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I have the following quantity:

$$ I = \int_{\mathbb{R}^3 \times \mathbb{R}^3} d^3r_1 \, d^3 r_2 \, \frac{F(\vec{r}_1) F^*(\vec{r}_2)}{|\vec{r}_1 - \vec{r}_2|} $$

$I$ is obviously real by symmetry. How can I show that it is also positive semidefinite?

Jolyon
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    The solution to Poisson's equation $\nabla^2 J = F$ is $J(r_2) = -\frac{1}{4\pi}\int \frac{F(r_1){\rm d}r_1}{|r_1-r_2|}$ so the integral can be written $-\frac{1}{4\pi}\int J(r_2)\nabla^2 J^*(r_2){\rm d}r_2$. Integration by parts. – Winther Dec 22 '18 at 20:36
  • Very suave! If you want to write that up as an answer, I'll accept it. – Jolyon Dec 22 '18 at 20:40

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Recall that the solution to Poisson's equation $\nabla^2 J = F$ is given by $J(r_2) = -\frac{1}{4\pi}\int \frac{F(r_1){\rm d}r_1}{|r_1-r_2|}$. Using this the integral can be written $$I = -4\pi\int J(r_2)\nabla^2 J^*(r_2){\rm d}r_2 = 4\pi\int |\nabla J(r_2)|^2 {\rm d}r_2 \geq 0$$ where we used integration by parts (as usual assuming boundary terms vanish) to arrive at the last equality.

Winther
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