## To Converge or Not to Converge – Hilbert Spaces Part 4/Banach Spaces 1

Last time I introduced the notion of the weak topology on a Banach space $X$. I defined the topology in terms of net convergence, but remarked that this was equivalent to the weak topology generated by $X^*$. 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 $Y$ be a topological space and $X$ be a set. If $\mathcal{F}=\{f_\alpha:X\to Y\}$ is a collection of functions, then the weak topology generated by $\mathcal{F}$ is the smallest topology on $X$ so that each $f_\alpha\in \mathcal{F}$ is continuous.

I’ll leave it as an exercise to show that this is equivalent to the topology generated by sets of the form $f_\alpha^{-1}(U)$, where $U\subseteq Y$ is open in the topology on $Y$. Now, suppose $X$ has the weak topology generated by $\mathcal{F}$. 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 $X,Y,\mathcal{F}$ be as in the above definition and endow $X$ with the weak topology generated by $\mathcal{F}$. Then a net $\{x_\beta\}$ in $X$ converges to $x\in X$ if and only if for every $f_\alpha\in\mathcal{F}$, the net $\{f_\alpha(x_\beta)\}$ converges to $f_\alpha(x)$ in $Y$.

Proof: If $\{x_\beta\}$ converges to $x$, then by the continuity of $f_\alpha$ (for all $\alpha$), we get the result. Conversely, suppose $\{x_\beta\}$ does not converge to $x\in X$. I’ll exhibit some $f_\alpha\in\mathcal{F}$ so that $f_\alpha(x_\beta)$ does not converge to $f(x)$. Since $x_\beta$ does not converge to $x$, there is some basic open set $U$ containing $x$ so that $x_\beta$ is not eventually in $U$. But now, by the comment above, $U=\bigcap_1^n f_i^{-1}(V)$, where $V\subseteq Y$ is open (technically we should take distinct sets $V_i$, but if we intersect these sets, the resulting smaller set works just fine). Now, since this is a finite intersection, the condition that $x_\beta$ is not eventually in $U$ implies that there must exist $i$ so that $x_\beta$ is not eventually in $f_i^{-1}(V)$. But this shows that $f_i(x_\beta)$ is not eventually in $V$, a neighborhood of $f_i(x)$, hence $f_i(x_\beta)$ does not converge to $f_i(x)$.

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 $\pi_\alpha:\prod_{j\in J} X_j\to X_\alpha$. 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 $\pi_\alpha^{-1}(U)$, for $U$ open in $X_\alpha$. However, let’s see what happens if we consider this in terms of nets. For a finite product, then this says a net $(x_\alpha,y_\alpha)$ converges to $(x,y)$ if and only if $\pi_1(x_\alpha,y_\alpha)=x_\alpha\to x$ and similarly $y_\alpha\to y$.

But whatt happens if we view $X\times Y$ as the set of all functions from $\{0,1\}$ to $X\cup Y$ with $f(1)\in X$ and $f(2)\in Y$? Well, this says that $f_\alpha\to f$ if and only if $f_\alpha(1)\to f(1)$ and $f_\alpha(2)\to f(2)$, in other words it’s a type of pointwise convergence. Suppose $X$ is any topological space, recall that $C(X)$ denotes the continuous functions $f:X\to \mathbb{R}$ (or $\mathbb{C}$, either works). We can view $C(X)$ as a subspace of the set of all functions from $X$ to $\mathbb{R}$, i.e. $\mathbb{R}^X=\prod_{x\in X}\mathbb{R}$. If we give this space the product topology, then a net $f_\alpha$ in $C(X)$ converges to $f$ (which need not be in $C(X)$!) if and only if $\pi_x(f_\alpha)\to\pi_x(f)$ for each $x\in X$. But under the natural identification of functions with products of sets, this says that $f_\alpha\to f$ if and only if $f_\alpha(x)\to f(x)$ for all $x\in X$. Thus the product topology is often referred to as the topology of pointwise convergence. Clearly $C(X)$ need not be a closed subspace of $\mathbb{R}^X$, since, for example, it’s easy to construct continuous functions on $\mathbb{R}$ 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 $X^*$ is itself a Banach space, we can consider the weak topology on $X^*$, i.e. the topology generated by $X^{**}$. However, a more natural topology to consider is the so called weak* topology. Recall that we can naturally embed $X$ in $X^{**}$ by $x\mapsto \hat x$, where $\hat x(f)=f(x)$, i.e. $\hat x$ is the so-called evaluation functional. The weak* topology on $X^*$ is the topology generated by the set $\{\hat x:x\in X\}$. In terms of nets then, a net in $X^*$ converges to $f$ if and only if for all $x\in X$, we have $\hat x(f_\alpha)\to \hat x(f)$, i.e. if and only if $f_\alpha(x)\to f(x)$ for all $x\in X$. In other words, this is precisely the topology of pointwise convergence on $X^*$! In general, the weak* topology is weaker than the weak topology on $X^*$, though of course if $X$ 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 $X^*$ (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 $B(X)$, the collection of bounded (i.e. continuous) operators $T:X\to X$. 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 $A_\alpha$ converges to $A$ in the weak operator topology if and only if for all $x\in X$, we have that $A_\alpha x$ converges weakly to $Ax$, 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 $X$ be a Banach space. Then the unit ball of $X^*$ is compact in the weak* topology.

