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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?

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  • $\begingroup$ What is $OPS$ here? $\endgroup$ – Andrew Jan 26 '18 at 20:17
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You can find a lot on this operator in the book Agranovich, Mikhail S. Sobolev spaces, their generalizations and elliptic problems in smooth and Lipschitz domains. Revised translation of the 2013 Russian original. Springer Monographs in Mathematics. Springer, Cham, 2015.

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