I ask this question out of curiosity, not a specific need. Euclidean spaces and manifolds. Are there examples of locally compact function spaces? Could (some?) Sobolev spaces be locally compact?

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    $\begingroup$ A Hausdorff topological vector space (over $\mathbb{R}$ or $\mathbb{C}$) is locally compact if and only if it is finite-dimensional. That's a celebrated theorem of Riesz. $\endgroup$ Jan 30, 2014 at 11:56

1 Answer 1


Theorem (Riesz Frigyes): A Hausdorff topological vector space over $\mathbb{R}$ or $\mathbb{C}$ is locally compact if and only if it is finite-dimensional.

Proof: For the implication that a finite-dimensional (real or complex) Hausdorff TVS is locally compact, I refer to the fact that every finite-dimensional (real or complex) vector space carries a unique Hausdorff TVS topology. That is (under any linear isomorphism) the product topology of $\mathbb{K}^n$, which we know is locally compact.

For the converse, suppose $E$ is a Hausdorff locally compact topological vector space. Let $K$ a compact neighbourhood of $0$ in $E$. Since $K$ is compact, there are finitely many $x_1,\dotsc,x_n$ such that

$$K \subset \bigcup_{i=1}^n \left(x_i + \frac{1}{2}\cdot \overset{\circ}{K}\right).\tag{1}$$

Let $F = \operatorname{span} \{ x_1,\dotsc,x_n\}$. Then $F$ is a finite-dimensional, hence complete, hence closed, subspace of $E$. Thus $E/F$ is a Hausdorff topological vector space. Let $\pi \colon E \to E/F$ the canonical projection. Since $\pi$ is open, $\tilde{K} = \pi(K)$ is a neighbourhood of $0$ in $E/F$, and since $\pi$ is continuous, it is compact. By $(1)$, we have

$$\tilde{K} \subset \pi\left(\frac{1}{2}\cdot K\right) = \frac{1}{2}\pi(K) = \frac{1}{2}\tilde{K}.$$

Inductively, we have $2^n\cdot \tilde{K}\subset \tilde{K}$ for all $n\in\mathbb{N}$, and therefore

$$E/F = \bigcup_{n\in\mathbb{N}} \left(2^n\cdot\tilde{K}\right) \subset \tilde{K}$$

is compact, hence $E/F = \{0\}$, whence $E = F$ is finite-dimensional.

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    $\begingroup$ By Montel's theorem, if $X$ and $Y$ are hyperbolic Riemann surfaces, $Hol(X,Y)$ is locally compact. Does it mean that $Hol(X,Y)$ is finite dimensional? $\endgroup$ Feb 12, 2021 at 13:51
  • $\begingroup$ @GiulioBinosi Cf. en.wikipedia.org/wiki/Montel_space I am definitely not an expert, but I think the idea is that, as long as the TVS is not a normed space (or a metric space), it can have the Heine-Borel property without being locally compact. If Hol(X,Y) is really locally compact, then if it is a TVS it is finite-dimensional. But again my understanding is that Montel's theorem doesn't necessarily imply local compactness. Cf. en.wikipedia.org/wiki/… $\endgroup$ Feb 8, 2022 at 3:42

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