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Let's define $T:L_2(0,\infty) \to L_2(0,\infty)$ as $$(Tf)(x) = \int_0^\infty \frac{f(y)\sqrt{xy}}{x^2y^2+1}dy.$$ I'm interested, if this operator is compact. $T$ is integral operator with kernel $K = \frac{\sqrt{xy}}{x^2y^2+1}$. The Holder inequality, which gives the usual bound $\|Tf\|_{L_2}\leq \|K\|_{L_2}\cdot \|f\|_{L_2}$, is not working here, because $K \notin L^2((0,\infty)\times(0,\infty))$. But I tried to use Schur test, and succeed to find this bound: $||T|| \le {\pi}/{\sqrt2}$. Then I tried to prove that $T$ is not compact. Because $L_2(0,\infty)$ is reflexive space, $T$ is compact iff it's completely continuous. So I tried to prove that $T$ is not completely continuous, that is to find weakly convergent sequence $f_n$ for which $Tf_n$ is not norm-convergent. But I cannot find such a sequence. Any suggestions? Maybe $T$ compact, but how to prove it then?

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