So I should explain: this is one of those weird weeks where I have actually finished all my homework for the week two days early, so I actually have some time to write on the blog! So, let’s talk about some more functional analysis. Last time we talked about , the set of bounded linear maps . In particular we defined to be the set of all bounded linear functionals on a normed space . We showed that this is a normed space (the reader should verify that an equivalent definition is ) and I claimed that is always complete with respect to this norm. Let’s prove this:
Proposition: Let be a normed space. Then is always complete with respect to the norm given in the previous post.
Proof: Suppose is a Cauchy sequence. If , then the sequence satisfies by assumption, and hence this sequence converges. Define . It is clear that is linear. Note that for all ,
But for each we have . Now, by the reverse triangle inequality, , so the numerical sequence is Cauchy, hence is (uniformly) bounded, say by . Then for all , we have , and taking limits gives , hence is bounded. Now given , choose so that implies . Then for all with , we have
But then taking to infinity, this shows that for all with we have , and taking the supremum over such preserves the inequality, thus for , we have .
I presented this proof because I realize in the last post I didn’t prove much and I wanted to give you a feel for how to work with these operator norms. Note above that this proof carries through mutatis mutandis if for any Banach space . Okay, let’s shift gears and talk about inner products. Recall that an inner product on a vector space is a map so that 1) for all with equality if and only if ; 2) If , , then , and 3) . There are several familiar facts from finite dimensional linear algebra whose proofs carry over almost verbatim to the arbitrary inner product space setting, so I will list them here without proof. Let be an inner product space, then
- The Cauchy-Bunyakovsky-Schwartz Inequality: For all , with equality if and only if for some , where .
- The map is a norm on . Unless stated otherwise, we always assume to be equipped with this norm.
- The Parallelogram Law (and its converse): For all , we have
Moreover, if is any normed space such that satisfies this expression, then there is an inner product so that for all . For a detailed explanation of the second result, check out this Wikipedia article.
- The Pythagorean Theorem: Let be an orthogonal set of vectors. That is, we have if . Then .
- Let be a subset. We say vectors are orthogonal and write if . We define . Then is a closed subspace of . If is a closed subspace, then is called the orthogonal complement of
This last definition is very important. In some sense this is what will allow us to prove the Riesz Representation Theorem. In fact, we have the following
Lemma/Proposition/Theorem: Let be a closed subspace of a Hilbert space . Then . Moreover, if we write with , , then are the unique elements of whose distance to is minimal.
Proof (sketch): Given , let and let be a sequence in so that . Apply the parallelogram law to show that is a Cauchy sequence (you may need to use that ) and let and let . Since is closed, we have and . To show that , let . By multiplying by an appropriate (complex) scalar, we may assume . Thus the function is real for all . Elementary calculus shows that has a minimum at , and so , which shows . I’ll leave it to the reader to verify the uniqueness statements as well as the statement about distance to .
Some remarks are in order. First, if is a Banach space and is a closed subspace of , we say is complemented if there exists a closed subspace such that . The above theorem tells us that in a Hilbert space, every closed subspace is complemented. It is a non-trivial theorem (due to Lindenstrauss and Tzafriri) that if is a Banach such that every closed subspace of is complemented, then is isomorphic to a Hilbert space. The question of which subspaces of a Banach space are complemented is open.
Given a closed subspace , we can define a map as follows: given , write uniquely with and . Then define . Note that , so . Moreover, we have and . A map so that is called a projection. For the special case of above, is called the orthogonal projection onto and is the orthogonal projection onto . It can be shown that in a Banach space, a subspace is complemented if and only there exists a continuous projection onto . For more information on projections, check out this link.
Alright, enough asides! The real reason I presented the above theorem was to prove the following
Corollary: Riesz Representation Theorem: Let be a Hilbert space and . Then there is a unique so that for all . Moreover, .
Proof: For uniqueness, suppose is another vector satisfying the theorem. Then for all we have
and so , i.e. . For existence, if , then clearly works. Otherwise is a proper subspace of , so its orthogonal complement contains a non-zero vector of norm one. Let and note that , so . Thus
so letting gives the existence. Finally, note that , so , and by Cauchy-Schwartz, for all we have , so , which completes the proof.
So what does this tell us? Well, note that if , then the map is a linear functional on . It’s easy to see that is linear, and if for all , then by the same argument as above, so this embeds into . What the above theorem tells us is that this correspondence is surjective, and so we have a conjugate-linear, isometric isomorphism between and . Repeating the construction again, we see that . It’s natural to ask if there are other classes of spaces where we can write down such a relation. I’ll write about these in later posts, but I think the next few posts I write will give some examples of applications of the Riesz representation theorem and perhaps a few more facts about Hilbert spaces. For example, how many different Hilbert spaces are there? Well, if you haven’t seen the result before, the answer may surprise you! Unfortunately I’ll probably have to start working on homework again soon, so I don’t know when I’ll get to write again, but until then, enjoy!