See Uniform convergence of Eigenfunction decomposition on Riemannian sphere? on MO. Unless I’m tripping, I feel like the following is a “softer” argument which only uses Sobolev embedding and elliptic estimates, rather than an explicit Weyl-law type of eigenvalue growth estimate.
Throughout, let $(M,g)$ be a smooth compact boundaryless Riemannian manifold of dimension $n$, and let $A$ be a self-adjoint second-order elliptic differential operator with smooth coefficients (like the Laplacian). Due to compactness of $M$, there are several equivalent ways of describing the function spaces below (e.g via local charts and partitions of unity, or by using the Levi-Civita connection to define these spaces). Now, we record the following estimates.
- (Sobolev Embedding) For all $\ell\in \Bbb{N}_0$ and all integer $m>\ell +\frac{n}{2}$, there is a $B_{\ell,m,n}>0$ such that for all $f\in C^{\infty}(M)$, we have
\begin{align}
\|f\|_{C^{\ell}}\leq B_{\ell,m,n}\cdot \|f\|_{H^m}.
\end{align}
Roughly speaking, we can quantiatively trade-in some super high Sobolev regularity for some uniform pointwise regularity.
- (Higher Order Elliptic Estimate for $A$) For all $m\in\Bbb{N}_0$, there is a constant $B_{m,n,A}>0$ such that for all $f\in C^{\infty}(M)$, we have
\begin{align}
\|f\|_{H^{m+2}}\leq B_{m,n,A}\left(\|Af\|_{H^{m}}+ \|f\|_{L^2}\right).
\end{align}
Before coming to the main result, let’s introduce some notation. Suppose $\{\phi_j\}_{j=0}^{\infty}$ is an orthonormal basis of $L^2(M)$ consisting of smooth functions which are eigenfunctions of $A$. For each $N\in\Bbb{N}_0$, let $P_N$ be the orthogonal projection of $L^2(M)$ onto the subspace spanned by $\{\phi_0,\dots, \phi_N\}$, i.e for all $f\in L^2(M)$,
\begin{align}
P_Nf:=\sum_{j=0}^N\langle f,\phi_j\rangle_{L^2}\,\phi_j.
\end{align}
We then have that for each $f\in L^2(M)$, $\|f-P_Nf\|_{L^2}\to 0$ because the $\phi_j$’s form an orthonormal basis. If $f$ is smooth, then we observe also that $Af$ makes sense, so we can commute $A(P_Nf)=P_N(Af)$.
Theorem.
With notation as above, for each $\ell\in\Bbb{N}_0$ and $f\in C^{\infty}(M)$, we have $\|f-P_Nf\|_{C^{\ell}}\to 0$ as $N\to\infty$.
To prove this, fix $\ell\in\Bbb{N}_0$, and an integer $m$ so large that $2m>\ell+ \frac{n}{2}$. Then, by the Sobolev embedding, and applying the elliptic estimate $m$ times, we see that for all $f\in C^{\infty}(M)$,
\begin{align}
\|f\|_{C^{\ell}}&\leq \widetilde{B}_{\ell,m,n}\cdot \|f\|_{H^{2m}}
\leq B_{\ell,m,n,A}\sum_{j=0}^m\|A^jf\|_{L^2}.
\end{align}
Now, apply this same estimate with $f-P_Nf$ instead (which is still a smooth function since the eigenfunctions are smooth), then commute with the $A$’s to get that for all $f\in C^{\infty}(M)$,
\begin{align}
\|f-P_Nf\|_{C^{\ell}}&\leq B_{\ell,m,n,A}\sum_{j=0}^m\|A^j(f-P_Nf)\|_{L^2}\\
&=B_{\ell,m,n,A}\sum_{j=0}^m\|(I-P_N)(A^jf)\|_{L^2}.
\end{align}
Since $f$ is smooth, we have that $f,Af,\dots, A^mf$ all lie in $L^2(M)$, so as $N\to\infty$, the RHS vanishes. This shows $P_Nf\to f$ in the $C^{\ell}$-norm, hence completes the proof.
In particular, this shows that for smooth $f$, the eigenfunction expansion is uniform, so if $x,y\in M$ are such that for all $j$, $\phi_j(x)=\phi_j(y)$, then we can indeed perform pointwise evaluations to get for all smooth $f$ that $f(x)=f(y)$. Hence, we deduce $x=y$ (we use bump functions to prove that smooth functions separate points on any manifold, compact or not).
