Compactness and Infinite-Dimensional Hilbert Spaces II

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 E of L^{2}[-\pi,\pi] formed by the functions

\displaystyle f_{n}(x):=\sin nx,\indent\forall x\in[-\pi,\pi],\forall n\in\mathbb{N}

define a closed and bounded subset of L^{2}[-\pi,\pi], 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 f_{n} are orthonormal.

This is an easy computation. Observe that, for integers n and m,

\begin{array}{lcl}\displaystyle\int_{-\pi}^{\pi}\sin(nx)\sin(mx)dx&=&\displaystyle\left(-\dfrac{i}{2}\right)^{2}\int_{-\pi}^{\pi}\left(e^{inx}-e^{-inx}\right)\left(e^{imx}-e^{-imx}\right)dx\\[.9 em]&=&\displaystyle-\dfrac{1}{4}\int_{-\pi}^{\pi}\left[e^{i(n+m)x}-e^{i(n-m)x}-e^{i(m-n)x}+e^{-i(n+m)x}\right]dx\\[.9 em]&=&\displaystyle\begin{cases}2\pi&{n-m=0,n\neq0}\\{-2\pi}&{n+m=0,n\neq0}\\{0} &{\text{otherwise}}\end{cases}\end{array}

Scaling by a factor of (2\pi)^{-1} completes the proof.

It’s worth mentioning that we do not need the hypothesis of a Hilbert space. In fact, a normed space (\mathbb{X},\left\|\cdot\right\|) 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 Y be a closed, proper subspace of a normed space (X,\left\|\cdot\right\|). For any \alpha\in(0,1), there exists an element x\in\mathbb{X} such that

\left\|x\right\|=1\text{ and }d(x,Y):=\inf_{y\in Y}\left\|x-y\right\|>\alpha

Proof. If Y=\left\{0\right\}, then there is nothing to prove, so assume otherwise. Fix \alpha\in(0,1). Since Y is a proper subspace of \mathbb{X}, there exists a nonzero element z\notin Y such that \delta:=d(z,Y)>0. Let y_{1}\in Y be a nonzero element satisfying \left\|z-y_{1}\right\|<\delta+\varepsilon, where \varepsilon>0 is some fixed arbitrarily small quantity. Define an element with unit norm by

\displaystyle x:=\frac{z-y_{1}}{\left\|z-y_{1}\right\|}

Then, for all y\in Y,

\displaystyle\left\|x-y\right\|=\left\|\frac{z-y_{1}}{\left\|z-y_{1}\right\|}-y\right\|=\dfrac{\left\|z-\left(y_{1}+\left\|z-y_{1}\right\|\cdot y\right)\right\|}{\left\|z-y_{1}\right\|}>\dfrac{\delta}{\delta+\varepsilon}

Since \varepsilon can be taken arbitrarily small, and therefore \frac{\delta}{\delta+\varepsilon} can be made arbitrarily close to 1 from below, we obtain the desired conclusion. \Box

We inductively construct an most countable sequence of linearly independent vectors e_{n} with unit norm as follows. If \mathbb{X}=\left\{0\right\}, then there is nothing to prove; assume otherwise. Choose x_{1}\neq0 in \mathbb{X} and define

e_{1}:=\dfrac{1}{\left\|x_{1}\right\|}\cdot x_{1}

If V_{1}:=\text{span}_{\mathbb{F}}\left\{e_{1}\right\}=\mathbb{X}, then we’re done. Otherwise choose a nonzero vector e_{2}\notin V_{1} such that

\displaystyle\left\|e_{2}\right\|=1\text{ and }d(e_{2},V_{1})>\alpha

Suppose we have chosen e_{1},\cdots,e_{n} as above. If

\displaystyle V_{n}:=\text{span}_{\mathbb{F}}\left\{e_{1},\ldots,e_{n}\right\}=\mathbb{X},

then we’re done. Otherwise, choose a vector e_{n+1}\notin V_{n} such that

\left\|e_{n+1}\right\|=1\text{ and }d(e_{n+1},V_{n})>\alpha

Suppose this process produces a countable sequence (e_{n})_{n=1}^{\infty}. The continuity of the norm implies that the closed unit ball


is closed and bounded, hence compact by the Heine-Borel property. Let (e_{n_{k}})_{k=1}^{\infty} be a convergent subsequence with limit e\in\mathbb{X}A fortiori, the sequence (e_{n_{k}})_{k=1}^{\infty} is Cauchy, and therefore for all k,j sufficiently large,

\displaystyle\frac{\alpha}{2}>\left\|e_{n_{k}}-e_{n_{j}}\right\|\geq d(e_{n_{k}},V_{n_{j}})>\alpha,

which is a contradiction.

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