1
$\begingroup$

I am wondering about a way to construct a base or subbase for the weak* topology on $\ell_2$. I am fairly new to topology and functional analysis, so I apologize if the question is not precisely formulated and/or stupid.

This is how far I got: Because $\ell_2$ is a Hilbert space, it holds that the dual space $\ell_2^\ast$ equals $\ell_2$ itself (i.e., $\ell_2$ is reflexive). Moreover, linear functionals on the space have the simple form of the dot product. Thus, for $a\in \ell_2$, any linear function takes the form $<a,x >$ for a fixed coefficient sequence $x$. (That is to say, any linear operator $T:\ell_2 \rightarrow \mathbb R$ takes the form $T(a) = <a,x >$.) So I need to construct the set of all such functions $T$ that are continuous - any idea how I would do that? Any kind of help is much appreciated, including books/online sources/...

$\endgroup$

1 Answer 1

1
$\begingroup$

(1). For a Banach space $B,$ with scalars $S=\mathbb R$ or $S=\mathbb C,$ the weak$^*$ topology on $B^*$ is defined to be the weakest topology on the set $B^*$ such that for every $x\in B$ the function $$\bar x:B^*\to S$$ is continuous, where $$\bar x (f)=f(x).$$ So it is necessary and sufficient that $\bar x^{-1}U=\{f\in B^*:f(x)\in U\}$ is a weak$^*$-open set whenever $x\in B$ and $U$ is open in $S.$

(2a).The family $F$ of all $\{f\in B^*:f(x)\in U\},$ over all open $U\subset S$ and all $x\in B,$ is a sub-base for the weak$^*$ topology. The family of intersections of finite subsets of $F$ is a base for the weak$^*$ topology.

(2b).Instead of taking all open $U\subset S,$ it suffices to take just those $U$ that belong to a given base for $S.$ In particular let $\mathbb B$ be the set of non-empty open balls (or open intervals ) of $S.$ Let $F_{\mathbb B}=\{\bar x^{-1}U: x\in B \land U\in \mathbb B\}.$ Then $F_{\mathbb B}$ is a sub-base for the weak$^*$ topology.

(3). For Hilbert space the weak$^*$ topology is often called the weak topology because $l_2^*=l_2.$ For $s\in S$ and $r>0$ let $B(s,r)=\{t\in S:|t-s|<r\}.$ For $x,y\in l_2$ and $r>0$ we have $$\{z\in l_2: |<x,z>-<x,y>|<r\}=\bar x^{-1} B(<x,y>,r)\in F_{\mathbb B},$$ and it is easily seen that every member of the sub-base $F_{\mathbb B}$ is of the above form.

(4). A good introductory text on this, and more, is A Hilbert Space Problem Book by Paul Halmos. Basic material is presented as problems. Solutions ARE included.

$\endgroup$
3
  • $\begingroup$ Thanks a million times! This is very helpful indeed! What I don't understand yet is why is $F_\mathbb{B}$ only a subbase (and not also a base)? I don't seem to see it, unfortunately. $\endgroup$
    – Jeremias K
    Jun 16, 2016 at 9:36
  • $\begingroup$ It would be a base if $B$ only had one point $x,$ which won't happen.... Generally if $ z\in T= (\bar x_1^{-1}U_1)\cap$ $ (\bar x_2^{-1}U_2),$ there may fail to be $x_3, U_3$ with $z\in \bar x_3^{-1}U_3\subset T.$ So one of the necessary conditions (for being a base ) fails. $\endgroup$ Jun 16, 2016 at 15:04
  • $\begingroup$ Thanks man! This explanation makes total sense. You're my hero - and I found the book in our library, and it is reaaaally well written. $\endgroup$
    – Jeremias K
    Jun 16, 2016 at 21:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .