## 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

$B_{1}(0):=\left\{x\in\mathbb{X}:\left\|x\right\|\leq1\right\}$

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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