Local nonlinear constraint construction with explicit analytic obligations

Uniform contraction bound. Define $T_{ϵ}(x):=x-L^{-1}F(x,ϵ)$ wherever $(x,ϵ)\in U$. Then $D_x{T}_0(0)=I-L^{-1}L=0$. By continuity of $D_{x}F$, choose $\rho >0$, ${ϵ}_1>0$, and $q\in (0,1)$ such that the closed product ${\bar{B}}_X(0,\rho )\times [-{ϵ}_1,{ϵ}_1]$ lies in $U$ and $\mathrm{\parallel }{D}_x{T}_{ϵ}(x)\mathrm{\parallel }\le q$ there. For fixed $ϵ$ and $x,y$ in the ball, the curve $s\mapsto T_{ϵ}(y+s(x-y))$ is continuous and differentiable with derivative norm at most $q\mathrm{\parallel }x-y\mathrm{\parallel }$; [F5] therefore gives $\mathrm{\parallel }{T}_{ϵ}(x)-T_{ϵ}(y)\mathrm{\parallel }\le q\mathrm{\parallel }x-y\mathrm{\parallel }$.

Fixed points and local uniqueness. Since $T_0(0)=0$, continuity of $F$ allows ${ϵ}_0\in (0,{ϵ}_1]$ such that $\mathrm{\parallel }{T}_{ϵ}(0)\mathrm{\parallel }<(1-q)\rho$ for $\mathrm{\mid }ϵ\mathrm{\mid }<{ϵ}_0$. Step 1.1 then gives $\mathrm{\parallel }{T}_{ϵ}(x)\mathrm{\parallel }\le q\rho +\mathrm{\parallel }{T}_{ϵ}(0)\mathrm{\parallel }<\rho$ whenever $\mathrm{\parallel }x\mathrm{\parallel }\le \rho$, so $T_{ϵ}$ maps the nonempty closed ball ${\bar{B}}_X(0,\rho )$ into itself and its fixed point lies in the open ball. It is complete by [F1] and [F6], and is a contraction by step 1.1. By [F3] it has exactly one fixed point $x(ϵ)$, and $T_{ϵ}(x)=x$ is equivalent to $F(x,ϵ)=0$ because $L^{-1}$ is injective. Thus the zero set in the product neighborhood is the graph of $x$, and $x(0)=0$.

Invertibility of nearby partial derivatives. Let $B(X)$ be the bounded linear operators on $X$, with the norm of [F7]. It is complete in the operator norm: an operator-norm Cauchy sequence converges pointwise to a linear map, the uniform norm bounds make that limit bounded, and the same Cauchy estimate gives convergence in operator norm. Shrinking the product neighborhood if necessary, continuity of $D_{x}F$ gives $\mathrm{\parallel }{L}^{-1}{D}_{x}F(x,ϵ)-I\mathrm{\parallel }\le q<1$. Put $E=L^{-1}{D}_{x}F-I$. The series ${\sum }_{j=0}^{\mathrm{\infty }}(-E{)}^j$ converges in $B(X)$ because ${\sum }_j\mathrm{\parallel }E{\mathrm{\parallel }}^j<\mathrm{\infty }$, and its finite partial sums multiplied on either side by $I+E$ converge to $I$; thus $D_{x}F(x,ϵ)$ is invertible. The inverse is continuous, since $A^{-1}-B^{-1}=A^{-1}(B-A)B^{-1}$ for invertible nearby operators with uniformly bounded inverses.

Continuity and a local Lipschitz bound. For $ϵ,\eta$ in a smaller interval, the fixed-point equations and step 1.1 yield $\mathrm{\parallel }x(ϵ)-x(\eta )\mathrm{\parallel }\le (1-q{)}^{-1}\mathrm{\parallel }{T}_{ϵ}(x(\eta ))-T_{\eta }(x(\eta ))\mathrm{\parallel }$. The derivative $D_{ϵ}F$ is bounded on a sufficiently small neighborhood of $(0,0)$, since it is continuous there; applying [F5] to the parameter segment bounds the right-hand side by $C\mathrm{\mid }ϵ-\eta \mathrm{\mid }$. Hence $x$ is locally Lipschitz, in particular continuous.

