4

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? 

thomas.en
  • 313
  • Yes if you can show something about ${G(x,y) : G(x,y)=g(x)h(y), g,h\in C_c^\infty(\mathbb{R}^n)}$, like density if true. Right now nothing come to my mind, but maybe someone else will think about this. – Bertrand R Sep 03 '13 at 20:17
  • 1
    Why not use the lemma on $\widetilde{F}(y) := \int_{\mathbb{R}^n} F(x,y)g(x)dx$ first. – Shuhao Cao Sep 03 '13 at 20:28
  • For using the lemma on $\tilde{F}(y)$, we need $\tilde{F}$ to be in $L^1_{loc}$. I am not sure, whether this holds. – thomas.en Sep 04 '13 at 07:06

2 Answers2

2

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 $(x, y)\in\mathbb{R}^n\times\mathbb{R}^n$.

Community
  • 1
Giuseppe Negro
  • 32,319
  • Interesting! In Lemma 1, the function $f$ is real-valued, does it change something, if we let $F, g, h$ be complex-valued? – thomas.en Sep 05 '13 at 05:54
  • No, because if $F$ is complex valued then you can just apply the whole argument to its real and imaginary part. The same thing applies if $F$ takes values in $\mathbb{R}^n$ or $\mathbb{C}^n$. As a side note, observe that you only need to check that $$\iint_{\mathbb{R}^n\times\mathbb{R}^n} F(x, y)g(x)h(y), dxdy=0$$ for all $g, h\in C^{\infty}_c(\mathbb{R}^n\to [0, 1]).$ – Giuseppe Negro Sep 05 '13 at 09:30
  • 1
    @thomas.en: I edited the answer to add an alternative solution, you might want to see if it is useful to you. – Giuseppe Negro Sep 06 '13 at 08:38
1

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....

Bertrand R
  • 1,084