Symmetry and Fubini representation of the Newtonian linkage pairing

The supports $K_u:=\mathrm{supp}u$ and $K_v:=\mathrm{supp}v$ are compact by [F8], hence closed and bounded in $({\mathbb{R}}^3,d_2)$ by [F11] and [F15]; by [F16] there are $x_u,{\rho }_u$ and $x_v,{\rho }_v$ with $K_u\subseteq B(x_u,{\rho }_u)$ and $K_v\subseteq B(x_v,{\rho }_v)$ (for an empty support any pair works, since the empty set is bounded). Put $R:=\max \{\mathrm{\mid }{x}_u\mathrm{\mid }+{\rho }_u,\mathrm{\mid }{x}_v\mathrm{\mid }+{\rho }_v\}>0$. For $x\in K_u$ the triangle inequality gives $\mathrm{\mid }x\mathrm{\mid }\le \mathrm{\mid }{x}_u\mathrm{\mid }+\mathrm{\mid }x-x_u\mathrm{\mid }<\mathrm{\mid }{x}_u\mathrm{\mid }+{\rho }_u\le R$, and likewise on $K_v$, so $K_u\cup K_v\subseteq B_R(0)=B(0,R)$; moreover $u(x)=0$ for $x\mathrm{\notin }{K}_u$ and $v(y)=0$ for $y\mathrm{\notin }{K}_v$ by [F8].

For $z\mathrm{\ne }0$, [F2] gives $\mathrm{\Phi }(z)=\mathrm{\mid }z{\mathrm{\mid }}^{2-3}\mathrm{/}((3-2){\omega }_2)=\mathrm{\mid }z{\mathrm{\mid }}^{-1}\mathrm{/}{\omega }_2$, and [F3] gives ${\omega }_2=\mathrm{\mid }{S}^2\mathrm{\mid }=3\mathrm{\mid }{B}_1\mathrm{\mid }$ with $\mathrm{\mid }{B}_1\mathrm{\mid }=V_3(1)$. By [F4], $V_3(1)={\pi }^{3\mathrm{/}2}\mathrm{/}\mathrm{\Gamma }(5\mathrm{/}2)$; by [F5] applied twice and [F6], $\mathrm{\Gamma }(5\mathrm{/}2)=\frac{3}{2}\cdot \frac{1}{2}\cdot \mathrm{\Gamma }(1\mathrm{/}2)=\frac{3}{4}\sqrt{\pi }$, hence $V_3(1)={\pi }^{3\mathrm{/}2}\mathrm{/}(\frac{3}{4}\sqrt{\pi })=\frac{4}{3}\pi$ and ${\omega }_2=4\pi$. Therefore $4\pi \mathrm{\Phi }(z)=\mathrm{\mid }z{\mathrm{\mid }}^{-1}=g(z)$ for $z\mathrm{\ne }0$; assigning the permitted pole value $\mathrm{\Phi }(0)=0$ by [F2] gives $\mathrm{\Phi }=g\mathrm{/}(4\pi )$ on all of ${\mathbb{R}}^3$, so by [F1] the transform of the Statement is $Nf(x)=\frac{1}{4\pi }{\int }_{{\mathbb{R}}^3}f(y)g(x-y)d{\lambda }_3(y)$ at each $x$ where this integral is absolutely finite.

For $i,j\in \{0,1,2\}$, by [F9] the components $u_i,v_j$ are smooth, so by [F10] they are continuous, and by [F27] they are Borel measurable; likewise $g$ is Borel measurable because assigning the value $0$ at the pole of the continuous function $z\mapsto \mathrm{\mid }z{\mathrm{\mid }}^{-1}$ on ${\mathbb{R}}^3\setminus \{0\}$ gives the function whose superlevel sets $\{g>t\}$ are $B_{1\mathrm{/}t}(0)\setminus \{0\}$ for $t>0$, ${\mathbb{R}}^3\setminus \{0\}$ for $t=0$ and ${\mathbb{R}}^3$ for $t<0$, all open and hence Borel. Applying [F28] to $g$ and $v_j$ shows $(x,y)\mapsto g(x-y)v_j(y)$ is Borel on ${\mathbb{R}}^6$; the maps $(x,y)\mapsto u_i(x)$ are Borel as compositions with the continuous projection, and for each pair $(i,j)$ the map ${\mathrm{\Phi }}_{ij}(x,y):=(u_i(x),g(x-y)v_j(y))$ has Borel preimages of open rectangles by [F29]; multiplication ${\mathbb{R}}^2\to \mathbb{R}$ is continuous, so [F30] makes $(x,y)\mapsto u_i(x)v_j(y)g(x-y)$ Borel, and the finite sum $H(x,y):=⟨u(x),v(y)⟩g(x-y)$ is Borel as well.