Differentiation of the branch. Write $\delta x=x(ϵ+h)-x(ϵ)$. Step 3.1 gives $\mathrm{\parallel }\delta x\mathrm{\parallel }=O(\mathrm{\mid }h\mathrm{\mid })$. Fréchet differentiability at $(x(ϵ),ϵ)$ yields $0=D_{x}F(x(ϵ),ϵ)\delta x+D_{ϵ}F(x(ϵ),ϵ)h+o(\mathrm{\parallel }\delta x\mathrm{\parallel }+\mathrm{\mid }h\mathrm{\mid })$. Divide by $h\mathrm{\ne }0$; the remainder divided by $\mathrm{\mid }h\mathrm{\mid }$ tends to zero because $\mathrm{\parallel }\delta x\mathrm{\parallel }=O(\mathrm{\mid }h\mathrm{\mid })$. Multiplying by the bounded inverse from step 2.2 gives $\delta x\mathrm{/}h\to -D_{x}F(x(ϵ),ϵ{)}^{-1}{D}_{ϵ}F(x(ϵ),ϵ)$. This proves differentiability and the displayed formula. The formula is continuous in $ϵ$ because $x$, $DF$, and the nearby operator inverse are continuous. Therefore $x$ is $C^1$ on a possibly smaller interval, which proves the statement.

Extend $v\in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ by zero. For fixed $(x_2,x_3)$, the fundamental theorem of calculus gives $v(x_1,x_2,x_3)={\int }_{-R}^{x_1}{\mathrm{\partial }}_{1}v(s,x_2,x_3)ds$. Cauchy--Schwarz bounds its square by $2R{\int }_{-R}^R\mathrm{\mid }{\mathrm{\partial }}_{1}v{\mathrm{\mid }}^{2}ds$. Integrating over the cube gives $\mathrm{\parallel }v{\mathrm{\parallel }}_2^2\le (2R{)}^2\mathrm{\parallel }{\mathrm{\partial }}_{1}v{\mathrm{\parallel }}_2^2\le (2R{)}^2\mathrm{\parallel }\mathrm{\nabla }v{\mathrm{\parallel }}_2^2$. Pass to an $H^1$-convergent compactly supported smooth approximation to obtain the same inequality on $H_0^1$.

Suppose $u\in C^1(\bar{\mathrm{\Omega }})$ and $u=0$ on the boundary. Take a finite graph cover and a subordinate partition from [F3], with an additional interior term. In a boundary chart $\mathrm{\Psi }(y^{\mathrm{\prime }},t)=(y^{\mathrm{\prime }},t+\phi (y^{\mathrm{\prime }}))$, $t>0$ is the inward graph coordinate; its Jacobian is one and its differential and inverse differential are bounded on the compact chart. Each localized pullback $z$ is $C^1$, vanishes at $t=0$, and is supported away from artificial chart edges. The identity $z(y^{\mathrm{\prime }},t)={\int }_0^t{\mathrm{\partial }}_{t}z(y^{\mathrm{\prime }},s)ds$ gives $\mathrm{\mid }z\mathrm{\mid }\le Ct$. Choose a smooth ${\eta }_{\delta }(t)$ equal to zero for $t\le \delta$, one for $t\ge 2\delta$, with $\mathrm{\mid }{\eta }_{\delta }^{\mathrm{\prime }}\mathrm{\mid }\le C\mathrm{/}\delta$. The support of $z-z{\eta }_{\delta }$ is a layer of volume $O(\delta )$; both $Dz$ and $z{\eta }_{\delta }^{\mathrm{\prime }}$ are uniformly bounded there. Thus $\mathrm{\parallel }z-z{\eta }_{\delta }{\mathrm{\parallel }}_{H^1}=O({\delta }^{1\mathrm{/}2})$. Transport back and sum the finite partition. The resulting $C^1$ functions have compact support in $\mathrm{\Omega }$, and ordinary mollification at a radius smaller than their distance to the boundary approximates each of them in $H^1$ by $C_c^{\mathrm{\infty }}(\mathrm{\Omega })$. Hence $u\in H_0^1$.

