Compact-support variation: the fundamental lemma, integration by parts and differentiation under the integral sign

Zero extensions. If a continuous $u:\mathrm{\Omega }\to \mathbb{R}$ vanishes outside a compact $K\subseteq \mathrm{\Omega }$, its zero extension is continuous on ${\mathbb{R}}^4$: it is continuous at points of $\mathrm{\Omega }$, and every point outside $\mathrm{\Omega }$ has a neighbourhood disjoint from the closed set $K$, on which the extension is zero. Its support lies in $K$; [F1], [F2] and [F4] therefore define its finite integral independently of a bounding rectangle. The same argument gives a $C^1$ zero extension when $u$ is $C^1$, with derivatives extended by zero; outside $K$ both $u$ and its derivatives vanish. For the parameter family in (c), zero extension of $F$ and ${\mathrm{\partial }}_{s}F$ is jointly continuous on $(-s_0,s_0)\times {\mathbb{R}}^4$, and the extended parameter derivative is still ${\mathrm{\partial }}_{s}F$, because the common support $K$ gives the same zero neighbourhoods.

Fundamental lemma. Suppose $g(p)\mathrm{\ne }0$ and set $\sigma =1$ if $g(p)>0$, $\sigma =-1$ otherwise. Continuity and openness give a nondegenerate closed box $Q\subseteq \mathrm{\Omega }$ and $a>0$ with $\sigma g\ge a$ on $Q$. Choose a smaller nondegenerate closed box $Q_1\subseteq \mathrm{int}Q$. By [F9] and [F13] there is a smooth $\rho :{\mathbb{R}}^4\to [0,1]$, equal to $1$ on $Q_1$, whose support is a compact subset of $\mathrm{int}Q$; compactness follows because the support is closed and bounded. For $\psi =\sigma \rho {\mathrm{\mid }}_{\mathrm{\Omega }}$, the continuous product $g\psi$ is nonnegative, compactly supported, and at least $a$ on $Q_1$. On a bounding rectangle, [F3] and [F5] give ${\int }_{\mathrm{\Omega }}g\psi \ge a\mathrm{vol}(Q_1)>0$, contrary to the test hypothesis. Thus $g=0$, proving (a).

Integration by parts. For (b), put $K=\mathrm{supp}f$. The function $fg$ is $C^1$ on $\mathrm{\Omega }$, vanishes outside $K$, and has a $C^1$ zero extension $u$ by step 1.1. Its derivative ${\mathrm{\partial }}_{a}u$ is the zero extension of $({\mathrm{\partial }}_{a}f)g+f{\mathrm{\partial }}_{a}g$ by [F8]. Choose a nondegenerate bounding rectangle with $K$ in its interior. On each line in the $a$th coordinate, $u$ vanishes at both endpoints, so [F7] gives integral zero for ${\mathrm{\partial }}_{a}u$. By [F4] and [F6], integrating over the remaining coordinates gives $0={\int }_{\mathrm{\Omega }}({\mathrm{\partial }}_{a}f)g+{\int }_{\mathrm{\Omega }}f{\mathrm{\partial }}_{a}g$. This is (b).

Parameter differentiation. For (c), fix a closed interval $[u,v]\subseteq (-s_0,s_0)$ with $u<v$ and a nondegenerate rectangle $Q$ containing $K$ in its interior. Use the joint zero extensions from step 1.1 on $[u,v]\times Q$. Integrating in one coordinate produces a jointly continuous function of $s$ and the remaining coordinates: its difference at two parameter tuples is bounded by that coordinate's interval length times the uniform oscillation of $F$ on the compact product, which tends to zero by [F11]. Apply [F10] with $s$ as parameter to differentiate this one-coordinate integral. The resulting derivative, the integral of ${\mathrm{\partial }}_{s}F$, is jointly continuous by the same bound and [F11]. Repeat this argument in each of the four coordinates. [F4] and [F6] identify the repeated integrals with the four-dimensional integrals; hence $G(s):={\int }_{\mathrm{\Omega }}F(s,z)d^{4}z$ is $C^1$ on $[u,v]$ and $G^{\mathrm{\prime }}(s)={\int }_{\mathrm{\Omega }}{\mathrm{\partial }}_{s}F(s,z)d^{4}z$.

Every parameter in $(-s_0,s_0)$ lies in the interior of such a closed interval, so step 2.3 proves (c) on the entire open interval. Steps 2.1 and 2.2 prove (a) and (b). All selections are finite, the bump supplier and [F13] are choice-free, and no distributional or local-integrability hypothesis is used.