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Suppose $X$ is a normed space and $K$ is a subset of $X$ such that $K$ is weakly compact. Show that $K$ is norm-closed and norm-bounded.

I manage to show that $K$ is norm-bounded by using the Uniform Boundedness Theorem. However, I am not sure on how to show that $K$ is norm-closed. The following is my attempt:

Since $K$ is weakly compact, we have $x^*(K)$ is compact in the scalar field of $X$, say $F$. By the Heine-Borel Theorem, $x^*(K)$ is closed. Since $x^*$ is continuous, we have $K$ is norm-closed.

Is it correct?

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No, your reasoning is not correct. In fact, you only have shown that the preimage of $x^*(K)$ w.r.t. $x^*$ is closed. But $$K \subset (x^*)^{-1}(x^*(K))$$ might be a proper inclusion. Even $$K \subset \bigcap_{x^* \in X^*} (x^*)^{-1}(x^*(K))$$ might be proper.

However, the argument for closedness is quite simple. Assume that $K$ is not closed. Hence, there is a sequence $\{x_n\}$ such that ... Using the weak compactness of $K$ we find...

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  • $\begingroup$ If $K$ is not closed, then it does not contain all its limit points. In other words, if $x$ is a limit point of $K$, then there exists a sequence $\{ x_n \}$ such that $x_n \rightarrow x$ in the norm of $K$, but $x \notin K$. How to use weak compactness to obtain contradiction? Do I need the Eberlein-Smulian theorem? $\endgroup$ – Idonknow Oct 23 '15 at 7:24
  • $\begingroup$ By weak compactness, you get a subnet $\{x_i\}$ of the sequence $\{x_n\}$, such that $x_i \to \tilde x$ with $\tilde x \in K$. But every subnet already converges towards $x$. $\endgroup$ – gerw Oct 23 '15 at 8:41

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