The fundamental lemma of the calculus of variations

The vanishing statement follows from the sign statement. Assume part (b) proved and first take $g$ real-valued. Applying it to $g$ gives $g\ge 0$ almost everywhere; applying it to $-g$, whose pairing with every nonnegative $\phi$ equals $-{\int }_{\mathrm{\Omega }}g\phi =0\ge 0$, gives $-g\ge 0$ almost everywhere. Hence $g=0$ almost everywhere. For complex $g$, the vanishing pairing with every real test implies vanishing pairings for $\mathrm{Re}g$ and $\mathrm{Im}g$; applying this real argument to each gives part (a). So it suffices to prove the sign statement, and from the next step on we assume $g$ real-valued and ${\int }_{\mathrm{\Omega }}g\phi \ge 0$ for every nonnegative $\phi \in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$.

Exhaustion of $\mathrm{\Omega }$ and localisation. For integers $k\ge 1$ put ${\mathrm{\Omega }}_k:=\{x\in \mathrm{\Omega }:\mathrm{\mid }x\mathrm{\mid }<k\text{ and }\mathrm{dist}(x,{\mathbb{R}}^n\setminus \mathrm{\Omega })>1\mathrm{/}k\}$, with $\mathrm{dist}(x,\mathrm{\varnothing })=+\mathrm{\infty }$. Each ${\mathrm{\Omega }}_k$ is open (both conditions are open or strict), its closure is bounded and contained in $\mathrm{\Omega }$, so $\overline{{\mathrm{\Omega }}_k}$ is a compact subset of $\mathrm{\Omega }$; moreover ${\mathrm{\Omega }}_k\subseteq {\mathrm{\Omega }}_{k+1}$ and ${\bigcup }_k{\mathrm{\Omega }}_k=\mathrm{\Omega }$, because for $x\in \mathrm{\Omega }$ openness gives $\mathrm{dist}(x,{\mathbb{R}}^n\setminus \mathrm{\Omega })>0$ and one may take $k>\max \{\mathrm{\mid }x\mathrm{\mid },1\mathrm{/}\mathrm{dist}(x,{\mathbb{R}}^n\setminus \mathrm{\Omega })\}$.

The localised functions are integrable on ${\mathbb{R}}^n$. Let $g_k:=g{\mathbf{1}}_{{\mathrm{\Omega }}_k}$, extended by zero outside ${\mathrm{\Omega }}_k$. Since $\overline{{\mathrm{\Omega }}_k}$ is a compact subset of $\mathrm{\Omega }$ and $g\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(\mathrm{\Omega })$, one has ${\int }_{{\mathbb{R}}^n}\mathrm{\mid }{g}_k\mathrm{\mid }={\int }_{{\mathrm{\Omega }}_k}\mathrm{\mid }g\mathrm{\mid }<\mathrm{\infty }$, so $g_k\in L^1({\mathbb{R}}^n)$.

Almost every point is a Lebesgue point of every $g_k$. By [F3] applied to $g_k$ there is a Lebesgue null set $N_k\subseteq {\mathbb{R}}^n$ such that every $z\mathrm{\notin }{N}_k$ is a Lebesgue point of $g_k$; the union $N:={\bigcup }_k{N}_k$ is again null, being a countable union of null sets.

At points of $\mathrm{\Omega }\setminus N$ the Lebesgue averages are small. Fix $z\in \mathrm{\Omega }\setminus N$ and $k$ with $z\in {\mathrm{\Omega }}_k$. Then $z\mathrm{\notin }{N}_k$, so for $A_k(r):={\int }_{B(z,r)}\mathrm{\mid }{g}_k(y)-g_k(z)\mathrm{\mid }dy$ the Lebesgue point property gives $A_k(r)\mathrm{/}\lambda (B(z,r))\to 0$; combined with [F4] this yields $A_k(r)\mathrm{/}{r}^n\le 2^n{A}_k(r)\mathrm{/}\lambda (B(z,r))\to 0$, that is $A_k(r)=o(r^n)$. Moreover $g_k(z)=g(z)$, since $z\in {\mathrm{\Omega }}_k$.

Kernel convergence. By step 4.1 the point $z$ is a Lebesgue point of $g_k$ with $g_k(z)=g(z)$ and $A_k(r)=o(r^n)$, so applying [F2] with $f=g_k$, $x=z$, $a=g_k(z)$ and the kernel $K=\rho$ of [F1] gives, for $r\downarrow 0$, the limit $r^{-n}{\int }_{{\mathbb{R}}^n}\rho (y\mathrm{/}r)g_k(z-y)dy⟶g(z),$ since [F1] realises $\rho$ as a compactly supported kernel with the required bounded nonincreasing majorant.

Admissible test functions. Fix $z\in \mathrm{\Omega }\setminus N$ and $k$ with $z\in {\mathrm{\Omega }}_k$ as in step 4.1. For $0<r<\mathrm{dist}(z,{\mathbb{R}}^n\setminus {\mathrm{\Omega }}_k)$ the function ${\phi }_r(w):=r^{-n}\rho ((z-w)\mathrm{/}r)$ lies in $C_c^{\mathrm{\infty }}(\mathrm{\Omega })$: it is smooth in $w$, nonnegative, and has support in $B(z,r)\subseteq \mathrm{\Omega }$. Changing variables $w=z-y$ gives ${\int }_{\mathrm{\Omega }}g(w){\phi }_r(w)dw=\int r^{-n}\rho (y\mathrm{/}r)g(z-y)dy$, and $g=g_k$ on the support of this test. Thus the integral equals the mollified $g_k$ in step 5.1, and the hypothesis of part (b) gives ${\int }_{\mathrm{\Omega }}g{\phi }_r\ge 0$.

Conclusion of the sign statement. For $z\in \mathrm{\Omega }\setminus N$ and $r\downarrow 0$, step 6.1 keeps the quantities ${\int }_{\mathrm{\Omega }}g{\phi }_r$ nonnegative while step 5.1 identifies their limit as $g(z)$; hence $g(z)\ge 0$. Since $N$ is null, $g\ge 0$ almost everywhere on $\mathrm{\Omega }$, and by step 1.1 this also gives $g=0$ almost everywhere under the hypotheses of part (a).