Are $L^2$-closed subspaces of $C^\infty$ finite-dimensional?

Question

I have the following question: Given a closed Riemannian manifold $M$, let $V\subseteq C^\infty(M)$ be a linear subspace that is closed with respect to the $L^2$-norm. The question now is whether $V$ is finite-dimensional.

My idea is to use Rellich's embedding theorem together with the open mapping theorem:

Denote by $\iota : H^m(M) \rightarrow L^2(M)$, $m \in \mathbb{N}$ the canonical inclusion, by Rellich this is a compact injective map. The restriction of $\iota$ to $V$ gives a bijection $\iota\lvert_V: (V, \lvert\lvert\cdot\rvert \rvert_{H^m})\rightarrow (V, \lvert\lvert\cdot\rvert \rvert_{L^2})$. If I now can show that $V$ is also closed with respect to the $H^m$-norm, I can use the open mapping theorem to get that the inverse $(\iota\lvert_V)^{-1}:(V, \lvert\lvert\cdot\rvert \rvert_{L^2})\rightarrow (V, \lvert\lvert\cdot\rvert \rvert_{H^m})$ is continous. In this case, the identity of $V$ factors through $H^m$, i.e. $id_V = \iota\lvert_V \circ (\iota\lvert_V)^{-1}$ and hence it is a compact map. This then implies that $V$ is finite-dimensional.

The missing puzzle piece is therefore to show that $V$ is closed as a subspace of $H^m$ (doesn't matter for which $m$).

I'm grateful for any hint on how to prove this, less grateful for a counterexample, but at least then I can stop thinking about this ;)

Thanks!

Closed as a subspace of $L^2$ or closed in the subspace topology of $C^\infty$? — Sassatelli Giulio, Apr 24 '23 at 10:18
I don't know too much about Riemannina manifolds, but would you be okay with $M = [-1, 1]$? Then $V := \left \lbrace u \in C^\infty([-1, 1]): \int_{\mathbb{R}} u = 0\right \rbrace$ is $L^2$-closed and $x, x^3, x^5, ...$ are all in $V$. — Hyperbolic PDE friend, Apr 24 '23 at 10:18
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. — José Carlos Santos, Apr 24 '23 at 10:19
@Meowdog $M=[-1,1]$ is not a closed manifold, since it has the boundary points $-1$ and $1$. However I don't think that the vector space you gave me there is L^2-closed, the closure is $\bar{V}={u\in L^2([-1,1]) : \int_{-1}^1 u =0}$. — pizzalberto, Apr 24 '23 at 10:36

score 1 · Answer 1 · answered Apr 24 '23 at 12:12

I think I solved it myself, it was actually right before my eyes:

Let $(u_n)\subseteq V$ be a sequence converging in $H^m(M)$, let $u$ be the limit in $H^m(M)$. But then $(u_n)$ converges also against $u$ in $L^2(M)$ and since $V$ is closed in $L^2$ we have that $u\in V$. Hence $V$ is also closed in $H^m(M)$.

Are $L^2$-closed subspaces of $C^\infty$ finite-dimensional?

1 Answers1