Uniform boundary Schauder bounds and smooth interior domain approximants

Let $k=n-1\ge 1$, $0<\alpha <1$, and $P_t(z)=c_k\frac{t}{(\mathrm{\mid }z{\mathrm{\mid }}^2+t^2{)}^{(k+1)\mathrm{/}2}},Q_{j,t}(z)=c_k\frac{z_j}{(\mathrm{\mid }z{\mathrm{\mid }}^2+t^2{)}^{(k+1)\mathrm{/}2}},$ where $c_k$ is the reciprocal of ${\int }_{{\mathbb{R}}^k}(1+\mathrm{\mid }z{\mathrm{\mid }}^2{)}^{-(k+1)\mathrm{/}2}dz$. This integral is finite by polar integration: its radial behavior is $r^{k-1}$ at zero and $r^{-2}$ at infinity. Thus $P_t$ is nonnegative, has mass one by scaling, and its usual differentiation verifies harmonicity in $(z,t)$. The formulas apply also when $n=2$, $k=1$; no $n\ge 3$ half-space theorem is assumed.

We first prove the elementary singular-kernel estimate required below. If $a\in C_c^{0,\alpha }({\mathbb{R}}^k)$ has support in a fixed ball, then on every fixed bounded cylinder $\mathrm{\mid }x\mathrm{\mid }\le R$, $0\le t\le R$ the functions $T_{j,t}a(x)={\int }_{{\mathbb{R}}^k}{Q}_{j,t}(z)a(x-z)dz(t>0)$ extend continuously to $t=0$ and satisfy $\mathrm{\parallel }{T}_{j}a{\mathrm{\parallel }}_{C^{0,\alpha }(\overline{{\mathcal{C}}_R})}\le C\mathrm{\parallel }a{\mathrm{\parallel }}_{C^{0,\alpha }({\mathbb{R}}^k)}\mathrm{.}$ Here the boundary integral is the absolutely convergent cancellation expression $T_{j,0}a(x)={\int }_{\mathrm{\mid }z\mathrm{\mid }<1}{Q}_{j,0}(z)(a(x-z)-a(x))dz+{\int }_{\mathrm{\mid }z\mathrm{\mid }\ge 1}{Q}_{j,0}(z)a(x-z)dz\mathrm{.}$ Indeed the kernels are odd in $z$, their integrals over every centered annulus vanish, and uniformly in $t\ge 0$, $\mathrm{\mid }{Q}_{j,t}(z)\mathrm{\mid }\le C\mathrm{\mid }z{\mathrm{\mid }}^{-k},\mathrm{\mid }{D}_z{Q}_{j,t}(z)\mathrm{\mid }+\mathrm{\mid }{\mathrm{\partial }}_t{Q}_{j,t}(z)\mathrm{\mid }\le C\mathrm{\mid }z{\mathrm{\mid }}^{-k-1}(z\mathrm{\ne }0)\mathrm{.}$ The near part of the displayed cancellation integral is bounded by $C[a{]}_{\alpha }{\int }_0^1{r}^{\alpha -1}dr$. For bounded $x$, the far integral lies in a fixed annulus and is bounded by $C_R\mathrm{\parallel }a{\mathrm{\parallel }}_{\mathrm{\infty }}$; this proves the supremum bound uniformly in $t$.

Here are the complete difference estimates. For a vertical change $\mathrm{\mid }t-s\mathrm{\mid }=\delta <1\mathrm{/}8$, subtract $a(x)$ under the integral on $\mathrm{\mid }z\mathrm{\mid }<1$ using oddness. The part $\mathrm{\mid }z\mathrm{\mid }<2\delta$ is bounded by $C[a{]}_{\alpha }{\delta }^{\alpha }$. On $2\delta <\mathrm{\mid }z\mathrm{\mid }<1$, the mean value estimate for ${\mathrm{\partial }}_{t}Q$ bounds the difference by $C[a{]}_{\alpha }\delta {\int }_{2\delta }^1{r}^{\alpha -2}dr\le C_{\alpha }[a{]}_{\alpha }{\delta }^{\alpha }\mathrm{.}$ The far part costs $C_R\mathrm{\parallel }a{\mathrm{\parallel }}_{\mathrm{\infty }}\delta$, hence also $C_R\mathrm{\parallel }a{\mathrm{\parallel }}_{\mathrm{\infty }}{\delta }^{\alpha }$. The same argument with $s=0$ proves convergence to the boundary expression.

