Newtonian potential of radial compact data

The profile $\rho (s)=f(s{e}_1)$ is continuous on $[0,\mathrm{\infty })$ because $f$ is continuous and $e_1$ exists by [F22]; the support $\mathrm{supp}f$ is compact by hypothesis, hence bounded by [F18], so there is $R>0$ with $\rho (s)=0$ for $s\ge R$. Also $\mathrm{\parallel }\rho {\mathrm{\parallel }}_{\mathrm{\infty }}\le {\sup }_{{\mathbb{R}}^n}\mathrm{\mid }f\mathrm{\mid }<\mathrm{\infty }$: $f$ is continuous and vanishes off the compact set $\mathrm{supp}f$, so it is bounded by [F18]. By [F3] the defining integral for $Nf$ is absolutely finite at every $x$, and by [F4] the potential satisfies $Nf\in C^2({\mathbb{R}}^n)$ with $-\mathrm{\Delta }Nf=f$ pointwise.

We prove that $u=Nf$ is radial. Fix $x,x^{\mathrm{\prime }}\in {\mathbb{R}}^n$ with $\mathrm{\mid }x\mathrm{\mid }=\mathrm{\mid }{x}^{\mathrm{\prime }}\mathrm{\mid }$; if $x=x^{\mathrm{\prime }}$ there is nothing to prove, so assume $x\mathrm{\ne }{x}^{\mathrm{\prime }}$ and put $v:=x-x^{\mathrm{\prime }}\mathrm{\ne }0$ and $Qz:=z-2\frac{⟨z,v⟩}{⟨v,v⟩}v$. Bilinearity and symmetry of the inner product [F16] make $Q$ linear, and the expansion $⟨Qz,Qw⟩=⟨z,w⟩-4\frac{⟨z,v⟩⟨w,v⟩}{⟨v,v⟩}+4\frac{⟨z,v⟩⟨w,v⟩}{⟨v,v⟩}=⟨z,w⟩$ shows that $Q$ preserves the inner product; also $⟨Qz,v⟩=-⟨z,v⟩$, so $Q^{2}z=z$ and $Q$ is a bijective linear isometry, that is, an orthogonal operator [F16]. Since $2⟨x,v⟩=2(\mathrm{\mid }x{\mathrm{\mid }}^2-⟨x,x^{\mathrm{\prime }}⟩)=\mathrm{\mid }v{\mathrm{\mid }}^2$ and $x\mathrm{\ne }{x}^{\mathrm{\prime }}$, we get $Qx=x-v=x^{\mathrm{\prime }}$. The identity $Q(z+h)-Qz-Qh=0$ shows $DQ(z)=Q$ for every $z$, so $Q$ is $C^1$ with derivative the invertible map $Q$; as $Q^{-1}=Q$, it is a $C^1$ diffeomorphism of ${\mathbb{R}}^n$ with $\mathrm{\mid }\det DQ(z)\mathrm{\mid }=\mathrm{\mid }\det Q\mathrm{\mid }=1$ [F16]. Choose the representative of $\mathrm{\Phi }$ with $\mathrm{\Phi }(0)=0$; then $\mathrm{\mid }Qw\mathrm{\mid }=\mathrm{\mid }w\mathrm{\mid }$ gives $\mathrm{\Phi }(Qw)=\mathrm{\Phi }(w)$ for every $w$, because the two profiles in [F1] depend only on the modulus, and likewise $f(Qz)=\rho (\mathrm{\mid }Qz\mathrm{\mid })=\rho (\mathrm{\mid }z\mathrm{\mid })=f(z)$. Put $g(y):=\mathrm{\Phi }(x^{\mathrm{\prime }}-y)f(y)$; by [F3] the integral ${\int }_{{\mathbb{R}}^n}\mathrm{\mid }g\mathrm{\mid }d{\lambda }_n$ is finite, so $g\in L^1({\lambda }_n)$ and the change-of-variables formula [F16] gives $u(x^{\mathrm{\prime }})={\int }_{{\mathbb{R}}^n}gd{\lambda }_n={\int }_{{\mathbb{R}}^n}g(Qz)d{\lambda }_n(z)$. For every $z$ the transformed integrand equals $\mathrm{\Phi }(x^{\mathrm{\prime }}-Qz)f(Qz)=\mathrm{\Phi }(Q(x-z))f(z)=\mathrm{\Phi }(x-z)f(z)$, so $u(x^{\mathrm{\prime }})=u(x)$. Applying this with $x^{\mathrm{\prime }}=\mathrm{\mid }x\mathrm{\mid }{e}_1$ (and trivially for $x=0$) gives $u(x)=u(\mathrm{\mid }x\mathrm{\mid }{e}_1)$ for every $x$, so $u$ is radial.

