Resolvent of the Laplacian as a pseudodifferential operator and its single layer potential

In M.Taylor's book "Partial differential equations II" it is shown that the fundamental solution $E(x,y)$ of the Laplacian equation gives rise to an elliptic pseudodifferential operator $S$ on the boundary $\partial \Omega$ of a domain $\Omega$ (for simplicity, let $\Omega \subset \mathbb R^3$ and $\partial \Omega$ be smooth) by the formula: $$Sf (x) = \int_{\partial \Omega} E(x,y)f(y) \, dS(y), \\ E(x,y) = \frac{1}{4\pi|x-y|},$$ where $dS(y)$ is the surface measure. The fact $S \in OPS^{-1}(\partial \Omega)$ follows from $E \in OPS^{-2}(\mathbb R^3)$ and from the representation formula $$Sf(x) = \lim_{y \in \Omega, y\to x} \int_{\mathbb R^3} E(x,y)(f\delta_{\partial \Omega}) dy, \quad x \in \partial \Omega,$$ where $\delta_{\partial \Omega}$ is the delta-function concentrated on $\partial \Omega$. Ellipticity follows from the explicit expression for the symbol of $S$ (the symbol of $E$ restricted to $\partial \Omega$ and integrated along conormal direction).

Here is my question. Consider, instead of $E$, the limit value of resolvent $R_+(\lambda)=R(\lambda+i0)$, $\lambda > 0$, for the operator $-\Delta$. Its integral kernel is $$R_+(x,y,\lambda) = \frac{e^{i\sqrt \lambda |x-y|}}{4\pi |x-y|}.$$ In analogy with $S$ we can define $$S_+(\lambda)f(x) = \int_{\partial \Omega} R_+(x,y,\lambda)f(y) \,dS(y).$$ Is it true that $R_+(\lambda) \in OPS^{-2}(\mathbb R^3)$? If yes, what is its symbol? Is it true that $S_+(\lambda) \in OPS^{-1}(\partial \Omega)$ and is elliptic?

• What is $OPS$ here? Jan 26 '18 at 20:17