$\newcommand\Id{\mathrm{Id}}$Assumptions and Notations :
$\Omega$ is a bounded Lipschitz domain in $\mathbb R^2$, $\Gamma$ denotes its boundary and $n$ is the normal vector to the boundary $\Gamma$,
$K'_0$ is the normal derivative of the single layer potential associated to the Laplace equation:
$$K'_0 u(x)=\int_{\Gamma}u(y) \partial_{n(x)} g_0(x-y) ds(y)$$
with $g_0(x-y)=\frac{1}{2\pi} \ln(\frac{1}{|x-y|})$.
The question:
If we choose the real number $\lambda$ such that the operator $\lambda\Id - K'_0$ is Fredholm on the Sobolev space $H^{-\frac 1 2}(\Gamma)$, can we then find an explicit expression of a regularizer for that operator, which will be denoted by $(\lambda\Id - K'_0)'$? (at least, for some particular domains as those with corners (polyhedral domain), using the local Mellin transformation I suppose).
Here, $A'$ a regularizer of $A$ means a bilateral regularizer, i.e. such that
$$A' A = \Id + K_l,\qquad A A' = \Id + K_r,$$
$K_l$ and $K_r$ are compact operators and $\Id$ is the identity map).
Note that if $\Gamma$ is smooth, then $K'_0$ defines a compact operator on $H^{-\frac 1 2}(\Gamma)$ and the answer is trivial.
