Let $\Omega\subset\mathbb{R}^{d}$ be a bounded Lipschitz domain. Let $f\in H^{s}(\Omega)$, for $s\geq -1$. Then Lax-Milgram guarantees a unique solution to the Poisson problem

$$\begin{cases}\Delta u=f & \Omega \\ u=0 & \partial\Omega \end{cases} \tag{Poisson}$$

in $H_{0}^{1}(\Omega)$, where the boundary condition is satisfied in the sense of trace.

It is known that if we impose the stronger assumption that $\Omega$ is a $C^{k,1}$ (i.e. $k$ times continuously differentiable with $k^{th}$ derivative Lipschitz) domain, for some nonnegative integer $k$, and $f\in H^{k}(\Omega)$, then we have the *elliptic regularity* result $u\in H^{k+2}(\Omega)$ and moreover,

$$\|u\|_{H^{k+2}(\Omega)} \lesssim_{\Omega,k} \|f\|_{H^{k}(\Omega)}.$$

Now let $0<s<1$. If we assume that $\Omega$ is a $C^{2,1}$ domain, then by interpolation between the estimates $$\|u\|_{H^{2}(\Omega)} \leq C(\Omega,0)\|f\|_{L^{2}(\Omega)}$$ and $$\|u\|_{H^{3}(\Omega)} \leq C(\Omega,1)\|f\|_{H^{1}(\Omega)},$$

we have the estimate $$\|u\|_{H^{2+s}(\Omega)} \leq C(\Omega,s)\|f\|_{H^{s}(\Omega)}.$$

In the preceding argument, it seems that we're assuming much more regularity on the domain $\Omega$ then we need to. Heuristically interpolating between $C^{1,1}$ and $C^{2,1}$, my guess is that the natural regularity assumption on the domain $\Omega$ is $C^{2,s}$, but I do not know how to prove this, nor can I find a reference.

Question.Let $0<s<1$. Is it true that if $\Omega$ is $C^{2,s}$ domain and $f\in H^{s}(\Omega)$, then the solution $u$ to (Poisson) is in $H^{2+s}(\Omega)$, and moreover, $u$ satisfies the estimate $$\|u\|_{H^{2+s}(\Omega)} \leq C(\Omega,s) \|f\|_{H^{s}(\Omega)}?$$