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Let $\Omega \subset \mathbb R^2$ be a bounded domain with a smooth boundary. Let $$\bar{\partial}= \frac{1}{2} ( \partial_{x^1} + i \,\partial_{x^2}),$$ and let $G: L^2(\Omega)\to H^2(\Omega)$ be the Dirichlet Green's function on $\Omega$ that is defined by $GF:=u$ where $u\in H^2(\Omega)$ is the unique solution to $-\Delta u =F$ on $\Omega$ with $u|_{\partial \Omega}=0$.

Let us define the compact operator $T:L^2(\Omega)\to L^2(\Omega)$ via $$ T F = \bar{\partial}GF.$$

My question is to try to understand the spectrum of the operator $T$ and whether for instance it admits a discrete spectrum that provides a basis for $L^2(\Omega)$.

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  • $\begingroup$ The only thing I see is compactness: $G$ is bounded form $L^2$ to $H^2$ and then $T$ is bounded form $L^2$ to $H^1$, hence compact from $L^2$ into itself. $\endgroup$
    – Giorgio Metafune
    Commented Apr 29, 2023 at 14:17
  • $\begingroup$ It actually turns out that the spectrum contains only zero and nothing else. So I guess the best next question to ask is what is the best resolvent estimates for T near zero. $\endgroup$
    – Ali
    Commented Apr 29, 2023 at 23:03
  • $\begingroup$ Nice. How can you prove this? In vol 1 of Dunford Schwartz there is a result saying that the linear span of the generalized eigenfunctions is dense if the resolvent decays like $1/|\lambda|$ on certain arcs. Then this is not your case. $\endgroup$
    – Giorgio Metafune
    Commented Apr 30, 2023 at 8:11

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I claim that the spectrum contains only zero, which amounts to proving that $T-\lambda$ has a bounded inverse for $\lambda$ away from zero. To see this, we need to construct a bounded inverse which I will denote by $A$. Let us first define the linear operator $A: C^{\infty}_c(\Omega)\to L^2(\Omega)$ by $u:=AF$ where $u$ is the unique solution to $$ -e^{\frac{1}{4\lambda}z} \Delta(e^{-\frac{1}{4\lambda}z}u) =\frac{1}{\lambda}\bar \partial u - \Delta u = \frac{1}{\lambda} \Delta F,\quad \text{on $\Omega$},$$ subject to $u|_{\partial \Omega}=0$. Note that $$ \|AF\|_{L^2(\Omega)} \leq C_\lambda \|F\|_{L^2(\Omega)}.$$ Subsequently, we can define $A:L^2(\Omega)\to L^2(\Omega)$ by continuous extension and density of $C^{\infty}_c(\Omega)$ in $L^2(\Omega)$.

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  • $\begingroup$ I see. Writing $Tf-\lambda f=F$ and setting $u=Gf$, then $u=-\frac 1 \lambda e^{z/4 \lambda} G(F e^{-z/4\lambda})$ which shows that the resolvent satisfies an exponential estimate near zero (in 1d this is in fact the Volterra operator). Since 0 is not an eigenvalue ($Tf=0$ gives $f=\partial Tf=0$), 0 is not a pole of the resolvent, otherwise would be an eigenvalue. $\endgroup$
    – Giorgio Metafune
    Commented May 1, 2023 at 14:36

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