Let $f$ and $g$ be real functions such that $\int_0^\infty(f(x))^2dx<\infty$ and $\int_0^\infty(g(x)^2dx<\infty$. Prove that:
$$\left(\int_0^\infty\int_0^\infty\frac{f(x)g(y)}{x+y}dxdy \right)^2\leq C\int_0^\infty(f(x))^2dx\int_0^\infty(g(y))^2dy$$
where $C$ is an universal constant (independent of $f$ and $g$)
My attempt: using Cauchy-Schwarz inequality I evaluate:
$$LHS\leq(\int_0^\infty (g(y))^2dy)\times (\int_0^\infty (\int_0^\infty \frac{f(x)}{x+y}dx)^2dy)\leq$$ $$\leq(\int_0^\infty(g(y))^2dy)\times(\int_0^\infty((\int_0^\infty(f(x))^2dx)\times(\int_0^\infty\frac{dx}{(x+y)^2}))dy)\leq$$ $$\leq\left(\int_0^\infty(f(x))^2dx\int_0^\infty(g(y))^2dy\right)\times\int_0^\infty\int_0^\infty\frac{dxdy}{(x+y)^2}$$
which is useless since $\int_0^\infty\int_0^\infty\frac{dxdy}{(x+y)^2}$ is not finite.
Assume without loss of generality that $f,g$ are nonnegative (in order to use safely Fubini's Theorem), and denote by $I$ the double integral in the left-hand side. We have \begin{eqnarray} I&=&\int_0^\infty g(y)\left(\int_y^\infty f(u-y)\,\frac{du}u \right)dy\\ &=&\int_0^\infty g(y)\left(\int_1^\infty f(y(v-1))\,\frac{dv}v \right)dy\\ &=&\int_0^\infty g(y)\left(\int_0^\infty f(yu)\frac{du}{u+1} \right)dy\\ &=&\int_0^\infty\left(\int_0^\infty g(y)f(uy)\, dy\right)\frac{du}{u+1}\cdot \end{eqnarray} Now, apply Cauchy-Scharz to the inner integral: this gives \begin{eqnarray} \int_0^\infty g(y)f(uy)\, dy&\leq& \Vert g\Vert_2\times \left(\int_0^\infty f(uy)^2 dy\right)^{1/2}\\ &=&\Vert g\Vert_2\times \left(\int_0^\infty f(t)^2 \frac{dt}{u}\right)^{1/2}\\ &=&\Vert g\Vert_2\times \Vert f\Vert_2\times \frac{1}{\sqrt u}\cdot \end{eqnarray} Altogether, we obtain $$I\leq \Vert g\Vert_2\times \Vert f\Vert_2\times\int_0^\infty\frac{du}{(u+1)\sqrt u}\, , $$ which gives the result since the integral in the right-hand side is finite.