A compact test variation and gauge-source sign check

With $j\equiv 0$ and $A=0$ one has $F(A)=0$ by [F1], so the coefficient $-{\mu }_0^{-1}{F}^{\mu \nu }{\mathrm{\partial }}_{\mu }{h}_{\nu }+j^{\nu }{h}_{\nu }$ of [F2] vanishes identically on $U$ for every $h\in C_c^{\mathrm{\infty }}(U;{\mathbb{R}}^4)$; the first variation is $c^{-1}$ times the integral of that coefficient by [F1], hence it is zero for every $h$ and $A=0$ is stationary. This is part (a).

For the potential $A_0=x^1$, $A_1=A_2=A_3=0$ of part (b), [F15] gives ${\mathrm{\partial }}_1{A}_0=1$ and ${\mathrm{\partial }}_0{A}_1=0$, so the dictionary [F5] yields $F_{01}={\mathrm{\partial }}_0{A}_1-{\mathrm{\partial }}_1{A}_0=-1$; raising indices with ${\eta }^{00}{\eta }^{11}=(-1)(1)=-1$ gives $F^{01}=+1$ and $F^{10}=-1$, and all remaining components vanish because the other $A_{\nu }$ are constant; the inverse dictionary of [F5] then gives $E_1=-c{F}_{01}=c$ with $E_2=E_3=0$ and $B=0$, so $E=(c,0,0)$ and ${\mathrm{\partial }}_{\mu }{F}^{\mu \nu }=0$ on $U$ because every component of $F$ is constant and only the two nonzero components $F^{01}$, $F^{10}$ occur, whose coordinate derivatives vanish.

For part (c), apply [F8] with $\mathrm{\Omega }=(-1\mathrm{/}4,1\mathrm{/}4)$ and $K=\{0\}$, with $\mathrm{\Omega }=(-1,1)$ and $K=[-1\mathrm{/}2,1\mathrm{/}2]$, and with $\mathrm{\Omega }=(-1,1{)}^3$ and $K=\{0\}$; this gives nonzero nonnegative $\rho \in C_c^{\mathrm{\infty }}((-1\mathrm{/}4,1\mathrm{/}4))$, $\zeta \in C_c^{\mathrm{\infty }}((-1,1))$ with $\zeta =1$ near $[-1\mathrm{/}2,1\mathrm{/}2]$, and nonzero nonnegative $\psi \in C_c^{\mathrm{\infty }}((-1,1{)}^3)$, all extended by zero. Set ${\phi }_{\pm }(u):=(2\pm u)\zeta (u)$ and ${\chi }_{\pm }(x):={\phi }_{\pm }(x^0)\psi (x^1,x^2,x^3)$; then ${\phi }_{\pm }$ is a nonnegative compactly supported smooth function with ${\phi }_{\pm }^{\mathrm{\prime }}(u)=\pm \zeta (u)+(2\pm u){\zeta }^{\mathrm{\prime }}(u)$, which equals $+1$ and $-1$ respectively on a neighbourhood of $[-1\mathrm{/}2,1\mathrm{/}2]$ and in particular on the whole support of $\rho$.

With $j\equiv 0$ the equation ${\mathrm{\partial }}_{\mu }{F}^{\mu \nu }=-{\mu }_0{j}^{\nu }=0$ of step 1.2 is exactly the coefficient equation of [F3], so [F3] gives that $A$ is stationary; directly, for every $h\in C_c^{\mathrm{\infty }}(U;{\mathbb{R}}^4)$ the first variation is $c^{-1}{\int }_U(-{\mu }_0^{-1}{F}^{\mu \nu }{\mathrm{\partial }}_{\mu }{h}_{\nu })d\lambda =-(c{\mu }_0{)}^{-1}{\int }_U(F^{01}{\mathrm{\partial }}_0{h}_1+F^{10}{\mathrm{\partial }}_1{h}_0)d\lambda =-(c{\mu }_0{)}^{-1}{\int }_U({\mathrm{\partial }}_0{h}_1-{\mathrm{\partial }}_1{h}_0)d\lambda$ by [F2] and step 1.2. Each of ${\int }_U{\mathrm{\partial }}_0{h}_{1}d\lambda$ and ${\int }_U{\mathrm{\partial }}_1{h}_{0}d\lambda$ vanishes: Fubini [F10] writes the first as the iterated integral whose inner $x^0$ integral is over a continuous compactly supported one-variable function, whose Lebesgue integral equals its Riemann integral by [F17] and is $0$ by the fundamental theorem [F9], and symmetrically for the second; hence the first variation is $0$ for every $h$ and the direct computation confirms stationarity with no boundary term. This is part (b).

