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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.

1 Answers 1

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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.2012-06-13