The cancelled representation of the second derivatives of Newtonian potentials

Put $r=\mathrm{\mid }z\mathrm{\mid }$ for $z\mathrm{\ne }0$. Differentiating the profiles of [F1] with the rules of [F2] gives, in both cases $n\ge 3$ and $n=2$, ${\mathrm{\partial }}_i\mathrm{\Phi }(z)=-\frac{z_i}{{\omega }_{n-1}{r}^n},k_{ij}(z)={\mathrm{\partial }}_i{\mathrm{\partial }}_j\mathrm{\Phi }(z)=-\frac{1}{{\omega }_{n-1}}\cdot \frac{1}{r^n}({\delta }_{ij}-n\frac{z_i{z}_j}{r^2})\mathrm{.}$ Both formulas are smooth in $z\mathrm{\ne }0$, and $k_{ij}(\lambda z)={\lambda }^{-n}{k}_{ij}(z)$ for every $\lambda >0$, so $k_{ij}$ is homogeneous of degree $-n$.

Estimates. Directly from step 1.1, $\mathrm{\mid }{k}_{ij}(z)\mathrm{\mid }\le (1+n){\omega }_{n-1}^{-1}{r}^{-n}$. Since $k_{ij}$ is smooth and homogeneous of degree $-n$ on ${\mathbb{R}}^n\setminus \{0\}$, each partial derivative ${\mathrm{\partial }}_l{k}_{ij}$ is homogeneous of degree $-n-1$, and its supremum over the compact unit sphere is a finite constant $C_{n,j,l}$, so $\mathrm{\mid }\mathrm{\nabla }{k}_{ij}(z)\mathrm{\mid }\le C_n{r}^{-n-1}$. If $\mathrm{\mid }z\mathrm{\mid }\ge 2\mathrm{\mid }h\mathrm{\mid }$, every point of the segment from $z-h$ to $z$ has norm at least $r\mathrm{/}2$, so the mean value theorem bound in [F2], applied along that segment, gives $\mathrm{\mid }{k}_{ij}(z-h)-k_{ij}(z)\mathrm{\mid }\le C_n\mathrm{\mid }h\mathrm{\mid }(r\mathrm{/}2{)}^{-n-1}$, which is the stated pointwise estimate with a constant doubled to the power $n+1$.

Zero spherical means. On the unit sphere $k_{ij}(\theta )=-{\omega }_{n-1}^{-1}({\delta }_{ij}-n{\theta }_i{\theta }_j)$. By [F3] the measure $\sigma$ is invariant under coordinate reflections, so $\int {\theta }_i{\theta }_{j}d\sigma =0$ for $i\mathrm{\ne }j$; it is also invariant under coordinate permutations, so all $n$ numbers $\int {\theta }_i^{2}d\sigma$ are equal, and ${\sum }_i{\theta }_i^2=1$ gives $\int {\theta }_i^{2}d\sigma ={\omega }_{n-1}\mathrm{/}n$. Hence ${\int }_{S^{n-1}}{\theta }_i{\theta }_{j}d\sigma ={\delta }_{ij}{\omega }_{n-1}\mathrm{/}n$ and therefore ${\int }_{S^{n-1}}{k}_{ij}(\theta )d\sigma (\theta )=-{\omega }_{n-1}^{-1}({\delta }_{ij}{\omega }_{n-1}-n{\delta }_{ij}{\omega }_{n-1}\mathrm{/}n)=0$. By the homogeneity of step 1.1 the mean over every centred sphere vanishes: ${\int }_{S(0,\rho )}{k}_{ij}dS={\rho }^{n-1}{\rho }^{-n}{\int }_{S^{n-1}}{k}_{ij}(\theta )d\sigma =0$.

