Newtonian cancellation and the three-dimensional Holder bound

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.