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I am practicing some elementary problems from functional analysis. I am trying to show when X is a normed space then the closed unit ball of its dual $ X^* $ is weak-star closed.

I can show whenever a sequence weak-star converges in the closed unit ball of X^* then the weak-star limit is also in the closed unit ball. Is this enough to show that the closed unit ball of $ X^* $ is weak-star closed?

Thanks!

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  • $\begingroup$ You have to be a little careful using sequences in a general topological space. In general, being closed implies that the limits of a sequence from the set all lie in the set itself, but the converse does not necessarily hold. $\endgroup$ – Theo Bendit Apr 7 '18 at 3:37
  • $\begingroup$ Yeah, that is why I am a bit cautious on this approach. I'm not sure whether or not a set containing all of its weak-star limit points implies that set is weak-star closed. $\endgroup$ – Carey Washington Apr 7 '18 at 5:06
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By Banach-Alaoglu Theorem, the closed unit ball in $X^{\ast}$ is weak-star compact. Since the weak-star topology is Hausdorff, the closed unit ball is weak-star closed.

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  • $\begingroup$ I am aware of this way to show that the closed ball of $ X^* $ is weak-star closed but I am trying to prove this w/o Alaoglu's theorem! Thanks! $\endgroup$ – Carey Washington Apr 7 '18 at 3:28
  • $\begingroup$ Have to think about it how to avoid that. $\endgroup$ – user284331 Apr 7 '18 at 3:29
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We can just write $$\tag{1}\overline{B_{X^*}(0)} = \bigcap_{x \in \overline{B_1(0)}} \{x^* \in X^* : |x^*(x)| \leq 1 \}.$$ By definition of the weak*-star topology, we know that $x^* \mapsto x^*(x)$ is continuous. So all sets are closed and thus also the intersection. Hence the $X^*$-unit ball is closed.

In order to see the equality of the sets in (1), we note that "$\subset$" is clear by definition of the operator norm on $X^*$; we awlays have $|x^*(x)| \leq |x^*| |x| \leq 1$. On the other hand, if $|x^*(x)| \leq 1$ for all $x \in \overline{B_1(0)}$, then $|x^*| \leq 1$.

For separable banach spaces it is true that any weak*-sequentially convex closed set is already closed.

However, there are sequentially closed sets in $l^\infty$, which are not closed. Example: Let $\delta_n$ denote the operator $\delta_n(f) = f(n)$, then $$C:=\{\delta_n : n \in \mathbb{N}\} \subset (l^\infty)^*$$ is sequentially closed. Let $\delta_{n_k} \rightarrow h$ in weak*-topology. If $(n_k)$ is unbounded, we may pass over to a subsequence with $n_k \rightarrow \infty$. In this case let $f \in l^\infty$ such that $f(n_k)$ is not convergent. Then $\lim_{n \rightarrow \infty} f(n_k) = h(f)$. That is a contradiction! So $(n_k)$ have to be already eventually constant.

Since $C \subset \overline{B_1(0)}$ is a subset of compact set, the sequence $(\delta_n)_{n \in \mathbb{N}}$ has limit points!

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