The retarded point-history pair solves wave, Lorenz and Maxwell equations distributionally, with locally integrable field representatives off the worldline

Fix $\psi \in \mathcal{D}(Q)$ and put $F:=K⊛\psi$, so that $F(y,s)=(4\pi {)}^{-1}\int \psi (y+w,s+\mathrm{\mid }w\mathrm{\mid }\mathrm{/}c)\mathrm{\mid }w{\mathrm{\mid }}^{-1}dw$. Then $F$ is of class $C^1$ on $Q$ and ${\mathrm{\partial }}_{y_i}F=K⊛({\mathrm{\partial }}_{x_i}\psi )$, ${\mathrm{\partial }}_{s}F=K⊛({\mathrm{\partial }}_t\psi )$. Indeed, fix a bounded parameter neighbourhood $\mathrm{\mid }y\mathrm{\mid }\le R$, $\mathrm{\mid }s\mathrm{\mid }\le R$ and choose $M$ with $\mathrm{supp}\psi \subseteq {\bar{B}}_M(0)\times [-M,M]$; every nonzero contribution then has $w\in B_{M+R}$, on which $\mathrm{\mid }w{\mathrm{\mid }}^{-1}$ is integrable by the polar formula. For fixed $(y,s)$ the difference quotients $[\psi (y+h{e}_i+w,s+\mathrm{\mid }w\mathrm{\mid }\mathrm{/}c)-\psi (y+w,s+\mathrm{\mid }w\mathrm{\mid }\mathrm{/}c)]\mathrm{/}h$ converge as $h\to 0$ to the partial derivative ${\mathrm{\partial }}_{x_i}\psi (y+w,s+\mathrm{\mid }w\mathrm{\mid }\mathrm{/}c)$ by definition of the partial derivative, and the mean value inequality bounds them by $\sup \mathrm{\mid }{\mathrm{\partial }}_{x_i}\psi \mathrm{\mid }{\mathbf{1}}_{B_{M+R}}\mathrm{\mid }w{\mathrm{\mid }}^{-1}$, an integrable majorant; dominated convergence gives ${\mathrm{\partial }}_{y_i}F=K⊛({\mathrm{\partial }}_{x_i}\psi )$ on the parameter neighbourhood, and the same argument with the $s$-difference quotients and the $t$-partial derivatives gives ${\mathrm{\partial }}_{s}F=K⊛({\mathrm{\partial }}_t\psi )$; the resulting integrals are continuous in $(y,s)$ by the same dominated convergence, so $F$ is $C^1$.

For every test $\psi$ and every $(y,s)\in Q$, $(K⊛W\psi )(y,s)=\psi (y,s)$. Indeed, with the translated test $\chi (z,\tau ):=\psi (y+z,s+\tau )$ one has $(W\psi )(y+\cdot ,s+\cdot )=W\chi$ by the chain rule and the constant coefficients of $W$, so $(K⊛W\psi )(y,s)=⟨K,W\chi ⟩=⟨WK,\chi ⟩=⟨{\delta }_{(0,0)},\chi ⟩=\chi (0,0)=\psi (y,s)$, where the second equality is the definition of the distributional operator $W$ applied to $K$ and the third is $WK={\delta }_{(0,0)}$.

