I'm studying Conway's a course in Functional Analysis by myself. The following is corollary 6.12.7 of this book.

If $X$ is a separable Banach space and $A$ is a convex subset of $X^*$ that is weak* sequentially closed, then $A$ is weak* closed.

Proof: Because X is separable, $r(ball X^*)$ is weak* metrizable for every $r>0$. So if A is weak* sequentially closed, $A\cap (r(ball X^*))$ is weak* closed for every $r>0$. Hence the Krein-Smulian theorem applies.

My problem: I do not know how A is weak* sequentially closed implies $A\cap (r(ball X^*))$ is weak* closed for every $r>0$.


As you write $A \cap r\cdot B_{X^*}$ is weak$^*$ly metrizable as a subspace of the metrizable space $r\cdot B_{X^*}$, now let $x_i^* \in A \cap r \cdot B_{X^*}$ be any net converging weakly$^*$ to $x^* \in X^*$. As $rB_{X^*}$ is weakly$^*$ closed (the norm is weakly$^*$ lower semicontinuous), we have $x^* \in rB_{X^*}$. Now, as $rB_{X^*}$ is metrizable in the weak$^*$ topology, there is a subsequence $(x_{i(n)}^*)$ of the net $(x_i^*)$ converging to $x^*$. As $A$ is weakly$^*$ sequentially closed and $x_{i(n)}^* \in A$, we have $x^* \in A$. So $x^* \in A \cap rB_{X^*}$. As $(x_i^*)$ was an arbitrary net, $A \cap rB_{X^*}$ is weakly$^*$ closed.

Addendum: Let $(U_n)$ be a basis of weakly$^*$ neighbourhoods of $x^* \in rB_{X^*}$, for example the $1/n$-balls in a metric generating the topology, such that $U_n \supseteq U_{n+1}$. For any $n$, by convergence, there is an $i(n)>i(n-1)$ with $x_{i}^* \in U_n$ for $i \ge i(n)$. By construction, hence $x^*_{i(n)} \in U_N$ for $n \ge N$. Now let $U$ any neighbourhood of $x^*\in rB_{X^*}$. There is an $N$ such that $U_N \subseteq U$. Hence $x^*_{i(n)} \in U$ for $ n \ge N$. So $x_{i(n)}^* \to x^*$.

  • $\begingroup$ martini@ Thanks. But I do not know how metrizability $rB_{X^*}$ implies converging subsequence in $rB_{X^*}$ $\endgroup$ – niki Aug 7 '14 at 7:11
  • $\begingroup$ @niki Added something. Does this help? $\endgroup$ – martini Aug 7 '14 at 7:16

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