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.

Advertisements
This entry was posted in Textbook Solutions and tagged , , . Bookmark the permalink.

One Response to Compactness and Infinite-Dimensional Hilbert Spaces

  1. Pingback: Compactness and Infinite-Dimensional Hilbert Spaces II | Math by Matt

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s