## Compactness and Infinite-Dimensional Hilbert Spaces

It is well-known that any finite-dimensional normed space $\mathbb{X}$ is isomorphic to $\mathbb{R}^{n}$, where $n=\dim(\mathbb{X})$. Therefore, the Heine-Borel theorem tells us that closed and bounded subsets of $\mathbb{X}$ are compact. We can say that $\mathbb{X}$ has the Heine-Borel property. Perhaps less well-known to beginning students is that an infinite-dimensional normed space $\mathbb{X}$ need not have the Heine-Borel property. Taking our cue from Big Rudin Chapter 4 Exercise 7, we prove in this post that no infinite-dimensional Hilbert space has the Heine-Borel property.

Let $\mathbb{H}$ be an infinite-dimensional Hilbert space, and let $E:=\left\{e_{n}\right\}_{n=1}^{\infty}$ be an orthonormal system. Then $E$ is closed and bounded but not compact (thus, disproving the Heine-Borel theorem in infinite-dimensions).

Proof. Clearly, $E$ is bounded by the constant $1$. To see that $E$ is closed, suppose $\left\{e_{n_{k}}\right\}_{k=1}^{\infty}$ is some sequence in $E$ such that $e_{n_{k}}\rightarrow x\in\mathbb{H}$. I claim that $e_{n_{k}}=e_{N}$ for all $k$ sufficiently large. Otherwise, it follows from Bessel’s inequality that

$\displaystyle 0=\lim_{k\rightarrow\infty}\langle{x,e_{n_{k}}}\rangle=\left\|x\right\|^{2}=\lim_{k\rightarrow\infty}\left\|e_{n_{k}}\right\|^{2}=1$,

which is a contradiction. This proves that $x\in E$. To see that $E$ is not compact, observe that the sequence $(e_{n})_{n=1}^{\infty}$ is not Cauchy since by orthonormality,

$\displaystyle\left\|e_{n}-e_{m}\right\|^{2}=\langle{e_{n}-e_{m},e_{n}-e_{m}}\rangle=\langle{e_{n},e_{n}}\rangle+\langle{e_{m},e_{m}}\rangle=2$,

so that $(e_{n})$ contains no convergent subsequences. $\Box$

Let $(\delta_{n})_{n=1}^{\infty}$ be a sequence of positive numbers, and let $S$ be the closure of the set

$\displaystyle\left\{x\in\mathbb{H}:x=\sum_{n=1}^{N}c_{n}e_{n}, \left|c_{n}\right|\leq\delta_{n}\right\}$

$S$ is compact if and only if $\sum_{n=1}^{\infty}\delta_{n}^{2}<\delta$.

Proof. If $S$ is compact, then $S$ is bounded and therefore there exists some positive constant $M > 0$ such that

$\displaystyle\left\|\sum_{n=1}^{N}\delta_{n}e_{n}\right\|^{2}=\sum_{n=1}^{N}\left|\delta_{n}\right|^{2}\leq M,\indent\forall N\in\mathbb{Z}^{\geq 1}$,

which implies that $\sum_{n=1}^{\infty}\delta_{n}^{2}\leq M$.

Now suppose that $\sum_{n=1}^{\infty}\delta_{n}^{2}<\infty$. Let $(x_{n})_{n=1}^{\infty}$ be some sequence in $S$. I claim that $\left|\langle{x_{n},e_{k}}\rangle\right|\leq\delta_{k}$, for each $k,n\in\mathbb{Z}^{\geq 1}$. Indeed, fix $n\in\mathbb{Z}^{\geq 1}$. By definition of $S$, there exists a finite linear combination $\sum_{k=1}^{N}c_{k}e_{k}$ such that $\left|c_{k}\right|\leq\delta_{k}$ and $\left\|\sum_{j=1}^{N}c_{j}e_{j}-x_{n}\right\|<\epsilon$. By the triangle inequality and Cauchy-Schwarz,

$\displaystyle\left|\langle{x_{n},e_{k}}\rangle\right|\leq\left|\left\langle{x_{n}-\sum_{j=1}^{N}c_{j}e_{j},e_{k}}\right\rangle\right|+\left|\left\langle{\sum_{j=1}^{N}c_{j}e_{j},e_{k}}\right\rangle\right|<\epsilon+\delta_{k}$

Since $\epsilon>0$ was arbitrary, we see that $\left|\langle{x_{n},e_{k}}\rangle\right|\leq\delta_{k}$, $\left\|x_{n}\right\|\leq\sum_{k=1}^{\infty}\delta_{k}^{2}$, and $(\langle{x_{n},e_{k}}\rangle)_{n=1}^{\infty}$ is a bounded sequence in $\mathbb{C}$, for each $k\geq 1$. By the Bolzano-Weierstrass theorem, there exists a covergent subsequence $(\langle{x_{n,1},e_{1}}\rangle)_{n=1}^{\infty}$. By induction, we can construct nested subsequences $(x_{n,1})_{n=1}^{\infty}\supset\cdots\supset (x_{n,k})_{n=1}^{\infty}$ such that

$\displaystyle(\langle{x_{n,j},e_{j}}\rangle)_{n=1}^{\infty}$ converges to limit $\displaystyle\alpha_{j}$, for each $\displaystyle 1\leq j\leq k$

Set $y_{n}:=x_{n,n}$. Since $\sum_{k=1}^{\infty}\left|\alpha_{k}\right|^{2}\leq\sum_{k=1}^{\infty}\delta_{k}^{2}$, the series $x:=\sum_{k=1}^{\infty}\alpha_{k}e_{k}$ converges in $\mathbb{H}$. I claim that the sequence $(y_{n})_{n=1}^{\infty}$ converges to $x$. Indeed, by uniform convergence we can evaluate the limit term-by-term to obtain

$\displaystyle\lim_{n\rightarrow\infty}\sum_{k=1}^{\infty}\langle{y_{n},e_{k}}\rangle e_{k}=\sum_{k=1}^{\infty}\left(\lim_{n\rightarrow\infty}\langle{y_{n},e_{k}}\rangle \right)e_{k}=\sum_{k=1}^{\infty}\alpha_{k}e_{k}=x$,

where we use the continuity of the inner product to obtain the penultimate equality. $\Box$

Lastly, we remark that $\mathbb{H}$ is not a locally compact space by Riesz’s lemma.