Proof: Let $B$ denote the unit ball of $X^*$. Given $x\in X$, define $D_x=\{z\in\mathbb{C}:|z|\leq ||x||\}$, i.e. to each $x\in X$ we associate a scaled copy of the closed complex unit disk. Let $D=\prod_{x\in X} D_x$ and note that $D$ is compact by Tychonoff’s theorem. Now, $D$ is the collection of all complex valued functions $\psi$ on $X$ so that $|\psi(x)|\leq||x||$ for all $x\in X$, and so $B$ is simply the subset of $D$ whose elements are linear. Moreover, as noted above, the product topology on $D$ is the topology of pointwise convergence, which implies that the topology $B$ inherits as a subspace of $D$ 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 $B$ is a compact subset of $D$, and since $D$ is compact, it suffices to prove that $B$ is closed. If $f_\alpha$ is a net in $B$ converging to $f\in D$, all the remains to show is that $f$ is linear. But for any $x,y\in X$ and $a,b\in\mathbb{C}$, we have

$f(\alpha x+\beta y)=\lim f_\alpha(ax+by)=\lim\left(af_\alpha(x)+bf_\alpha(y)\right)=af(x)+bf(y)$

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 $X$ be a Banach space. Then there exists a compact space $K$ so that $X$ is isometrically isomorphic to a closed subspace of $C(K)$ equipped with the supremum norm.

Proof: Let $K=B$ be the unit ball of $X^*$ with the weak* topology, and identify $X$ with its natural image $\hat X$ in $X^{**}$, viewed as functions on $K$ simply by restriction of domain. I claim that $\hat X\subset C(K)$ (it’s not immediately clear that $\hat x$ remains a continuous map when $X^*$ is given the weak* topology instead of the norm topology). Fix $x\in X$. If $f_\alpha$ is a net in $K$ converging to $f$. In the weak* topology, this simply says $f_\alpha\to f$ pointwise, so for any $x\in X$, we have $f_\alpha(x)\to f(x)$. But this is precisely the statement that $\hat x(f_\alpha)\to\hat x(f)$, so indeed $\hat x\in C(K)$. Clearly the map $x\mapsto \hat x$ is injective, and we have

$||\hat x||_K=\sup_{f\in B}||\hat x(f)||=||\hat x||_{X^{**}}=||x||_{X}$

by definition of the operator norm on $X^{**}$ and using the fact that $x\mapsto \hat x$ preserves the norm on $X$. This shows that $f$ is an isometry (which implies that $T^{-1}$ is also bounded, hence $T$ is actually an isometric isomorphism onto its image). The last thing to prove is that $\hat X$ is actually a closed subspace of $C(K)$. Suppose $\hat x_\alpha$ is a net in $C(K)$ converging to $\Psi\in C(K)$. Then $x_\alpha$ is a Cauchy net in $X$, so it converges, say to $x$ (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 $f\in K$, we have

$\Psi(f)=\lim \hat x_\alpha(f)=\lim f(x_\alpha)=f(\lim x_\alpha)=f(x)=\hat x(f)$

so indeed $\Psi\in\hat X$ and the proof is complete.

Actually, one can make the result a bit better. The Banach-Mazur theorem states that if $X$ is a real separable Banach space, then we can take $K$ in the above proposition to be the interval $[0,1]$ 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.