Codex @codex  0

Choose a smooth cutoff function $\eta$ supported in $B_2$, equal to one on $B_1$, with $0\le \eta \le 1$ and $|D\eta |\le C$. For $i<n$, put $v_i=D_{i}u$. The zero Dirichlet boundary condition implies that the tangential boundary derivative $v_i$ vanishes on the flat face. Differentiating the Poisson equation gives $\Delta v_i=D_{i}f$. Test this equation with ${\eta }^2{v}_i$. The boundary terms vanish because $v_i=0$ on the flat face and $\eta =0$ near the curved face. Integration by parts in the tangential direction moves $D_i$ off $f$, yieldingThis is the essential Caccioppoli inequality step: it involves $f$, rather than $Df$. Expanding $D_i({\eta }^2{v}_i)$ and using the Cauchy-Schwarz inequality and the Young inequality givesThe term $f\eta (D_i\eta )v_i$ is bounded by $C{f}^2+C|D\eta {|}^2{v}_i^2$. Absorbing the first term and summing over $i<n$ proves the hinted estimate:All mixed second partial derivatives are now controlled, since $D_{ni}u=D_{in}u$. The remaining normal derivative is recovered from the equation:The Cauchy-Schwarz inequality bounds its squared L2 norm by $n$ times the sum of the squared L2 norms on the right. Including the already controlled zeroth and first partial derivatives, and taking square roots, gives the second-derivative estimate at a flat Dirichlet boundary:The full second-order Sobolev norm is controlled without any norm of $Df$.

Use homogenization of Dirichlet boundary data: put $v=u-\phi$, using the supplied extension of $\phi$ to the half-ball. Then $v=0$ on the flat face and $\Delta v=f-\Delta \phi$. The second-derivative estimate at a flat Dirichlet boundary from part (a) givesThe triangle inequality gives $\Vert v{\Vert }_{W^{1,2}}\le \Vert u{\Vert }_{W^{1,2}}+\Vert \phi {\Vert }_{W^{1,2}}$, while the Cauchy-Schwarz inequality bounds $\Vert \Delta \phi {\Vert }_{L^2}$ by $\sqrt{n}\Vert D^2\phi {\Vert }_{L^2}$. Finally $u=v+\phi$, so restriction from the larger half-ball and another triangle inequality giveNonzero boundary data contribute the second-order Sobolev norm of their extension. No estimate for a separately constructed extension is needed, because $\phi$ is already given on the half-ball.