For a horizontal change $\mathrm{\mid }x-y\mathrm{\mid }=\delta <1\mathrm{/}8$ at fixed height, write both operators in the physical integration variable $\zeta$, subtract respectively $a(x)$ and $a(y)$ in unit balls centered at their poles, and split at the union of the two radius-$2\delta$ balls. The near pieces cost $C[a{]}_{\alpha }{\delta }^{\alpha }$, since each pole is at most $3\delta$ from the other near ball. On the common far region inside the unit balls, use $\mathrm{\mid }{Q}_t(x-\zeta )-Q_t(y-\zeta )\mathrm{\mid }\le C\delta \mathrm{\mid }x-\zeta {\mathrm{\mid }}^{-k-1}$ and $\mathrm{\mid }a(\zeta )-a(x)\mathrm{\mid }\le [a{]}_{\alpha }\mathrm{\mid }x-\zeta {\mathrm{\mid }}^{\alpha }$. Its integral is the same $C[a{]}_{\alpha }\delta {\int }_{2\delta }^2{r}^{\alpha -2}dr$ as above. The remaining constant-difference term is $a(x)-a(y)$ times an integral of $Q_t$ over the far integration region. That integral is uniformly bounded: compare its domain to a centered annulus, whose integral is zero. At the inner boundary the symmetric difference is contained in a shell of radius at most $4\delta$ and outside radius $\delta$, of volume $C{\delta }^k$ and kernel bound $C{\delta }^{-k}$; its integral is $O(1)$. At the outer boundary the symmetric difference has volume $C\delta$ and bounded kernel, also $O(1)$. These bounds use only the stated kernel estimates, uniformly in $t$. Multiplying by $\mathrm{\mid }a(x)-a(y)\mathrm{\mid }\le [a{]}_{\alpha }{\delta }^{\alpha }$ gives the required bound. Outside the unit neighborhoods the kernels and their first derivatives are bounded on the fixed support region, so that part is Lipschitz with constant $C_R\mathrm{\parallel }a{\mathrm{\parallel }}_{\mathrm{\infty }}$; the moving unit-boundary shells have the same bound after the cancellation constants are included. This proves horizontal $\alpha$-Hölder control. Larger increments are covered by the supremum bound. Combining a horizontal and a vertical change proves the asserted joint Hölder estimate.

The Poisson operator itself maps bounded $C^{\alpha }$ data to $C^{\alpha }$ on the closed half-space, locally with controlled norm. Horizontal differences follow by translation and its mass one. For vertical differences, differentiating the explicit kernel and scaling gives $\int \mathrm{\mid }{\mathrm{\partial }}_t{P}_t(z)\mathrm{\mid }\mathrm{\mid }z{\mathrm{\mid }}^{\alpha }dz\le C_{\alpha }{t}^{\alpha -1},\int {\mathrm{\partial }}_t{P}_t=0.$ The moment is finite since $\alpha <1$. Thus $\mathrm{\mid }{\mathrm{\partial }}_t(P_t\ast a)\mathrm{\mid }\le C[a{]}_{\alpha }{t}^{\alpha -1}$; integration from $s$ to $t$ gives $C[a{]}_{\alpha }\mathrm{\mid }t-s{\mathrm{\mid }}^{\alpha }$. The same moment estimate for $P_t$ proves its convergence to $a$ at $t=0$ with error $C[a{]}_{\alpha }{t}^{\alpha }$.

Now let $\varphi \in C_c^{2,\alpha }({\mathbb{R}}^k)$. Set $H(x,t)=P_t\ast \varphi (x)$. Differentiation under the integral for $t>0$, followed by tangential integration by parts, gives $D_{a}H=P_t\ast D_a\varphi$ and $D_{ab}H=P_t\ast D_{ab}\varphi$. The explicit identities ${\mathrm{\partial }}_t{P}_t=-{\sum }_{j=1}^k{\mathrm{\partial }}_j{Q}_{j,t},{\mathrm{\Delta }}_{x,t}{P}_t=0$ give $D_t{D}_{a}H=-{\sum }_j{Q}_{j,t}\ast D_{ja}\varphi ,D_{tt}H=-{\sum }_a{P}_t\ast D_{aa}\varphi \mathrm{.}$ The same identity without $D_a$ bounds $D_{t}H$ by the $T_j$ estimates applied to $D_j\varphi$. Each datum on the right is compactly supported $C^{\alpha }$. The just-proved $P$ and $T$ bounds show that $H$, its gradient and Hessian extend to the boundary, with Hessian $\alpha$-Hölder there and $\mathrm{\parallel }H{\mathrm{\parallel }}_{C^{2,\alpha }(\overline{{\mathcal{C}}_R})}\le C\mathrm{\parallel }\varphi {\mathrm{\parallel }}_{C^{2,\alpha }({\mathbb{R}}^k)}\mathrm{.}$ Classical derivatives up to the boundary mean the continuous interior derivative extensions; their compatibility follows from the fundamental theorem of calculus along interior coordinate segments and then passage to the boundary. No Riesz-transform boundedness theorem or unproved mixed-derivative claim has been cited.

