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Given a normed vector space $X$, consider a subspace $Z$ of its continuous dual $X^{\ast}$. I want to show that if the annihilator of $Z$ is zero, then $Z$ is $weak^{\ast}ly$ dense in $X^{\ast}$. How should I prove this?

I'd like to use Hahn Banach, probably a version of it for locally convex topological space, since we are not considering the norm topology on $X^{\ast}$. Also, I should probably prove by contradiction.

Keith
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  • In any locally convex space $Y$ a subspace $L$ is dense if and only if its annihilator $L^\perp={y^\in Y^: y^|_L=0}$ is $0$. Apply this to $Y=X^$ with the weak$^*$ topology whose dual is $X$ (more precisely, every element of the dual is an evaluation in a point of $X$). – Jochen Nov 08 '17 at 07:55

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