$k_{ij}$ is a standard Hölder Calderón–Zygmund kernel. The annular size condition follows from step 2.1 and [F3]: ${\int }_{R\le \mathrm{\mid }z\mathrm{\mid }\le 2R}\mathrm{\mid }{k}_{ij}\mathrm{\mid }dz={\int }_R^{2R}{\rho }^{-1}d\rho {\int }_{S^{n-1}}\mathrm{\mid }{k}_{ij}(\theta )\mathrm{\mid }d\sigma \le (1+n)\log 2=:A_1$. Hörmander's condition follows from the difference estimate of step 2.1 and polar integration: for $y\mathrm{\ne }0$, ${\int }_{\mathrm{\mid }x\mathrm{\mid }\ge 2\mathrm{\mid }y\mathrm{\mid }}\mathrm{\mid }{k}_{ij}(x-y)-k_{ij}(x)\mathrm{\mid }dx\le C_n\mathrm{\mid }y\mathrm{\mid }{\int }_{2\mathrm{\mid }y\mathrm{\mid }}^{\mathrm{\infty }}{\rho }^{-n-1}{\omega }_{n-1}{\rho }^{n-1}d\rho =C_n{\omega }_{n-1}\mathrm{/}2=:A_2$. The pointwise estimate of step 2.1 is exactly condition (1) of [F9] with $\delta =1$ and constant $C_n{2}^{n+1}$, so $k_{ij}$ is standard $1$-Hölder in the sense of Standard (Hölder) Calderón–Zygmund kernels.

The principal value exists. Let $\phi \in \mathcal{S}({\mathbb{R}}^n)$ and $0<\epsilon <1$. Using the vanishing spherical means of step 2.2 on the annulus, ${\int }_{\mathrm{\mid }z\mathrm{\mid }\ge \epsilon }{k}_{ij}(z)\phi (z)dz={\int }_{\epsilon \le \mathrm{\mid }z\mathrm{\mid }<1}{k}_{ij}(z)(\phi (z)-\phi (0))dz+{\int }_{\mathrm{\mid }z\mathrm{\mid }\ge 1}{k}_{ij}(z)\phi (z)dz\mathrm{.}$ The second integral is absolutely convergent because $\mathrm{\mid }{k}_{ij}(z)\phi (z)\mathrm{\mid }\le C_n\mathrm{\mid }z{\mathrm{\mid }}^{-n}\mathrm{\mid }\phi (z)\mathrm{\mid }$ and $\phi$ is Schwartz, and the first integrand is dominated by $C_n\mathrm{\parallel }\mathrm{\nabla }\phi {\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }z{\mathrm{\mid }}^{1-n}$, which is integrable on $\mathrm{\mid }z\mathrm{\mid }<1$ for $n\ge 2$; hence the first integral converges as $\epsilon \downarrow 0$ by dominated convergence. Therefore $W_{ij}(\phi ):={\lim }_{\epsilon \downarrow 0}{\int }_{\mathrm{\mid }z\mathrm{\mid }\ge \epsilon }{k}_{ij}(z)\phi (z)dz$ exists for every Schwartz $\phi$; the bound $\mathrm{\mid }{W}_{ij}(\phi )\mathrm{\mid }\le C_n(\mathrm{\parallel }\mathrm{\nabla }\phi {\mathrm{\parallel }}_{\mathrm{\infty }}+{\sup }_z(1+\mathrm{\mid }z\mathrm{\mid }{)}^2\mathrm{\mid }\phi (z)\mathrm{\mid })$ proves continuity in Schwartz seminorms, so the map is a tempered distribution agreeing with $k_{ij}$ off the origin, and taking ${\delta }_j=1\mathrm{/}j$ exhibits it as a principal-value distribution for $k_{ij}$ in the sense of [F9].

