# Weak topologies and weak convergence - Looking for feedbacks

I am currently trying to get exactly what the weak and the weak* topologies are, in particular in connection to the concept of weak convergence in measure, however I am not completely sure on what I have got so far. So here, there is a summary of what I understood. I disseminated the text with some numbered questions, to point out what I am not sure about this summary.
Any feedback will be most welcome!

Edit: The answers I received have been quite useful and stimulating (e.g. I ended up looking at the Kelley-Namioka text), but I would love to get something more specific. Hence, I am adding a bounty. Thus, I slightly changed the text, because I actually changed my "opinion" regarding some issues.

Edit 2: I know that editing put a question again on the top, however I did not edit with this purpose. I actually noticed that I did not get any new answer, and, considering I actually changed my view on some issues (I am studying the topic intensively these days), I changed the text improving (IMO) the content, without affecting any new answer. Moreover, I add some straightforward questions that arose out of these hours of study, in order to make clear what I am looking for, and I put the rest of the text as block, for those who want to see what I ended up with.

Questions:

1. Isn’t it problematic to use the inner product notation when we deal with dual pairs and weak topologices, considering that we work in this case mainly with Banach spaces, and not all norm norms are associated to an inner product?

2. Does it ever enter in the discussion on weak topologies the concept of algebraic dual?

3. If by definition $X^*$ is the topological dual of $X$, hence it is the set of all continuous functionals on $X$, isn’t by definition that each $x^* \in X^*$ is actually continuous? Why do we need the weak topology on $X$?
[The same problem does not really apply to the weak* topology on $X^*$ that makes each evaluation functional $e_x: X \to \mathbb{R}$ continuous, because those evaluation functionals are not continuous by definition.]

4. I see in which way the fact that the bidual $X^{**}$ can be way bigger than $X$ when we deal with weak topologies can be relevant, namely if $X = X^{**}$ then weak and weak* topologies coincide. However, I don’t see how can be that $X \subset X^{**}$.

5. Related to the previous problem is the fact that when we move from dual pairs and weak topologies to weak convergence in measure, I don’t see why the weak convergence in measure coincides with endowing the reference metric space $(X,d)$ with the weak* topology, and not simply the weak topology. Thus the question should be, how do we decide what is the main space, what is the dual one, and which one is the one endowed with a topology, in order to come up with the concept of weak convergence? Is it through the Riesz Representation theorem?
(...and now I should end up looking exactly what that representation thoerem does...)

6. I read that what is called weak convergence should be called weak* convergence, because it is based on the weak* topology. Why, actually?
[True, this question can be considered a repetition of the previous]

In the following, there are my thoughts on weak topologies and weak convergence. I hope they are correct or interesting. I put them as a spoiler in order not to scare who would like to answer.

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.
Notice that, if $\mathcal{F} := \{ f | f: X \to \mathbb{R} \}$, then $X$ can be seen as set of real valued functions on $\mathcal{F}$, where each $x \in X$ is an evaluation functional $e_x : X^* \to \mathbb{R}$ with $e_x (f) = f(x)$. Thus, 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.

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. $X^* := \{ x^* | x^*: X \to \mathbb{R}, x^* \text{ continuous} \}$ is, or – in plain english – the vector space of all continuous linear functionals on $X$, with $x^* \in X^*$.

Notice that, for the following, for example Rudin points out that $\langle x , x^* \rangle \equiv x^*(x)$, thus we should have that $\langle x^* , x \rangle \equiv e_x (x^*)$. [a. Right?]

Thus, $X$ 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 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 $$x^{*}_\alpha \overset{w^*}{\to} x^* \in X \Longleftrightarrow \forall x \in X , \langle x^{*}_a, x \rangle \to \langle x^* , x \rangle.$$

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. Indeed, 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 $e_x$ from $X$ to $\mathbb{R}$ continuous.

