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 is orthonormal if . The set is called an orthonormal basis if, in addition, it satisfies one of the following equivalent properties:
- For all , we have , 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.
- If for all , then .
- Given any , we have .
In the last post on Hilbert spaces, I proved that every Hilbert space has an orthonormal basis by considering a maximal orthonormal set in (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 and are orthonormal bases for the same Hilbert space , then . The cardinal number is called the dimension of the Hilbert space and is denoted by .
Proof: The case where either or is trivial since any orthonormal spanning set in a finite dimensional space is a (Hamel) basis and so is the vector space dimension of . Thus we consider only the case where and are infinite. For fixed , we define . By definition of an ONB, is countable. Given any , there must exist some with , otherwise . But so such must exist. Thus . But then since is infinite. The proof of the reverse inequality is symmetric, hence .
Recall that if and are two Hilbert spaces, then a linear map is called an isometry if for all . Note that an isometry is bounded with norm one, since for any , we have . Moreover, if , then , so , hence an isometry is always injective and so is an isomorphism onto its range. If is surjective, then we say is an isometric isomorphism between and and the spaces and are isometrically isomorphic and we write . With these definitions in hand, we can prove the following important theorem:
Theorem Two Hilbert spaces and 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 be any set and define
Then is a Hilbert space under the given inner product and if and only if .
Proof sketch: It is relatively straightforward to show that is a Hilbert space (see, for example, Folland or Conway). For the isomorphism, given , define by and for . It is clear and that the set is an orthonormal set in . To show that it is a basis, suppose that for all . By definition of , . Since this holds for all , , and so is indeed an ONB for . Define similarly. If an isomorphism exists, then is an orthonormal basis for , and so by the well-definedness of dimension, this basis has the same cardinality as . But the former set is indexed by and the latter by , which implies . For the converse, use Folland’s proposition 5.30 to get the desired map.
I remark that the space is best understood when . In this case, is simply the set of all -tuples that are square summable. In other words, with the Euclidean inner product. If , then is the set of all square-summable sequences and is often denoted by simply and is the natural extension of to countably infinitely many entries. Finally, we have:
Proof of Theorem: Suppose and let be an isometric isomorphism. Let be an orthonormal basis for and let . It is clear that is an orthonormal set in . If is such that for all , then write for some . Then for all , we have
so since is an ONB for , it follows that , and so , and so indeed is an ONB for . Moreover, clearly and so . Conversely, suppose . In the proof above we showed that , so indeed from the lemma we have . As isometric isomorphism is a transitive relation (check this), it suffices to show that (and similarly that ). For , define . Then
where I used property three (Parseval’s identity) in the definition of an ONB. Thus the map 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 such that for all but finitely many is dense in . If is such a function, let be the collection of such that . Then if we let , we have that , and so has dense range. Note that if is any isometry of Hilbert spaces and , then so is Cauchy, hence converges to (as is complete). As is continuous, , so , and it follows that has closed range. It follows from these two facts that the map 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 , is a compact subset of the Hilbert space (I say Hilbert space since any finite dimensional Banach space is isometrically isomorphic to , 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 . After that, I’ll introduce some different topologies one can consider on , 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 . 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.