Uniform convergence of an expansion into an orthonomal basis of Laplacian eigenfunctions on a manifold Let $(M,g)$ be a compact Riemannian manifold $h \in C^{\infty}(M).$ For simplicity, I will assume $M$ is an embedded submanifold of $\mathbb R^N$ with the induced metric. Consider the elliptic operator for the form,
$$ L = \Delta + h, $$
where $\Delta$ is the Laplace-Beltrami operator on $M.$ Standard results tell us that there exists an orthonormal basis for $L^2(M)$ of eigenfunctions $(\phi_k)_{k=1}^{\infty} \in C^{\infty}(M)$ for $L$ with associated real eigenvalues $\lambda_k$ tending to infinity as $k \rightarrow \infty.$

Question If $f \in C^{0,\alpha}(M)$ for some $\alpha > 0,$ does the series,
  $$ \sum_{k=1}^{\infty} f_k \phi_k = \sum_{k=1}^{\infty} \left( \int_M f(x)\phi_k(x)\ dx\right) \phi_k $$
  converge uniformly to $f$ on $M$? If $f$ is moreover differentiable, is the convergence absolute in $C^0(M)$?

In the case $M$ is the unit circle and $h=0,$ this is a classical result about Fourier series. I'm wondering if this can be generalized.
Even if my general formulation doesn't hold, I would be interested in partial results (e.g. whether this holds when $h=0,$ higher regularity assumptions on $f$, further conditions on $M,$ etc).

Since this question has been unanswered for a long time, I will update it with a partial answer. One can show that if $f$ lies in an appropriate Besov space, then we obtain absolute uniform convergence. This is detailed in the following paper,
Peetre, J. Absolute convergence of eigenfunction expansions. Math. Ann. (1967) 169: 307. https://doi.org/10.1007/BF01362354
I unfortunately no longer have access to the paper to provide the precise details. The rough idea is that we can prove the eigenvalues grow sufficiently quickly, which provides an $L^{\infty}$ estimate in terms of some Sobolev norm. It turns out that this estimate is very weak however, so an interpolation argument gives an estimate in terms of a Besov norm.
