Browsing through Chapter 11 of Baby Rudin, which I previously hadn’t read, I realized that a past post of mine, hereafterto referred to as my first post, answers exercise 11.13. This exercise asks the reader to show that the subset of formed by the functions
define a closed and bounded subset of , which is not compact. In my first post, we showed that a countable orthonormal system in a Hilbert space forms a closed and bounded set, it remains for us to show that the functions are orthonormal.
This is an easy computation. Observe that, for integers and ,
Scaling by a factor of completes the proof.
It’s worth mentioning that we do not need the hypothesis of a Hilbert space. In fact, a normed space is finite-dimensional if and only if it has the Heine-Borel property. I will only prove sufficiency. Necessity is an easy consequence of the Bolzano-Weierstrass theorem, which I leave to the reader.
Riesz’s Lemma. Let be a closed, proper subspace of a normed space . For any , there exists an element such that
Proof. If , then there is nothing to prove, so assume otherwise. Fix . Since is a proper subspace of , there exists a nonzero element such that . Let be a nonzero element satisfying , where is some fixed arbitrarily small quantity. Define an element with unit norm by
Then, for all ,
Since can be taken arbitrarily small, and therefore can be made arbitrarily close to from below, we obtain the desired conclusion.
We inductively construct an most countable sequence of linearly independent vectors with unit norm as follows. If , then there is nothing to prove; assume otherwise. Choose in and define
If , then we’re done. Otherwise choose a nonzero vector such that
Suppose we have chosen as above. If
then we’re done. Otherwise, choose a vector such that
Suppose this process produces a countable sequence . The continuity of the norm implies that the closed unit ball
is closed and bounded, hence compact by the Heine-Borel property. Let be a convergent subsequence with limit . A fortiori, the sequence is Cauchy, and therefore for all sufficiently large,
which is a contradiction.