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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 $A \subset Y'$, such that $Y'$ 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 !

Integral
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Theorem
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2 Answers2

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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.

David Mitra
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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
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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).

Nate Eldredge
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  • A minor question: I think we always have $x_n^* \in X^$ is weak$^$ convergent iff it is pointwise convergent.

    Bounded is not necessary here .

     – Red shoes May 31 '17 at 04:09
  • @nonlinearthought: Here I am viewing the elements of $Y' = \ell^\infty$ as bounded functions on $\mathbb{N}$ rather than as linear functionals on $Y$. So a sequence ${f_n} \in \ell^\infty$ converges weak-* to $f$ iff (1) $\sup_{n,x} |f_n(x)| < \infty$ and (2) $\lim_{n \to \infty} f_n(x) = f(x)$ for each $x \in \mathbb{N}$. (1) and (2) are what I mean by "bounded" and "converges pointwise". (1) is necessary by the uniform boundedness principle, and (2) alone is not sufficient. – Nate Eldredge May 31 '17 at 05:29
  • But the point is that for any $f \in \ell^\infty$, you can find a sequence $f_n \in c_0$ that not only converges pointwise to $f$ (1), but moreover is uniformly bounded (2). That proves weak-* density. – Nate Eldredge May 31 '17 at 05:30
  • Ok, Now it makes sense. By points I thought you meant $x \in X or Y$. That was a little confusing, because the definition of weak$^*$ convergence is exactly pointwise convergence in general. – Red shoes May 31 '17 at 05:42