Hölder data give a classical Newtonian solution

Put $r=\mathrm{\mid }z\mathrm{\mid }$. Differentiating the two profiles in [F1] using [F2] and [F11] gives, for every $z\mathrm{\ne }0$ and all $i,j$, ${\mathrm{\partial }}_i\mathrm{\Phi }(z)=-\frac{z_i}{{\omega }_{n-1}{r}^n},K_{ij}(z):={\mathrm{\partial }}_i{\mathrm{\partial }}_j\mathrm{\Phi }(z)=\frac{n{z}_i{z}_j{r}^{-n-2}-{\delta }_{ij}{r}^{-n}}{{\omega }_{n-1}}\mathrm{.}$ A further differentiation and the displayed formulas give $\mathrm{\mid }D\mathrm{\Phi }(z)\mathrm{\mid }\le C_n{r}^{1-n}$, $\mathrm{\mid }{K}_{ij}(z)\mathrm{\mid }\le C_n{r}^{-n}$, and $\mathrm{\mid }D{K}_{ij}(z)\mathrm{\mid }\le C_n{r}^{-n-1}$. The trace of the Hessian formula is zero away from $0$. Thus the same estimates hold in the logarithmic and power cases; in $n=2$ [F3] gives the needed ${\omega }_1=2\pi$.

By continuity and compact support, [F7] gives $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}<\mathrm{\infty }$. The bounded-data potential lemma [F8] makes the integral for $Nf(x)$ absolutely finite at every point and locally bounded, including when $f=0$ or its support is empty.

Choose the smooth bump $\rho$ from [F4] and set ${\eta }_{\epsilon }(z)=1-\rho (z\mathrm{/}\epsilon )$ and $q_{\epsilon }(z)=\mathrm{\Phi }(z){\eta }_{\epsilon }(z)$, assigning $q_{\epsilon }(0)=0$. Then $q_{\epsilon }$ is smooth: it is zero near $0$ and equals the smooth kernel off $B_{2\epsilon }$. Its derivatives of orders one and two in the transition annulus are bounded by the product rule and [F4]. Define $u_{\epsilon }(x)=\int q_{\epsilon }(x-y)f(y)dy\mathrm{.}$ Differentiation under the integral sign [F5] applies on every compact $x$-set: the support of $f$ is compact and, for fixed $\epsilon$, the kernel derivatives are bounded on the corresponding difference set.

Orthogonal invariance in [F3] gives ${\int }_{S^{n-1}}{\theta }_i{\theta }_{j}d\sigma =0$ for $i\mathrm{\ne }j$ by coordinate reflection. Coordinate permutations make all diagonal integrals equal, and ${\sum }_i{\theta }_i^2=1$ makes each equal ${\omega }_{n-1}\mathrm{/}n$. Therefore, on the sphere centered at $x$, ${\int }_{\mathrm{\partial }{B}_R(x)}{\mathrm{\partial }}_i\mathrm{\Phi }(x-y){\nu }_j(y)dS(y)=\frac{{\delta }_{ij}}{n},$ where $\nu (y)=(y-x)\mathrm{/}R$ is its outward normal. This follows by substituting $y=x+R\theta$ and the first formula of step 1.1. The same calculation in the variable $z$ gives ${\int }_{\mathrm{\partial }{B}_R(0)}{\mathrm{\partial }}_i\mathrm{\Phi }(z)z_j\mathrm{/}RdS(z)=-{\delta }_{ij}\mathrm{/}n$.

The difference $u_{\epsilon }-Nf$ is supported in the kernel variable $\mathrm{\mid }x-y\mathrm{\mid }<2\epsilon$, so [F7] and polar integration [F3] give a uniform-in-$x$ bound $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}{\int }_{B_{2\epsilon }}\mathrm{\mid }\mathrm{\Phi }(z)\mathrm{\mid }dz$, which tends to zero like $O({\epsilon }^2)$ for $n\ge 3$ and $O({\epsilon }^2(1+\mathrm{\mid }\log \epsilon \mathrm{\mid }))$ for $n=2$. Also $G_i(x):=\int {\mathrm{\partial }}_i\mathrm{\Phi }(x-y)f(y)dy$ is absolutely finite, since $D\mathrm{\Phi }$ is locally integrable by step 1.1 and $f$ is bounded with compact support. From the product rule, the difference ${\mathrm{\partial }}_i{u}_{\epsilon }-G_i$ is bounded uniformly in $x$ by $C_n\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}({\int }_{B_{2\epsilon }}\mathrm{\mid }z{\mathrm{\mid }}^{1-n}dz+{\epsilon }^{-1}{\int }_{B_{2\epsilon }}\mathrm{\mid }\mathrm{\Phi }(z)\mathrm{\mid }dz),$ which tends to zero (the second term is $O(\epsilon (1+\mathrm{\mid }\log \epsilon \mathrm{\mid }))$ when $n=2$). Thus $u_{\epsilon }\to Nf$ and ${\mathrm{\partial }}_i{u}_{\epsilon }\to G_i$ uniformly on compact sets. The limit $G_i$ is continuous as a locally uniform limit of continuous functions.

