0
$\begingroup$

I was studying functional analysis a few months ago and asked a similar question (On the definition of weak and weak-* topologies). I thought I had understood weak* topologies rather well, until I started thinking about them some more today. I realized I am unable to even "visualize" the open sets.

Suppose $X$ is a Banach space over a field $\mathbb{F}$ and let $X^*$ be its dual. The weak topology is by definition the weakest topology making all $f \in X^*$ continuous. From this definition it is easy to visualize the open sets in the weak topology on $X$; simply take any open $U \subset \mathbb{F}$ then we define $f^{-1}(U)$ to be open in the weak topology for any $f \in X^*$.

I am having trouble carrying out a similar picture for the weak* topology. For example, let $X^{**}$ denote the double dual of $X$ and again let $U \subset \mathbb{F}$ be open. Consider all linear functionals of the form $\Lambda_x(f) = f(x)$. Then what do the open sets in the weak* topology on $X^*$ look like? My current understanding is that they look like $f^{-1}(U) = \Lambda_x^{-1}(U)$ for any open set $U \subset \mathbb{F}$. Is this correct?

$\endgroup$

2 Answers 2

2
$\begingroup$

@cbbam this is too long for a comment the elements of the dual $X^{*}$ are already "there",i.e $X^{*}$ consists of all continuous linear forms $f:X\to \mathbb{K}$,this is already a normed space whose elements are forms,picking (a fixed) $x$, the map (evaluation at $x$) $ev_{x}:X^{*}\to \mathbb{K};f\to f(x)$ is continous,i.e $ev_{x}\in X^{**}$ (all this is proved befor introducing any weak or weak* topology),in fact the map $i:X\to X^{**},x\to ev_{x}$ is the natural embedding of $X$ into $X^{**}$ and a space is called reflexive if this embedding is surjective,So u can see $x$ as a vector on which elements of $X^{*}$ acts or as linear form acting on elements of $X^{*}$.Let's return now to the weak* topology on $X^{*}$,this is by definition the smallest (coresest) topology on $X^{*}$ making all $ev_{x}$ continous,So we want all our $ev_{x}$ to be continous but with the least amount of open subsets,this means that given an open subset $U\subset \mathbb{K}$,$ev_{x}^{-1}(U)$ must be open for every $x\in X$,Now $ev_{x}^{U}=\{f\in X^{*} \backslash f(x)\in U\}$, so an open subset of $X^{*}$ consists of all (continous since this is the definition of $X^{*}$) linear form that send ( a fixed) $x$ inside $U$.In general if you have topological space $X, Y$ and u pick a family of $f_{i}\in C(X,Y)$,then u can define a new topology on $X$ a the smallest topology on $x$ making all the $f_{i}$ continuous,how does open subsets of this new toplogy looks like?well let $U_{k},k=1,..n$ be open subsets of $Y$,a basic open subset has the form $\cap_{k=1}^{n}f_{i_{k}}^{-1}(U_{k})$,when u take $Y=\mathbb{K}$ and the family of continuous maps to be $X^{**}$, u get the weak topology on $X^{*}$,which as I mentioned before coarser that the weak* topology(remark here we require all linear forms on $X^{*}$ not only $ev_{x}$,that why we get "more" open subsets in general). u can look for Initial topology if you are intersted, hope this helps.

$\endgroup$
4
  • $\begingroup$ Thank you this is very helpful! I have one follow up question, so we require that $\Lambda_x(f) = f(x)$ be continuous. Does this require that $f$ also be continuous for all $y \neq x$? $\endgroup$
    – CBBAM
    Commented Nov 4, 2022 at 17:42
  • 1
    $\begingroup$ @CBBAM yes $f$ is an element of $X^{*}$, so (by definition) it is continous at every point of $X$.remember that we trying to put a topology on $X^{*}$ (which already ahs its elements, we don't change their nature).in conclusion $X^{*}$ has three topologies(of course its elements are the same ,continuous linear forms $f:X\to \mathbb{K}$),one is induced by the operator norm,$\lvert \lvert f\rvert \rvert$ $\endgroup$ Commented Nov 4, 2022 at 19:33
  • 1
    $\begingroup$ @CBBAM The second one is the topology making every evaluation continuous,where an open subset is given by $ev_{x}^{-1}(U)$,that is all linear forms $f$ that send a fixed vector $x$ inside an open subset $U$ (and is comming from $X$)and the one making every linear form on $X^{*}$,that is every element of $X^{**}$ continous ,the first one is the finer (with the most amount of open subsets ),while the second one is the coarser(with the least amount of open subsets) $\endgroup$ Commented Nov 4, 2022 at 19:34
  • $\begingroup$ I got it now, thank you for all your help! $\endgroup$
    – CBBAM
    Commented Nov 4, 2022 at 19:46
1
$\begingroup$

it is an open subset of $X^{*}$, so it need to consisit of linear forms, more precisely it consists of $\lambda_{x} ^{-1}(U)=\{f\in X^{*} \backslash f(x) \in U \}$ ($x$ is fixed and the open subset consists of forms that send $x$ inside $U$),reamark that if u restrict to the evaluations $$\lambda_{x}:X^{*}\to \mathbb{F},\lambda_{x}(f)=f(x)$$ then u get the weak* topology on $X^{*}$,which is coarser than the weak topology comming from the double dual,unless $X$ is reflexive.

$\endgroup$
1
  • $\begingroup$ I see, so here we require $f$ to be continuous for only some $x$ (namely $f(x)$), which in turn makes $\Lambda_x$ continuous for every $f$? $\endgroup$
    – CBBAM
    Commented Nov 4, 2022 at 16:48

You must log in to answer this question.

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