Write $B_r^{+}=B_r\cap \{x_n>0\}$ and its flat face $B_r^{\mathrm{\prime }}=B_r\cap \{x_n=0\}$. If $u\in C^2(B_1^{+})\cap C(\overline{B_1^{+}})$, $\mathrm{\Delta }u=f\in C^{0,\alpha }(\overline{B_1^{+}})$, and $u=0$ on $B_1^{\mathrm{\prime }}$, then $\mathrm{\parallel }u{\mathrm{\parallel }}_{C^{2,\alpha }(\overline{B_{1\mathrm{/}4}^{+}})}\le C(\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty };B_1^{+}}+\mathrm{\parallel }f{\mathrm{\parallel }}_{C^{0,\alpha }(\overline{B_1^{+}})})\mathrm{.}$ To prove this, extend $f$ evenly across the flat face on $B_1$, and multiply by a smooth cutoff supported in $B_1$ and equal to one on $B_{3\mathrm{/}4}$. The even extension is Hölder: if the two points are on opposite sides, the distance between their reflected upper-half-space points is no greater than their original distance. Thus the resulting $F\in C_c^{\alpha }({\mathbb{R}}^n)$ has controlled norm. The actual local theorem thm-newtonian-potential-for-holder-data-is-classical gives $v=-NF$, with $\mathrm{\Delta }v=f$ on $B_{3\mathrm{/}4}^{+}$ and $\mathrm{\parallel }v{\mathrm{\parallel }}_{C^{2,\alpha }({\bar{B}}_{3\mathrm{/}4})}\le C\mathrm{\parallel }f{\mathrm{\parallel }}_{C^{\alpha }}$. This theorem covers $n=2$ with its logarithmic kernel, and no boundedness of $v$ at infinity is required.

The continuous ball-Dirichlet supplier currently has dimension restriction $n\ge 3$. For $n=2$ the needed continuous-data disk solution follows from the actual $n\ge 2$ supplier lem-smooth-sphere-data-have-a-harmonic-replacement as follows. Parametrize the boundary circle by angle and extend its continuous datum to a continuous $2\pi$-periodic function $a(\theta )$. Convolve in the angle with a nonnegative smooth periodic unit-mass kernel supported, modulo $2\pi$, in $\mathrm{\mid }\theta \mathrm{\mid }<ϵ$. The result $a_{ϵ}$ is smooth and periodic, and $\mathrm{\parallel }{a}_{ϵ}-a{\mathrm{\parallel }}_{\mathrm{\infty }}\le {\sup }_{\mathrm{\mid }h\mathrm{\mid }\le ϵ,\theta }\mathrm{\mid }a(\theta -h)-a(\theta )\mathrm{\mid }\to 0$ by uniform continuity on the circle. The smooth-sphere supplier gives harmonic $H_{ϵ}$ continuous on the closed disk with these data. The weak maximum principle gives $\mathrm{\parallel }{H}_{ϵ}-H_{\eta }{\mathrm{\parallel }}_{\mathrm{\infty };\bar{B}}\le \mathrm{\parallel }{a}_{ϵ}-a_{\eta }{\mathrm{\parallel }}_{\mathrm{\infty }}$, so $H_{ϵ}$ converges uniformly on the closed disk to a continuous function with the prescribed boundary datum. Interior derivative estimates applied to the differences show that all derivatives converge uniformly on smaller disks; the fundamental theorem of calculus identifies them as the derivatives of the uniform limit, which is therefore harmonic. Uniqueness follows from the weak maximum principle. This supplies continuous ball data in dimension two without extending a dimension-restricted theorem silently.

