## Moving Around a Sphere – Hilbert Spaces Part 3

Well, it’s been a long time since I’ve written a post, but I have a bit of time tonight and I really don’t feel like doing homework tonight, so let’s talk about some mathematics! Last time I wrote, I promised some discussion of the unit ball in infinite dimensional Hilbert spaces. Many of the definitions and even some of the facts I’ll state carry over to arbitrary Banach spaces, but some of the proofs are a bit easier with the inner product structure, so I’ll stick to the nicer Hilbert space case. Here we go:

Last time I remarked that if $H$ is a finite dimensional Hilbert space, then the unit ball of $H$ is compact in the norm topology. The proof of this is a bit messy, so I won’t give it, but I’ll sketch the basic idea, which is straightforward: in $\mathbb{R}^n$ this is obviously true, since the unit ball is closed and bounded. Given any finite dimensional Hilbert space, it is isometrically isomorphic to $\mathbb{R}^{\dim{H}}$. Under this isomorphism, the unit ball of $\mathbb{R}^{\dim{H}}$ maps onto the unit ball of $H$, and so the unit ball of $H$ is the continuous image of a compact set, so is itself compact. The messy part is writing down the correct isomorphism, but that’s left to the reader. It’s a fun exercise to use this fact to prove that any two norms on a finite dimensional Hilbert space are equivalent: i.e. if $||\cdot||_1,||\cdot||_2$ are two norms on $H$, there exist non-zero, positive constants $A,B$ so that $A||x||_1\leq ||x||_2\leq B||x_1||$ for all $x\in H$, and as an easy corollary to this we see that any two norms generate the same topology on $H$.

Now let’s suppose $H$ is infinite dimensional. I’ll leave $H$ is an arbitrary infinite dimensional case, but if you want a concrete example, just take $H=l^2(\mathbb{N})$, the set of square summable sequences of complex (or real) numbers with the natural inner product. Let $\{e_\alpha\}_{\alpha\in J}$ be an orthonormal basis for $H$. For $l^2$, the easiest example is to take $e_i=(0,0,\dots,1,0,\dots)$, where the $1$ occurs in the $i$‘th slot. For arbitrary $H$, let $\{e_i\}_1^\infty$ be a countable subset of the orthonormal basis. Now, for $i\neq j$, we have

$||e_i-e_j||^2=\langle e_i-e_j,e_i-e_j\rangle=||e_i||^2+||e_j||^2=2$

so the set $\{e_i\}_1^\infty$ is a $\sqrt{2}$-separated sequence sitting in the unit ball, and so it has no convergent sub-sequence. For a metric spaces, sequential compactness and compactness are equivalent, so this shows the unit ball is not compact.

The fact that the unit ball of a Hilbert space is not compact is unfortunate. We like compact spaces, they tend to make life easier! One thing we might try to do is define a topology on $H$ where the unit ball is compact. In fact, we can do something similar to this for any Banach space. Actually, we have to work in the dual space for an arbitrary Banach space, but since Hilbert spaces are naturally (anti)isomorphic to their dual via the Riesz representation theorem, we can think of it as a topology on $H$. For an arbitrary Banach space, this topology will be called the weak* topology (read “weak star”) on $X^*$. In this post, I’ll define the so called weak topology on $X$. In the Hilbert space case, these two will coincide, but for an arbitrary Banach space they are different topologies on two distinct spaces. Finally, I’ll prove a cool fact about the weak topology on a Hilbert space. My next post will contain the definition of the weak* topology and Alaoglu’s theorem, which states that the unit ball in $X^*$ is weak* compact.

Definition: Let $X$ be a Banach space. The weak topology on $X$ is defined as follows: a net $\{x_\alpha\}$ converges to $x$ if and only if for all $f\in X^*$, the net $f(x_\alpha)$ converges to $f(x)$.

Remark: The weak topology depends on the norm on $X$, it’s not intrinsic to the set $X$. We use the norm to define the continuous linear functionals that make up $X^*$, and in turn use these to define the weak topology on $X$. Also, note that we could have defined the weak topology to be the weak topology generated by $X^*$, i.e. the weakest (smallest, coarsest) topology on $X$ so that every $f\in X^*$ is continuous. In particular, since the norm topology on $X$ satisfies this condition, it follows immediately that the weak topology on $X$ is in fact weaker than the norm topology: i.e. $||x_n-x||\to 0$ implies $x_n$ converges to $x$ in the weak topology (in this case we often we say $x_n$ converges to $x$ weakly).

Now, moving back to the world of Hilbert spaces, what does it mean for a sequence (technically a net, but we’ll just think sequences for the sake of simplicity) to converge weakly in $H$? Well, recall that by the Riesz Representation theorem, any bounded linear functional $f\in H^*$ is given by $f(x)=\langle x,y_f\rangle$ for some fixed vector $y_f\in H$ (and conversely, the map $x\mapsto\langle x,y\rangle$ defines a bounded linear functional on $H$). Translating this to weak convergence, we have $x_n\to x$ weakly in $H$ if and only if for all $y\in H$, we have $\langle x_n,y\rangle\to \langle x,y\rangle$. Thus for Hilbert spaces, we can think of weak convergence as “convergence of inner products”. One might ask if this is actually a genuinely weaker topology than the norm topology on $H$. That is, can we find a sequence $x_n$ that converges weakly in $H$, but does not converge in the norm topology? I’ll end my post by proving the following cool

Proposition: Let $\{e_i\}_1^\infty$ be an orthonormal set in $H$. Then $\{e_i\}$ converges weakly to $0$. Moreover, $\{e_n\}_1^\infty$ does not converge in the norm topology on $H$.

Proof: The statement about norm convergence was shown above, so we need only prove the statement about weak convergence. To this end, let $y\in H$. By Bessel’s inequality (or, if you want, extend $\{e_i\}$ to an ONB and apply Parseval’s identity), we have

$\sum_{i=1}^\infty |\langle y,e_i\rangle|^2\leq ||y||<\infty$

and so in particular the infinite sum converges. But this is just a sum of real number, so we must have that the terms of the sequence go to zero as $i\to \infty$. But this is precisely the statement that $\lim_{i\to\infty} |\langle y,e_i\rangle|^2=0$. It is left to the reader to verify that this implies $\langle e_i,y\rangle\to 0$ as $i\to\infty$, i.e. $e_i\to 0$ weakly in $H$.