Quasinilpotent vectors of Newton potential vanish Suppose $\Omega$ is a smooth bounded domain in $\mathbb{R}^3$. Consider the Newton potential
\begin{equation}
T [\phi](x) = \int_{\Omega} \frac{1}{|x-y|} \phi(y)dy.  
\end{equation}
It is well know that $T$ is a bounded linear operator from $L^2(\Omega)$ to $H^2(\Omega)$. Hence it is a self adjoint compact operator defined on $L^2(\Omega)$. Suppose that it has the following spectral decomposition:
$$T \phi = \sum^\infty_{j = 1}\lambda_j (\phi,e_j) e_j,$$
where $(\lambda_j,\phi_j)$ is the eigenpair counting multiplicity. And we can see $ker T = \{0\}$ from the following observation: $\Delta T[\phi] = C\phi$ on $\Omega$ for some positive constant $C$. 
We say that a vector $q$ in $L^2(\Omega)$ is a quasinilpotent vector if 
$$
\lim_{n \to \infty}||T^n q||^{\frac{1}{n}} = 0.
$$
Then from above spectral decomposition and fact that $\lambda_j > 0$, we can claim all the quasinilpotent vectors of $T$ vanish. Indeed, if $\phi$ is a quasinilpotent vector, then 
$$ \lim_{n \to \infty}|(e_j,T^n \phi)|^{\frac{1}{n}} = \lambda_j |(e_j,\phi)|^{1/n} = 0 ,$$
which gives us $(e_j,\phi)$ vanishes for all $j$. 
I would like to prove the same result (all the quasinilpotent vectors vanish) for the following operator, 
$$ T_k[\phi] = \int_{\Omega} \frac{e^{ik|x-y|}}{|x-y|}\phi(y)dy,$$
which is a also a compact operator on $L^2(\Omega)$. But we may not expect the above arguments work in our case since the spectral structure of $T_k$ is not clear. Perhaps we need turn to elliptic PDE theory for help.
Thank you very much in advance for any insight or suggestions.
 A: This is really no different from $k=0$. Your kernel is the kernel of the resolvent $(-\Delta-k^2)^{-1}$ on $L^2(\mathbb R^3)$. This is a standard fact, though I'm having trouble now locating a useful reference; see this question perhaps and especially the comment, except that there's a typo in the key formula, it should be $\sqrt{-z}$ in the exponent, not $-\sqrt{z}$.
Also, it's maybe not completely appropriate to call this operator the resolvent since $k^2\ge 0$ is in the spectrum, but the inverse exists (and is unbounded on $L^2(\mathbb R^3)$) since there is no point spectrum.
You're compressing this to $L^2(\Omega)$, so if $P$ denotes the corresponding projection, then $T_k=P(-\Delta-k^2)^{-1}P$, and this operator is also self-adjoint (edit: this claim is wrong, but $T_k$ is normal, which is enough; see the comments for clarification), and it doesn't have a kernel. Now we can complete the argument as above.
(It is true that $T_k$ is compact, but we don't need this since we can also in general compute
$$
\|T^n q\|^{1/n} = \left( \int t^{2n}\, d\rho(t) \right)^{1/(2n)} ,
$$
and here $\rho\not= 0$ is the spectral measure of $q$. Since $\rho(\{ 0\})=0$, the expression cannot go to zero.)
