# regularity of solution of linear elliptic PDE

I am interested in the boundary regularity of solutions of $L(u) = f(x) \ge 0$ in $\Omega$ with zero Dirichlet boundary conditions, here $L(u) = (-\Delta)^\frac{\alpha}{2}$ where $0 < \alpha <2$.

I have found results like:
- if $f$ bounded and $\alpha<1$ then $u \in C^{t}$ (to the boundary) for all $t < \alpha$.

• if $\alpha=1$ and $f$ is smooth with $f=0$ on the boundary then $u$ is $C^{2,\gamma}$ (to the boundary) for some $\gamma >0$.

My question is related to the following calculation which seems to contradict the above results. We let $G(x,y)$ denote the Greens function associated to $L$. One can show that

$G(x,y)= \frac{ \delta(x)^\frac{\alpha}{2} \delta(y)^\frac{\alpha}{2} }{ |x-y|^{N-\alpha} \left( \max\{ |x-y|^2, \delta(x)\delta(y) \} \right)^\frac{\alpha}{2} }$

or at least is bounded above and below by constant multiples of this and where $\delta(x)$ is the distance from $x$ to the boundary of $\Omega$. Here $N$ is the space dimension of $\Omega$.

So using the integral representation and taking $f(x) \ge 0$ smooth and zero in a neighborhood of the boundary of $\Omega$ i seem to be able to show that $u(x)$ cannot be Holder continuous of order $> \frac{\alpha}{2}$ at the boundary. To do this let $x_m$ be a sequence that converges to $x_0$ which lies on the boundary and assume that $x_m$ approaches the boundary at right angles. Use the above representation to write out

$u(x_m)$ and note that one can calculate the maximum for big enough $m$ since $f$ is identially zero near the boundary. Then one uses this to get a lower bound on the Holder quotient of $u$ at $x_m$ and $x_0$ and arrives at a contradiction. (I will add more details of the exact calculation if this would help).

In anycase I cannot spot the error in my logic.

Any comments would be apprecaited. thanks