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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.

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  • $\begingroup$ Just a small observation: suppose $\phi$ is a quasinilpotent vector, to prove the vanishing of $\phi$, it suffices to prove the vanishing of $T^l \phi$ for some $l$. Hence, we may further assume the quasinilpotent vector $\phi$ is continuous. $\endgroup$ – Bowen Mar 26 at 12:42
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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.)

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  • $\begingroup$ Thank you very much for your answer. You wrote that Tk is a self-adjoint operator (unbounded). But, if we use the original definition and check directly, we will find Tk is not self-adjoint. The kernel of the adjoint operator is $\frac{e^{-ik|x-y|}}{|x-y|}$. Of course, both of them are the kernels of $(\Delta-k^2)^{-1}$. Maybe I miss some basic and important fact. Could you explain a bit more on this point? Thanks. $\endgroup$ – Bowen Mar 28 at 1:28
  • $\begingroup$ @Bowen: Right, that was a bit sloppy. What I denoted by $(-\Delta-k^2)^{-1}$ is really something like $\lim_{y\to 0+} (-\Delta-(k^2+iy))^{-1}$ (without worrying now about in what sense exactly this limit is supposed to be taken), and taking the adjoint of this operator changes $iy$ to $-iy$, so we're now approaching from the lower half plane. The operators are normal, though (again, not thinking carefully about domain issues), and that's enough for the argument to work. $\endgroup$ – Christian Remling Mar 28 at 18:32
  • $\begingroup$ Thank you very much for your comment. Here are some of my own understandings. The resolvent $(-\Delta - z)^{-1}$ ($ z \in \mathbb{C}\backslash [0,+\infty)$) are normal operators since up the $L^2$ Fourier transform they can be represented as the multiplication operators. Normal is directly from the commutativity of scalar. Also, if we write it as the convolution of the kernel(well-defined since exponential decay at infinity and locally weak singularity), then we can directly check they are normal operators by the change of variable. $\endgroup$ – Bowen Mar 29 at 8:50
  • $\begingroup$ for $[0,+\infty)$ continuous spectrum, $-\Delta - z$ (definition domain is $H^2$) is self-adjoint and injective which implies its inverse with dense domain is also self-adjoint. This should be a standard result on self-adjoint operator. (I am still not clear about the contracdition with the integral form). For the domain issue, it seems to me that it is the key point. Denote by $P$ the projection, then its adjoint $P^*$ is the zero extension. But we can not argue $P(-\Delta - z)^{-1}P^*$ is also normal from normality of $(-\Delta - z)^{-1}$. Any corrections would be very helpful. $\endgroup$ – Bowen Mar 29 at 9:21
  • $\begingroup$ In fact, I am thinking whether there is a possibility that based on the assumption that $\phi$ is smooth we do some (local) estimate to argue that $\phi$ must vanish pointwise. And the estimate should be strongly related to the singularity $\frac{1}{|x-y|}$, or the fact that the operator is pseudodifferential operators of order $2$, or 'some coercivity' from $-\Delta$. $\endgroup$ – Bowen Mar 29 at 9:43

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