Let $f$ be an arbitrary function in $L^2(0,\infty)$ and consider the function


$$(g\_f)(y) = \frac{1}{y-x\_0} \int\_{0}^{\infty} f(x) \left(\frac{xy}{(x^2+y^2+1)}\right)^2 \ dx$$


where $x\_0$ is an arbitrary point in $(0,\infty).$


I ask: Is the function $g\_f$ ever a function in $L^2(0,\infty)?$


In other words, fix $x\_0$ as above, does there exist a square-integrable $f$ such that $g\_f$ is also in $L^2(0,\infty)$ [other than zero, of course]?



Function square-integrable
Yes. Indeed set
$$
G\_f(y) = (y-x\_0) g\_f(y)
 = \int\_0^\infty f(x) \, \left(\frac{xy}{(x^2+y^2+1)}\right)^{\!2} \, dx.
$$
It is easy to find linearly independent $f\_1,f\_2 \in L^2(0,\infty)$
such that each $G\_{f\_i}(y)$ is also in $L^2(0,\infty)$ and is differentiable
on $(0,\infty)$; for instance, $f\_1,f\_2$ can be characteristic functions of
different intervals. Let $f$ be a nonzero linear combination 
$c\_1 f\_1 + c\_2 f\_2$ such that $G\_f(x\_0) = 0$. Then $g\_f \in L^2(0,\infty)$.


