Stationary functional need not be a minimum

In the source-free case $j=0$, [F3] turns the density of [F1] into $-(4{\mu }_0{)}^{-1}{F}_{\mu \nu }{F}^{\mu \nu }=\mathrm{\mid }E{\mathrm{\mid }}^2\mathrm{/}(2{\mu }_0{c}^2)-\mathrm{\mid }B{\mathrm{\mid }}^2\mathrm{/}(2{\mu }_0)$, so linearity of the integral gives $S[A]=(2c{\mu }_0{)}^{-1}{\int }_U(\mathrm{\mid }E{\mathrm{\mid }}^2\mathrm{/}{c}^2-\mathrm{\mid }B{\mathrm{\mid }}^2)d\lambda$; writing the time coordinate as $t=x^0\mathrm{/}c$ multiplies the one-dimensional time integral by $c$ and carries $(-1,1)$ to $(-1\mathrm{/}c,1\mathrm{/}c)$ by [F6] and [F8], while the spatial coordinates are unchanged, so $S[A]=\frac{1}{2}{\int }_{U_t}({\epsilon }_0\mathrm{\mid }E{\mathrm{\mid }}^2-\mathrm{\mid }B{\mathrm{\mid }}^2\mathrm{/}{\mu }_0)dt{d}^{3}x$ by [F5]; and since $A\mapsto F(A)$ is linear in $A$ and $S$ is quadratic in $F$, $F(\lambda A)=\lambda F(A)$ gives $S[\lambda A]={\lambda }^{2}S[A]$ for every real $\lambda$.

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

By [F7] with $\mathrm{\Omega }=U$ and $K=\{(0,0,0,0)\}$ there is $\chi \in C_c^{\mathrm{\infty }}(U)$ with $0\le \chi \le 1$ and $\chi =1$ near the origin; then $\theta :=x^1\chi$ lies in $C_c^{\mathrm{\infty }}(U)$ and ${\mathrm{\partial }}_1\theta (0)=\chi (0)+0\cdot {\mathrm{\partial }}_1\chi (0)=1$, so ${\mathrm{\nabla }}_x\theta \equiv ̸0$. With covariant components $h_0=\theta$, $h_i=0$ the dictionary [F4] gives $F_{0i}=-{\mathrm{\partial }}_i\theta$, hence $E=c{\mathrm{\nabla }}_x\theta$ and $B=0$; step 1.1 then gives $S[h]=(2c{\mu }_0{)}^{-1}{\int }_U\mathrm{\mid }{\mathrm{\nabla }}_x\theta {\mathrm{\mid }}^{2}d\lambda$. The integrand $g:=\mathrm{\mid }{\mathrm{\nabla }}_x\theta {\mathrm{\mid }}^2$ is continuous, nonnegative and nonzero at the origin; if ${\int }_{U}gd\lambda =0$, then for every ball $V$ compactly contained in $U$ monotonicity [F14] would give ${\int }_{V}g\le {\int }_{U}g=0$ with ${\int }_{V}g\ge 0$, so [F15] would force $g=0$ on $U$, a contradiction; hence ${\int }_{U}g>0$ and $S[h]>0$, one positive direction.

Apply [F7] with $\mathrm{\Omega }=(-1\mathrm{/}2,1\mathrm{/}2)$ and $K=\{0\}$ four times (or once, taking all four bumps equal) to obtain nonzero nonnegative $\eta ,\phi ,{\psi }_2,{\psi }_3\in C_c^{\mathrm{\infty }}((-1\mathrm{/}2,1\mathrm{/}2))$, extended by zero to $\mathbb{R}$; define $f_N(x):=\eta (x^0)\phi (N{x}^1){\psi }_2(x^2){\psi }_3(x^3)$ for integers $N\ge 1$ and let $h^{(N)}$ have $h^{(N)2}=f_N$ and all other contravariant components $0$. Since $\mathrm{supp}\phi (N\cdot )\subseteq (-1\mathrm{/}(2N),1\mathrm{/}(2N))\subseteq (-1\mathrm{/}2,1\mathrm{/}2)$, every $h^{(N)}$ is smooth with support in the fixed compact set $K_{\ast }=[-1\mathrm{/}2,1\mathrm{/}2{]}^4\subset U$ by [F22]. The product and chain rules [F12] and [F13] give ${\mathrm{\partial }}_0{f}_N={\eta }^{\mathrm{\prime }}(x^0)\phi (N{x}^1){\psi }_2{\psi }_3$, ${\mathrm{\partial }}_1{f}_N=N\eta (x^0){\phi }^{\mathrm{\prime }}(N{x}^1){\psi }_2{\psi }_3$ and ${\mathrm{\partial }}_3{f}_N=\eta (x^0)\phi (N{x}^1){\psi }_2{\psi }_3^{\mathrm{\prime }}$; since $f_N=h^{(N)2}$ and $h^{(N)0}=h^{(N)1}=h^{(N)3}=0$, the dictionary [F4] gives $E=-c({\mathrm{\partial }}_0{f}_N)e_2$ and $B=-({\mathrm{\partial }}_3{f}_N)e_1+({\mathrm{\partial }}_1{f}_N)e_3$.