Local integrability of the classical fields. On $V:=Q\setminus \mathrm{\Gamma }$ the map $g(s,w):=(z(s)+w,s+\mathrm{\mid }w\mathrm{\mid }\mathrm{/}c)$ is a $C^1$ diffeomorphism from $U:=\mathbb{R}\times ({\mathbb{R}}^3\setminus \{0\})$ onto $V$ with $\mathrm{\mid }\det Dg(s,w)\mathrm{\mid }=\kappa (s,w)\in [1-v_{\ast }\mathrm{/}c,1+v_{\ast }\mathrm{/}c]$: the inverse is $(x,t)\mapsto (s(x,t),x-z(s(x,t)))$ by the root of [F1], injectivity follows from $\mathrm{\mid }s-s^{\mathrm{\prime }}\mathrm{\mid }=\mathrm{\mid }\mathrm{\mid }{w}^{\mathrm{\prime }}\mathrm{\mid }-\mathrm{\mid }w\mathrm{\mid }\mathrm{\mid }\mathrm{/}c\le (v_{\ast }\mathrm{/}c)\mathrm{\mid }s-s^{\mathrm{\prime }}\mathrm{\mid }$, and the determinant expansion along the last row gives $-1+z^{\mathrm{\prime }}(s)\cdot w\mathrm{/}(c\mathrm{\mid }w\mathrm{\mid })=-\kappa$. For a compact observation window $Q_0\subseteq Q$ contained in $\{\mathrm{\mid }x\mathrm{\mid }\le M,t_0\le t\le t_1\}$ put $Z:=\mathrm{\mid }z(0)\mathrm{\mid }$, $S:=\max \{0,(M+Z-c{t}_0)\mathrm{/}(c-v_{\ast })\}$, $L:=1+M+Z+v_{\ast }\max \{S,\mathrm{\mid }{t}_1\mathrm{\mid }\}$ and $T:=\max \{0,t_1+S\}$; then every $(s,w)$ with $g(s,w)\in Q_0$ has $s\in [-S,t_1]$ and $\mathrm{\mid }w\mathrm{\mid }<L$, so by [F9] ${\int }_{Q_0\setminus \mathrm{\Gamma }}{r}^{-p}dxdt={\int }_U{\mathbf{1}}_{Q_0}(g(s,w))\mathrm{\mid }w{\mathrm{\mid }}^{-p}\kappa (s,w)dsdw\le (1+v_{\ast }\mathrm{/}c)T{\int }_{B_L}\mathrm{\mid }w{\mathrm{\mid }}^{-p}dw$, which is finite for $p=1,2$ by the polar formula ($\sigma (S^2)L^2\mathrm{/}2$ and $\sigma (S^2)L$ respectively). On the other hand the explicit formulas of [F1] show $\mathrm{\mid }\phi \mathrm{\mid }+\mathrm{\mid }A\mathrm{\mid }+\mathrm{\mid }\mathrm{\nabla }\phi \mathrm{\mid }+\mathrm{\mid }\mathrm{\nabla }A\mathrm{\mid }+\mathrm{\mid }{\mathrm{\partial }}_t\phi \mathrm{\mid }+\mathrm{\mid }{\mathrm{\partial }}_{t}A\mathrm{\mid }+\mathrm{\mid }\mathrm{curl}A\mathrm{\mid }\le C_0(r^{-2}+r^{-1})$ on $Q_0$ with $C_0$ depending on $q,{\epsilon }_0,c,v_{\ast }$ and the supremum of $\mathrm{\mid }\dot{\beta }\mathrm{\mid }$ over the compact history interval. Hence the zero extensions of $\phi ,A$ and of the classical fields $E_0:=-\mathrm{\nabla }\phi -{\mathrm{\partial }}_{t}A$, $B_0:=\mathrm{curl}A$, and of the first partial derivatives of $\phi ,A$ are locally integrable on $Q$; no local-integrability claim for derivatives of $E_0,B_0$ is needed or made. The graph $\mathrm{\Gamma }$ is closed and Borel and has spatial singleton sections, so [F11] and Tonelli give spacetime measure zero; zero extension therefore does not affect these integrals.

The history function $s\mapsto F(z(s),s)$ is $C^1$ with compact support contained in a finite interval, and its derivative is ${\mathrm{\nabla }}_{y}F(z(s),s)\cdot z^{\mathrm{\prime }}(s)+{\mathrm{\partial }}_{s}F(z(s),s)$. Indeed $F$ is $C^1$ by step 1.1, $z$ is $C^1$, and the chain rule gives the derivative; the support bound $\{s:g(s,w)\in \mathrm{supp}\psi \text{ for some }w\}\subseteq [-S,t_1]$ of the same cylinder estimate shows compact support, so the total derivative integrates to zero over $\mathbb{R}$.