Conversely, let $u\in C^1(\bar{\mathrm{\Omega }})\cap H_0^1$ and choose $u_j\in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ converging in $H^1$. In a fixed smaller graph chart, for a $C^1$ pullback $w$ and $0<t<h$, the identity $w(y^{\mathrm{\prime }},0)=w(y^{\mathrm{\prime }},t)-{\int }_0^t{\mathrm{\partial }}_{t}w(y^{\mathrm{\prime }},s)ds$ gives $\mathrm{\mid }w(y^{\mathrm{\prime }},0){\mathrm{\mid }}^2\le 2\mathrm{\mid }w(y^{\mathrm{\prime }},t){\mathrm{\mid }}^2+2h{\int }_0^h\mathrm{\mid }{\mathrm{\partial }}_{t}w(y^{\mathrm{\prime }},s){\mathrm{\mid }}^{2}ds$. Average over $t\in (0,h)$ and integrate over a smaller fixed tangential base. Bounded chart Jacobians and differentials give $\mathrm{\parallel }w(\cdot ,0){\mathrm{\parallel }}_{L^2}^2\le C_h\mathrm{\parallel }w{\mathrm{\parallel }}_{H^1}^2$. Apply this to the localized pullback of $u-u_j$; its boundary value is the localized boundary value of $u$. The right side tends to zero, so that boundary value vanishes almost everywhere. Continuity makes it vanish everywhere: a nonzero value would persist on an open tangential disk of positive area. The finite cover proves $u{\mathrm{\mid }}_{\mathrm{\partial }\mathrm{\Omega }}=0$.

Steps 1.2 and 2.1 identify zero classical boundary values with the zero-boundary Sobolev closure for the stated $C^1$ representatives. Together with step 1.1 they establish both claims; no unproved trace-to-classical regularity upgrade has been used.

By scaling it suffices to prove the estimate for $R=1$. Direct differentiation gives $D_i\mathrm{\Phi }(z)=-z_i\mathrm{/}(4\pi \mathrm{\mid }z{\mathrm{\mid }}^3)$ and $K_{ij}(z)=D_{ij}\mathrm{\Phi }(z)=(3z_i{z}_j\mathrm{\mid }z{\mathrm{\mid }}^{-5}-{\delta }_{ij}\mathrm{\mid }z{\mathrm{\mid }}^{-3})\mathrm{/}(4\pi )$. Therefore $\mathrm{\mid }D\mathrm{\Phi }\mathrm{\mid }\le C\mathrm{\mid }z{\mathrm{\mid }}^{-2}$, $\mathrm{\mid }K\mathrm{\mid }\le C\mathrm{\mid }z{\mathrm{\mid }}^{-3}$ and $\mathrm{\mid }DK\mathrm{\mid }\le C\mathrm{\mid }z{\mathrm{\mid }}^{-4}$. The spherical integrals satisfy ${\int }_{S^2}{\theta }_i{\theta }_{j}d\sigma =4\pi {\delta }_{ij}\mathrm{/}3$ by coordinate reflections, permutations and ${\sum }_i{\theta }_i^2=1$. The divergence theorem consequently gives zero integral of $K_{ij}$ on every concentric annulus and the correction $-{\delta }_{ij}\mathrm{/}3$ for the inner-sphere Hessian flux.

To justify the second derivatives for merely Hölder data, choose from [F5] a smooth radial cutoff ${\chi }_{\epsilon }$ equal to zero in $B_{\epsilon }$ and one outside $B_{2\epsilon }$, with derivatives bounded by $C{\epsilon }^{-j}$, and put $q_{\epsilon }=\mathrm{\Phi }{\chi }_{\epsilon }$. For each fixed $\epsilon >0$, [F3] differentiates $q_{\epsilon }\ast g$. On any fixed compact set choose one large containing radius $M$ for all translated supports. Splitting off $g(x)$ inside $B_M(x)$ and using the boundary flux from step 1.1 gives $D_{ij}(q_{\epsilon }\ast g)(x)={\int }_{B_M(x)}{D}_{ij}{q}_{\epsilon }(x-y)(g(y)-g(x))dy-\frac{{\delta }_{ij}}{3}g(x)\mathrm{.}$ The integral converges uniformly on that compact set to the same expression with $D_{ij}{q}_{\epsilon }$ replaced by $K_{ij}$: the omitted ball and transition annulus are bounded by $C_{\alpha }[g{]}_{\alpha }{\epsilon }^{\alpha }$, using $\mathrm{\mid }g(y)-g(x)\mathrm{\mid }\le [g{]}_{\alpha }\mathrm{\mid }x-y{\mathrm{\mid }}^{\alpha }$ and ${\int }_0^{2\epsilon }{r}^{\alpha -1}dr$. The potentials and first derivatives likewise converge uniformly, with errors $O({\epsilon }^2\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }})$ and $O(\epsilon \mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }})$. Apply [F3] first to the functions and their first derivatives, then to the first and second derivatives on coordinate segments. This proves $Ng\in C^2$ and the absolutely convergent cancellation formula $D_{ij}Ng(x)={\int }_{B_M(x)}{K}_{ij}(x-y)(g(y)-g(x))dy-\frac{{\delta }_{ij}}{3}g(x)\mathrm{.}$

