Let $(M,g)$ be a compact rimannian manifold. It is well known that we can diagonalyse the Green kernel as a $L^2$ operator acting on functions. Moreover we have the convergence of the following series, viewed as $L^2$ operators :
 \begin{equation} \underset{k \in \mathbb{N}}{\sum} \frac{f\_k(x) \otimes f\_k(y)}{\lambda\_k} \end{equation}
where $f\_k$ is an orthormormal Hilbert basis of $ \ker \Delta^{\perp} $ composed of eigenvectors of $\Delta$. 


My question is : what kind of convergence could we expect if we look at the series as a series of functions, $ \underset{k \in \mathbb{N}}{\sum} \frac{f\_k(x) f\_k(y)}{\lambda\_k} $ ?


More precisely have we got an uniform $C^1$ convergence away from the diagonal ?


I don't ask for a $C^2$ convergence, because in the case of the circle is already false.


Any help will be appreciated.



Does the green kernel converge as a series of functions?
The answer should depend, among other things, on the dimension of your manifold. Let's look at tori.


One first observation is that $\mathcal{C}^1$ convergence is too strong to ask for, since computing eigenfunctions gives us


$$\frac{f\_k(x) f\_k(y)}{\lambda\_k} = cst \frac{e^{ik \cdot (x+y)}}{|k|^2}$$


which is not the general term of an absolutely convergent series in $\mathcal{C}^1$, even in dimension $1$. On the other hand, still in dimension $1$, we have convergence in $H^1$ (and actually a bit better).


Now, if your torus has dimension $\geq 3$, you even lose convergence of your series in $L^2$.