Zero-extension representatives of the first derivatives. Let $f$ be one of the components of $\phi$ and $A$ on $V$, with zero extension $\tilde{f}$, and let $D\in \{{\mathrm{\partial }}_t,{\mathrm{\partial }}_1,{\mathrm{\partial }}_2,{\mathrm{\partial }}_3\}$. Since $\tilde{f}$ and $Df$ (zero extended) are locally integrable by step 1.3, both $u_{\tilde{f}}$ and $u_{(Df{)}^{\sim }}$ are distributions. Then $D{u}_{\tilde{f}}=u_{(Df{)}^{\sim }}$: choose the spatial cutoff ${\chi }_{\epsilon }(x,t):=\chi ((x-z(t))\mathrm{/}\epsilon )$ built from a smooth cutoff $\chi$ that is $0$ on ${\bar{B}}_1(0)$ and $1$ outside $B_2(0)$; it is $C^3$ and its first derivatives satisfy $\mathrm{\mid }D{\chi }_{\epsilon }\mathrm{\mid }\le C^{\mathrm{\prime }}\mathrm{/}\epsilon$ with support in the shell $\epsilon \le d\le 2\epsilon$, where $d:=\mathrm{\mid }x-z(t)\mathrm{\mid }$ and $(1-v_{\ast }\mathrm{/}c)r\le d\le (1+v_{\ast }\mathrm{/}c)r$; the product ${\chi }_{\epsilon }\tilde{f}$ is $C^1$ on $Q$ because it vanishes on a neighbourhood of the singular set. Classical integration by parts in each coordinate over a large box containing $\mathrm{supp}\psi$ (boundary terms vanish, Riemann-Fubini iterates the one-dimensional identities, and the Riemann-Lebesgue comparison converts to Lebesgue integrals) gives ${\int }_Q{\chi }_{\epsilon }\tilde{f}D\psi =-{\int }_Q{\chi }_{\epsilon }(Df{)}^{\sim }\psi -{\int }_Q\tilde{f}(D{\chi }_{\epsilon })\psi$, and dominated convergence with the locally integrable bounds of step 1.3 lets $\epsilon \to 0$ in the first two integrals; the third is $O(\epsilon )$ because $\mathrm{\mid }\tilde{f}\mathrm{\mid }\le C_0^{\mathrm{\prime }}\mathrm{/}\epsilon$ on the shell of volume $O({\epsilon }^3)$ in space and $O(1)$ in time. Hence $-{\int }_Q\tilde{f}D\psi ={\int }_Q(Df{)}^{\sim }\psi$, that is, $⟨D{u}_{\tilde{f}},\psi ⟩=⟨u_{(Df{)}^{\sim }},\psi ⟩$ for every test, proving the claim. Consequently $\mathcal{E}=-\mathrm{\nabla }\mathrm{\Phi }-{\mathrm{\partial }}_t\mathrm{{\rm Y}}=u_{E_0^{\sim }}$ and $\mathcal{B}=\mathrm{curl}\mathrm{{\rm Y}}=u_{B_0^{\sim }}$ by linearity of the regular functional, with $E_0,B_0$ the classical fields of step 1.3.

The wave equations. For every test $\psi$, $⟨\mathrm{\Phi },W\psi ⟩=\frac{q}{{\epsilon }_0}{\int }_{\mathbb{R}}(K⊛W\psi )(z(s),s)ds=\frac{q}{{\epsilon }_0}{\int }_{\mathbb{R}}\psi (z(s),s)ds=\frac{1}{{\epsilon }_0}⟨\rho ,\psi ⟩$, the first equality by the definition of $\mathrm{\Phi }$ in [F1], the second by the pointwise identity of step 1.2 applied under the history integral (whose integrand is bounded by the cylinder estimate of step 2.1, so the interchange is legitimate), and the last by [F3]. Since $⟨W\mathrm{\Phi },\psi ⟩=⟨\mathrm{\Phi },W\psi ⟩$ for the even-order constant-coefficient operator $W$, this gives $W\mathrm{\Phi }=\rho \mathrm{/}{\epsilon }_0$. The same computation componentwise with the prefactor ${\mu }_{0}q$ and the weight $z_j^{\mathrm{\prime }}$ gives $⟨{\mathrm{{\rm Y}}}_j,W\psi ⟩={\mu }_{0}q{\int }_{\mathbb{R}}{z}_j^{\mathrm{\prime }}(s)\psi (z(s),s)ds={\mu }_0⟨J_j,\psi ⟩$, hence $W\mathrm{{\rm Y}}={\mu }_{0}J$.

The Lorenz identity. For every test $\psi$, using the definitions of the distributional derivatives and the identities of step 1.1, $⟨\mathrm{div}\mathrm{{\rm Y}}+c^{-2}{\mathrm{\partial }}_t\mathrm{\Phi },\psi ⟩=-{\mu }_{0}q{\int }_{\mathbb{R}}[z^{\mathrm{\prime }}(s)\cdot (K⊛\mathrm{\nabla }\psi )(z(s),s)+(K⊛{\mathrm{\partial }}_t\psi )(z(s),s)]ds=-{\mu }_{0}q{\int }_{\mathbb{R}}[z^{\mathrm{\prime }}(s)\cdot {\mathrm{\nabla }}_{y}F(z(s),s)+{\mathrm{\partial }}_{s}F(z(s),s)]ds$, that is, $-{\mu }_{0}q$ times the integral of the total derivative of $s\mapsto F(z(s),s)$, where $c^{-2}{\epsilon }_0^{-1}={\mu }_0$; by step 2.1 that total derivative is continuous with compact support, so Newton-Leibniz gives ${\int }_{\mathbb{R}}\frac{d}{ds}[F(z(s),s)]ds=0$ and the pairing vanishes for every test. Hence $\mathrm{div}\mathrm{{\rm Y}}+c^{-2}{\mathrm{\partial }}_t\mathrm{\Phi }=0$.