Classical formula for Hölder data. If $0<\alpha <1$ put $\beta :=\alpha$, and if $\alpha =1$ put $\beta :=1\mathrm{/}2$; $f\in C_c^{0,\beta }({\mathbb{R}}^n)$ in either case: for $\alpha =1$, if $\mathrm{\mid }x-y\mathrm{\mid }\le 1$ then $\mathrm{\mid }f(x)-f(y)\mathrm{\mid }\mathrm{/}\mathrm{\mid }x-y{\mathrm{\mid }}^{\beta }\le [f{]}_{0,1}$, while if $\mathrm{\mid }x-y\mathrm{\mid }>1$ it is at most $2\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$. Thus $[f{]}_{0,\beta }\le \max \{[f{]}_{0,1},2\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\}<\mathrm{\infty }$, so [F7] gives $Nf\in C^2({\mathbb{R}}^n)$ together with the ball formula. Fix $x\in {\mathbb{R}}^n$ and choose $r>1$ with $\mathrm{supp}f\subset B_r(x)$; substituting $z=x-y$ and splitting $\{\mathrm{\mid }z\mathrm{\mid }<r\}$ into $\{\mathrm{\mid }z\mathrm{\mid }<1\}$ and $\{1\le \mathrm{\mid }z\mathrm{\mid }<r\}$ gives ${\int }_{B_r(x)}{k}_{ij}(x-y)(f(y)-f(x))dy={\int }_{\mathrm{\mid }x-y\mathrm{\mid }<1}{k}_{ij}(x-y)(f(y)-f(x))dy+{\int }_{\mathrm{\mid }x-y\mathrm{\mid }\ge 1}{k}_{ij}(x-y)f(y)dy,$ because the pure multiple term $-f(x){\int }_{1\le \mathrm{\mid }z\mathrm{\mid }<r}{k}_{ij}(z)dz$ vanishes by the annular zero-mean property of step 2.2, while the remaining $\{1\le \mathrm{\mid }z\mathrm{\mid }<r\}$-integral equals the integral over $\{\mathrm{\mid }z\mathrm{\mid }\ge 1\}$: indeed $f(x+z)=0$ for $\mathrm{\mid }z\mathrm{\mid }\ge r$, so the two extended integrands agree almost everywhere. Hence ${\mathrm{\partial }}_i{\mathrm{\partial }}_{j}Nf(x)={\int }_{\mathrm{\mid }x-y\mathrm{\mid }<1}{k}_{ij}(x-y)(f(y)-f(x))dy+{\int }_{\mathrm{\mid }x-y\mathrm{\mid }\ge 1}{k}_{ij}(x-y)f(y)dy-\frac{{\delta }_{ij}}{n}f(x)$ for the fixed $x$.

The principal value and the two integrals. With $0<\epsilon <1$, the vanishing of ${\int }_{\epsilon \le \mathrm{\mid }x-y\mathrm{\mid }<1}{k}_{ij}(x-y)dy$ by step 2.2 gives ${\int }_{\mathrm{\mid }x-y\mathrm{\mid }\ge \epsilon }{k}_{ij}(x-y)f(y)dy={\int }_{\epsilon \le \mathrm{\mid }x-y\mathrm{\mid }<1}{k}_{ij}(x-y)(f(y)-f(x))dy+{\int }_{\mathrm{\mid }x-y\mathrm{\mid }\ge 1}{k}_{ij}(x-y)f(y)dy\mathrm{.}$ The first integrand is bounded by $C_n[f{]}_{0,\alpha ;{\mathbb{R}}^n}\mathrm{\mid }x-y{\mathrm{\mid }}^{\alpha -n}$, whose integral over $\mathrm{\mid }x-y\mathrm{\mid }<1$ is finite because $\alpha -n>-n$, so dominated convergence lets $\epsilon \downarrow 0$ pass to the absolutely convergent first integral of the statement; the other integral is absolutely convergent because $\mathrm{\mid }{k}_{ij}(x-y)f(y)\mathrm{\mid }\le C_n\mathrm{\mid }f(y)\mathrm{\mid }$ on $\mathrm{\mid }x-y\mathrm{\mid }\ge 1$ and $f$ is integrable. Hence the principal value $\mathrm{p}\mathrm{.}\mathrm{v}\mathrm{.}\int k_{ij}(x-y)f(y)dy$ exists for every $x$ and equals the sum of the two integrals of the displayed formula.

