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I have a question about the following statement:
*"Every bounded net in a Banach space $Y$ has a subnet that converges in $Y^{**}$ with respect to the weak-$\ast$ topology."

Is this statement true? To clarify, $Y^{**}$ denotes the bidual of the Banach space $Y$, and the weak-$\ast$ topology refers to the weak-$\ast$ topology on $Y^{**}$.

Is this really a general result?

Thank you in advance for any clarification or references!

vinipenalty27
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  • It may be a good idea to provide some more context for this question. – Dean Miller Mar 25 '25 at 22:54
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    Yes…? This is basically just Banach-Alaoglu theorem. – David Gao Mar 25 '25 at 22:56
  • @DavidGao Thank you! I see your point. But just to clarify — the Banach-Alaoglu theorem is originally about the weak-* compactness of the closed unit ball in the dual space $Y^*$.

    What confuses me is how exactly this extends to the bidual $Y^{\ast}$.

     – vinipenalty27 Mar 25 '25 at 23:03
  • @vinipenalty27 The bidual is also a dual space, namely, the dual space of $Y^\ast$. – David Gao Mar 25 '25 at 23:07
  • @DavidGao Thank you for the explanation! But I’m still a bit confused — my net is originally in $Y$.

    Could you please give a bit more detail on how exactly the Banach-Alaoglu theorem applies here?

    I’d really appreciate it!

     – vinipenalty27 Mar 25 '25 at 23:17
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    @vinipenalty27 Any net in $Y$ is also a net in $Y^{\ast\ast}$ since $Y \subset Y^{\ast\ast}$. The net is uniformly bounded, so it is contained in a weak$^\ast$ compact subset of $Y^{\ast\ast}$, namely, the unit ball of $Y^{\ast\ast}$ multiplied by some constant. By compactness, a subnet must converge weak$^\ast$ to some element of $Y^{\ast\ast}$. – David Gao Mar 25 '25 at 23:42
  • @DavidGao Just to clarify, when you say $Y\subseteq Y^{\ast\ast}$, are you referring to the canonical embedding $J_Y$ ?

    Thank you so much for your attention and help — I really appreciate it!

     – vinipenalty27 Mar 25 '25 at 23:48
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    @vinipenalty27 That is what I meant, yes. – David Gao Mar 26 '25 at 00:25
  • @vinipenalty27 If these comments helped answer your question, it can be a good idea to type this information up as an answer to your question. – Dean Miller Mar 26 '25 at 02:27

1 Answers1

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Here is the answer from the comments with details. We start off by noting that the statement "Every bounded net in a Banach space $Y$ has a subnet that converges in $Y^{**}$ with respect to the weak-$\ast$ topology." is identifying the Banach space $Y$ with its image under the canonical embedding into $Y^{**}$. With this in mind, a precise formulation of the above statement is the following.

Let $Y$ be a Banach space and let $(y_{\alpha})_{\alpha \in I}$ be a bounded net in $Y$. Let $J_{Y}\colon Y \to Y^{**}$ be the canonical embedding. Then the net $(J_{Y}y_{\alpha})_{\alpha \in I}$ has a subnet which converges with respect to the weak$^{*}$ topology on $Y^{**}$.

Since the net $(y_{\alpha})_{\alpha \in I}$ is bounded, there is some $c > 0$ such that $\|y_{\alpha}\|_{Y} \leq c$ for all $\alpha \in I$. Then as the map $J_{Y}$ is an isometry, we have that $\|J_{Y} y_{\alpha} \|_{Y^{**}} \leq c$ for all $\alpha \in I$. Hence the net $(J_{Y}y_{\alpha})_{\alpha \in I}$ is contained in the bounded set $\{y^{**}\in Y^{**} : \|y^{**}\|_{Y^{**}} \leq c\}$, which is compact in the weak$^{*}$ topology by the Banach-Alaoglu theorem. By compactness, the net $(J_{Y}y_{\alpha})_{\alpha \in I}$ has a subnet which converges with respect to the weak$^{*}$ topology on $Y^{**}$ as desired.

Note that the proof did not use the completeness of $Y$ anywhere. So the result still holds if $Y$ is only assumed to be a normed space.

Dean Miller
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