I have tried to solve this problem several times, and I still cannot find a good answer. Maybe I'm too naive, and the problem is really complicated.
Consider a nonnegative function $g$ defined on $(0,\infty)$. Assume additionally that $$ \int_0^\infty \frac{g(s)}{\sqrt{s}}ds <\infty. $$ Consider now the operator for $x>0$, $$ Tf(x) := \int_0^\infty g(xs)f(s)ds. $$ The question now is simple: show that $Tf$ is in $L^2$ provided $f$ is also in $L^2$. Could you help me with ideas to prove this result? Thanks a lot.
\begin{align} &\int_{0}^{\infty}\left|\int_{0}^{\infty}g(xs)f(s)ds\right|^2dx \\ & = \mbox{( let s=u/x ) }\int_{0}^{\infty}\left|\int_{0}^{\infty}g(u)f(u/x)\frac{du}{x}\right|^2dx \\ & = \int_{0}^{\infty}\left|\int_{0}^{\infty}\frac{g(u)^{1/2}}{u^{1/4}} \frac{u^{1/4}g(u)^{1/2}f(u/x)}{x}du\right|^2dx \\ & \le \int_{0}^{\infty}\int_{0}^{\infty}\frac{g(u)}{\sqrt{u}}du\int_{0}^{\infty}\frac{\sqrt{u}g(u)f(u/x)^2}{x^2}du\,dx \\ & = C\int_{0}^{\infty}\int_{0}^{\infty}\frac{\sqrt{u}g(u)f(u/x)^2}{x^2}dudx, \;\;\; C=\int_{0}^{\infty}\frac{g(u)}{\sqrt{u}}du \\ & = C\int_{0}^{\infty}\sqrt{u}g(u)\int_{0}^{\infty}\frac{f(u/x)^2}{x^2}dxdu \\ & = \mbox{ ( v=u/x ) } C\int_{0}^{\infty}\sqrt{u}g(u)\frac{1}{u}\int_{0}^{\infty} f(v)^2 dv du \\ & = C^2\int_{0}^{\infty}f(v)^2dv \end{align} Therefore, $$ \left\|\int_{0}^{\infty}g(xs)f(s)ds\right\|_{L^2} \le C\|f\|_{L^2}, \;\;\; C=\int_{0}^{\infty}\frac{g(u)}{\sqrt{u}}du. $$