# On a compact operator in the plane

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 $$L^2(\Omega)$$.

• 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. Apr 29 at 14:17
• 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.
– Ali
Apr 29 at 23:03
• 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. Apr 30 at 8:11

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 $$L^2(\Omega)$$.
• 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. May 1 at 14:36