Choose a tangential cutoff equal to one on $B_{2\mathrm{/}3}^{\mathrm{\prime }}$ and supported in $B_{3\mathrm{/}4}^{\mathrm{\prime }}$, and put $\varphi$ equal to that cutoff times $v{\mathrm{\mid }}_{x_n=0}$, extended by zero to ${\mathbb{R}}^{n-1}$. the Poisson-extension calculation above gives a harmonic $H=P\varphi$ with controlled $C^{2,\alpha }$ norm on $\overline{B_{2\mathrm{/}3}^{+}}$. The remainder $w=u-v+H$ is harmonic in $B_{2\mathrm{/}3}^{+}$, continuous on its closure and zero on its flat face. Its odd extension is harmonic in $B_{1\mathrm{/}2}$. Here is a proof requiring no boundary derivatives of $u$: on each ball centered on the flat face and compactly contained in $B_{2\mathrm{/}3}$, solve the ordinary ball Dirichlet problem for the continuous odd boundary data obtained from $w$; uniqueness on the ball makes its solution odd, and uniqueness by the weak maximum principle on the upper half-ball identifies it with $w$. Thus the odd extension is harmonic near every plane point, as well as off the plane.

The actual harmonic interior derivative estimates on $B_{1\mathrm{/}2}$ bound its derivatives through order three on $B_{1\mathrm{/}4}$ by $C\mathrm{\parallel }w{\mathrm{\parallel }}_{\mathrm{\infty };B_{1\mathrm{/}2}^{+}}$. A third derivative bound gives the required second derivative $\alpha$-Hölder bound on that ball, since $\alpha <1$. Combining this with the $v,H$ bounds proves the half-ball estimate, and at the same time proves its boundary regularity upgrade noncircularly for the Laplacian.

Rescaling gives the corresponding estimate between $B_r^{+}$ and $B_{r\mathrm{/}4}^{+}$ in scaled norms. It holds for every constant symmetric positive definite principal matrix $A_0$ with $\lambda I\le A_0\le \mathrm{\Lambda }I$, with constants depending only on $n,\alpha ,\lambda ,\mathrm{\Lambda }$ and a fixed target-radius fraction. To see that the linear normalization preserves the boundary plane, take $S=A_0^{1\mathrm{/}2}$, choose an orthogonal matrix $O$ whose last column is $S{e}_n\mathrm{/}\mathrm{\mid }S{e}_n\mathrm{\mid }$, and set $T=SO$. Then $T{T}^T=A_0$ and $(Ty{)}_n=\mathrm{\mid }S{e}_n\mathrm{\mid }{y}_n$. Therefore $A_0:D_x^2$ becomes ${\mathrm{\Delta }}_y$ under $x=Ty$, and upper half-spaces map onto upper half-spaces. The singular values of $T$ lie in $[\sqrt{\lambda },\sqrt{\mathrm{\Lambda }}]$, so contained/containing half-balls have fixed comparable radii. Restricting the estimate to the contained smaller half-ball gives the assertion.

This section is intentionally an a priori statement: let $V$ have smooth boundary but a uniformly controlled finite $C^{2,\alpha }$ graph atlas, let $A,\beta ,c$ be smooth near $\bar{V}$ with uniform $C^{\alpha }$ bounds and ellipticity, and let $u$ be smooth to the boundary, zero on the boundary, with $Lu=F$. Then $\mathrm{\parallel }u{\mathrm{\parallel }}_{C^{2,\alpha }(\bar{V})}\le C(\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty }}+\mathrm{\parallel }F{\mathrm{\parallel }}_{C^{\alpha }(\bar{V})}),$ where $C$ depends on the uniform atlas, the coefficient Hölder bounds, ellipticity, dimension and exponent, but not on higher smoothness norms.

A smooth graph flattening $x=\mathrm{\Psi }(y)$ transforms the equation to $\tilde{A}:D_y^{2}w+\tilde{\beta }\cdot D_{y}w+\tilde{c}w=\tilde{F}$ with $w=u\circ \mathrm{\Psi }$ and zero flat trace. The chain rule expresses $\tilde{A}=D{\mathrm{\Psi }}^{-1}A(D{\mathrm{\Psi }}^{-1}{)}^T$ and the drift as the transformed original drift plus $A:D_x^2{\mathrm{\Psi }}^{-1}$. The uniform $C^{2,\alpha }$ chart bounds give uniform ellipticity and $C^{\alpha }$ bounds for every transformed coefficient. Thus no derivative of the original coefficients beyond the stated Hölder bound enters this estimate.