Fix a sequence ${\epsilon }_k\downarrow 0$, for example ${\epsilon }_k=2^{-k}$. Restrict to any closed coordinate segment inside an open box. For each real and imaginary component, [F5] and step 2.2 give convergence of the smooth restrictions at one point and uniform convergence of their derivatives in the chosen coordinate. Hence the uniform derivative limit theorem [F5], applied separately to both components, shows that the corresponding partial derivative of $Nf$ exists and equals $G_i$. The continuous-partials theorem [F5], applied to the real two-component map $(\mathrm{Re}Nf,\mathrm{Im}Nf)$, makes $Nf\in C^1$ with ${\mathrm{\partial }}_{i}Nf=G_i$.

Fix a compact box $Q$ and choose $R>0$ so that $\mathrm{supp}f\subset B_R(x)$ for every $x\in Q$, with a positive margin. For $\epsilon <R\mathrm{/}2$, differentiating $G_{i,\epsilon }:={\mathrm{\partial }}_i{u}_{\epsilon }$ and using $f=0$ off its support gives ${\mathrm{\partial }}_j{G}_{i,\epsilon }(x)={\int }_{B_R(x)}{\mathrm{\partial }}_i{\mathrm{\partial }}_j{q}_{\epsilon }(x-y)(f(y)-f(x))dy+f(x){\int }_{B_R(x)}{\mathrm{\partial }}_i{\mathrm{\partial }}_j{q}_{\epsilon }(x-y)dy\mathrm{.}$ The second integral equals $-{\delta }_{ij}\mathrm{/}n$: change variables $z=x-y$, apply the divergence theorem [F6] to the smooth field ${\mathrm{\partial }}_i{q}_{\epsilon }(z)e_j$ on $B_R(0)$, and use the last sphere integral in step 2.1, since $q_{\epsilon }=\mathrm{\Phi }$ near that boundary.

Set $H_{ij}(x):={\int }_{B_R(x)}{K}_{ij}(x-y)(f(y)-f(x))dy-\frac{{\delta }_{ij}}{n}f(x)\mathrm{.}$ The singular integral is absolutely convergent: its absolute integrand is at most $C_n[f{]}_{\alpha ;{\mathbb{R}}^n}\mathrm{\mid }x-y{\mathrm{\mid }}^{\alpha -n}$, whose polar radial integral at zero is a constant times ${\int }_0^R{r}^{\alpha -1}dr<\mathrm{\infty }$. In the omitted inner ball $\mathrm{\mid }z\mathrm{\mid }<\epsilon$, $D^2{q}_{\epsilon }=0$, so the error from replacing it by $K_{ij}$ after multiplication by $\mathrm{\mid }f(x-z)-f(x)\mathrm{\mid }$ is at most $C_{n,\alpha }[f{]}_{\alpha ;{\mathbb{R}}^n}{\int }_0^{\epsilon }{r}^{\alpha -1}dr\le C_{n,\alpha }[f{]}_{\alpha ;{\mathbb{R}}^n}{\epsilon }^{\alpha }\mathrm{.}$ On the transition annulus $\epsilon <\mathrm{\mid }z\mathrm{\mid }<2\epsilon$, the product rule, [F4], and the kernel estimates of step 1.1 bound the same error by $C_{n,\alpha }[f{]}_{\alpha ;{\mathbb{R}}^n}({\epsilon }^{\alpha }+{\epsilon }^{\alpha }(1+\mathrm{\mid }\log \epsilon \mathrm{\mid }){\mathbf{1}}_{\{n=2\}})\mathrm{.}$ Outside $B_{2\epsilon }$ the kernels agree. The total error tends to zero by [F2], uniformly for $x\in Q$. Therefore ${\mathrm{\partial }}_j{G}_{i,\epsilon }\to H_{ij}$ uniformly on $Q$. Each $H_{ij}$ is continuous as this uniform limit.

