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Today I learnt something about the weak-star topology, but I don't know what the use of weak-star topology is. I hope someone can tell me what we can do with the weak-star topology. Thanks in advance!

Martin
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Brain Zhang
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    We have a nice compactness result, for starters: http://en.wikipedia.org/wiki/Banach%E2%80%93Alaoglu_theorem – Zach L. Jun 04 '13 at 03:22
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    Also, forget about what it's useful for at the moment, it's only natural to consider the topology of pointwise convergence for elements of the dual. On top of that, as Zack L mentioned, there are compactness results, such as Alaoglu and Krein-Milman. – Michael Jun 04 '13 at 03:26
  • thanks. I have one more question. In class, the teacher said that the dual space of $L_{p}$ is $L_{q}$ while $1/p+1/q=1$, while when p is infinte there is an exception. We know that the dual of a space is the set of all the bounded fuctionals on it. How can we say that the dual space of $L_{p}$ is $L_{q}$ ,since, in my opinion, $L_{q}$ is not a space containing the functionals of $L_{p}$. – Brain Zhang Jun 04 '13 at 03:30
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    @Martin: You're absolutely right. People will often say things are "equal" or "identical" when they mean "canonically isomorphic". Depending on how comfortable you are with this sloppiness, it can either a convenience or an annoyance. – Zev Chonoles Jun 04 '13 at 03:52
  • It's not. Notice that if we take $L_p$ and a function $g \in L_q$, then we can define an operator on $L_p$ by $f \mapsto \int fg$. This is a bounded linear operator and so can be thought of as an element in the dual of $L_p$. As it turns out, every linear operator arises this way. – Zach L. Jun 04 '13 at 03:53
  • «In your opinion»?! Are the elements of $L_q$ or not linear functionals on $L_p$? You can certainly answer this :-) – Mariano Suárez-Álvarez Jun 04 '13 at 03:54
  • @Zach L. yup! that is ensured by Riesz Representation Thm, am I right? I think Zev has answered my question. I am just wondering why the teacher was saying that thet are 'equal'. Thanks a lot!! – Brain Zhang Jun 04 '13 at 03:57
  • @Zev Chonoles Thank you so much!! – Brain Zhang Jun 04 '13 at 03:58
  • @Mariano Suárez-Alvarez♦ I think they are not~ – Brain Zhang Jun 04 '13 at 04:00
  • As Zach L. mentioned, the weak-$\ast$ topology gives rise to the Banach--Alaoglu theorem. A very nice application of the weak-$\ast$ topology and of Banach--Alaoglu can be found in the proof of Gelfand--Naimark duality (http://en.wikipedia.org/wiki/Gelfand_representation). The point is that if $A$ is a commutative unital $C^\ast$-algebra (http://en.wikipedia.org/wiki/C*-algebra), then the set $M(A)$ of characters of $A$ (unital $\ast$-homomorphisms $A \to \mathbb{C}$) is a closed subset of the unit ball of the dual of $A$ as a Banach space, with the weak-$\ast$ topology... – Branimir Ćaćić Jun 04 '13 at 06:22
  • ...so that by Banach--Alaoglu, $M(A)$ is compact. The upshot then is that $A$ is therefore $\ast$-isomorphic to the commutative unital $C^\ast$-algebra $C(M(A))$ of continuous complex-valued functions on the compact Hausdorff space $M(A)$, so that every commutative unital $C^\ast$-algebra, up to $\ast$-isomorphism, is of the form $C(X)$ for a unique (up to homeomorphism) compact Hausdorff space $X$. – Branimir Ćaćić Jun 04 '13 at 06:23
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    Usually, if you think something your teacher said is "wrong", you will learn a lot by discussing it with her/him. – GEdgar Jun 04 '13 at 11:59
  • @GEdgar yesBut sometimes it takes some time before the teacher reply me and I feel uncomfortable to leave question unsolved I have just found this site days ago and I really really enjoy discussing with the masters here! – Brain Zhang Jun 04 '13 at 15:16

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The main use of weak* topology is to provide a topolgy on $V^*$ for a normed space(or a TVS) $V$ such that the unit ball in $V^*$ is compact, which is the Banach-Alaoglu Theorem. There are numerous places in analysis where we use this topology, for example from Riesz Representation theorem, for any locally compact Hausdorff space $X$, if $\mathcal{M}(X) $ is the space of complex Radon measures, then the dual $ C_0(X)^* \cong \mathcal{M}(X) $, So from usual sup norm topology of $ C_0(X) $ you can induce the weak* topology on Radon measures, thus you can have convergence, compactness results in measures. If you are familiar with distribution theory, you find that given the inductive limit topology on $ C^\infty_c(\Omega) $, weak* topology is induced on the space of distributions. This is a fundamental requirement in PDEs.

smiley06
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