# Is a specific product function orthogonal to all harmonic functions

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?

• 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 $u$. – Giorgio Metafune Sep 23 at 18:52
• 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 $w \in H^2_0(\Omega)$. – Ali Sep 23 at 19:15
• 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? – Giorgio Metafune Sep 23 at 20:33

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 $$g$$.