Resolvent of the Laplacian

Question

We know that the Laplacian $-\Delta$ on $\mathbb{R}^n$ is a radial operator. I was wondering if its resolvent $(\lambda - \Delta)^{-1}$ is also a radial operator on $L^2(\mathbb{R}^n)$. I strongly suspect it should be, but cannot come up with a rigorous proof. Help, please? Thanks!!

Answer 1

$$ f = (\lambda - \Delta)^{-1} g \Leftrightarrow g = (\lambda - \Delta) f .$$ Now if we define $R_Af(x) = f(Ax)$ for any matrix $A$, then since $\Delta$ is a radial operator, for any orthogonal $A$ we have $$ g = (\lambda - \Delta) f \Leftrightarrow R_A g = (\lambda - \Delta) R_A f .$$ Go from there.