I have a question on the solvability of Neumann boundary problems with singular data. To state my question, let $\Omega$ be a bounded Lipschitz domain (open and connected) in $\mathbb{R}^n$.
In the case of the Dirichlet problem $$ \triangle u =0\quad \text{in } \Omega \quad \text{and} \quad u=g\quad \text{on } \partial \Omega.$$ it is proved by Dahlberg (1979) that for any $g\in L^{2}(\partial\Omega)$, there exists a unique harmonic function $u$ such that $u\rightarrow g$ nontangentially a.e. on $\partial\Omega$. Also, $u\in H^{1/2}(\Omega)$.
In the case of Neumann problem, $$ \triangle u =0\quad \text{in } \Omega \quad \text{and} \quad \frac{\partial u}{\partial \nu}=g\quad \text{on } \partial \Omega.$$ it is proved by Jerison and Kenig (1981) that for any $g\in L^{2}(\partial\Omega)$, there exists a unique harmonic function $u$ such that $\nabla u \cdot \nu \rightarrow g$ nontangentially a.e. on $\partial\Omega$. Also, $u\in H^{3/2}(\Omega)$.
My question is motivated by these results and the result of Fabes-Mendez-Mitrea (1998).
For $0<s<1$ and $1<p<\infty$, let $B_{s}^{p}(\partial\Omega)$ be a Besov space and define $B_{-s}^{p'}(\partial\Omega) = (B_{s}^{p}(\partial\Omega))^{*}$. They proved that under some condition on $(s,p)$, for any $f\in L^{p}_{s+1/p-2,0}(\Omega)$ and $g \in B^{p}_{-(1-s)}(\partial\Omega)$ satisfying the compatibility condition $\left<f,1 \right>=\left<g,1 \right>$, there exists a solution $u\in L^{p}_{s+1/p}(\Omega)$ of $$ \triangle u =f\quad \text{in } \Omega \quad \text{and} \quad \frac{\partial u}{\partial \nu}=g\quad \text{on } \partial \Omega$$ in the sense that $$-\left<\nabla u, \nabla \phi \right> = \left<f,\phi \right>-\left<g,\mathrm{Tr}\, \phi \right> $$ for all $\phi \in L^{p'}_{1-s+1/p'}(\Omega)$.
My question is the following.
Question. Given $g \in B^{p}_{-1}$, can we find a harmonic function $u$ satisfying the Neumann problem in 'some sense'? Or just $(H^1(\partial\Omega))^*$, i.e., the case when $p=2$.
I cannot find any relevant papers and any comments on this question.
Thank you for your reading.
