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The problem is:

Let $E$ be a normed space, let $x\in E$, let $\left(x_{n}\right)_{n\in\mathbb{N}}\subset E$ be a sequence. I need to show that $\left(x_{n}\right)$ converges weakly to $x$ if and only if $\left(x_{n}\right)$ converges to $x$ in $\left(E,\,\sigma\left(E,\, E^{*}\right)\right)$, $\sigma\left(E,\, E^{*}\right)$ denoting the weak topology on $E$.

My efforts:

I have that $x_{n}\rightarrow x$ in $\left(E,\,\sigma\left(E,\, E^{*}\right)\right)$ is equivalent to: $\forall U\in\sigma\left(E,\, E^{*}\right),\, x\in U\,:\quad\left\{ n\in\mathbb{N}:\, x_{n}\notin U\right\}$ is finite.

My question:

It is unclear to me how to pass from the weak convergence in $E$ to the convergence in the weak topology on $E$.

How can I prove both directions?

Thanks, Franck.

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What is $X$ (capital $X$) The same as $E$? And what is $\sigma$? Not everyone has access to the index of notation to your book or lecture. –  user20266 Jun 10 '12 at 15:54
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Assume $x_n \to 0$ weakly. This means that for fixed $\varepsilon$ you have for every $f \in E^\star$ an index $N_f$ such that $\lvert f(x_n)\rvert <\varepsilon$ for every $n > N_f$. This is what you have. Now let $U(f_1 \ldots f_k; \varepsilon)$ denote a weak neighborhood of the origin like this: $$\{x\in E \ :\ \lvert f_j(x)\rvert < \varepsilon,\ j=1, 2\ldots k\}.$$ Playing a bit with what you have you can prove that there exists an index $N$ such that $x_n \in U(f_1 \ldots f_k; \varepsilon)$ for every $n\ge N$. (...) –  Giuseppe Negro Jun 10 '12 at 17:47
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So $x_n\in U(f_1 \ldots f_k; \varepsilon)$ except, at most, for a finite number of indices. This is enough to infer that $x_n \to 0$ in $\sigma (E; E^\star)$, since every weakly open set $U$ containing the origin of $E$ contains a neighborhood of the form $U(f_1 \ldots f_k; \varepsilon)$ for some $f_1\ldots f_k \in E^\star$ and $\varepsilon >0$. (This is what "base of neighborhoods" means). –  Giuseppe Negro Jun 10 '12 at 17:51
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Try reasoning in a similar way to prove the other implication, namely: $x_n \to 0$ in $\sigma(E;E^\star) \Rightarrow x_n \to 0$ weakly. –  Giuseppe Negro Jun 10 '12 at 17:54
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What you have is not "almost equivalent" to $x_n \to 0$ weakly: indeed, it is stronger. $x_n \to 0$ weakly means exactly that for every $f \in E^\star$ and $\varepsilon > 0$ we have $x_n \in U(f; \varepsilon)$ except at most for a finite number of indices. And you already have that: just set $k=1$. –  Giuseppe Negro Jun 10 '12 at 21:15

1 Answer 1

up vote 1 down vote accepted

WLOG, set $x=0$ (weak translation is continuous):

We have to show that $x_{n}\rightharpoonup0\Leftrightarrow x_{n}\rightarrow0$ in $\left(E,\sigma\left(E,E^{\star}\right)\right).$

$$ \Rightarrow\forall\epsilon\quad\exists N_{\epsilon}\in\mathbb{N}\quad\forall f\in E^{\star}:\quad\left|f(x_{n})\right|<\epsilon\quad\forall n\geq N_{\epsilon}. $$

Let $U\left(f_{1},...,f_{k};\epsilon\right)$ denote a weak neighborhood of the origin:

$$ U\left(f_{1},...,f_{k};\epsilon\right):=\left\{ x\in E:\,\left|f_{i}\left(x\right)\right|<\epsilon,\, i=1,...,k\right\} . $$

So $x_{n}\in U\left(f_{1},...,f_{k};\epsilon\right)$, except at most for a finite number of indices. This is enough to infer that $x_{n}\rightarrow0$ in $\left(E,\sigma\left(E,E^{\star}\right)\right)$, since every weakly open set $U$ containing the origin of $E$ contains a neighborhood of the form $U\left(f_{1},...,f_{k};\epsilon\right)$ for some $f_{1},...,f_{k}\in E^{\star}$ and $\epsilon>0$.

For the reverse:

$x_{n}\rightarrow0$ in $\left(E,\sigma\left(E,E^{\star}\right)\right)\Rightarrow x_{n}\in U\left(f_{1},...,f_{k};\epsilon\right)$, except for at most a finite number of indices, $f_{i}\in E^{\star},\,\epsilon>0$.

$\Rightarrow f_{i}(x_{n})\rightarrow0$ for $f_{i}\in E^{\star},\, i=1,...,k$.

Set $k=1\Rightarrow\forall f\in E^{\star}:\, f\left(x_{n}\right)\rightarrow f\left(0\right)=0\Rightarrow x_{n}\rightharpoonup0$

$\square$

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Good, thank you! I should have made our comments into an answer but I'm busy right now, you saved me some precious time. –  Giuseppe Negro Jun 13 '12 at 17:13

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