I consider an analytic bounded domain $\Omega\subset \mathbb R^3$ and an the operator $L_a=-\Delta +a$ where $a$ is a function from $\Omega$ to $\mathbb R$. I assume the operator to be coercive, in particular there exists a Green function $G_a :\Omega\times \Omega \setminus D \rightarrow \mathbb R$ symmetric such that $$ L_a G_a(x,y)= \delta_y \text{ on } \Omega,$$ and $$G_a(x,y)=0 \text{ on } \partial \Omega,$$ where $D=\{ (x,x) \,\vert\, x \in \Omega \}$. I can write $$G_a(x,y)= \frac{c}{\vert x-y\vert} + H_a(x,y),$$ where $H_a \in L^\infty( \Omega \times \Omega, \mathbb R)$. We use to call $\rho_a(x)= H_a(x,x)$ the Robin function.
My question is, if $a$ analytic, is $\rho_a$ analytic ?
First it has to be noticed, it is false for $H_a$, since it is not hard to prove that locally $H(x,y)=d +d'\vert x-y\vert +...$.
But it is true for $\rho_a$ when $a$ is constant. The idea is quite simple, we first build a solution $S_a$ of $$ L_a S_a =-\frac{ac}{\vert x-y\vert}$$ by setting $$S_0(x,y)=\frac{ac \vert x-y\vert}{2}$$ which solves $$ -\Delta S_ 0 =-\frac{ac}{\vert x-y\vert}$$ then $$ -\Delta S_{n+1}= -ac S_n$$ it is not hard to show that $S_a =\sum_{n=0}^\infty S_n$ converges and solves the desired equations. Moreover $S_a(x,y)= \sum_{n=0}^\infty c_n \vert x-y \vert^{1+2n}$ is such that $S_a(x,y)$ is symmetric and $S_a(x,x)=0$. Hence $F_a = H_a -S_a$ solves $$L_a F_a =0$$ Hence it is analytic in $x$ in the interior( we just need $a$ analytic here) and symmetric in $x$ and $y$( because $H_a$ and $S_a$ are), hence $\rho_a(x)=F_a(x,x)$ is analytic.
The argument work well, since when $a$ is constant, we can almost explicitly solves
$$ L_aS_a =-\frac{ac}{\vert x-y\vert}$$
What we only need in the general case is to find any symmetric solution in $x$ and $y$ which vanishes on the diagonal. The end of the proof should work the same.
Despite the problem of the regularity of the Robin function seems quite natural, I found only few references... any idea or reference is welcome.