The four Maxwell equations. Using the distributional algebra for finite combinations of coordinate derivatives (which commutes in the sense ${\mathrm{\partial }}^{\alpha }{\mathrm{\partial }}^{\beta }={\mathrm{\partial }}^{\alpha +\beta }$ and is linear) together with the classical identities [F6] applied to smooth tests, one computes, with $G:=\mathrm{div}\mathrm{{\rm Y}}+c^{-2}{\mathrm{\partial }}_t\mathrm{\Phi }=0$ from step 3.2: $\mathrm{div}\mathcal{E}=-\mathrm{\Delta }\mathrm{\Phi }-{\mathrm{\partial }}_t\mathrm{div}\mathrm{{\rm Y}}=W\mathrm{\Phi }-c^{-2}{\mathrm{\partial }}_t^2\mathrm{\Phi }-{\mathrm{\partial }}_t\mathrm{div}\mathrm{{\rm Y}}=\rho \mathrm{/}{\epsilon }_0-{\mathrm{\partial }}_{t}G=\rho \mathrm{/}{\epsilon }_0$ by step 3.1; $\mathrm{div}\mathcal{B}=\mathrm{div}\mathrm{curl}\mathrm{{\rm Y}}=0$; $\mathrm{curl}\mathcal{E}=-\mathrm{curl}\mathrm{\nabla }\mathrm{\Phi }-{\mathrm{\partial }}_t\mathrm{curl}\mathrm{{\rm Y}}=-{\mathrm{\partial }}_t\mathcal{B}$; and $\mathrm{curl}\mathcal{B}=\mathrm{curl}\mathrm{curl}\mathrm{{\rm Y}}=\mathrm{\nabla }\mathrm{div}\mathrm{{\rm Y}}-\mathrm{\Delta }\mathrm{{\rm Y}}={\mu }_{0}J-c^{-2}{\mathrm{\partial }}_t^2\mathrm{{\rm Y}}+\mathrm{\nabla }\mathrm{div}\mathrm{{\rm Y}}={\mu }_{0}J+c^{-2}{\mathrm{\partial }}_t\mathcal{E}+\mathrm{\nabla }G={\mu }_{0}J+c^{-2}{\mathrm{\partial }}_t\mathcal{E}$, where the fourth step uses $-\mathrm{\Delta }\mathrm{{\rm Y}}=W\mathrm{{\rm Y}}-c^{-2}{\mathrm{\partial }}_t^2\mathrm{{\rm Y}}$ and ${\mathrm{\partial }}_t\mathcal{E}=-\mathrm{\nabla }{\mathrm{\partial }}_t\mathrm{\Phi }-{\mathrm{\partial }}_t^2\mathrm{{\rm Y}}$.

Off the worldline. For a test $\psi$ supported in $V$ one has $⟨\rho ,\psi ⟩=q\int \psi (z(s),s)ds=0$ and likewise $⟨J,\psi ⟩=0$, since $\psi$ vanishes on $\mathrm{\Gamma }$. By step 2.2 the restrictions of $\mathcal{E},\mathcal{B}$ to $V$ are the regular distributions of the $C^1$ fields $E_0,B_0$ of [F1]. The classical-compatibility clause of [F4], applied on the open set $V$, identifies their first distributional derivatives there with the regular distributions of their continuous classical first derivatives; hence the four equations of step 4.1, restricted to $V$, read $\mathrm{div}{E}_0=0$, $\mathrm{div}{B}_0=0$, $\mathrm{curl}{E}_0=-{\mathrm{\partial }}_t{B}_0$, $\mathrm{curl}{B}_0=c^{-2}{\mathrm{\partial }}_t{E}_0$ as identities of continuous regular distributions. If a continuous function on $V$ has zero regular distribution, then it vanishes almost everywhere and hence, by continuity and the positivity of ball measure, at every point; applying this to each component of the classical residuals gives the source-free Maxwell equations pointwise on $V$.