Because all integrands below are smooth with support in the fixed compact subset $K_{\ast }\subset U$, [F9] lets each Lebesgue integral be computed as the equal Riemann integral, and Fubini [F10] together with linearity [F11] factors each product of one-variable functions into a product of one-dimensional integrals: ${\int }_U\mathrm{\mid }E{\mathrm{\mid }}^2\mathrm{/}{c}^{2}d\lambda =\int ({\eta }^{\mathrm{\prime }}{)}^2\cdot \int \phi (N{x}^1{)}^{2}d{x}^1\cdot \int {\psi }_2^2\cdot \int {\psi }_3^2$ and ${\int }_U\mathrm{\mid }B{\mathrm{\mid }}^{2}d\lambda =\int {\eta }^2\cdot \int \phi (N{x}^1{)}^{2}d{x}^1\cdot \int {\psi }_2^2\cdot \int {\psi }_3^{\mathrm{\prime }2}+\int {\eta }^2\cdot \int (N{\phi }^{\mathrm{\prime }}(N{x}^1){)}^{2}d{x}^1\cdot \int {\psi }_2^2\cdot \int {\psi }_3^2$. By [F8] with the affine map $y=N{x}^1$ on $[-1,1]$, ${\int }_{-1}^1\phi (N{x}^1{)}^{2}d{x}^1=N^{-1}{\int }_{-N}^N\phi (y{)}^{2}dy=N^{-1}{\int }_{-1}^1\phi (y{)}^{2}dy$, the last equality because $\phi$ vanishes outside $(-1\mathrm{/}2,1\mathrm{/}2)$; the same identity holds for ${\phi }^{\mathrm{\prime }}$. Hence ${\int }_U\mathrm{\mid }E{\mathrm{\mid }}^2\mathrm{/}{c}^{2}d\lambda =C_E\mathrm{/}N$ and ${\int }_U\mathrm{\mid }B{\mathrm{\mid }}^{2}d\lambda =C_{B}N+C_3\mathrm{/}N$ with the stated constants, and $C_B>0$: if ${\phi }^{\mathrm{\prime }}$ vanished identically then [F16] would make $\phi$ constant on $\mathbb{R}$, hence zero by compact support, contradicting the nonzero choice, so ${\phi }^{\mathrm{\prime }2}$ is a continuous nonnegative nonzero square, and together with the nonzero bumps $\eta ,{\psi }_2,{\psi }_3$ the same argument as in step 3.1 with [F14] and [F15] gives each of the four one-dimensional integrals in $C_B$ strictly positive.

Step 1.1 with $j=0$ and [F3] expresses $S[h^{(N)}]=(2c{\mu }_0{)}^{-1}{\int }_U(\mathrm{\mid }E{\mathrm{\mid }}^2\mathrm{/}{c}^2-\mathrm{\mid }B{\mathrm{\mid }}^2)d\lambda$, so step 5.1 gives $S[h^{(N)}]=(2c{\mu }_0{)}^{-1}((C_E-C_3)\mathrm{/}N-C_{B}N)$; since $C_B>0$, the bound $(C_E-C_3)\mathrm{/}N-C_{B}N\le \mathrm{\mid }{C}_E-C_3\mathrm{\mid }-C_{B}N$ for $N\ge 1$ tends to $-\mathrm{\infty }$ by [F20], proving the stated limit. To fix one negative direction, choose $N\ge 1$ with $C_{B}N>(C_E-C_3)\mathrm{/}N$: if $C_E-C_3\le 0$ every $N$ works, while if $C_E-C_3>0$ then $\epsilon :=C_B\mathrm{/}(C_E-C_3)>0$ and [F20] provides $n\ge 1$ with $1\mathrm{/}n<\epsilon$, that is $n>(C_E-C_3)\mathrm{/}{C}_B$; setting $N:=n$ gives $C_{B}N>C_E-C_3\ge (C_E-C_3)\mathrm{/}N$ because $N\ge 1$. For that fixed $N_0$ one has $S[h^{(N_0)}]<0$, giving one fixed-support negative direction $h_{-}:=h^{(N_0)}$.

For every real $\lambda$ the scaling law of step 1.1 gives $S[\lambda h_{+}]={\lambda }^{2}S[h_{+}]$ and $S[\lambda h_{-}]={\lambda }^{2}S[h_{-}]$, so the signs of the two directions are preserved for nonzero $\lambda$ and both action magnitudes tend to $0$ as $\lambda \to 0$. In the fixed-support space [F17] each seminorm satisfies $p_m(\lambda h)=\mathrm{\mid }\lambda \mathrm{\mid }{p}_m(h)$ by definition of the maximum and supremum, and in the LF topology [F18] every seminorm satisfies $q(\lambda h)=\mathrm{\mid }\lambda \mathrm{\mid }q(h)$ and a zero-neighbourhood contains finitely many bounds $q<\epsilon$; hence $\lambda h_{\pm }\to 0$ in both topologies, and in any usual topology with continuous scalar multiplication the map $\lambda \mapsto \lambda h_{\pm }$ is continuous at $0$. Therefore every such zero-neighbourhood contains configurations $\lambda h_{+}$ with $S>0$ and $\lambda h_{-}$ with $S<0$, with $\mathrm{\mid }S\mathrm{\mid }={\lambda }^2\mathrm{\mid }S[h_{\pm }]\mathrm{\mid }$ arbitrarily small: the stationary configuration $0$ is neither a local minimum nor a local maximum.

Every object above is smooth and compactly supported in the box $U$, all integrals are the declared finite-region Lebesgue integrals, and the arguments used exactly [F7], [F8], [F9], [F14], [F15], [F16] and [F20]; Countable Choice [F19] enters only through the measure interface of [F9] and [F15], as inherited from the field-action sector. The computation concerns the adopted functional [F21] and asserts nothing about physical fields, measurements, existence of stationary configurations beyond the exhibited zero, or any other region or variation class.