I know (after quite some bookreading) that what can be found in the following text as the old point of view is the correct way of looking at the problem, however I keep on feel that it is the new point of view the correct way. In particular, I find what I wrote in the old point of view misleading. Indeed, I implicitly state that we have as dual pair $\langle X, C(X) \rangle$ and every $\mu$ is on $X$, but this does not make sense. Indeed, we should have that the dual pair is $\langle \Delta (X), C(X) \rangle$ and every $\mu$ is on $\Delta (X)$. But then, if the dual pair is really $\langle \Delta (X), C(X) \rangle$, according to the definition of weak and weak* topology I wrote down, here we are actually dealing with the weak topology! Hence, the dual pair should be $\langle C(X), \Delta(X) \rangle$, but why?

Old point of view:

Thus, if we take all these definitions and we use them to get what the weak convergence in measures is, we actually have that weak convergence in measure is the same as convergence in the weak* topology. [b. Right?]

Indeed, given a metric space $(X,d)$ a sequence of measures $(\mu_n)$ converges weakly to $\mu$ in $X$ if and only if for every $\phi \in C(X)$ (where $C(X)$ is the space of all continuous functions on $X$), $$\lim_{n \to \infty} \int_X \phi d\mu_n = \int_X \phi d\mu,$$ and here the $\mu_n$ should make the same job of our $x^{*}_n$ in our definition of the weak* topology. [c. Right?].

New point of view:

Thus, if we take all these definitions and we use them to get what the weak convergence in measures is, we actually have that weak convergence in measure is the same as convergence in the weak topology. [d. Right?]

Indeed, given a metric space $(X,d)$ a sequence of measures $(\mu_n)$ converges weakly to $\mu$ in $\Delta (X)$, where $\Delta (X)$ denotes the set of all probability measures on $X$, if and only if for every $\phi \in C(X)$ (where $C(X)$ is the space of all continuous functions on $X$), $$\lim_{n \to \infty} \int_X \phi d\mu_n = \int_X \phi d\mu.$$ Thus, in this setting the dual pair of spaces is $\langle \Delta (X) , C (X) \rangle$ and indeed the definition of weak convergence in $\Delta (X)$ ends up to be equivalent to endow $\Delta (X)$ with the weak topology (and not the weak* topology!). [e. Right?].

• Do you know initial topologies? Do you know locally convex spaces? Do you know dual pairs? (All of these come together in the weak topologies.) – C-Star-W-Star Jan 19 '15 at 0:19
• Yes, in fact another name for the weak topology is really initial topology generated by the chosen duals. Wiki gives this definition for example. This way one guarantees their continuity. You're welcome. ;) – C-Star-W-Star Jan 19 '15 at 10:25
• Next the questions on the set-up: One has to make clear two things namely which space to consider and which topology to endow it with. For example the original Banach space endowed with the norm-topology. Another one would be to take the initial topology induced by all linear functionals. But that again does something on the original Banach space and not the space of linear functionals. So what you study is the structure of the original Banach space and you do that with for example the help of functionals. – C-Star-W-Star Feb 5 '15 at 15:51
• A property that original Banach space then possesses with that new topology is that these linear functionals become continuous w.r.t. that new topology. As another example consider now the initial topology on the original Banach space induced by the zero functional only. Then the Banach space is depicted with the trivial topology and suddenly the earlier continuousy functionals (w.r.t. the norm-topolog) become now all discontinuous. – C-Star-W-Star Feb 5 '15 at 15:55
• Now the problem about to consider the functionals as maps on the elements of the Banach space or vice versa: It doesn't matter wether $\langle x',x\rangle:=x'(x)$ or $\langle x',x\rangle:=\varepsilon_x(x')$. They both have the same effect and therefore give rise to the same topology on the dual $X'$. Sometimes one can't even decide which one was first as for example is it $f(\mu)=\int f\mathrm{d}\mu$ or is it $\mu(f)=\int f\mathrm{d}\mu$. That is a hint why one doesn't distinguish so strictly and simply writes $\langle x,u\rangle$. – C-Star-W-Star Feb 5 '15 at 16:09

I would say "Yes" to all your questions.

The so-called weak* topology is an example of a weak topology.

Weak convergence of measures can be considered a weak* topology.