The closed ball ${\bar{B}}_{2R}(0)$ is closed by [F12] and bounded by [F16], hence compact in $({\mathbb{R}}^3,d_2)$ by [F13]. Since $\mathrm{\Phi }\ge 0$ in dimension three and $\mathrm{\Phi }=g\mathrm{/}(4\pi )$ by step 1.2, [F7] and [F18] give $K_R:={\int }_{B_{2R}(0)}gd{\lambda }_3={\int }_{B_{2R}(0)}4\pi \mathrm{\Phi }d{\lambda }_3\le 4\pi {\int }_{{\bar{B}}_{2R}(0)}\mathrm{\Phi }d{\lambda }_3<\mathrm{\infty }\mathrm{.}$

Fix $x\in B_R(0)$. The translation-dilation map $T_x:{\mathbb{R}}^3\to {\mathbb{R}}^3$, $T_x(y):=x-y$, is smooth with $T_x^{-1}=T_x$ and $\mathrm{\mid }\det D{T}_x(y)\mathrm{\mid }=1$ for every $y$, so it is a $C^1$ diffeomorphism; and $T_x(B_R(0))=x-B_R(0)\subseteq B_{2R}(0)$, since $\mathrm{\mid }x-y\mathrm{\mid }\le \mathrm{\mid }x\mathrm{\mid }+\mathrm{\mid }y\mathrm{\mid }<2R$ for $\mathrm{\mid }y\mathrm{\mid }<R$. The function $g$ is nonnegative and ${\lambda }_3$-measurable, and ${\int }_{B_{2R}(0)}g<\mathrm{\infty }$ by step 2.1, so $g{\mathbf{1}}_{B_{2R}(0)}\in L^1({\lambda }_3)$; applying [F22] to it and using [F18] and $T_x(B_R(0))\subseteq B_{2R}(0)$ yields ${\int }_{B_R(0)}g(x-y)d{\lambda }_3(y)={\int }_{T_x(B_R(0))}g(z)d{\lambda }_3(z)\le {\int }_{B_{2R}(0)}gd{\lambda }_3=K_R\mathrm{.}$

By step 1.1, $\mathrm{\mid }{u}_i\mathrm{\mid }\le M_u$ and $\mathrm{\mid }{v}_j\mathrm{\mid }\le M_v$ on all of ${\mathbb{R}}^3$ for the finite constants $M_u:={\max }_{{\bar{B}}_R(0)}\mathrm{\mid }u\mathrm{\mid }$ and $M_v:={\max }_{{\bar{B}}_R(0)}\mathrm{\mid }v\mathrm{\mid }$, which exist by [F14] applied to the continuous functions $\mathrm{\mid }u\mathrm{\mid },\mathrm{\mid }v\mathrm{\mid }$ on the nonempty compact set ${\bar{B}}_R(0)$ of steps 1.1 and 2.1 (a vanishing field has $M=0$). The measure space is sigma-finite because ${\mathbb{R}}^3={\bigcup }_{m=1}^{\mathrm{\infty }}{B}_m(0)$ and each ball has finite measure by [F21]. Hence, using [F17] and [F23] for the nonnegative Borel function $\mathrm{\mid }{u}_i(x)\mathrm{\mid }\mathrm{\mid }{v}_j(y)\mathrm{\mid }g(x-y)$ on ${\mathbb{R}}^3\times {\mathbb{R}}^3$, [F18] and step 3.1 give ${\int }_{{\mathbb{R}}^3}{\int }_{{\mathbb{R}}^3}\mathrm{\mid }{u}_i(x)\mathrm{\mid }\mathrm{\mid }{v}_j(y)\mathrm{\mid }g(x-y)d{\lambda }_3(y)d{\lambda }_3(x)={\int }_{B_R(0)}{\int }_{B_R(0)}\mathrm{\mid }{u}_i(x)\mathrm{\mid }\mathrm{\mid }{v}_j(y)\mathrm{\mid }g(x-y)d{\lambda }_3(y)d{\lambda }_3(x)\le M_u{M}_v{\lambda }_3(B_R(0))K_R<\mathrm{\infty },$ with ${\lambda }_3(B_R(0))<\mathrm{\infty }$ by [F21] and $K_R<\mathrm{\infty }$ by step 2.1. Summing over the three pairs $(i,j)=(0,0),(1,1),(2,2)$ and using [F26] and $\mathrm{\mid }⟨u,v⟩\mathrm{\mid }\le {\sum }_i\mathrm{\mid }{u}_i\mathrm{\mid }\mathrm{\mid }{v}_i\mathrm{\mid }$, the double integral of $⟨u(x),v(y)⟩g(x-y)\mathrm{/}(4\pi )$ converges absolutely; by [F25] and [F29] the Euclidean and product-measure structures on ${\mathbb{R}}^6$ agree on Borel sets, so [F24] applies to this $L^1$ function and permits either order of integration.

