 Solution

First perform homogenization of Dirichlet boundary data: put $v=u-\psi \in H_0^1(B_1^{+})$ and $g=f-\Delta \psi \in L^2(B_1^{+})$. Then $\Delta v=g$ weakly. Extend both functions oddly across the flat face:The zero Sobolev trace of $v$ makes this odd reflection an $H^1(B_1)$ function. This can also be seen by approximating $v$ by smooth compactly supported functions in $B_1^{+}$ and reflecting the approximants. Its $H^1$ norm is $\sqrt{2}$ times that of $v$, and $\Vert \tilde{g}{\Vert }_{L^2(B_1)}=\sqrt{2}\Vert g{\Vert }_{L^2(B_1^{+})}$.

For $\varphi \in C_c^{\infty }(B_1)$, the function $\varphi (x^{\prime },t)-\varphi (x^{\prime },-t)$ on the upper half-ball has zero trace on the flat face and vanishes near the curved face, hence is an admissible $H_0^1$ test. Splitting the integrals over the two half-balls and changing $t$ to $-t$ showsThus $\Delta \tilde{v}=\tilde{g}$ weakly throughout the full ball; there is no extra distribution on the reflecting face.

Apply part (i) to $B_{1/2}⋐B_1$, restrict to the upper half-ball, and then add $\psi$. Since $\Vert \Delta \psi {\Vert }_2\le C(n)\Vert \psi {\Vert }_{H^2}$, this gives $u\in W^{2,2}(B_{1/2}^{+})$ and the second-derivative estimate at a flat Dirichlet boundaryThe fixed radii leave distance $1/2$ for the interior estimate, so the constant depends only on dimension.