Hilbert Spaces – Part 2

Recall that in the finite dimensional world, two vector spaces are isometrically isomorphic if and only if they have the same dimension. In the last post, I mentioned that, by using the appropriate definition of a basis and dimension, one can make the same statement about infinite dimensional (complete) inner product spaces, commonly known as Hilbert spaces. This post will be devoted to proving this fact.

First, let’s recall the definition of an orthonormal basis from last time. Recall that a set of vectors $\{e_\alpha\}_{\alpha\in A}$ is orthonormal if $\langle e_\alpha,e_\beta\rangle=\delta_{\alpha,\beta}$. The set is called an orthonormal basis if, in addition, it satisfies one of the following equivalent properties:

1. For all $x\in H$, we have $x=\sum_{\alpha\in A} \langle x,e_\alpha\rangle e_\alpha$, where the sum on the right has only countably many non-zero terms and converges in the norm topology no matter how the terms are ordered.
2. If $\langle x,e_\alpha\rangle=0$ for all $\alpha\in A$, then $x=0$.
3. Given any $x\in H$, we have $||x||^2=\sum_{\alpha\in A} |\langle x,e_\alpha\rangle|^2$.

In the last post on Hilbert spaces, I proved that every Hilbert space has an orthonormal basis by considering a maximal orthonormal set in $H$ (and so naturally had to invoke Zorn’s lemma) and showed that such a set must be a basis. The following proof is adapted from J.B. Conway’s A Course in Functional Analysis.

Proposition/Definition If $E$ and $F$ are orthonormal bases for the same Hilbert space $H$, then $|E|=|F|$. The cardinal number $|E|$ is called the dimension of the Hilbert space $H$ and is denoted by $\dim H$.
Proof: The case where either $|E|$ or $|F|$ is trivial since any orthonormal spanning set in a finite dimensional space is a (Hamel) basis and so $|E|$ is the vector space dimension of $H$. Thus we consider only the case where $|E|$ and $|F|$ are infinite. For fixed $e\in E$, we define $F_e=\{f\in F:\langle e,f\rangle\neq0\}$. By definition of an ONB, $F_e$ is countable. Given any $f\in F$, there must exist some $e\in E$ with $\langle e,f\rangle\neq 0$, otherwise $f=\sum_{e\in E}\langle e,f\rangle e=0$. But $||0||=0\neq1=||f||$ so such $e$ must exist. Thus $F=\bigcup_{e\in E} F_e$. But then $|F|\leq |E|\cdot \aleph_0=|E|$ since $E$ is infinite. The proof of the reverse inequality is symmetric, hence $|F|=|E|$.

Recall that if $H$ and $K$ are two Hilbert spaces, then a linear map $T:H\to K$ is called an isometry if $\langle x,y\rangle_H=\langle Tx,Ty\rangle_K$ for all $x,y\in H$. Note that an isometry is bounded with norm one, since for any $x\in H$, we have $||Tx||^2=\langle Tx,Tx\rangle=\langle x,x\rangle=||x||^2$. Moreover, if $Tx=0$, then $||x||=||Tx||=0$, so $x=0$, hence an isometry is always injective and so is an isomorphism onto its range. If $T$ is surjective, then we say $T$ is an isometric isomorphism between $H$ and $K$ and the spaces $H$ and $K$ are isometrically isomorphic and we write $H\simeq K$. With these definitions in hand, we can prove the following important theorem:

Theorem Two Hilbert spaces $H$ and $K$ are isometrically isomorphic if and only if they have the same dimension.

To prove the theorem, I’ll follow a standard technique of finding a sort of canonical object and showing that any Hilbert space is isomorphic to this object which behaves the way we want it to behave.

Lemma Let $A$ be any set and define

$\displaystyle l^2(A):=\{f:A\to\mathbb{C}:\sum_{a\in A}|f(a)|^2<\infty\},\hspace{3mm} \langle f,g\rangle=\sum_{a\in A}f(a)\overline{g(a)}$

Then $l^2(A)$ is a Hilbert space under the given inner product and $l^2(A)\simeq l^2(B)$ if and only if $|A|=|B|$.
Proof sketch: It is relatively straightforward to show that $l^2(A)$ is a Hilbert space (see, for example, Folland or Conway). For the isomorphism, given $a\in A$, define $f_a$ by $f_a(a)=1$ and $f_a(b)=0$ for $b\neq a$. It is clear $f_a\in l^2(A)$ and that the set $\{f_a\}_{a\in A}$ is an orthonormal set in $l^2(A)$. To show that it is a basis, suppose that $\langle x,f_a\rangle=0$ for all $a\in A$. By definition of $f_a$, $0=\langle x,f_a\rangle=x(a)$. Since this holds for all $a\in A$, $x=0$, and so $\{f_a\}_{a\in A}$ is indeed an ONB for $l^2(A)$. Define $\{g_b\}_{b\in B}$ similarly. If an isomorphism $T:l^2(A)\to l^2(B)$ exists, then $\{Tf_a\}_{a\in A}$ is an orthonormal basis for $B$, and so by the well-definedness of dimension, this basis has the same cardinality as $\{g_b\}_{b\in B}$. But the former set is indexed by $A$ and the latter by $B$, which implies $|A|=|B|$. For the converse, use Folland’s proposition 5.30 to get the desired map.

