A have a question related to the boundary regularity of a solution of a Poisson equation on a bounded domain. But to make the question easier to pose I will state it on $ R\_+^2:=\{ x \in R^2:x\_2>0\}$ but now the statement may not be true, but keep in mind i am only concerned about local issues near the boundary. 


So consider $$ \Delta \phi(x) = \frac{f(x)}{x\_2^{2-\gamma}}$$ in $ R\_+^2$ with $ \phi=0$ on $ x\_2=0$ and where $f$ is bounded and where $ 0<\gamma$ is small. Using a scaling argument one can show that 


$$ \sup\_{R\_+^2} x\_2^{-\gamma} | \phi | + \sup\_{R\_+^2} x\_2^{1-\gamma} | \nabla \phi | \le C \sup\_{R\_+^2} |f|$$. (This is the statement that may not be exactly correct since really i am on a bounded domain and $x\_2=dist(x, \partial \Omega)$). 


In any case my question is in obtaining more regularity for the gradient of $ \phi$ in the tangential direction. So suppose $ f$ is $C^1$ in $ x\_1$ direction then we have something like $$ \Delta \phi\_{x\_1}= \frac{f\_{x\_1}(x)}{x\_2^{2-\gamma}}$$ in $ R\_+^2$ with $ \phi\_{x\_1}=0$ on $ x\_2=0$ (at least morally). We can then see that we get added regularity in $x\_1$ direction. 


QUESTION. So assuming $f$ is only bounded i would like to obtain some added regularity in $x\_1$ direction. I assume the proof is quite trivial and a non-research level question but I have attempted for a while and I am having difficulty. 


thanks



elliptic boundary regularity, tangential regularity
Boundedness of $f$ isn't enough to get extra regularity for $\phi\_1$. Take for example $\phi = r^{\gamma}\sin(\theta)$, which solves an equation of the desired form with $f = (\gamma^2-1)\sin(\theta)$, which is discontinuous at the origin. Then 
$$\phi\_1(1/{\sqrt{2}},1/{\sqrt{2}}) = \frac{1}{2}(\gamma-1),$$
so by homogeneity
$$\phi\_1(\epsilon,\epsilon) \sim \epsilon^{\gamma-1}$$
and the estimate you got from scaling can't be improved.