For every smooth potential $A$ on a neighbourhood of $\bar{U}$, [F4] gives $S[A+{\mathrm{\partial }}^{\mathrm{♯}}{\chi }_{\pm }]-S[A]=c^{-1}{\int }_U{j}^{\mu }{\mathrm{\partial }}_{\mu }{\chi }_{\pm }d\lambda$; with the source $j^0=c\rho (x^0)$, $j^i=0$ of step 1.3 and the convention [F7], the integrand is $j^0{\mathrm{\partial }}_0{\chi }_{\pm }=c\rho (x^0){\phi }_{\pm }^{\mathrm{\prime }}(x^0)\psi (x^1,x^2,x^3)$, whose support is a closed subset of $\mathrm{supp}\rho \times \mathrm{supp}\psi$, a compact subset of $U$ by [F18]; Fubini [F10] factors the integral as $c^{-1}\cdot c\left({\int }_{-1}^1\rho {\phi }_{\pm }^{\mathrm{\prime }}d{x}^0\right)\left({\int }_{(-1,1{)}^3}\psi d^{3}x\right)$, the factors being finite Lebesgue integrals of continuous compactly supported functions; and since ${\phi }_{\pm }^{\mathrm{\prime }}=\pm 1$ on $\mathrm{supp}\rho$ by step 1.3, $\int \rho {\phi }_{\pm }^{\mathrm{\prime }}=\pm \int \rho$ with $\int \rho >0$, while $\int \psi >0$: each positivity follows because the integrand is continuous, nonnegative and nonzero, so if the integral over the ambient open box vanished then monotonicity [F13] would give vanishing over every compactly contained ball and [F14] would force the integrand to vanish identically, a contradiction. Hence $S[A+{\mathrm{\partial }}^{\mathrm{♯}}{\chi }_{\pm }]-S[A]=\pm (\int \rho )(\int \psi )$, strictly positive for $+$ and strictly negative for $-$.

For the same data, the second form of [F4] gives $S[A+{\mathrm{\partial }}^{\mathrm{♯}}{\chi }_{\pm }]-S[A]=-c^{-1}{\int }_U({\mathrm{\partial }}_{\mu }{j}^{\mu }){\chi }_{\pm }d\lambda$; here ${\mathrm{\partial }}_{\mu }{j}^{\mu }={\mathrm{\partial }}_0(c\rho (x^0))=c{\rho }^{\mathrm{\prime }}(x^0)$ by [F7] and [F15], so the right-hand side is $-{\int }_U{\rho }^{\mathrm{\prime }}(x^0){\phi }_{\pm }(x^0)\psi d\lambda =-\left({\int }_{-1}^1{\rho }^{\mathrm{\prime }}{\phi }_{\pm }d{x}^0\right)(\int \psi )$ by Fubini [F10]; by [F17] the one-variable integrals equal their Riemann integrals, and integration by parts [F12] applied to $\rho$ and ${\phi }_{\pm }$ has vanishing endpoint terms because both are compactly supported, giving $\int {\rho }^{\mathrm{\prime }}{\phi }_{\pm }=-\int \rho {\phi }_{\pm }^{\mathrm{\prime }}=\mp \int \rho$, which reproduces exactly the value $\pm (\int \rho )(\int \psi )$ of step 2.2. If instead $\rho$ is constant on the whole neighbourhood, then ${\mathrm{\partial }}_{\mu }{j}^{\mu }=c{\rho }^{\mathrm{\prime }}=0$, and [F4] (b) gives $S[A+{\mathrm{\partial }}^{\mathrm{♯}}\chi ]=S[A]$ for every compact gauge variation $\chi$; no globally monotone compact function occurs anywhere.

The computation used exact substitutions in the declared box functional, the one-variable fundamental theorem for compactly supported functions and the factoring of the four-dimensional integral by Fubini; it used no existence of sources or fields, no numerical value of ${\mu }_0$, no global gauge statement and no globally monotone compact function. All integrals are the finite-region Lebesgue integrals of the definition, and Countable Choice [F16] enters only through the inherited measure statements [F14] and the cited theorems; this is part (d).