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Let us define $E^*=\{f:E\rightarrow \mathbb{R} \mid f $ is linear and continuous$\}$, where $E$ is separable space equipped with a norm. Let $A = \{x_1,x_2,\ldots\}$ be a countable dense subset of $E$. Suppose that $(f_n)_{n\in\mathbb{N}}$ is a sequence of functions from $E^*$ such that $$\lim_{n\to\infty}f_n(x_i)=f(x_i)$$for all $i\in\{1,2,\ldots\},$ i.e. $(f_n)$ converges pointwise to $f$ on $A$. My teacher writes then:

Consequently, $f_n(x) \rightarrow f(x) $ $\forall x \in E$.

My question is: why is it true?

I am trying to demonstrate this as follows:
Choose any $x_i$ close to $x$. Then $f_n(x - x_i) = 0 = f_n(x) -f_n(x_i)$. We have $$\lim_{n\rightarrow \infty}|f_n(x) - f(x_i)| = 0 \iff f_n(x) \rightarrow f(x_i)$$ as $n\rightarrow \infty$. By hypothesis $f_n(x_i) \rightarrow f(x_i)$ whilst by continuity $f(x_i) \rightarrow f(x)$ and then I will be able to conclude that $f_n(x) \rightarrow f(x) $ $\forall x \in E$. Is it correct?

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Why $f_n(x-x_i)=0$ ? –  no identity Aug 14 '12 at 22:17
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Is $f$ supposed to be in $E^*$? (you should write the sequence $\{f_n\}$, not $\{f_n(x_i)\}$, and that $f_n(x_i)\to f(x_i)$ for each $i$. And you have to assume that the norms of $f_n$ are uniformly bounded, otherwise take $E=L^2(\Bbb R)$ and $f_n(x):=n\int_{[n,n+1]}f(x)dx$, and $A$ which consists of functions with compact support. –  Davide Giraudo Aug 14 '12 at 22:18

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up vote 3 down vote accepted

Your proof is quite strange because it starts from the assumption that $f_n(x-x_i)=0$ for all $n\in\mathbb{N}$. This is not true in general.

I'll additionally assume that $\{f_n:n\in\mathbb{N}\}\subset \mathrm{Ball}_{E^*}(0,C)$ for some $C>0$, i.e. norms of functionals $\{f_n:n\in\mathbb{N}\}$ is uniformly bounded by some constant $C$.

Take arbitrary $x\in E$. Fix $\varepsilon>0$, and find $i\in\mathbb{N}$ such that $$ \Vert x- x_i\Vert\leq\frac{\varepsilon}{3\max\{ C,\Vert f\Vert\}} $$ Since $\lim\limits_{n\to\infty} f_n(x_i)=f(x_i)$, then there exist $N\in\mathbb{N}$ such that $n>N$ implies $|f_n(x_i)-f(x_i)|\leq\varepsilon/3$. Now we use $\varepsilon/3$-trick, i.e. for all $n>N$ we have $$\eqalign{ |f_n(x)-f(x)|&\leq |f_n(x)-f_n(x_i)|+|f_n(x_i)-f(x_i)|+|f(x_i)-f(x)|\phantom{A\over B}\cr &= |f_n(x-x_i)|+|f_n(x_i)-f(x_i)|+|f(x_i-x)|\phantom{A\over B}\cr &\leq \Vert f_n\Vert\Vert x-x_i\Vert+\varepsilon/3+\Vert f\Vert\Vert x-x_i\Vert\phantom{A\over B}\cr &\leq C\frac{\varepsilon}{3\max\{ C,\Vert f\Vert\}}+\varepsilon/3+\Vert f\Vert\frac{\varepsilon}{3\max\{ C,\Vert f\Vert\}}\cr &\leq \varepsilon/3+\varepsilon/3+\varepsilon/3\phantom{A\over B}\cr &=\varepsilon} $$ Thus for all $x\in E$ and $\varepsilon>0$ we found $N\in\mathbb{N}$ such that $n>N$ implies $|f_n(x)-f(x)|<\varepsilon$. This means that for all $x\in E$ we have $\lim\limits_{n\to\infty} f_n(x)=f(x)$.

P.S.

If we won't require uniform boundedness of norms of $\{f_n:n\in\mathbb{N}\}$ we can construct a counterexample. See Davide's comment.

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And actually a simpler counter-example in $\ell^2(\Bbb Z)$: $f_n(x):=nx_n$ and $D$ the set of sequence which are eventually $0$ and rational entries. –  Davide Giraudo Aug 14 '12 at 22:37
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$f_n$ is linear so if equality $f_n(x-x_i)=0$ holds, then $f_n(x)=f_n(x_i)$. But how do you ensure that the last equality is true? I think you can't because there is a simple counterexample: take $f_n=f$. –  no identity Aug 15 '12 at 0:56
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That was wrong, I've edited answer to get correct inequlities –  no identity Aug 15 '12 at 20:53
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Juan in the case $3\max\{C,\Vert f\Vert\}=\Vert f\Vert$ you will get $\Vert f\Vert\frac{\varepsilon}{3\max\{C,\Vert f\Vert\}}=\frac{\varepsilon}{3}$ so I don't see a problem –  no identity Aug 15 '12 at 21:31
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Nope, if $\Vert f\Vert>C$, then $\max\{C, \Vert f\Vert\}=\Vert f\Vert$ and $\Vert f\Vert\frac{\varepsilon}{3\max\{C,\Vert f\Vert\}}=\frac{\varepsilon}{3}$ –  no identity Aug 16 '12 at 0:34

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