Fix a small common chart radius $r$. On each boundary patch freeze $\tilde{A}$ at its flat-face center and apply the constant-coefficient half-ball estimate on an ambient half-ball of radius comparable to $r$; on interior balls use the corresponding constant-coefficient full-ball estimate, which follows directly from the local Newtonian potential and harmonic remainder argument. For the frozen forcing use ${\tilde{A}}_0:D^{2}w=\tilde{F}+({\tilde{A}}_0-\tilde{A}):D^{2}w-\tilde{\beta }\cdot Dw-\tilde{c}w\mathrm{.}$ There is no regularity issue in this application, since $w$ is already smooth on the closed half-patch.

Use the common scaled global norm $N_r(u)=\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty }}+r\mathrm{\parallel }Du{\mathrm{\parallel }}_{\mathrm{\infty }}+r^2\mathrm{\parallel }{D}^{2}u{\mathrm{\parallel }}_{\mathrm{\infty }}+r^{2+\alpha }[D^{2}u{]}_{\alpha }\mathrm{.}$ The product inequality $[pq{]}_{\alpha }\le \mathrm{\parallel }p{\mathrm{\parallel }}_{\mathrm{\infty }}[q{]}_{\alpha }+[p{]}_{\alpha }\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }}$, the freezing bounds $\mathrm{\parallel }\tilde{A}-{\tilde{A}}_0{\mathrm{\parallel }}_{\mathrm{\infty }}\le K{r}^{\alpha }$, $[\tilde{A}{]}_{\alpha }\le K$, and the convex-half-ball estimate $[Dw{]}_{\alpha }\le C{r}^{1-\alpha }\mathrm{\parallel }{D}^{2}w{\mathrm{\parallel }}_{\mathrm{\infty }}$ show that the scaled frozen error is at most $C(K{r}^{\alpha }+r\mathrm{\parallel }\tilde{\beta }{\mathrm{\parallel }}_{\mathrm{\infty }}+r^{1+\alpha }[\tilde{\beta }{]}_{\alpha }+r^2\mathrm{\parallel }\tilde{c}{\mathrm{\parallel }}_{\mathrm{\infty }}+r^{2+\alpha }[\tilde{c}{]}_{\alpha })N_r(u)=:\eta (r)N_r(u)\mathrm{.}$ Chart norm comparison changes $C$ only by uniform atlas constants; the terms from second chart derivatives times $Du$ carry a factor $r$ and are controlled in the same way. In particular $\eta (r)\to 0$ uniformly over the stated family.

For precision, gluing the patch estimates costs a fixed constant independent of the number of patches. Take a finite cover by their smaller inner patches and interior balls, with a Lebesgue number at least $c_{0}r$, using the common graph radii and a lattice of balls of radius a fixed multiple of $r$; the constant $c_0>0$ depends only on the uniform atlas distortion. Supremum derivative norms are bounded by the maximum of the corresponding local norms, with the uniform chart comparisons. If $\mathrm{\mid }x-y\mathrm{\mid }<c_{0}r$, the pair lies in one inner patch and its Hessian quotient is controlled there. If $\mathrm{\mid }x-y\mathrm{\mid }\ge c_{0}r$, its contribution to $r^{2+\alpha }[D^{2}u{]}_{\alpha }$ is at most $2c_0^{-\alpha }{r}^2\mathrm{\parallel }{D}^{2}u{\mathrm{\parallel }}_{\mathrm{\infty }}$. Thus $N_r(u)\le C_0(\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty }}+r^2\mathrm{\parallel }F{\mathrm{\parallel }}_{\mathrm{\infty }}+r^{2+\alpha }[F{]}_{\alpha })+C_1\eta (r)N_r(u)\mathrm{.}$ The constants $C_0,C_1$ are uniform; no sum over all covering balls occurs. Choose the common radius so $C_1\eta (r)\le 1\mathrm{/}2$ and absorb the final term. Since $r$ is now fixed in terms of the permitted data, the unscaled asserted global estimate follows. This argument proves the estimate only for the smooth functions to which it was applied; a consuming existence proof obtains the original solution's regularity by passing this estimate to a limit.