Use the same sequence ${\epsilon }_k$ from step 3.1 and apply [F5]'s uniform derivative limit theorem separately to the real and imaginary components on coordinate segments in $Q$, now for $G_{i,\epsilon }$ and their $j$-derivatives. Steps 2.2 and 4.1 supply the function and derivative limits, so ${\mathrm{\partial }}_j{G}_i=H_{ij}$. Apply the continuous-partials theorem [F5] to the real two-component map as in step 3.1 to obtain $Nf\in C^2$ on the box and ${\mathrm{\partial }}_i{\mathrm{\partial }}_{j}Nf=H_{ij}$ there; use mixed partial symmetry on the same two real components as in Continuous mixed partials of order $k$ are invariant under permutations. Since every point lies in such a box, this holds throughout ${\mathbb{R}}^n$. The radius $R$ was any sufficiently large containing radius. Comparing the expression for two such radii shows independence: their difference is the integral of $K_{ij}$ over an annulus against the constant $-f(x)$, and the divergence theorem turns it into the difference of the two outer sphere fluxes, both $-{\delta }_{ij}\mathrm{/}n$ in the $z=x-y$ orientation. Thus the formula in the Statement holds for every $r>0$ whose centered ball contains the support.

We prove the local Hölder estimate. Take distinct $x,x^{\mathrm{\prime }}$ in a compact box $Q$; the case $x=x^{\mathrm{\prime }}$ is trivial. Put $m=(x+x^{\mathrm{\prime }})\mathrm{/}2$ and $d=\mathrm{\mid }x-x^{\mathrm{\prime }}\mathrm{\mid }>0$. Choose one radius $R$ so large that for every pair in $Q$, $\mathrm{supp}f\subset B_R(m)$, the ball $B_R(m)$ has positive distance from both poles to its boundary, and $R\ge 2d$. For any bounded $C^1$ domain $\mathrm{\Omega }$ with $\mathrm{supp}f\cup \{x\}\subset \mathrm{\Omega }$, choose $\epsilon$ so small that $B_{2\epsilon }(x)\subset \mathrm{\Omega }$ and $q_{\epsilon }=\mathrm{\Phi }$ near $\mathrm{\partial }\mathrm{\Omega }$. Since $f$ vanishes outside $\mathrm{\Omega }$ and $u_{\epsilon }$ is defined by convolution, ${\mathrm{\partial }}_j{\mathrm{\partial }}_i{u}_{\epsilon }(x)={\int }_{\mathrm{\Omega }}{\mathrm{\partial }}_j{\mathrm{\partial }}_i{q}_{\epsilon }(x-y)f(y)dy={\int }_{\mathrm{\Omega }}{\mathrm{\partial }}_j{\mathrm{\partial }}_i{q}_{\epsilon }(x-y)(f(y)-f(x))dy-f(x){\int }_{\mathrm{\partial }\mathrm{\Omega }}{\mathrm{\partial }}_i{q}_{\epsilon }(x-y){\nu }_j(y)dS(y),$ where the last equality is the divergence theorem in $y$ and uses ${\mathrm{\partial }}_{y_j}{\mathrm{\partial }}_i{q}_{\epsilon }(x-y)=-{\mathrm{\partial }}_{x_j}{\mathrm{\partial }}_i{q}_{\epsilon }(x-y)$. By the inner-ball and transition-annulus estimates of step 4.1, the first integral tends to ${\int }_{\mathrm{\Omega }}{K}_{ij}(x-y)(f(y)-f(x))dy$; the boundary integral equals $g_{\mathrm{\Omega }}(x)$ for these small $\epsilon$. The left side tends to $H_{ij}(x)$ by steps 3.2 and 4.1. Thus $H_{ij}(x)={\int }_{\mathrm{\Omega }}{K}_{ij}(x-y)(f(y)-f(x))dy-f(x)g_{\mathrm{\Omega }}(x),g_{\mathrm{\Omega }}(x)={\int }_{\mathrm{\partial }\mathrm{\Omega }}{\mathrm{\partial }}_i\mathrm{\Phi }(x-y){\nu }_j(y)dS(y)\mathrm{.}$ We use this formula with the fixed domain $\mathrm{\Omega }=B_R(m)$ for both $x$ and $x^{\mathrm{\prime }}$. Splitting the integral difference into $B_d(m)$ and its complement, the inner part is at most $C_{n,\alpha }[f{]}_{\alpha ;{\mathbb{R}}^n}{\int }_{B_d(m)}(\mathrm{\mid }x-y{\mathrm{\mid }}^{\alpha -n}+\mathrm{\mid }{x}^{\mathrm{\prime }}-y{\mathrm{\mid }}^{\alpha -n})dy\le C_{n,\alpha }[f{]}_{\alpha ;{\mathbb{R}}^n}{d}^{\alpha }\mathrm{.}$ On $\mathrm{\Omega }\setminus B_d(m)$ write the integrand difference as $[K_{ij}(x-y)-K_{ij}(x^{\mathrm{\prime }}-y)](f(y)-f(x))-(f(x)-f(x^{\mathrm{\prime }}))K_{ij}(x^{\mathrm{\prime }}-y)\mathrm{.}$ The mean-value theorem [F12] and $\mathrm{\mid }D{K}_{ij}(z)\mathrm{\mid }\le C_n\mathrm{\mid }z{\mathrm{\mid }}^{-n-1}$ bound the first term by $C_n[f{]}_{\alpha }d\mathrm{\mid }y-m{\mathrm{\mid }}^{\alpha -n-1}$. Its integral is bounded by $C_{n,\alpha }[f{]}_{\alpha }{d}^{\alpha }$, because $d{\int }_d^{\mathrm{\infty }}{r}^{\alpha -2}dr=d^{\alpha }\mathrm{/}(1-\alpha )$. For the second term, apply the divergence theorem to the annulus $B_R(m)\setminus {\bar{B}}_d(m)$; the two boundary fluxes of $D\mathrm{\Phi }$ are bounded by $C_n$ using $\mathrm{\mid }D\mathrm{\Phi }(z)\mathrm{\mid }\le C_n\mathrm{\mid }z{\mathrm{\mid }}^{1-n}$ and $R\ge 2d$. Its contribution is at most $C_n[f{]}_{\alpha }{d}^{\alpha }$. Finally, reflection through $m$ and the oddness of $D\mathrm{\Phi }$ show $g_{\mathrm{\Omega }}(x)=g_{\mathrm{\Omega }}(x^{\mathrm{\prime }})$, while the same outer-sphere bound gives $\mathrm{\mid }{g}_{\mathrm{\Omega }}(x)\mathrm{\mid }\le C_n$. Hence $\mathrm{\mid }f(x)g_{\mathrm{\Omega }}(x)-f(x^{\mathrm{\prime }})g_{\mathrm{\Omega }}(x^{\mathrm{\prime }})\mathrm{\mid }\le C_n[f{]}_{\alpha }{d}^{\alpha }$. Together these bounds give $[H_{ij}{]}_{\alpha ;Q}\le C_{n,\alpha }[f{]}_{\alpha ;{\mathbb{R}}^n}\mathrm{.}$ The estimates use the real exponent strictly below $1$ in the convergent outer radial integral.