For every $x\in {\mathbb{R}}^3$, $x-B_R(0)\subseteq B_{\mathrm{\mid }x\mathrm{\mid }+R}(0)$; the change of variables $z=x-y$ of [F22] and local integrability [F7] give $\int \mathrm{\mid }{v}_j(y)\mathrm{\mid }\mathrm{\Phi }(x-y)dy\le M_v(4\pi {)}^{-1}{\int }_{{\bar{B}}_{\mathrm{\mid }x\mathrm{\mid }+R}(0)}g(z)dz<\mathrm{\infty }$. The same argument applies to $u$, so $Nu,Nv$ are defined everywhere. Their components are measurable: apply [F23] to the nonnegative Borel integrands $\mathrm{\Phi }(x-y)v_j^{\pm }(y)$ and take the difference of the finite section integrals, and likewise for $u$. By [F20], $⟨u(x),Nv(x)⟩=(4\pi {)}^{-1}\int ⟨u(x),v(y)⟩g(x-y)dy$. Pointwise, its absolute value is at most $(4\pi {)}^{-1}{\sum }_j\int \mathrm{\mid }{u}_j(x)\mathrm{\mid }\mathrm{\mid }{v}_j(y)\mathrm{\mid }g(x-y)dy$; integrating this bound and using step 4.1 proves absolute integrability. Integration of the preceding equality therefore gives the displayed double-integral representation in the Statement by [F24]. Every fraction with denominator $\mathrm{\mid }x-y\mathrm{\mid }$ is interpreted using $g(0)=0$ as stipulated there.

The same computation with the roles of $u$ and $v$ exchanged, using [F24] to interchange the two orders of integration of the common absolutely convergent double integral of step 4.1, gives ${\int }_{{\mathbb{R}}^3}⟨Nu(x),v(x)⟩d{\lambda }_3(x)=\frac{1}{4\pi }{\int }_{{\mathbb{R}}^3}{\int }_{{\mathbb{R}}^3}\frac{⟨u(x),v(y)⟩}{\mathrm{\mid }x-y\mathrm{\mid }}d{\lambda }_3(y)d{\lambda }_3(x)\mathrm{.}$

For bilinearity let $u^{(1)},u^{(2)},v,\alpha ,\beta$ be given as in the Statement, with $w:=\alpha u^{(1)}+\beta u^{(2)}$. The componentwise linearity of the Newtonian integral [F20] and step 1.1's support bound for the union of the supports show $Nw=\alpha N{u}^{(1)}+\beta N{u}^{(2)}$ at every point, hence $B(w,v)=\alpha B(u^{(1)},v)+\beta B(u^{(2)},v)$ by [F20] and the absolute finiteness of step 4.1 (the argument of steps 1.1-6.1 applies verbatim to $w$, whose support is contained in the union of the two supports and hence bounded). The second-slot computation is the same with the roles exchanged.

Steps 5.1, 6.1 and the definition of $B$ show that $B(u,v)$ is well defined and real-valued, that all three integrals of the Statement are absolutely convergent with that common value, and that the exchanged pairing satisfies $B(v,u)=\frac{1}{4\pi }\int \int \frac{⟨v(y),u(x)⟩}{\mathrm{\mid }y-x\mathrm{\mid }}d{\lambda }_3(x)d{\lambda }_3(y)$; since $\mathrm{\mid }x-y\mathrm{\mid }=\mathrm{\mid }y-x\mathrm{\mid }$ and $⟨v(y),u(x)⟩=⟨u(x),v(y)⟩$ by [F26], [F24] identifies this with $B(u,v)$, giving symmetry.

If $u=0$ or $v=0$ then all integrands above vanish identically, all three integrals are $0$, and the symmetry and bilinearity identities hold trivially; the empty support case was already admitted in step 1.1. The proof uses Countable Choice exactly through [A1] in the cited change-of-variables, product-measure, ball-measure and Borel-measurability interfaces, and no full Axiom of Choice.