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Prove that the following operator is bounded on $L^{2}(0, \infty)$:

$Af(x)$ = $\frac{1}{\pi} \int_{0}^{\infty} \frac{f(y)}{x+y}dy$

with $||A|| \le 1$.

Attempt at Solution

It can be shown that:

$\int_{0}^{\infty} \frac{|f(y)|}{|x+y|}dy \le [\omega(x)\int_{0}^{\infty} \frac{|f(y)|^{2}}{|x+y|}\omega(y)^{-1}dy]^{\frac{1}{2}}$

I am trying to show that:

$\int_{0}^{\infty}\frac{1}{\pi}\omega(x)\int_{0}^{\infty} \frac{|f(y)|^{2}}{|x+y|}\omega(y)^{-1}dy = \frac{1}{\pi}\int_{0}^{\infty} \frac{1}{(1+y)y^{\frac{1}{2}}}\int_{0}^{\infty}|f(y)|^{2} = \int_{0}^{\infty}|f(y)|^{2}$

I can't seem to find a suitable choice of $\omega(x)$ such that result holds.

Martin
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Heisenberg
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    Surely what you want to do is apply Minkowski's integral inequality to $$ |Af|_2 = \left[\int_0^\infty \left|Af(x)\right|^2 dx\right]^{1/2} = \left[\int_0^\infty \left| \int_0^\infty \frac{f(y)}{x+y} dy \right|^2 dx\right]^{1/2}, $$ isn't it? – Branimir Ćaćić Mar 21 '13 at 07:34
  • @BranimirĆaćić, after applying Minkowski inequality, what do you do to finish? – Tomás Mar 21 '13 at 13:45

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