Flat Dirichlet Holder estimate by normal-jet lifting and reflection

Scale first to $R=1$. Write $x=(x^{\mathrm{\prime }},t)$, $t=x_3$. The boundary function $g(x^{\mathrm{\prime }})=f(x^{\mathrm{\prime }},0)$ on the closed unit disk is bounded by $M=\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$ and has seminorm at most $H=[f{]}_{\alpha }$. Extend it to all ${\mathbb{R}}^2$ by taking ${\inf }_z(g(z)+H\mathrm{\mid }{x}^{\mathrm{\prime }}-z{\mathrm{\mid }}^{\alpha })$ over the original disk and clamping to $[-M,M]$. The inequality $(a+b{)}^{\alpha }\le a^{\alpha }+b^{\alpha }$ shows that this infimum is $H$-Hölder; the original Hölder inequality and $z=x^{\mathrm{\prime }}$ show agreement on the disk. Clamping is one-Lipschitz and preserves this agreement. For $H=0$ take the constant extension. Thus no differentiability of the boundary forcing is assumed.

Choose a two-dimensional mollifier $\rho$ from [F2], set $g_t(x^{\mathrm{\prime }})=\int \rho (z)g(x^{\mathrm{\prime }}-tz)dz$ for $t>0$, and put $L(x^{\mathrm{\prime }},t)=t^2{g}_t(x^{\mathrm{\prime }})\mathrm{/}2$. A derivative of total order $k\ge 1$ in $(x^{\mathrm{\prime }},t)$ of the kernel $t^{-2}\rho ((x^{\mathrm{\prime }}-y^{\mathrm{\prime }})\mathrm{/}t)$ has integral zero and has the form $t^{-2-k}P((x^{\mathrm{\prime }}-y^{\mathrm{\prime }})\mathrm{/}t)$, with $P$ smooth and compactly supported. Subtract $g(x^{\mathrm{\prime }})$ under that differentiated integral to obtain $\mathrm{\mid }{D}^k{g}_t\mathrm{\mid }\le C_{k}H{t}^{\alpha -k}$ for $k=1,2,3$, and $\mathrm{\mid }{g}_t-g\mathrm{\mid }\le CH{t}^{\alpha }$. Differentiation is legitimate on every compact positive-$t$ interval because the differentiated kernels are bounded and compactly supported there.

Set $L=DL=0$ on the face. For $i,j<3$, direct differentiation gives $L_{ij}=t^2(g_t{)}_{ij}\mathrm{/}2$, $L_{it}=t(g_t{)}_i+t^2(g_t{)}_{it}\mathrm{/}2$ and $L_{tt}=g_t+2t(g_t{)}_t+t^2(g_t{)}_{tt}\mathrm{/}2$. Consequently $D^{2}L(x^{\mathrm{\prime }},t)$ differs by at most $CH{t}^{\alpha }$ from the boundary matrix whose sole nonzero entry is its normal-normal entry $g(x^{\mathrm{\prime }})$, and $\mathrm{\mid }{D}^{3}L\mathrm{\mid }\le CH{t}^{\alpha -1}$. These are actual boundary jets: $L\mathrm{/}{t}^2\to g\mathrm{/}2$, $L_t\mathrm{/}t\to g$, and $L_i\mathrm{/}t\to 0$. For points $X=(x^{\mathrm{\prime }},t),Y=(y^{\mathrm{\prime }},s)$ at distance $d$, if $d<\min (t,s)\mathrm{/}2$, integrate the third-derivative bound along their segment to get $CHd\min (t,s{)}^{\alpha -1}\le CH{d}^{\alpha }$. Otherwise $\min (t,s)\le 2d$ and $\max (t,s)\le 3d$; compare both Hessians to their boundary jets and use the Hölder bound of $g$. Thus $\mathrm{\parallel }L{\mathrm{\parallel }}_{C^{2,\alpha }({\mathbb{R}}^2\times [0,1])}\le C_{\alpha }(M+H)$ and $(\mathrm{\Delta }L)(x^{\mathrm{\prime }},0)=g(x^{\mathrm{\prime }})$.