A bounded $C^{2,\alpha }$ domain $U$ has a global $C^{2,\alpha }$ defining function $\rho$ on ${\mathbb{R}}^n$, equal to $1$ outside a large ball and to $-1$ away from a collar inside $U$, with $U=\{\rho <0\}$ and $\mathrm{\mid }D\rho \mathrm{\mid }\ge c>0$ near $\mathrm{\partial }U$. Here is a construction. On a finite boundary atlas use the signed graph functions ${\rho }_{\mathrm{\ell }}=x_n-h_{\mathrm{\ell }}(x^{\mathrm{\prime }})$, after rigid motions, each negative inside. Choose a smooth partition of unity on a collar subordinate to those chart neighborhoods and put ${\rho }_0=\sum {\chi }_{\mathrm{\ell }}{\rho }_{\mathrm{\ell }}$. On the boundary, the gradients of the signed graph functions are positive scalar multiples of the same outward unit normal; the terms $D{\chi }_{\mathrm{\ell }}{\rho }_{\mathrm{\ell }}$ vanish there. Therefore $D{\rho }_0\mathrm{\ne }0$ on the boundary, with a uniform lower bound by compactness. Every active signed graph function has the same sign on each side, so ${\rho }_0$ has exactly the correct sign in a sufficiently small collar. Multiply by a smooth collar cutoff equal to one in a smaller collar, and outside use the sign function $-1$ on $U$, $+1$ outside. The latter is multiplied by a function identically zero near the boundary, so this gluing is $C^{2,\alpha }$ globally and introduces no new zeros. Shrinking the collar preserves the gradient lower bound. All derivatives and their Hölder bounds are finite because only finitely many compactly supported chart pieces occur.

Let ${\rho }_j=\rho \ast {\rho }_{s_j}$ for $s_j\downarrow 0$, and put $a_j=\max \{j^{-1},\mathrm{\parallel }{\rho }_j-\rho {\mathrm{\parallel }}_{C^1}\}$. Mollification gives $a_j\to 0$, uniform $C^{2,\alpha }$ bounds on ${\rho }_j$ and $C^2$ convergence to $\rho$; the latter follows from uniform continuity of $D^2\rho$. Define $V_j=\{{\rho }_j<-2a_j\}\mathrm{.}$ For large $j$, its boundary lies in the fixed collar, where $\mathrm{\mid }D{\rho }_j\mathrm{\mid }\ge c\mathrm{/}2$. Hence $V_j$ has smooth boundary, ${\bar{V}}_j\subset U$, and every compact subset of $U$ eventually lies inside $V_j$. It is bounded since ${\rho }_j=1$ outside a fixed large ball. No regular-value selection is needed for this particular smooth-boundary approximation.

These domains have a common finite atlas with uniformly bounded $C^{2,\alpha }$ graph norms. Indeed, after shrinking the original charts, choose in each a coordinate with $\mathrm{\mid }{\mathrm{\partial }}_n\rho \mathrm{\mid }\ge c_1>0$. Uniform $C^1$ convergence gives $\mathrm{\mid }{\mathrm{\partial }}_n{\rho }_j\mathrm{\mid }\ge c_1\mathrm{/}2$, so the level equations ${\rho }_j(x^{\mathrm{\prime }},h_j(x^{\mathrm{\prime }}))=-2a_j$ have graphs on fixed smaller chart cylinders, by monotonicity in the normal coordinate and the intermediate value theorem. Their derivative formulas are $D_a{h}_j=-\frac{{\rho }_{j,a}}{{\rho }_{j,n}},$ $D_{ab}{h}_j=-\frac{{\rho }_{j,ab}+{\rho }_{j,an}{D}_b{h}_j+{\rho }_{j,bn}{D}_a{h}_j+{\rho }_{j,nn}{D}_a{h}_j{D}_b{h}_j}{{\rho }_{j,n}}\mathrm{.}$ The denominator stays bounded away from zero, the second derivatives of ${\rho }_j$ have uniform Hölder bounds, and the first derivative formulas give uniform Lipschitz bounds for the graphs and their first derivatives. Product, quotient and composition Hölder estimates then give the uniform $C^{2,\alpha }$ bounds. Compactness and convergence place all of $\mathrm{\partial }{V}_j$ in these fixed smaller cylinders. This proves every geometric property needed by the freezing and gluing argument, rather than invoking an unproved smooth-domain-approximation theorem.

The freezing and gluing argument proves the uniform a priori estimate for precisely the smooth solutions specified in the Statement, and the defining-function construction supplies the required smooth interior approximating domains with the uniform atlas controls needed by that estimate. The half-ball estimate and the Poisson mixed derivative bounds were derived by the kernel calculation and reflection above; they were not assumed as external analytic results. Both conclusions follow.