I've been reading through chapter 3, Rudin's Functional Analysis, and an important point is the one of weak topology. From the theorems it seems to me weak topologies are somehow the result of introducing a topology of a vector space by using the dual space, however I believe there must be some special and useful property of such topologies that I might be missing. Can anyone explain why weak and weak* topologies are actually useful?

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    $\begingroup$ Strong convergence implies weak convergence. The converse is not true. Sometimes things that aren't strongly convergent are weakly convergent. This gives you a broader notion of "convergence" and lets you work with things that are normally divergent. Working with things that are normally divergent is a key thing in many areas of PDEs, physics, etc. $\endgroup$ – user608030 Dec 28 '18 at 7:41
  • $\begingroup$ Because they contain more compact sets. $\endgroup$ – Will M. Jan 4 at 21:00
  • $\begingroup$ @WillM. Why is that useful? Also I'm not sure I understand, if having weak topology means to have to coarsest topology shouldn't be the opposite? $\endgroup$ – user8469759 Jan 9 at 15:08
  • $\begingroup$ Less open sets means less open coverings which means more compacts. Compact set play the rôle, in analysis, of that of finite sets in set theory. They are the easiest sets to handle and to work with. Also, the “underlying space” is the same when you change the topologies but there are certain properties that can be obtained using topologies and not pure sets or algebra. Anyway, I am not an expert, I, like I'm guessing you, are a student of mathematics. $\endgroup$ – Will M. Jan 9 at 16:01
  • $\begingroup$ Your guess is right, I'm not an expert either. I'm interested in your insight anyway. Why does less open sets imply more compact sets? What exactly do you mean with the role of finite set in set theory. $\endgroup$ – user8469759 Jan 9 at 16:05

There's a couple of ways to motivate these concepts. I'll provide some general reasons why the weak* topology is useful, but there are so many other reasons I don't have the time to mention.

Let $X$ be a topological vector space. Then we can define the topological dual space, $$ X^* = \{ f : X \rightarrow \mathbb R\, \text{ is linear and continuous} \}. $$ This makes sense as a set, and moreover it has a natural vector space structure by defining addition and scalar multiplication pointwise. But is it itself a topological vector space? Do we have a natural topology to endow it with?

One natural solution is to endow it with the weak* topology; whatever topology we endow $X^*$ with, we surely expect the evaluation functional $\hat x : X^* \rightarrow \mathbb R$ sending $f \mapsto f(x)$ is continuous for all $x \in X.$ Since the weak* topology is precisely the weakest topology for which this holds, it seems like a good starting point.

Also by the universal property, if $\tau$ is another topology on $X^*$ such that evaluations are continuous (e.g. $X$ is Banach and $\tau$ is the strong topology on the dual), then the identity map $(X,\tau) \rightarrow (X, w^*)$ is continuous; so any such topology $\tau$ also contains the weak* topology as a subset. So if we agree that evaluation functionals ought to be continuous, this is really the weakest topology we can put on $X^*$ that's reasonable.

Another interpretation of weak topologies is as the topology of pointwise convergence. It is known that for a general topological space $X,$ we have the space of maps $X \rightarrow \mathbb R,$ identified as the product ${\mathbb R}^X$ naturally admits the product topology. Moreover in this topology, $f_n \rightarrow f$ if and only if $f_n(x) \rightarrow f(x)$ for all $x \in X.$

The weak* topology is similar, but we only require this to hold amount continuous linear functionals. Going back to the case when $X$ is a TVS, the product space ${\mathbb R}^X$ is huge and difficult to work with, but $X^* \subset {\mathbb R}^X$ is a bit more manageable. But we do get a natural topology on $X^*$; we simply equip it with the subspace topology inherited from ${\mathbb R}^X.$ It turns out that this coincides with the weak* topology (essentially because the product topology itself is the weakest topology such that the projections are continuous), and so $X^*$ is also the topology of pointwise convergence (in the sense described above).

Admittedly this may convince you that this makes the weak* topology a natural choice to endow $X^*$ with, but it's not obvious why it's useful. This interpretation as a product topology has the benefit however, that you get nice compactness properties.

Tychonoff's theorem says that the product space $A^X$ is compact if $A$ is. Of course we can't simply take $A = \mathbb R$ as it is non-compact, but if we can find a subset $Y \subset X^*$ such that $Y \subset [a,b]^X$ and is closed with respect to suitable topologies, we get $Y$ is (appropriately) compact. It turns out that because the weak topology is suitably compatible with the product topology in $\mathbb R^X,$ we get nice compactness properties. In particular if $X$ is Banach, weak* closed and bounded subsets of $X^*$ are weak* compact.

Going back to a more concrete setting, in PDEs for example a common approach to finding solutions is to start with some sort of approximation, and extract some kind of limit. For this compactness is extremely useful, because you know any sequence will have a limit point. From there a common approach is to try and show said limit solves the PDE you were considering.

Other motivations I haven't included include their use in distribution theory (and more applications to PDEs), the relation between weak topologies and reflexivity, and probably a lot more that I can't think of right now.

  • $\begingroup$ Very good answer. Well thought out, clear, helpful. +1 $\endgroup$ – user608030 Dec 28 '18 at 9:47

One way of understanding the weak-$\ast$ topologies is by trying to figure out what the operation taking closures looks like. For instance, if you take $L^1(\mathbb{T})^\ast =L^\infty(\mathbb T)$ and consider $P \subset L^\infty(\mathbb T)$ the set of all trigonometric polynomials, then (by Stone-Weisstrass) its norm-closure will be $C(\mathbb{T})$, the continuous functions over $\mathbb T$, but its weak-$\ast$ closure gives the whole $L^\infty(\mathbb{T})$.

In general, if $f_\alpha$ are functions on $L^\infty(\mathbb{T})$ they converge in the weak-$\ast$ topology to $f$ whenever $f_\alpha \to f$ almost everywhere and the supremum $\sup_{\alpha} |f_\alpha|$ is essentially bounded. This is a routine application of Lebesgue's dominated convergence theorems (and almost a rephrasing of it).

Alternatively, a way of seeing the importance of weak (and weak-$\ast$) topologies is by analyzing important theorems that use those notions. I can mention:

  1. Von Neumann's bicommutant theorem. If $A \subset B(H)$ is a subalgebra of the bounded operators in a Hilbert space, then its bicommutant $A''$ is another algebra containing $A$, where the commutant of a set is given by $$ S' = \{ T \in B(H) : [T,R] = 0, \forall R \in S \}. $$ Von Neumann proved that the purely algebraic notion of $A''$ coincides with the weak-$\ast$ closure of $A$ (equivalently with its weak operator or strong operator closures).

  2. Central Limit Theorem. The central limit theorems asserts that for independent, $L^2$, identically distributed random variables their "normalized" sum $$ S_N = \frac1{\sqrt{N}} \bigg( \sum_{k = 1}^N X_k - N \, p \bigg), $$ where $p = \mathbb{E}[X_1]$. Then, the distributions of the $S_N$ are measures on the real line that converge in the weak-$\ast$ topology to a Gaussian (this is calles vague convergence in probability https://en.wikipedia.org/wiki/Vague_topology)


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