Let $f(u,v)$ be a "nice" complex-valued function defined on the region $u,v\geq 0$, e.g., $f$ is continuous and decays rapidly at infinity.
Problem: For any such function $f$, prove that
$$\int_{[0,\infty)^4} dx_1dx_2dx_3dx_4 \: f(x_1+x_2, x_3+x_4)^* f(x_1+x_3, x_2+x_4) \geq0.$$
Here the asterisk denotes the complex conjugation. The variables are "twisted" in $f^*$ and $f$.
Motivation: This problem is motivated from physics. By a physical argument, a certain quantity must be non-negative, and I found that it can be represented as an integral in the above form.
Note: One may assume that $f$ is symmetric, i.e., $f(u,v) = f(v,u)$. To see this, note that a generic function $f$ can be written as $f = f_s + f_a$, the sum of the symmetric and antisymmetic component. Enough to show that, e.g.,
$$A := \int f_s(x_1+x_2, x_3+x_4)^* f_a(x_1+x_3, x_2+x_4) =0$$
and
$$\int f_a(x_1+x_2, x_3+x_4)^* f_a(x_1+x_3, x_2+x_4) =0.$$
As an example, we prove the first equality. By changing the variables $x_1\leftrightarrow x_2$ and $x_3 \leftrightarrow x_4$, we obtain
$$A = \int f_s(x_1+x_2, x_3+x_4)^* f_a(x_2+x_4, x_1+x_3) = -A$$
by the antisymmetry $f_a(x_2+x_4, x_1+x_3) = – f_a(x_1+x_3, x_2+x_4)$. This implies $A=0$.
Best Answer
We consider the change of variables to $x_1, a, b, y$, where $$a = x_1 + x_2, b = x_1 + x_3, y = x_1 + x_2 + x_3 + x_4.$$ The Jacobian of this change of variable is $1$. The inverse of this change of variables is $$x_2 = a - x_1, x_3 = b - x_1, x_4 = y - a - b + x_1.$$ So the region $[0, \infty)^4$ is mapped to the new region $$R = \{(x_1, a, b, y): \max(0, a + b - y) \leq x_1 \leq \min(a, b)\}.$$ And $$\int_{[0,\infty)^4} dx_1dx_2dx_3dx_4 \: f(x_1+x_2, x_3+x_4)^* f(x_1+x_3, x_2+x_4)$$ is equal to $$\int_{R} dx_1da db dy \: f(a, y - a)^* f(b, y - b).$$ Now we can integrate in $x_1$. For each $a,b,y$ for which there is a satisfying $x_1$, the length of the integral in $x_1$ is $$\min(a, b) - \max(0, a + b - y) = \min(a, b, y - a, y - b).$$ So the integral is equal to $$\int_{[0,\infty)^3} da db dy \: \max(0, \min(a, b, y - a, y - b)) f(a, y - a)^* f(b, y - b).$$ Now it is familiar that $$\max(0, \min(a, b, y - a, y - b)) = \int_0^{\min(a, b, y - a, y - b)} dx$$ where we define the RHS to be $0$ if $\min(a, b, y - a, y - b) < 0$. So the integral becomes $$\int_{[0,\infty)^3} da db dy \: \int_0^{\min(a, b, y - a, y - b)} dx f(a, y - a)^* f(b, y - b).$$ Now we swap the order of integration so the $x$ variable is integrated first. If we set $R'(x)$ to be $$\{(a,b,y): a \geq x, b \geq x, y - a \geq x, y - b \geq x\}$$ then the integral is equal to $$\int_{[0, \infty)} dx \int_{R'(x)} da db dy \: f(a, y - a)^* f(b, y - b).$$ The inner integral decouples into $$\int_{R'(x)} da db dy \: f(a, y - a)^* f(b, y - b) = \int_{2x}^\infty dy \left\vert \int_{x}^{y - x} da f(a, y - a)\right\vert^2$$ and the non-negativity of the integral follows.