On a compact operator in the plane

Question

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 $L^2(\Omega)$.

Answer 1

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 $L^2(\Omega)$.