This might help http://math.stackexchange.com/questions/204157/pointwise-limit-on-a-complete-metric-space2012-09-30
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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.