To clarify a new example of convergence of this kind, you should master that $\sigma (\mathcal{F}, X)$ notation. (I think it goes back to Mackey, and was popularized by Bourbaki.)

• First of all, thanks a lot! Two questions. First one, where can I find some references regarding the $\sigma (\mathcal{F}, X)$ notation? – Kolmin Jan 18 '15 at 19:00
• The second question concerns weak convergence of measures as a form of a weak* topology. Why don't we use the expression "weak* convergence of measure"? Is there something like weak* convergence of measure? If it does exists, how does it actually look like? [This all sound quite misleading...] – Kolmin Jan 18 '15 at 19:02
• Kelley-Namioka, Topological Vector Spaces uses that $\sigma (\mathcal{F}, X)$ idea. But maybe tries to change the notation to $w(\mathcal{F}, X)$ which never caught on. – GEdgar Jan 18 '15 at 21:40
• Bratteli & Robinson apply them extensively but I'm not sure wether they actually define them. – C-Star-W-Star Jan 19 '15 at 0:23

Dual Pair

Two vector spaces with a bilinear form satisfying: $$\langle\cdot,\cdot\rangle:X\times Y\to\mathbb{C}:\quad X^\perp=(0),\,Y_\perp=(0)$$ (Informally, this way one guarantees that it separates points.)

Weak Topology

The initial topology generated by the dual: $$\sigma(X;Y):=\tau\left(\bigcup_{y\in Y}\varepsilon_y^{-1}(\mathcal{T}_\mathbb{C})\right)\quad(\varepsilon_y:=\langle\cdot,y\rangle)$$ (This way one guatantees that they become continuous.)

Example

Consider a Banach space with its continuous dual: $$\langle\cdot,\cdot\rangle:E\times E'\to\mathbb{C}:\quad\langle x,f\rangle:=f(x)$$ By the initial topology weak convergence boils down to: $$x_n\rightharpoonup x\iff f(x_n)\to f(x)\quad(f\in E')$$ (That is in terms of product topology convergence by components.)

Remark

The difference between the weak and weak* topology lie in the chosen dual: $$\langle x,f\rangle:=f(x):\quad\sigma(E';E)$$ $$\langle F,f\rangle:=F(f):\quad\sigma(E';E'')$$ (The comparison becomes solid on embedding the Banach space into its double dual.)

• Editing my question I was reading again your answer, and I have the feeling that partly answered my doubts. In particular, without your comment on the question regarding the definition of initial topology I would have not see (...my shame!) how the standard definition of initial topology applies to dual pairs with weak and weak* topology. However, there is something that is not answered explicitly here, most probably because it was only fuzzily – and implicitly – present in my original question... – Kolmin Jan 22 '15 at 16:24
• ...In particular, having the numbered questions at the beginning of the text as a reference, question 5 is the less addressed, which is at the same time the most connected to your insights regarding dual pairs (something I completely – and stupidly – overlooked at the beginning), when applied to weak convergence in measure. I am just pointing out because as it stands now, the answer is – even if a bit cryptic for my level... :) – the closest to an explanation of what is going on here, and it would be amazing if it would be expanded. But, thanks anyway for your insights! :) – Kolmin Jan 22 '15 at 16:27
• @Kolmin: I'm sorry I couldn't answer so far; I'm still very busy these days. Moreover there were some really interesting questions among; let me answer them another day if you're still interested. ;) (I'll mark it is favorite question anyway.) – C-Star-W-Star Jan 27 '15 at 18:30
• @Kolmin: In fact I want to leave this answer as the basic one and rather adress your specific problems in another answer; again still if interested. ;) – C-Star-W-Star Jan 27 '15 at 18:36
• Regarding the bounty, IMO your answer (as I mentioned in a previous comment) was a very interesting starting point, and your comments were very important in order to shape my understanding of the topic (that is still quite rough. Of course it would be great if you would expand it. In the meantime, the bounty is yours for the effort and the hints you already gave me. :) – Kolmin Jan 27 '15 at 18:49