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I'm reading the definition of weak topology in Banach Algebra Techniques in Operator Theory by Douglas:

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According to an article about the product topology in Wikipedia, the product topology is also called topology of pointwise convergence. I'm confused with the underscored sentence. According to the answer by @Brian M. Scott to the question What is the Topology of point-wise convergence?, I was expecting

$ \lim_{\alpha\in A}f_\alpha=f $ if and only if $\lim_{\alpha\in A}f_\alpha(x)=f(x)$ for every $x\in X$ where $(f_\alpha)_{\alpha\in A}$ is a net in $\mathscr{F}$.

Could anyone clarify what is going on?

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  • $\begingroup$ The one you are referring to is the weak*-topology, while the other one is indeed the topology of point-wise convergence (weak topology) $\endgroup$
    – Kolmin
    Commented Sep 28, 2015 at 16:57
  • $\begingroup$ Related Wikipedia article: Initial topology. $\endgroup$ Commented Nov 19, 2015 at 11:37
  • $\begingroup$ Not sure if you already found the answer you were looking for by yourself or not, but self-studying (I hope the "self" does not scary you) things close to these, I came back to your question again (recently I also edited my answer a bit). Actually, today (better late than never) I finally saw what your problem is, and why my answer did not really answer your question. So, here there is the answer in very few words: it all depends on which space we put the topology of pointwise convergece (hence, my new edit – most probably you were expecting that topology on $X$ rather than $X^*$)! $\endgroup$
    – Kolmin
    Commented Nov 24, 2015 at 19:48

2 Answers 2

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In general we have the following definition of initial (or weak) topology: given a set $X \neq \varnothing$, a family of topological spaces $\{ (Y_i, \tau_i ) \}_{i \in I}$, and for every $i \in I$ a function $f_i : X \to Y_i$, the initial (or weak) topology is the weakest topology on $X$ that makes all the functions $f_i$ continuous.

Take the set $\mathcal{F} := \{ f \ | \ f \in \mathbb{R}^X \}$, and notice that the weak topology on $\mathcal{F}$, denoted by $\sigma (\mathcal{F}, X)$, is identical to the relative topology on $\mathcal{F}$ as subset of $\mathbb{R}^X$ endowed with the product topology. Moreover, notice that, given our definition of $\mathcal{F}$, the set $X$ can be seen as set of real valued functions on $\mathcal{F}$, where each $x \in X$ can be seen as an evaluation functional $e_x \in \mathbb{R}^\mathcal{F}$ such that $e_x (f) = f(x)$, i.e. $X = \{ \ e_x \ | \ e_x \in \mathbb{R}^\mathcal{F}, \ e_x (f) = f(x) \ \}$

Now, let $X$ a linear space with $x \in X$. Thus, let $X’$ be the algebraic dual of $X$, i.e. the vector space of all linear functionals on $X$, and let $X^*$ be the topological dual of $X$, i.e. $$X^* := \big\{ x^* \ \big| \ x^* \in \mathbb{R}^X, \ x^* \text{ continuous} \big\},$$ or – in plain english – the vector space of all continuous linear functionals on $X$, with $x^* \in X^*$.

Notationally wise, notice that, for the following, (for example) Rudin writes $\langle x , x^* \rangle \equiv x^*(x)$, thus (according to the notation employed above) we should have that $\langle x^* , x \rangle \equiv e_x (x^*)$.

Hence, $X$ can be seen (like in the second paragraph) as a vector subspace of $\mathbb{R}^{X^*}$, in which case $\sigma (X, X^*)$ is the weak topology on $X$ (denoted by $w$), defined as

$$ x_\alpha \overset{w}{\to} x \in X \Longleftrightarrow \forall x^* \in X^* , \langle x_a, x^* \rangle \to \langle x , x^* \rangle \in \mathbb{R}, $$

that corresponds to the topology of pointwise convergence on $X^*$.

In the same vein, $X^*$ can be seen as a vector subspace of $\mathbb{R}^X$, in which case $\sigma (X^*, X)$ is the weak* topology on $X^*$ (denoted by $w^*$), defined as $$ x^{*}_\alpha \overset{w^*}{\to} x^* \in X^* \Longleftrightarrow \forall x \in X , \langle x^{*}_a, x \rangle \to \langle x^* , x \rangle \in \mathbb{R}. $$

Notice that in both cases we are just declining in different ways the original definition of initial topology given at the beginning. What we are changing is simply the "reference" space that makes all functions continuous:

  • In the case of the weak topology, we have that $(X, \sigma(X, X^*))$ is the topological space that makes all the $x^* \in X^*$ continuous;
  • In the case of the weak* topology, we have that $(X^*, \sigma(X^*, X))$ is the topological space that makes all the evaluation functionals $e_x \in \mathbb{R}^{X^*}$ continuous.

In light of what above, the one you were expecting is the weak*-topology, and not the weak-topology. Interestingly, and here we get to the all problem of the topology of pointwise convergence, we can endow different spaces with that topology. In my answer (and in your book), the space endowed with that topology is $X^*$, while – most probably – your problem arose because instead you had in mind $X$ endowed with that topology.

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  • $\begingroup$ Could you explain what you mean by "topology of pointwise convergence on $X^*$"? What I understand by the topology of pointwise convergence on a certain space of functions from $X$ to $Y$ is that $f_n\rightarrow f$ if and only if $f_n$ converges to $f$ pointwise on $X$. In which sense then, does the weak topology on $X$ correspond to the topology of pointwise convergence on $X^*$ (which is a collection of functions from $X$ to $\mathbb{C}$)? I would have called it the topology of pointwise convergence on $X\subset X^{**}$. $\endgroup$
    – Not Euler
    Commented Sep 4, 2020 at 15:59
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I do not know, what Douglas wanted to point out with this sentence, as pointwise convergence is usally used in the way, Brian explains in his answer, but you can, if you want, adjust things:

Each $x \in X$ deinfes a function $\hat x$ on $\def\F{\mathscr F}\F$ by $\hat x\color \mathscr F \ni f \mapsto f(x)$. Now - by definition of $\mathscr T$ $x_\alpha \to x$ in $\mathscr T$ iff $\hat x_\alpha \to \hat x$. Thus $\mathscr T$ is (induced by) the topology of pointwise convergence on the subset $\{\hat x \mid x \in X\}$ of $Y^{\mathscr F}$.

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  • $\begingroup$ What does $\hat xF$ mean? $\endgroup$
    – user9464
    Commented Sep 29, 2015 at 17:35

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