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Let $X$ be a Banach space and $X^*$ be its dual space. Let $\phi_n\in X^\ast$ and for all $x\in X$ we have $\phi_n(x)\to c\in\mathbb{C}$ as $n\to\infty$. I want to show that the sequence $\phi_n$ has a weak$^*$ limit $\phi\in X^*$.

Also if $x_n$ is a sequence in $X$ and for all $\phi\in X^*$ we have $\phi(x_n)\to a\in\mathbb{C}$. I want to show $x_n$ converges weakly in $X$ if $X$ is reflexive.

Thank you!

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yeah I know.. $X^*=\{\phi:X\mapsto\mathbb{C},\phi\,{\text continuous}\}$ – wang830 Nov 8 '12 at 9:51
Did you mean to write $\phi_n \in X^\ast$ in your second sentence? – Rudy the Reindeer Nov 8 '12 at 9:52
yeah $\phi_n\in X^*$ – wang830 Nov 8 '12 at 9:54
do we use the fact that $X$ is reflexive at all? – wang830 Nov 8 '12 at 9:55
I don't think so. But I might be missing something. – Rudy the Reindeer Nov 8 '12 at 9:57
up vote 1 down vote accepted

As pointed out by commenter in the comments:

You want to show that $c = c(x)$ is in $X^\ast$, that is, that it's continuous.

Let $c: X \to \mathbb C$ be such that $\phi_n (x) \xrightarrow{n \to \infty} c(x)$ for all $x \in X$. Then, as pointed out in the comments, $\phi_n$ are by assumption such that $|\phi_n(x)| < \infty$ for all $n$ and all $x$, hence we may apply BS to get that $\sup_n \|\phi_n\| < \infty$ and hence that $c$ is bounded:

$$ \|c\| = \sup_{\|x\|=1} |c(x)| = \sup_{\|x\|=1} \lim_{n \to \infty}|\phi_n(x)| \leq \sup_{\|x\|=1} \sup_n \|\phi_n\| \|x\| = \sup_n \|\phi_n\| < \infty$$

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Types don't match: $c$ is a complex number so you can't have $\phi_n \to c$ weak*. The question is a bit poorly phrased but it seems that the subtlety here is that $c$ depends on $x$ (otherwise $c = 0$). So: $\phi_n(x) \to c(x)$. Then you have to show that $x \mapsto c(x)$ defines an element of $X^\ast$ (uniform boundedness) and after that weak*-convergence is automatic, as you say. Similarly for the second part: you build $a(\phi)$ in $X^{\ast\ast}$ as above and then you use reflexivity to show that $a = \iota_X(x)$ for some $x \in X$ and finally you have to check that $x_n \to x$ weakly. – commenter Nov 8 '12 at 10:16
Thank you for the comment, @commenter. When I wrote it I read $c$ as the constant function. Of course, in retrospect, a stupid mistake, since that's not linear. I will try to fix my answer. – Rudy the Reindeer Nov 8 '12 at 10:20
Beware: you are assuming norm convergence in "Let $\varepsilon \gt 0$ and $n$ large enough so that $\lVert c - \phi_n\rVert \lt \varepsilon$." Here's one way to repair the argument: you know $|c(x)| = \lim_n |\phi(x)| \leq C \lVert x\rVert$ for $C = \sup_n \lVert \phi_n\rVert$, so $\lVert c\rVert \leq C$. – commenter Nov 8 '12 at 10:52
Thanks guys! I just worked out the second part. – wang830 Nov 8 '12 at 11:07
Well done! @wang830 – Rudy the Reindeer Nov 8 '12 at 11:08

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