From now on write $u(r)$ for the common value of the radial $u$ at points of modulus $r$, and use $e_1$ for the standard basis vector with coordinate index $1$. Since $u\in C^2({\mathbb{R}}^n)$ by [F4], the profile is continuous on $[0,\mathrm{\infty })$ and, for $r>0$, differentiable with $u^{\mathrm{\prime }}(r)=Du(r{e}_1)\cdot e_1$: the chain rule applied to the affine map $r\mapsto r{e}_1$, whose total derivative is the constant map $h\mapsto h{e}_1$, with [F13] and [F16] gives $D(u\circ e_1)(r)=Du(r{e}_1)\circ e_1$. For each $i$ the one-variable function $h\mapsto u(h{e}_i)$ is even, because $\mathrm{\mid }h{e}_i\mathrm{\mid }=\mathrm{\mid }-h{e}_i\mathrm{\mid }$ and $u$ is radial; consequently ${\mathrm{\partial }}_{i}u(0)=0$, since the difference quotient at $h\mathrm{\ne }0$ equals the negative of the quotient at $-h$ and both tend to ${\mathrm{\partial }}_{i}u(0)$ as $h\to 0$. Thus $Du(0)=0$, and by continuity of $Du$ [F4] the profile's one-sided derivative at the origin is ${\lim }_{r\downarrow 0}{u}^{\mathrm{\prime }}(r)=Du(0)\cdot e_1=0$.

Fix $r>0$. The ball $B_r(0)$ is a bounded $C^1$ domain with outward unit normal $\nu (y)=y\mathrm{/}r$ on its boundary sphere $S_r$, which is the regular level set $\{y:\mathrm{\mid }y{\mathrm{\mid }}^2=r^2\}$ [F10, F12]. Since $u\in C^2({\mathbb{R}}^n)$, the field $\mathrm{\nabla }u$ lies in $C^1(\overline{B_r(0)};{\mathbb{R}}^n)$, so [F9] applies and, with [F14], gives ${\int }_{B_r(0)}\mathrm{\Delta }udy={\int }_{S_r}\mathrm{\nabla }u\cdot \nu dS={\int }_{S_r}{\mathrm{\partial }}_{\nu }udS$. For $y\mathrm{\ne }0$ the chain rule [F13] applied to the composition of the profile $u(\cdot )$ with $\mathrm{\mid }\cdot \mathrm{\mid }$, together with the derivative $\mathrm{\nabla }\mathrm{\mid }\cdot \mathrm{\mid }(y)=y\mathrm{/}\mathrm{\mid }y\mathrm{\mid }$ obtained from $\mathrm{\mid }y\mathrm{\mid }=(\mathrm{\mid }y{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}$, $\mathrm{\nabla }(\mathrm{\mid }y{\mathrm{\mid }}^2)=2y$ and the half-power rule, gives $\mathrm{\nabla }u(y)=u^{\mathrm{\prime }}(\mathrm{\mid }y\mathrm{\mid })y\mathrm{/}\mathrm{\mid }y\mathrm{\mid }$; hence ${\mathrm{\partial }}_{\nu }u(y)=u^{\mathrm{\prime }}(r)$ for every $y\in S_r$ [F11]. Since $u^{\mathrm{\prime }}(r)$ is constant on $S_r$ and $\sigma (S^{n-1})={\omega }_{n-1}$ with radius scaling $r^{n-1}$ [F6], the surface integral equals ${\omega }_{n-1}{r}^{n-1}{u}^{\mathrm{\prime }}(r)$ by [F8]. On the other hand $\mathrm{\Delta }u=-f$ pointwise by [F4], so ${\int }_{B_r(0)}\mathrm{\Delta }u=-{\int }_{B_r(0)}f$; the product $f{\chi }_{B_r(0)}$ and its positive and negative parts are Borel, because $f$ is continuous, the open ball is Borel and the operations preserve measurability [F19], so applying [F5] to those parts, noting ${\chi }_{B_r(0)}(s\omega )=1$ exactly for $s<r$, and subtracting with the signed-integral convention [F18] gives ${\int }_{B_r(0)}f={\omega }_{n-1}{\int }_0^r{s}^{n-1}\rho (s)ds$. Comparing the two values of ${\int }_{B_r(0)}\mathrm{\Delta }u$ and dividing by ${\omega }_{n-1}{r}^{n-1}$ yields $u^{\mathrm{\prime }}(r)=-r^{1-n}{\int }_0^r{s}^{n-1}\rho (s)ds$.