On a compact $K$, choose $R_0$ so $\mathrm{supp}f$ and $K$ lie in a fixed bounded ball. Then for all $x\in K$, the integrals for $Nf$ and $DNf$ are bounded by $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$ times the integrals of $\mathrm{\mid }\mathrm{\Phi }\mathrm{\mid }$ and $\mathrm{\mid }D\mathrm{\Phi }\mathrm{\mid }$ over a fixed ball, which are finite by polar integration [F3]. The formula of step 4.1 bounds $\mathrm{\mid }{D}^{2}Nf(x)\mathrm{\mid }$ by $C_{n,\alpha ,R_0}[f{]}_{\alpha ;{\mathbb{R}}^n}+n^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{.}$ Together with step 5.2 this is the displayed local $C^{2,\alpha }$ estimate.

By compact support and continuity, $f$ is integrable, so [F8] applies and gives $-\mathrm{\Delta }{T}_{Nf}=T_f$. Since $Nf\in C^2$, classical/distributional compatibility [F9] and the Laplacian convention [F10] give $T_{-\mathrm{\Delta }Nf-f}=0$. Injectivity in [F9] makes this continuous difference zero almost everywhere. If it were nonzero at a point, continuity would make its modulus bounded below by a positive number on some nonempty ball, which has positive measure by [F9], a contradiction. Hence $-\mathrm{\Delta }Nf=f$ pointwise.

If $f=0$ or $\mathrm{supp}f=\mathrm{\varnothing }$, the integral and every term in the cancellation formula vanish. The proof covers $n=2$ by the logarithmic profile and $n\ge 3$ by the power profile. Dimensions $n=1$ and $n=0$ are outside the theorem's explicit range; the estimates and Green normalization used here are stated for $n\ge 2$. The strict endpoint restrictions $0<\alpha <1$ are used in local integrability of $r^{\alpha -1}$ and convergence of ${\int }_d^{\mathrm{\infty }}{r}^{\alpha -2}dr$. All radii and boxes are chosen individually from bounded sets; the only stated choice axiom is ${\mathrm{A}\mathrm{C}}_{\omega }$ in [A1]. The result is one-way and asserts no converse. [A1, F1, F2, F3, step 1.1, step 1.2, step 4.1, step 5.2, step 6.2, cases] $\mathrm{\square }$