fundamental lemma for variational calculus Is it possible to use the fundamental lemma of calculus of variations in some way in the following case:
$F(x,y)$ is a locally integrable function on $\mathbb{R}^n \times \mathbb{R}^n$. We know that for all $g,h \in C^{\infty}_{c}(\mathbb{R^n})$
$
\int_{\mathbb{R}^n} \int_{\mathbb{R}^n} F(x,y) g(x) h(y) dxdy = 0.
$
Is it possible to get $F(x,y) = 0$ almost everywhere? 
 A: Yes. See Lemma 1 in this answer, which says that it is sufficient to prove 
$$\iint_{R_1\times R_2} F(x, y)\, dxdy=0$$
for any pair $R_1, R_2$ of rectangles in $\mathbb{R}^n$. Now the characteristic function $\chi_{R_1\times R_2}(x, y)$ factors out as $\chi_{R_1}(x)\chi_{R_2}(y)$ and it can be approximated by a product $g(x)h(y)$ where $g, h\in C^\infty_c(\mathbb{R}^n)$ (see Lemma 2 of the aforementioned answer for details on this). 
P.S.: I just noticed that we could equally use the Fourier transform approach by Zarrax. Let $\phi, \psi\in C^{\infty}_c(\mathbb{R}^n)$ be arbitrary and note that 
\begin{equation}
\begin{split}\mathcal{F}\big[F(x,y)\phi(x)\psi(y)\big]_{(x,y)\to(\xi, \eta)}(\xi, \eta)&=\iint_{\mathbb{R}^n\times\mathbb{R}^n}F(x, y)\left(\phi(x)e^{-ix\cdot\xi}\right)\left(\psi(y)e^{-iy\cdot\eta}\right)\, dxdy\\
&=0
\end{split}
\end{equation}
by assumption. Therefore the function $F(x,y)\phi(x)\psi(y)$ vanishes, and since $\phi$ and $\psi$ were arbitrary, we can conclude that $F(x, y)$ vanishes at almost all $(x, y)\in\mathbb{R}^n\times\mathbb{R}^n$.
A: I think that you can use the Stone Weierstrass theorem in this case : 
A being a subalgebra of $C(\mathbb{R}^n\times\mathbb{R}^n,\mathbb{R})$ which contain non zero constant functions it is dense if and only if it separates points.
In your case if you choose 
$$A=\{G(x,y)=\sum_{i\in I} f_i(x)g_i(y): f_i,g_i\in C_c^\infty(\mathbb{R}^n)\}$$
It obviously contains constant functions and it can be shown that it separates points. Then it is dense in $C(\mathbb{R}^n\times\mathbb{R}^n ,\mathbb{R})$, and I'm almost sure it's dense in $L^1_{loc}$.
Then $A^\perp=\{0\}$ and $F(x,y)=0$.
I'm really not sure about this, but this could be a starting point....
