Last time I introduced the notion of the weak topology on a Banach space . I defined the topology in terms of net convergence, but remarked that this was equivalent to the weak topology generated by . The purpose of this post will be to introduce a bunch of different topologies we can put on a Banach space, its dual, and even the set of linear operators on the Banach space. Before I do this however, I want to spend a bit of time reviewing the notion of the weak topology on a space generated by a collection of functions.
Definition: Let be a topological space and be a set. If is a collection of functions, then the weak topology generated by is the smallest topology on so that each is continuous.
I’ll leave it as an exercise to show that this is equivalent to the topology generated by sets of the form , where is open in the topology on . Now, suppose has the weak topology generated by . In analysis, it’s often useful to describe a topology in terms of convergent sequences, or in a more general setting in terms of convergent nets. I’ll now establish the following fact:
Proposition: Let be as in the above definition and endow with the weak topology generated by . Then a net in converges to if and only if for every , the net converges to in .
Proof: If converges to , then by the continuity of (for all ), we get the result. Conversely, suppose does not converge to . I’ll exhibit some so that does not converge to . Since does not converge to , there is some basic open set containing so that is not eventually in . But now, by the comment above, , where is open (technically we should take distinct sets , but if we intersect these sets, the resulting smaller set works just fine). Now, since this is a finite intersection, the condition that is not eventually in implies that there must exist so that is not eventually in . But this shows that is not eventually in , a neighborhood of , hence does not converge to .
Okay, so that was a bit of technical, not-very-enlightening proof, but at least it’s there. Now for an example: the most familiar example of a weak topology is the product topology on a Cartesian product of sets, generated by the natural projection maps . Typically in point-set topology we are concerned with the first characterization of the product topology, namely it’s generated by finite intersections of cylinders , for open in . However, let’s see what happens if we consider this in terms of nets. For a finite product, then this says a net converges to if and only if and similarly .
But whatt happens if we view as the set of all functions from to with and ? Well, this says that if and only if and , in other words it’s a type of pointwise convergence. Suppose is any topological space, recall that denotes the continuous functions (or , either works). We can view as a subspace of the set of all functions from to , i.e. . If we give this space the product topology, then a net in converges to (which need not be in !) if and only if for each . But under the natural identification of functions with products of sets, this says that if and only if for all . Thus the product topology is often referred to as the topology of pointwise convergence. Clearly need not be a closed subspace of , since, for example, it’s easy to construct continuous functions on that converge pointwise to a discontinuous function.
Now let’s get back to Banach spaces. I’ve already defined the weak topology in a previous post, so now I’ll define the weak* topology on the dual of a Banach space. Of course, since is itself a Banach space, we can consider the weak topology on , i.e. the topology generated by . However, a more natural topology to consider is the so called weak* topology. Recall that we can naturally embed in by , where , i.e. is the so-called evaluation functional. The weak* topology on is the topology generated by the set . In terms of nets then, a net in converges to if and only if for all , we have , i.e. if and only if for all . In other words, this is precisely the topology of pointwise convergence on ! In general, the weak* topology is weaker than the weak topology on , though of course if is reflexive, then the two coincide. As I hinted at in my previous post, the weak* topology is important because in this topology, the unit ball of (i.e. the set of all functionals with operator norm less than or equal to 1) is a compact set. I’ll state and prove this below, but first I just want to mention topologies on operators.
There are a few natural topologies to put on , the collection of bounded (i.e. continuous) operators . As usual, we have the norm topology (using the operator norm), and the topology of pointwise convergence, which is often called the strong operator topology. We can also consider the weak operator topology, which is best defined in terms of nets. We say converges to in the weak operator topology if and only if for all , we have that converges weakly to , so the weak operator topology might best be described as the topology of weak pointwise convergence. Now, without further wait, I give you (with proof adapted from Folland)
Alaoglu’s Theorem: Let be a Banach space. Then the unit ball of is compact in the weak* topology.
Proof: Let denote the unit ball of . Given , define , i.e. to each we associate a scaled copy of the closed complex unit disk. Let and note that is compact by Tychonoff’s theorem. Now, is the collection of all complex valued functions on so that for all , and so is simply the subset of whose elements are linear. Moreover, as noted above, the product topology on is the topology of pointwise convergence, which implies that the topology inherits as a subspace of is also the topology of pointwise convergence, and as I remarked this is the same as the weak* topology. Thus it suffices to prove that is a compact subset of , and since is compact, it suffices to prove that is closed. If is a net in converging to , all the remains to show is that is linear. But for any and , we have
and the result is proven.
As you can see, once the proper identifications are made (and Tychonoff’s theorem is applied), this is actually a very clean proof. I’ll end my post, as usual, with an application of this theorem to prove a result that I think is pretty cool. The following is an exercise from Conway that shows, if we want, we can always view Banach spaces as continuous functions on some compact topological space.
Proposition: Let be a Banach space. Then there exists a compact space so that is isometrically isomorphic to a closed subspace of equipped with the supremum norm.
Proof: Let be the unit ball of with the weak* topology, and identify with its natural image in , viewed as functions on simply by restriction of domain. I claim that (it’s not immediately clear that remains a continuous map when is given the weak* topology instead of the norm topology). Fix . If is a net in converging to . In the weak* topology, this simply says pointwise, so for any , we have . But this is precisely the statement that , so indeed . Clearly the map is injective, and we have
by definition of the operator norm on and using the fact that preserves the norm on . This shows that is an isometry (which implies that is also bounded, hence is actually an isometric isomorphism onto its image). The last thing to prove is that is actually a closed subspace of . Suppose is a net in converging to . Then is a Cauchy net in , so it converges, say to (it’s a bit of an exercise, but one can show that in an first countable topological vector space, i.e. a vector space with a first countable topology so that addition and scalar multiplication are continuous, which certainly applies to normed spaces, that if every Cauchy sequence converges then every Cauchy net converges as well). But then for every , we have
so indeed and the proof is complete.
Actually, one can make the result a bit better. The Banach-Mazur theorem states that if is a real separable Banach space, then we can take in the above proposition to be the interval and use real valued functions instead of complex valued functions. That’s pretty cool! So, as you can see, the idea of weak topologies is useful and in some sense quite natural when considering normed spaces. Anyways, that’s about all I’ve got for this post, hopefully you enjoyed it.