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