For the Hölder estimate take $x\mathrm{\ne }{x}^{\mathrm{\prime }}$ in a fixed compact box, $d=\mathrm{\mid }x-x^{\mathrm{\prime }}\mathrm{\mid }$, and $m=(x+x^{\mathrm{\prime }})\mathrm{/}2$. Choose a large ball $B_M(m)$ containing the support and both points with a positive margin and $M\ge 2d$. The same divergence calculation on this fixed ball writes each Hessian as ${\int }_{B_M(m)}{K}_{ij}(x-y)(g(y)-g(x))dy-g(x)b(x)$, where $b(x)={\int }_{\mathrm{\partial }{B}_M(m)}{D}_i\mathrm{\Phi }(x-y){\nu }_j(y)dS(y)$. Reflection in $m$ and oddness of $D_i\mathrm{\Phi }$ show $b(x)=b(x^{\mathrm{\prime }})$; its magnitude is bounded by a dimension constant since both poles remain a fixed fraction of $M$ away from the sphere.

Split the difference into $B_d(m)$ and its complement. The near integrals are at most $C_{\alpha }[g{]}_{\alpha }{d}^{\alpha }$. On the complement, $\mathrm{\mid }K(x-y)-K(x^{\mathrm{\prime }}-y)\mathrm{\mid }\le Cd\mathrm{\mid }y-m{\mathrm{\mid }}^{-4}$, so the part multiplied by $g(y)-g(x)$ is at most $C[g{]}_{\alpha }d{\int }_d^M{r}^{\alpha -2}dr\le C_{\alpha }[g{]}_{\alpha }{d}^{\alpha }$. The remaining part is $(g(x^{\mathrm{\prime }})-g(x)){\int }_{B_M(m)\setminus B_d(m)}{K}_{ij}(x^{\mathrm{\prime }}-y)dy$. This kernel integral is bounded uniformly: the divergence theorem expresses it as the difference of two $D_i\mathrm{\Phi }$ boundary fluxes; each sphere stays at least half its radius from the pole and has area proportional to its radius squared. The boundary correction from step 3.1 is also at most $C[g{]}_{\alpha }{d}^{\alpha }$. Hence $[D^{2}Ng{]}_{\alpha }\le C_{\alpha }[g{]}_{\alpha }$ on the box. The convergent far radial integral uses $\alpha <1$.

For $x\in B_1$, the support is in $B_3(x)$. Integrating $\mathrm{\mid }\mathrm{\Phi }\mathrm{\mid }$ and $\mathrm{\mid }D\mathrm{\Phi }\mathrm{\mid }$ over that ball bounds $\mathrm{\mid }Ng\mathrm{\mid }$ and $\mathrm{\mid }DNg\mathrm{\mid }$ by $C\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}$. Step 2.1 with $M=3$ bounds $\mathrm{\mid }{D}^{2}Ng\mathrm{\mid }$ by $C_{\alpha }([g{]}_{\alpha }+\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }})$; step 4.1 supplies the seminorm. These are precisely the unit-radius scaled bounds. The identity $Ng(Rx)=R^{2}N(g(R\cdot ))(x)$ proves the displayed estimate for arbitrary $R$.

By [F4] the distribution of $-\mathrm{\Delta }Ng-g$ vanishes. Step 2.1 makes this difference continuous, so it vanishes everywhere: a nonzero value would persist on a ball of positive measure. Thus $-\mathrm{\Delta }Ng=g$ classically. Empty support and zero data give the zero potential, and all bounds remain valid.

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.