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Let $X$ be a Banach space and let $X^{*}$ be its dual space. Let $M$ a subspace of $X^{*}$ on which a weak*-continuous linear functional $L$ is given. In the case where $M$ is weak*-closed, we know that there exists an element $x\in X/{M}_\perp$ which represents $L,$ i.e., $$L(m)=m(x)\qquad (m\in M),$$ where $M_\perp$ denotes the pre-annihilaor of $M$ in $X.$ I have the following question;

Is the conclusion above true even when $M$ is not necessarily weak*-closed?

Tomasz Kania
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Abelvikram
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1 Answers1

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You can extend $L$ uniquely (by weak* continuity) to the weak*-closure of $M$. And the preannihilator of $M$ coincides with the preannihilator of its weak* closure.

gerw
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  • No, I don't think we can extend. I mean if you choose net ${\varphi_{\alpha}}$ which converges to a $\varphi_0$ in weak* closure, how do we know that ${L(\varphi_{\alpha})}$ converges? – Abelvikram Nov 16 '15 at 04:20
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    Let $\varepsilon > 0$. Since $L$ is weak* cts, we find a weak* neighborhood $V$ of $0$, such that $L(V\cap M) \subset (-\varepsilon,\varepsilon)$. Since ${\varphi_\alpha}$ converges, there is $\gamma$ with $\varphi_\alpha - \varphi_\beta \in V$ for $\alpha, \beta \ge \gamma$. Hence, $L(\varphi_\alpha - \varphi_\beta) \subset (-\varepsilon, \varepsilon)$. Hence, ${L(\varphi_\alpha)}$ is a Cauchy net and converges. – gerw Nov 16 '15 at 07:45