I remark that the space $l^2(A)$ is best understood when $|A|=n<\infty$. In this case, $l^2(A)$ is simply the set of all $n$-tuples that are square summable. In other words, $l^2(A)=\mathbb{C}^n$ with the Euclidean inner product. If $A=\mathbb{N}$, then $l^2(\mathbb{N})$ is the set of all square-summable sequences and is often denoted by simply $l^2$ and is the natural extension of $\mathbb{C}^n$ to countably infinitely many entries. Finally, we have:

Proof of Theorem: Suppose $H\simeq K$ and let $T:H\to K$ be an isometric isomorphism. Let $E$ be an orthonormal basis for $H$ and let $F=\{Te\}_{e\in E}$. It is clear that $F$ is an orthonormal set in $K$. If $k\in K$ is such that $\langle k,Te\rangle=0$ for all $e\in E$, then write $k=Th$ for some $h\in H$. Then for all $e\in E$, we have

$\langle h,e\rangle=\langle Th,Te\rangle=\langle k,Te\rangle=0$

so since $E$ is an ONB for $H$, it follows that $h=0$, and so $k=Th=0$, and so indeed $F$ is an ONB for $K$. Moreover, clearly $|E|=|F|$ and so $\dim K=|F|=|E|=\dim H$. Conversely, suppose $\dim H=\dim K$. In the proof above we showed that $\dim l^2(H)=|E|=\dim l^2(K)$, so indeed from the lemma we have $l^2(H)\simeq l^2(K)$. As isometric isomorphism is a transitive relation (check this), it suffices to show that $H\simeq l^2(E)$ (and similarly that $K\simeq l^2(F)$). For $h\in H$, define $f_h(e)=\langle h,e\rangle$. Then

$\langle f_h,f_h\rangle=\sum_{e\in E} \langle h,e\rangle\overline{\langle h,e\rangle}=\sum_{e\in E}|\langle h,e\rangle|^2=||h||^2=\langle h,h\rangle$

where I used property three (Parseval’s identity) in the definition of an ONB. Thus the map $h\mapsto f_h$ is an isometry. It’s clearly linear, so it remains to show the correspondence is surjective. I’ll leave it as an exercise to verify that the set of $f\in E$ such that $f(e)=0$ for all but finitely many $e\in E$ is dense in $l^2(E)$. If $f$ is such a function, let $e_1,\dots e_n$ be the collection of $e\in E$ such that $f(e)\neq 0$. Then if we let $h=\sum_1^n f(e_i)e_i$, we have that $f=f_h$, and so $h\mapsto f_h$ has dense range. Note that if $T:H\to K$ is any isometry of Hilbert spaces and $Tx_n\to y$, then $||x_n-x_m||=||Tx_n-Tx_m||\to0$ so $x_n$ is Cauchy, hence converges to $x\in H$ (as $H$ is complete). As $T$ is continuous, $Tx_n\to Tx$, so $y=Tx$, and it follows that $T$ has closed range. It follows from these two facts that the map $h\mapsto f_h$ is surjective, and this completes the proof.

That’s all I want to say about dimension and orthonormal bases for Hilbert spaces. Up until now I’ve focused on showing the similarities between infinite dimensional Hilbert spaces and their familiar finite dimensional counterparts. In my next post in this series, I’ll switch gears and talk about how these spaces are different. It’s not terribly difficult to show that, in finite dimensions, the unit ball, i.e. the set $\{x\in H:||x||=1\}$, is a compact subset of the Hilbert space $H$ (I say Hilbert space since any finite dimensional Banach space is isometrically isomorphic to $\mathbb{R}^n$, which can be given the structure of a Hilbert space). I’ll show that in infinite dimensions, the unit ball is NEVER closed in the norm topology on $H$. After that, I’ll introduce some different topologies one can consider on $H$, and describe convergence in these topologies, and work towards proving a very cool result known as Alaoglu’s theorem, which states that the unit ball of a Hilbert space is compact in a special topology known as the weak-star topology on $H$. These results have some natural generalizations to Banach spaces (and even more general objects known as topological vector spaces), and once we prove Alaoglu’s theorem, I’ll use it to show that a very familiar Banach space can not be isomorphic to the dual space of any other Banach space. Until then, best of luck with quals/finals/start of summer classes.

Great post, Hotovy. Another cool fact about the $\ell^2$ unit sphere – unlike its finite dimensional counterparts – is that it contracts to a point.