Distributional identity. Let $\phi \in C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$ and choose $R$ with $\mathrm{supp}\phi \subset B_{R\mathrm{/}2}(0)$. For $0<\epsilon <R\mathrm{/}2$ apply the divergence theorem [F4] on the bounded $C^1$ domain ${\mathrm{\Omega }}_{\epsilon }:=\{\epsilon <\mathrm{\mid }x\mathrm{\mid }<R\}$ to the field $F:=\mathrm{\Phi }{\mathrm{\partial }}_j\phi e_i-\phi {\mathrm{\partial }}_i\mathrm{\Phi }{e}_j$. Since $\mathrm{div}F={\mathrm{\partial }}_i(\mathrm{\Phi }{\mathrm{\partial }}_j\phi )-{\mathrm{\partial }}_j(\phi {\mathrm{\partial }}_i\mathrm{\Phi })=\mathrm{\Phi }{\mathrm{\partial }}_i{\mathrm{\partial }}_j\phi -k_{ij}\phi$ and $F=0$ near $\mathrm{\mid }x\mathrm{\mid }=R$, while $\nu =-x\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$ on $\mathrm{\mid }x\mathrm{\mid }=\epsilon$, ${\int }_{{\mathrm{\Omega }}_{\epsilon }}\mathrm{\Phi }{\mathrm{\partial }}_i{\mathrm{\partial }}_j\phi dx={\int }_{{\mathrm{\Omega }}_{\epsilon }}{k}_{ij}\phi dx+{\int }_{\mathrm{\mid }x\mathrm{\mid }=\epsilon }(\mathrm{\Phi }{\mathrm{\partial }}_j\phi {\nu }_i-\phi {\mathrm{\partial }}_i\mathrm{\Phi }{\nu }_j)dS\mathrm{.}$ On $\mathrm{\mid }x\mathrm{\mid }=\epsilon$ the first boundary term tends to $0$ as $\epsilon \downarrow 0$, because $\mathrm{\Phi }=O({\epsilon }^{2-n})$ or $O(\log (1\mathrm{/}\epsilon ))$ and $dS$ contributes ${\epsilon }^{n-1}$; the second boundary term is $-{\int }_{\mathrm{\mid }x\mathrm{\mid }=\epsilon }\phi {\mathrm{\partial }}_i\mathrm{\Phi }{\nu }_{j}dS$, and by step 1.1, the outward normal $\nu =-x\mathrm{/}\epsilon$ and the scaling rule of [F3], ${\int }_{\mathrm{\mid }x\mathrm{\mid }=\epsilon }\phi {\mathrm{\partial }}_i\mathrm{\Phi }{\nu }_{j}dS=\frac{1}{{\omega }_{n-1}{\epsilon }^{n+1}}{\int }_{\mathrm{\mid }x\mathrm{\mid }=\epsilon }\phi (x)x_i{x}_{j}dS(x)⟶\frac{{\delta }_{ij}}{n}\phi (0),$ because ${\int }_{\mathrm{\mid }x\mathrm{\mid }=\epsilon }\phi x_i{x}_{j}dS={\epsilon }^{n+1}{\int }_{S^{n-1}}\phi (\epsilon \theta ){\theta }_i{\theta }_{j}d\sigma (\theta )$ and $\phi (\epsilon \theta )=\phi (0)+O(\epsilon )$. Hence the boundary integral in the identity tends to $-\frac{{\delta }_{ij}}{n}\phi (0)$. Passing to the limit, using local integrability of $\mathrm{\Phi }$ on the left and step 3.2 on the right, gives $⟨{\mathrm{\partial }}_i{\mathrm{\partial }}_j\mathrm{\Phi },\phi ⟩=⟨\mathrm{\Phi },{\mathrm{\partial }}_i{\mathrm{\partial }}_j\phi ⟩=W_{ij}(\phi )-\frac{{\delta }_{ij}}{n}\phi (0)$; by [F5] this is the distributional identity $D_{ij}\mathrm{\Phi }=\mathrm{p}\mathrm{.}\mathrm{v}\mathrm{.}{k}_{ij}-\frac{{\delta }_{ij}}{n}{\delta }_0$.

