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Let $(X, \|.\|)$ be a Banach space and $(X^*, \|.\|_{X^*})$ its dual space. Suppose that $E^*$ is convex and closed in the norm topology of $X^*$. Suppose that $X$ is not reflexive, I would like to ask whether $E^*$ is weak$^*$ closed in $X^*$.

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Not necessary. Let $x^{**} \in X^{**} \setminus X$ and set $E^* = \ker x^{**}$. Then $E^*$ is convex and norm closed, as $x^{**} \colon X^* \to \mathbb K$ is norm continuous. If $E^*$ were weakly$^*$ closed, $x^{**}$ were weakly$^*$ continuous, hence an element of $X$, as $(X^*, \text{weak}^*)^* \cong X$.

To give a concrete example let $X = c_0$ and $x^{**} = (1,1, \ldots) \in \ell^\infty$. Our $E^*$ is then given by \[ E^* = \left\{x \in \ell^1 \biggm| \sum_n x_n = 0 \right\} \] which is easily seen to be norm closed and convex. To see that it isn't weakly$^*$ closed, consider $x^k = e^1 - e^k$ ($e^i$ denoting the $i$-th unit sequence $e^i = (0, \ldots, 0, 1, 0, \ldots)$) then $x^k \in E^*$ and given $a \in c_0$, we have \begin{align*} (e^1 - e^k)(a) &= a_1 - a_k\\ &\to a_1 \\ &= e^1(a). \end{align*} So $x^k \to e^1$ weakly$^*$, but $e^1 \not\in E^*$.

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Dear Sir. Thank you very much for your interesting solution. –  blindman Oct 23 '12 at 0:35
    
Thanks! This example is so great! –  user67339 Mar 18 '13 at 18:00

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