# Quasinilpotent vectors of Newton potential vanish

Suppose $$\Omega$$ is a smooth bounded domain in $$\mathbb{R}^3$$. Consider the Newton potential $$$$T [\phi](x) = \int_{\Omega} \frac{1}{|x-y|} \phi(y)dy.$$$$ It is well know that $$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. Commented Mar 26, 2019 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. Commented Mar 28, 2019 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. Commented Mar 28, 2019 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. Commented Mar 29, 2019 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. Commented Mar 29, 2019 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 $-\Delta$. Commented Mar 29, 2019 at 9:43