Assembly. Combining steps 3.3 and 3.4 gives, for every $x\in {\mathbb{R}}^n$, ${\mathrm{\partial }}_i{\mathrm{\partial }}_{j}Nf(x)=\mathrm{p}\mathrm{.}\mathrm{v}\mathrm{.}\int k_{ij}(x-y)f(y)dy-({\delta }_{ij}\mathrm{/}n)f(x)$, and the two ordinary integrals in the equivalent displayed form converge absolutely by step 3.4; step 3.3 supplies the classical second derivatives. The subtraction of $f(x)$ is made only on the unit ball: the globally subtracted integrand obeys $\mathrm{\mid }{k}_{ij}(x-y)(f(y)-f(x))\mathrm{\mid }\ge \frac{1}{2}\mathrm{\mid }f(x)\mathrm{\mid }\mathrm{\mid }{k}_{ij}(x-y)\mathrm{\mid }$ for all $y$ outside a large ball when $f(x)\mathrm{\ne }0$, and ${\int }_{\mathrm{\mid }z\mathrm{\mid }\ge 1}\mathrm{\mid }{k}_{ij}(z)\mathrm{\mid }dz={\int }_1^{\mathrm{\infty }}{\rho }^{-1}d\rho {\int }_{S^{n-1}}\mathrm{\mid }{k}_{ij}(\theta )\mathrm{\mid }d\sigma =\mathrm{\infty }$ by [F3] and step 1.1, so no absolutely convergent Lebesgue integral over ${\mathbb{R}}^n$ can replace the principal value; the annular cancellation of step 2.2 is what makes the truncations converge. The case $\alpha =1$ was reduced to $\beta =1\mathrm{/}2<\alpha$ in step 3.3, where higher Hölder regularity than assumed is irrelevant to the formula; the positivity of $\alpha$ makes $\mathrm{\mid }z{\mathrm{\mid }}^{\alpha -n}$ integrable at the origin in step 3.4. All constructions are pointwise in $x$ and use only Countable Choice in the cited measure, polar, potential and distribution interfaces.

Distributional form for bounded compact data. Let $f\in L_c^{\mathrm{\infty }}({\mathbb{R}}^n)$ and $\phi \in C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$. The double integral $\iint \mathrm{\Phi }(x-y)f(y){\mathrm{\partial }}_i{\mathrm{\partial }}_j\phi (x)dydx$ is absolutely convergent: the integrand is supported in a bounded subset of ${\mathbb{R}}^n\times {\mathbb{R}}^n$ and is dominated there by a constant times $\mathrm{\mid }\mathrm{\Phi }(x-y)\mathrm{\mid }$, and $\mathrm{\Phi }$ is locally integrable. Fubini [F10] and the substitution $z=x-y$ therefore give $⟨D_{ij}Nf,\phi ⟩=\int f(y)⟨\mathrm{\Phi },D_{ij}\phi (\cdot +y)⟩dy=\int f(y)⟨D_{ij}\mathrm{\Phi },\phi (\cdot +y)⟩dy$, the last equality being the definition of the distributional derivative. Inserting the kernel identity proved in step 4.1, $⟨D_{ij}Nf,\phi ⟩=\int f(y)[W_{ij}(\phi (\cdot +y))-\frac{{\delta }_{ij}}{n}\phi (y)]dy$. For $y$ in the compact support of $f$, subtract $\phi (y)$ on $\epsilon \le \mathrm{\mid }z\mathrm{\mid }<1$ using step 2.2. The resulting near integrand is bounded by $C_n\mathrm{\parallel }\mathrm{\nabla }\phi {\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }z{\mathrm{\mid }}^{1-n}$; the far integral is bounded by $C_n\mathrm{\parallel }\phi {\mathrm{\parallel }}_{\mathrm{\infty }}{\int }_{1\le \mathrm{\mid }z\mathrm{\mid }\le R_0}\mathrm{\mid }z{\mathrm{\mid }}^{-n}dz$, where one $R_0>1$ contains all differences of the two supports. These bounds are uniform in $y$ and $\epsilon$; multiplying by $\mathrm{\mid }f(y)\mathrm{\mid }$ gives an integrable majorant on its compact support. Dominated convergence now passes the truncation limit through the $y$ integral; another application of Fubini [F10] (the truncated double integral is absolutely convergent on the bounded region) gives $\int f(y)W_{ij}(\phi (\cdot +y))dy={\lim }_{\epsilon \downarrow 0}{\iint }_{\mathrm{\mid }x-y\mathrm{\mid }\ge \epsilon }{k}_{ij}(x-y)f(y)\phi (x)dydx$. Hence $⟨{\mathrm{\partial }}_i{\mathrm{\partial }}_{j}Nf,\phi ⟩={\lim }_{\epsilon \downarrow 0}{\iint }_{\mathrm{\mid }x-y\mathrm{\mid }\ge \epsilon }{k}_{ij}(x-y)f(y)\phi (x)dydx-\frac{{\delta }_{ij}}{n}\int f\phi$ for every test function, which is clause (iii).