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Suppose $\Omega=[-1,1]^3$. Let $f:[-1,1]\to \mathbb R$ and $g:[-1,1]^2\to \mathbb R$ be smooth functions and suppose that given any harmonic function on $\Omega$ (i.e. $\Delta u =0$ on $\Omega$), with $u \in L^2(\Omega)$, there holds: $$ \int_{\Omega} u(x^1,x^2,x^2) f(x^1)g(x^2,x^3)\,dx=0.$$

Does it follow that $f$ and $g$ are identically zero?

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  • $\begingroup$ If $h=\Delta w$ with $w$ smooth and compactly supported, then $\int uh=\int u\Delta w=\int (\Delta u) w=0$ for every harmonic function $u$. $\endgroup$ – Giorgio Metafune Sep 23 at 18:52
  • $\begingroup$ Its true that $\Delta w$ is orthogonal to harmonic functions for all $w \in H^2_0(\Omega)$ but I don’t see how that is relevant at all. The function at hand, may not be writable as $\Delta w$ with $w \in H^2_0(\Omega)$. $\endgroup$ – Ali Sep 23 at 19:15
  • $\begingroup$ Yes, true. This gives only a counteraxample with a function which is the sum of 2 in the above form. What happens in 2 variables? $\endgroup$ – Giorgio Metafune Sep 23 at 20:33
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Looks like it is so (though the conclusion is rather that either $f$, or $g$ is identically $0$ (one of the two is enough).

Let $v$ be the solution of the problem $\Delta v=fg$ in $\Omega$, $v|_{\partial\Omega}=0$. Then, by Green's formula, the integral in question is (up to minus) $\int_{\partial\Omega}u\frac{\partial v}{\partial n}$. However, the boundary values of $u$ can be anything sufficiently nice we want, so this may hold only if $\frac{\partial v}{\partial n}$ is identically zero on the boundary, in which case $v$ can be extended by $0$ outside $\Omega$ and its Laplacian (in the sense of generalized functions, at least) is still $fg\chi_\Omega$.

Now it suffices to show that the Laplacian of a compactly supported function cannot be a product like above unless it is zero. Indeed, its Fourier transform would then be the product $F(z_1)G(z_2,z_3)$ of two entire functions and it should vanish whenever $z_1^2+z_2^2+z_3^2=0$. If there exist $a,b\in\mathbb C$ with $a^2+b^2=-c^2\ne 0$ such that $G(a,b)\ne 0$, then the function $F(cz)G(az,bz)$ of one complex variable vanishes identically and, since the second factor is not zero for $z=1$, we must have $F\equiv 0$, i.e., $f\equiv 0$. Otherwise $G(z_2,z_3)$ is zero on a dense set in $\mathbb C^2$, so it is identically $0$ and so is $g$.

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