Define $G(t):=t^{1-n}{\int }_0^t{s}^{n-1}\rho (s)ds$ for $t>0$ and $G(0):=0$. For $0\le t$ we have $\mid {\int }_0^t{s}^{n-1}\rho (s)ds\mid \le \mathrm{\parallel }\rho {\mathrm{\parallel }}_{\mathrm{\infty }}{t}^n\mathrm{/}n$ by [F18], so $\mathrm{\mid }G(t)\mathrm{\mid }\le \mathrm{\parallel }\rho {\mathrm{\parallel }}_{\mathrm{\infty }}t\mathrm{/}n$ and $G(t)\to 0$ as $t\downarrow 0$; on $(0,\mathrm{\infty })$ the integral $t\mapsto {\int }_0^t{s}^{n-1}\rho (s)ds$ is a primitive of the continuous function $t\mapsto t^{n-1}\rho (t)$ [F15], hence is differentiable and continuous there, and $G$ is a product of continuous functions on $(0,\mathrm{\infty })$ [F17]. Step 1.4 gives $u^{\mathrm{\prime }}(t)=-G(t)$ for every $t>0$, and step 1.3 gives the one-sided derivative $u^{\mathrm{\prime }}(0)=0=-G(0)$; thus the profile $u$ is a primitive of the continuous function $-G$ on the interval $[0,\mathrm{\infty })$ [F15]. Applying the evaluation clause of [F15] with $a=0$ and $b=r>0$ gives $u(r)-u(0)={\int }_0^r-G(t)dt=-{\int }_0^r{t}^{1-n}{\int }_0^t{s}^{n-1}\rho (s)dsdt$, which is the displayed identity; at $r=0$ both sides are $0$.

It remains to compute the value at the origin. By [F2], [F3] and $\mathrm{\Phi }(-y)=\mathrm{\Phi }(y)$, where both sides equal the profile evaluated at $\mathrm{\mid }y\mathrm{\mid }$ under the representative chosen in step 1.2, $u(0)={\int }_{{\mathbb{R}}^n}\mathrm{\Phi }(y)\rho (\mathrm{\mid }y\mathrm{\mid })dy$. The integrand is Borel [F1, F19] and bounded in modulus by $\mathrm{\parallel }\rho {\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }\mathrm{\Phi }(y)\mathrm{\mid }{\mathbf{1}}_{B_R(0)}(y)$, since $\rho$ vanishes outside that ball; hence it is integrable, because $\mathrm{\mid }\mathrm{\Phi }\mathrm{\mid }$ is locally integrable [F20] and the remaining region is bounded [F18]. Therefore [F5] applied to the positive and negative parts and the subtraction rule of [F18] give $u(0)={\int }_0^{\mathrm{\infty }}{\int }_{S^{n-1}}\mathrm{\Phi }(s\omega )\rho (s)s^{n-1}d\sigma (\omega )ds={\omega }_{n-1}{\int }_0^{\mathrm{\infty }}{q}_n(s)\rho (s)s^{n-1}ds,$ where $q_n$ is the radial profile of [F1] and we used $\sigma (S^{n-1})={\omega }_{n-1}$ [F6] and the fact that $q_n(s)\rho (s)$ is independent of $\omega$. If $n\ge 3$, then ${\omega }_{n-1}{q}_n(s)s^{n-1}=s\mathrm{/}(n-2)$, so $u(0)=\frac{1}{n-2}{\int }_0^{\mathrm{\infty }}s\rho (s)ds$. If $n=2$, then [F7] gives $\mathrm{\mid }{B}_1\mathrm{\mid }=V_2(1)=\pi \mathrm{/}\mathrm{\Gamma }(2)=\pi$ and, with [F6], ${\omega }_1=\mathrm{\mid }{S}^1\mathrm{\mid }=2\mathrm{\mid }{B}_1\mathrm{\mid }=2\pi$, so ${\omega }_1{q}_2(s)s=-s\log s$; the integral $-{\int }_0^{\mathrm{\infty }}s\log s\rho (s)ds$ is absolutely convergent because $\rho$ vanishes outside $[0,R]$ and ${\int }_0^{R}s\mathrm{\mid }\log s\mathrm{\mid }ds<\mathrm{\infty }$ by [F13] and [F21]. This gives the two displayed values of $u(0)$.

If $\rho \equiv 0$ (including the case of empty support), then $f=0$, the integrands in steps 1.4, 2.1 and 2.2 all vanish, and $u\equiv 0$, so every displayed identity reduces to $0=0$; the estimates in steps 1.4 and 2.2 are then trivially finite. The cases $n=2$ and $n\ge 3$ are exhaustive under $n\ge 2$ and are exactly the two alternatives in [F1]; dimension one is excluded by the stated hypothesis and is not silently included. The derivative formula is asserted only for $r>0$; at $r=0$ step 1.3 supplies the one-sided derivative and step 2.1 the integrated identity. Countable Choice is the sole set-theoretic assumption: it is inherited from the kernel, potential, polar, surface and divergence conventions of [F1]–[F5], [F9] and [F16], while the pointwise reflection, chain-rule and one-dimensional integration arguments use no further choice. For complex-valued radial data the result applies to $\mathrm{R}\mathrm{e}f$ and $\mathrm{I}\mathrm{m}f$, whose profiles are the real and imaginary parts of $\rho$.