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The Banach Alaoglu theorem states:

Let $\mathcal{Z}$ be a banach space. The closed unit ball $\{Z^{\ast}\in\mathcal{Z}^{\ast}:\|Z^{\ast}\|_{\ast}\leq1\}$ is compact in the weak$\ast$ topology of $\mathcal{Z}^{\ast}$.

Then in the lecture notes I am using there is the following consequence:

It follows that any bounded (in the dual norm $\|\cdot\|_{\ast}$) and weakly$\ast$ closed subset of $\mathcal{Z}^{\ast}$ is weakly$\ast$ compact.

Now, I do not understand why this follows:

1.) Why can we transfer it to any subset of $\mathcal{Z}^{\ast}$ (that is bounded and closed of course) when the theorem only gives us the compactness of the unit ball? I.e. we would need something more general like $\|Z^{\ast}\|_{\ast}\leq M$ for some $M\in\mathbb{R}$ instead of just $\|Z^{\ast}\|_{\ast}\leq1$

2.) Why do we need boundedness and closedness (and even in two different topologies)?

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    $\begingroup$ Scalar multiplication is a homeomorphism, so the closed ball of any radius in the dual space is weak* compact. $\endgroup$
    – While I Am
    Commented Aug 7 at 14:42
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    $\begingroup$ a closed subset of a compact space is compact $\endgroup$
    – user8268
    Commented Aug 7 at 14:58
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    $\begingroup$ @guest1 A closed subset of a compact set is compact. So if $A \subset \mathcal{Z}^\ast$ is weak$^\ast$ closed and bounded, then it is a closed subset of the compact set $\{Z^\ast \in \mathcal{Z}^\ast: \|Z^\ast\| \leq M\}$ for some $M > 0$, whence it is compact (as it is a closed subset of a compact set). $\endgroup$
    – David Gao
    Commented Aug 7 at 15:00
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    $\begingroup$ @guest1 The image of a compact set under a continuous map is compact. The closed ball of radius $M$ is the image of the unit ball (which is compact) under the map given by multiplication by $M$ (which is continuous), so it is compact. $\endgroup$
    – David Gao
    Commented Aug 7 at 15:25
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    $\begingroup$ @guest1 I’d suggest you review the basics of general topology if you find that confusing. That really should be obvious from basic topology knowledge. $\endgroup$
    – David Gao
    Commented Aug 7 at 15:27

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Here is a complete argument, as has already been sketched in the comments. I have broken it up into smaller subparts to be as clear as possible.

Let $X$ be a Banach space, and let $C$ be a weak* closed, bounded subset of $X^\ast$.

  • By the Banach-Alaoglu theorem, $B := \{\ell \in X^\ast : \|\ell\|_{op} \leq 1\}$ is weak* compact.

  • By the structure of a TVS, the map $s_\lambda : x \mapsto \lambda x$ is a homeomorphism for any $\lambda \neq 0$.

  • The continuous image of a compact set is compact (this is a general topological fact), so $$s_\lambda(B) = \{\ell \in X^\ast : \|\ell\|_{op} \leq |\lambda|\}$$ is weak* compact.

  • Since $C$ is bounded, there is some $R > 0$ so that $C \subseteq s_R(B)$. Thus $C$ is a subset of a weak* compact set.

  • A closed subset of a compact set is compact (this is a general topological fact), and as $C$ is weak* closed and is contained in a weak* compact set, it follows that $C$ is weak* compact.

As you can see, the only "new" things we need to add are these facts from general topology. I am a believer in learning-on-the-go, but working with TVS's requires these kinds of facts to be second nature, as any "trivial" fact about a TVS will probably require some nontrivial general topology.

EDIT: In response to your second question, I want to point out that we don't need two topologies; boundedness is not a topological invariant, so we are really enjoying the rich interplay of the metric structure of the norm, and the topological structure of the weak* topology. That this interplay is worth studying should be justified by the Banach-Alaoglu theorem.

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  • $\begingroup$ thank you four your exhaustive answer! I have three question thought: 1.) So you define the map $s_{\lambda}$ as a mapping that maps elements from $X$ to $X$ but in your third bullet point you use it as a mapping that maps a subset of $X^*$ to another subset of $X^*$. So that is unclear to me. 2.) what do we precisley shift with our homeomorphism? Do we take all the elements from the set $B$ and shift them by $\lambda$ and then we know somehow that their norms must be smaller than $\lambda$? $\endgroup$
    – guest1
    Commented Aug 8 at 9:03
  • $\begingroup$ 3.) In your edit you say that one does not need two topologies. But we need a normed spaced to invoke the Banach Alaoglu theorem in the first place and then we also need the weak* topology. So we do need two topologies here right? 4.) you say that boundedness is not a topological invariant. I am not sure what you mean by that? But I think it is important that the boundedness is w.r.t. the norm topology. sorry now these are 4 instead of 3 questions $\endgroup$
    – guest1
    Commented Aug 8 at 9:05
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    $\begingroup$ I do not mean to sound impolite, but the answers to these questions involve “elementary” (relative to the question) skills which should be background to studying the weak* topology. You should try answering these on your own, and if one is particularly confusing you should search Math.SE to see if it’s been asked, and if not, post a question with demonstrated effort. $\endgroup$
    – While I Am
    Commented Aug 8 at 12:28
  • $\begingroup$ Well you do sound impolite but I guess that you dont care about that anyway $\endgroup$
    – guest1
    Commented Aug 8 at 12:32
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    $\begingroup$ In fact I don’t think it’s impolite to indicate that certain topics are considered “basic background” for a subject, as it can be a good signal about whether one should return to those topics for review. This is constructive feedback, not criticism. $\endgroup$
    – While I Am
    Commented Aug 8 at 12:50

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