Let $v=u-L$ and $F=f-\mathrm{\Delta }L$. Then $v=0$ and $F=0$ on the flat face, $\mathrm{\Delta }v=F$, and $\mathrm{\parallel }F{\mathrm{\parallel }}_{C^{0,\alpha }}+\mathrm{\parallel }L{\mathrm{\parallel }}_{C^{2,\alpha }}\le C_{\alpha }\mathrm{\parallel }f{\mathrm{\parallel }}_{C^{0,\alpha }}$. Because $v$ is $C^2$ up to the face and has zero face restriction, its tangential first and second derivatives vanish there. The continuous equation then gives $v_{tt}=0$ there. Its odd reflection $V(x^{\mathrm{\prime }},-t)=-v(x^{\mathrm{\prime }},t)$ is therefore $C^2$ across the face: tangential-tangential and normal-normal Hessian entries are odd with zero trace, and mixed entries are even with matching traces. Its equation is $\mathrm{\Delta }V=F_o$, with $F_o$ the odd reflection of $F$. For opposite-side points $(x^{\mathrm{\prime }},t),(y^{\mathrm{\prime }},-s)$, the zero trace gives $\mathrm{\mid }{F}_o(x^{\mathrm{\prime }},t)-F_o(y^{\mathrm{\prime }},-s)\mathrm{\mid }\le [F{]}_{\alpha }(t^{\alpha }+s^{\alpha })\le 2^{1-\alpha }[F{]}_{\alpha }\mathrm{\mid }(x^{\mathrm{\prime }},t)-(y^{\mathrm{\prime }},-s){\mathrm{\mid }}^{\alpha }$. Same-side pairs use the original bound. Hence $F_o\in C^{0,\alpha }(B_1)$ with controlled norm. The original forcing has not been reflected without first removing its possibly nonzero trace.

Choose a fixed radial $\eta$ equal to one on $B_{3\mathrm{/}4}$ and supported in $B_{7\mathrm{/}8}$, extend $G=\eta F_o$ by zero, and put $W=-NG$. The Hölder product inequality and the support buffer give $\mathrm{\parallel }G{\mathrm{\parallel }}_{C^{0,\alpha }({\mathbb{R}}^3)}\le C\mathrm{\parallel }{F}_o{\mathrm{\parallel }}_{C^{0,\alpha }(B_1)}$: a point in $B_{7\mathrm{/}8}$ is at distance at least $1\mathrm{/}8$ from the exterior of $B_1$. By [F1], $\mathrm{\Delta }W=G$ and $\mathrm{\parallel }W{\mathrm{\parallel }}_{C^{2,\alpha }(B_{3\mathrm{/}4})}\le C_{\alpha }\mathrm{\parallel }f{\mathrm{\parallel }}_{C^{0,\alpha }}$. The radial kernel and cutoff are even in the normal coordinate, so $W$ is odd, and on the upper half-space it is exactly the reflected Green potential $-{\int }_{y_3>0}(\mathrm{\Phi }(x-y)-\mathrm{\Phi }(x-y^{\ast }))\eta (y)F(y)dy$, with $y^{\ast }=(y^{\mathrm{\prime }},-y_3)$. This reflection uses the corrected zero-trace Hölder datum, not a discontinuous odd extension of $f$.

The remainder $H_0=V-W$ is harmonic in $B_{3\mathrm{/}4}$ with $\mathrm{\parallel }{H}_0{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty }}+C_{\alpha }\mathrm{\parallel }f{\mathrm{\parallel }}_{C^{0,\alpha }}$. By [F3], its sphere averages equal its center value. Average those identities against a fixed smooth radial kernel supported in $B_{1\mathrm{/}16}$. On $B_{5\mathrm{/}8}$ this writes $H_0=\kappa \ast H_0$ using only values in $B_{3\mathrm{/}4}$; differentiation falls on $\kappa$, giving $\mathrm{\parallel }{D}^j{H}_0{\mathrm{\parallel }}_{\mathrm{\infty };B_{1\mathrm{/}2}}\le C_j\mathrm{\parallel }{H}_0{\mathrm{\parallel }}_{\mathrm{\infty }}$ for $j\le 3$. Integration along segments then bounds $[D^2{H}_0{]}_{\alpha ;B_{1\mathrm{/}2}}$. Combining these bounds with $u=L+W+H_0$ on the upper half-ball proves the unit-radius estimate, including its flat-face limits.

For general $R$, apply the unit result to $u_R(x)=u(Rx)$ and $f_R(x)=R^{2}f(Rx)$. Derivatives scale by $R^j$ and the Hessian Hölder seminorm by $R^{2+\alpha }$. Thus the displayed estimate follows with the same constant. Only the given $C^2$ boundary regularity was used to justify reflection; no bound on higher derivatives of $u$ or $f$ has entered.