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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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What is $\varepsilon$? And in the supremum, you should write you take it over the linear continuous functionals of norm $\leqslant 1$. –  Davide Giraudo Dec 29 '12 at 14:43
    
$\epsilon$ is arbitrary small number? Will the choice of $\ell$ affect my result? –  Johan Dec 29 '12 at 14:55
    
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 $\varepsilon$). –  Davide Giraudo Dec 29 '12 at 14:59

2 Answers 2

up vote 4 down vote accepted

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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the strong topology is finer than the weak –  user52188 Dec 29 '12 at 14:49
    
why is weak topology hausdorff? –  Koushik Dec 29 '12 at 14:53
3  
you can use Hahn–Banach (second geometric form) to prove that weak topology is hausdorff –  user52188 Dec 29 '12 at 15:02

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 soln –  Koushik Dec 30 '12 at 8:16

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