Is every sufficiently smooth function on a compact manifold approximated by a linear combination of a 'few' Laplacian eigenfunctions? Let $M$ be some smooth, compact manifold. Let $\Delta$ be the usual Laplacian, and $f_0, f_1, \ldots, $ its eigenfunctions in order of increasing eigenvalues.
I know that minimizing Dirichlet energy subject to being norm 1 and orthogonal to the first $k$ eigenfunctions gives the $k+1$st eigenfunction: https://en.wikipedia.org/wiki/Dirichlet_eigenvalue
This seems to be a sense in which the canonical smooth functions on $M$ are the Laplacian eigenfunctions.
Along those lines, I'm interested in whether there is a sense in which every sufficiently smooth (low Dirichlet energy) function is 'near' to a linear combination of the first $n$ eigenfunctions, where $n = n(K,\epsilon)$ will be a function of the Dirichlet energy and the desired approximation.
More precisely:
Question: Is there a function $n(K, \epsilon)$, such that for any $f \in C^{\infty}(M)$ with $ \int || \nabla f||^2 \leq K$ and $||f||_2 = 1$, there are $a_1,\ldots, a_{n(K, \epsilon)}$ such that $|| \sum_{i = 0}^{n(K, \epsilon)} a_i f_i - f ||_2 < \epsilon$?
 A: I think the answer is yes, although manifold stuff is pretty far out of my wheelhouse so I could easily be missing something.
First, we'll write $f = \sum_{i = 0}^{\infty} a_i f_i$, where the $f_i$ are the orthonormal eigenfunctions ($\lambda f_i = \lambda f_i$) normalized so $||f_i||_2 = 1$. In particular, we have $\sum a_i^2 = 1$ by the assumption $\| f\|_2 = 1$.
Additionally, if $d\omega$ is the volume element on the manifold, we have $$K \geq \int_M || \nabla f||_2^2 d\omega = \int_M f \Delta f d\omega = \sum_{i = 0}^{\infty} a_i^2 \lambda_i \tag{1}$$
We have that $||f - \sum_{i = 0}^{N - 1} a_i f_i ||_2 = \| \sum_{i = N}^{\infty} a_i f_i\|_2 = \sum_{i = N}^{\infty} a_i^2$.
So, mainly the question is how large $N$ has to be so that $\epsilon > \sum_{i = N}^{\infty} a_i^2$.
If $\epsilon \leq \sum_{i = N}^{\infty} a_i^2$, since $\lambda_i \leq \lambda_{i + 1}$, from $(1)$ we have that $K \geq \epsilon \lambda_N$.
Since $N \to \infty$, we know that we can pick $N$ large enough so that $\lambda_N > K/\epsilon$ and obtain a contradiction to  $K/\epsilon \geq \lambda_N$. Moreover, we can estimate how large $N$ needs to be:
We know, by Weyl's law, that $\lambda_j \sim \frac{ (2 \pi)^2 }{ ( w_n Vol(M))^{2/n} } j^{2/n} = C_M j^{2/n}$. (Throughout I'll set $C_M =  \frac{ (2 \pi)^2 }{ ( w_n Vol(M))^{2/n} }$.)
Thus, we want $\inf \{ N : \lambda_N > K/\epsilon \} \sim \inf \{ N : C_M N^{2/n} > K/\epsilon\}$.
Modulo the handwaving in the previous line comparing the two infimums, this tells us that taking $N = \lceil (\frac{ K}{ C_M \epsilon})^{n/2} \rceil$ suffices so that  $||f - \sum_{i = 0}^{N - 1} a_i f_i ||_2 < \epsilon$ when  $K \geq \int_M || \nabla f||_2^2 d\omega$ and $||f||_2 = 1$.
