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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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    I don't think so. But I might be missing something.2012-11-08

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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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    You are right commenter, thank you for pointing it out!2012-11-08