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Let $X$ be a Banach space and let $x_n \overbrace{\rightarrow}^w x$ and $x_n \overbrace{\rightarrow}^s z$ can we then say that $x = z$? My try:

$\| x- z\| = \sup_{\ell \leq 1} |\ell(x-z)| = \sup_{\ell \leq 1} |\ell x -\ell z| \leq \epsilon$

Where $\ell$ is a continuous functional in $X'$ Is this correct? is there any easier way? Thanks Btw if this already is correct, should I delete the post or what do I do?

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    No problem as the norm is bounded by $1$. But the las inequality deserves an additional step introducing the sequence $\{x_n\}$ (hence you don't really need $\varepsilon$).2012-12-29

2 Answers 2

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Yes. If $ x_{n} \to z $ strongly, then $ x_{n} \to z $ weakly also. Since the weak topology is Hausdorff, we have uniqueness of the limit. Therefore, $ z = x $.

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    you can use Hahn–Banach (second geometric form) to prove that weak topology is hausdorff2012-12-29
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If $x_n\to x$ weakly and $x_n\to z$ weakly, it follows that $f(x)=f(z)$ for any $f\in X^{*}$. Hence $f(x-z)=0$ for any $f\in X^{*}$. An immediate application of Hahn-Banach implies $x=z$. Indeed, otherwise we can build a functional $g_0$ on span$\{x-z\}$ such that $g_0(x-z)=1$, and use Hahn-Banach extend it to a functional $g$ on $X$. For this $g\in X^{*}$ we would have $g(x-z)=1$.

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    this is a nice soln2012-12-30