I were doing this problem in Functional Analysis of Erwin Kreyszig(part 4.9, problem 10, page 269), but got stuck in the last point to come to the conclusion. Can anyone give me some hint to move on? Thanks.

Let $X$ be a separable Banach space and $M \subset X'$ a bounded set. Show that every sequence of elements of $M$ contains a subsequence which is weak* convergent to an element of $X'$.

Here $X'$ is the space contains all bounded linear functionals on $X$. Let $\{T_n\} \subset X'$, then we said $\{T_n\}$ weak* converges to $T \in X'$ if $\lim_{n \rightarrow \infty}{T_n(x)} = T(x)$ for all $x \in X$.

What I tried so far: Suppose we have a given sequence $\{T_n\} \subset M$. From the assumption, we've already have the sequence $\{||T_n||\}$ bounded, so from Corollary 4.9-7, we only need to find a subsequence $\{T_{n_i}\}$ of $\{T_n\}$such that the sequence $\{T_{n_i}(x)\}$ is Cauchy for every x in a total subset of $X$.

Because $X$ is separable, there exists a dense countable subset of $X$, we call $S$. For each $x \in X$, because $\{||T_n||\}$ bounded, we must have $\{||T_n(x)||\}$ bounded for each $x \in S$, by Bozzano-Weirstrass, there exists convergent subsequence of this sequence, and obviously, the subsequence is Cauchy sequence.

Here is the place I got stuck. If $S$ is finite, then we can repeat that process for the subsequence we just found, and finally, we have the subsequence $\{T_{n_i}\}$ of $\{T_n\}$ such that the $\{T_{n_i}(x)\}$ is Cauchy for all $x \in S$. But because $S$ is not finite, but countable, we can't do that.

Is there any trick to overcome this difficulty? I really appreciate any help.

  • $\begingroup$ Alternatively, you could use the fact that since $X$ is separable, $B(X^*)$ is weak* metrizable. Then, $B(X^*)$ is weak* sequentially compact as it is weak* compact. (I imagine you could diagonolize with your approach.) $\endgroup$ – David Mitra May 26 '16 at 15:36
  • $\begingroup$ I got you idea, but can you explain why if $X$ is separable, then $B(X^*)$ is weak* metrizable? $\endgroup$ – le duc quang May 27 '16 at 1:59
  • $\begingroup$ It should be in your book somewhere. Here is a post here, though. $\endgroup$ – David Mitra May 27 '16 at 6:46

The idea is to diagonalize, as mentioned earlier, but you have to do it carefully: Let $\{x_n\}$ be a countable dense subset of $X$. Not $\{T_n(x_1)\}$ has a convergent subsequence by Bolzano-Weierstrass, which you index with an increasing sequence $\{s(1,n) : n\in \mathbb{N}\} \subset \mathbb{N}$.

Now $\{T_{s(1,n)}(x_2)\}$ has a convergent subsequence, which you index by an increasing $\{s(2,n) : n\in \mathbb{N}\}$. Proceed inductively to obtain strictly increasing sequences $\{s(j,n) : n\in \mathbb{N}\}$ for each $j \in \mathbb{N}$ such thath

  • $\{s(j+1,n) : n \in \mathbb{N}\}$ is a subsequence of $\{s(j,n) : n \in \mathbb{N}\}$
  • $\{T_{s(j,n)}(x_j)\}$ is convergent for each $j \in \mathbb{N}$

Now consider the subsequence $T_{s(n,n)}$, and this converges pointwise at each $x_j$. As you mention, this completes your proof.

  • $\begingroup$ I got your idea. Thanks a lot! $\endgroup$ – le duc quang May 27 '16 at 2:10
  • $\begingroup$ @Prahlad Vaidyanathan Can you tell me why $||T_n(x)||$ is bounded because you are using the Bolzano Weierstrass theorem. There is some confusion with me because it seems to me that $||T_n(x)||$ is bounded for every x in the closed unit ball by definition of operator norm. But the dense subset you are using might not lie the closed unit ball. Can you clarify my confusion? Thanks a lot! $\endgroup$ – Riju Apr 18 '17 at 0:12
  • $\begingroup$ @Riju Since $M$ is bounded, there is an $R>0$ be such that $\|T\| \leq R$ for all $T\in M$. Now if $x\in X$ is fixed, then $\|T(x)\| \leq R\|x\|$ for all $T\in M$. In particular, the sequence $\{\|T_n(x)\| : n\in \mathbb{N}\}$ is bounded by $R\|x\|$. $\endgroup$ – Prahlad Vaidyanathan Apr 18 '17 at 14:33
  • $\begingroup$ @PrahladVaidyanathan Thank a lot!! I was missing that I am fixing an x I can use that for getting a pointwise boundedness. $\endgroup$ – Riju Apr 18 '17 at 16:47

Are you familiar with a so-called "diagonal argument"? In this way you can construct the subsequence you are looking for.

As you use Kreyszig as a reference, you could have a look into the proof of Theorem 8.1-5 of Kreyszig's book. The first part of this proof consists of a diagonal argument.


The dual unit ball is weak$^*$ compact by Alaoglu. If $S$ is a countable dense subset of $X$ then the topology on $X^*$ describing pointwise convergence on $S$ is metrizable (in particular, Hausdorff) and coarser that the weak$^*$ topology. But a compact space does not have strictly coarser Hausdorff topologies so that the dual unit ball with the weak$^*$ topology is metrizable. Hence it is sequentially compact.


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