# 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.

• 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. – Bowen Mar 26 at 12:42

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.)
• 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. – Bowen Mar 28 at 1:28
• @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. – Christian Remling Mar 28 at 18:32
• 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. – Bowen Mar 29 at 8:50
• 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. – Bowen Mar 29 at 9:21
• 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 $-\Delta$. – Bowen Mar 29 at 9:43