For $x,y\ge 0$, let $$ k(x,y)= \frac {J_1(2\sqrt{xy})}{\sqrt{xy}}, $$ where $J_1$ is the the Bessel function of the first kind $$ J_{1}(z)=\sum_{k=0}^{\infty}(-1)^{k} \frac{\left(\frac{z}{2}\right)^{2 k+1}}{k !(k+1) !}. $$ Does this kernel give a bounded linear operator on $L^2(0,+\infty)$: $$ Af(x) =\int_{0}^\infty k(x,y) f(y) dy. $$ Note that the kernel $k(x,y)$ is not square-integrable since $$ \int_0^\infty \left( \frac {J_1(2\sqrt{x})}{\sqrt{x}}\right)^2 dx= 1. $$
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