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I am trying to figure out if the statement holds true, the literature i am following says that its not true but i don't seem to understand,

If $Y$ is a Banach space and let subspace $A \subset Y'$ $Y'$ is a dual . $A$ is norm closed if and only if $A$ is weak star closed ?

Looks like reflexivity comes into play to argue this statement . Thank you for your hints !

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2 Answers 2

up vote 6 down vote accepted

From Goldstine's Theorem, it follows that $X$ is weak* dense in $X^{**}$. Now $X$ is norm closed in $X^{**}$. If $X$ were weak* closed in $X^{**}$, we would have $X=X^{**}$. Of course, this is not always the case.

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E.g., $c_0$ is norm-closed but not weak* closed in $\ell^\infty$, and $K(H)$ is norm-closed but not weak* closed in $B(H)$ (when $H$ is an infinite dimensional Hilbert space). –  Jonas Meyer Dec 13 '12 at 18:57
    
@David Mitra : you used homogeneity of scalar multiplication to say that the whole $X$ is weak* dense $X^{**}$ right ? I find this fact cool about topological vector spaces :D –  Theorem Dec 13 '12 at 19:14

Take $Y = \ell^1$, so $Y' = \ell^\infty$, and let $A = c_0$ be the space of all sequences which converge to 0. It is easy to see that $A$ is norm closed, but in fact $A$ is weak-* dense in $Y'$ (since a sequence converges weak-* in $Y'$ iff it is bounded and converges pointwise).

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