Could you help me to show the following:
The operator $$ T(f)(x) = \int _0^\infty \frac{f(y)}{x+y}dy $$ satisfies $$\Vert T(f)\Vert_p \leq C_p \Vert f\Vert_p $$ for $1 <p< \infty$ where $$ C_p = \int_0^\infty \frac{t^{-1/p}}{t+1}dt $$
Thanks a lot!
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Note that substitution $y=tx$ gives $$ \int\limits_0^\infty\frac{f(y)}{x+y}dy=\int\limits_0^\infty\frac{f(tx)}{1+t}dt $$ then from Minkowski's integral inequality you get $$ \begin{align} \Vert T(f)\Vert_p &=\left(\int\limits_0^\infty\left|\int\limits_0^\infty\frac{f(tx)}{1+t}dt\right|^p dx\right)^{1/p}\\ &\leq\int\limits_0^\infty\left(\int\limits_0^\infty\left|\frac{f(tx)}{(1+t)}\right|^pdx\right)^{1/p}dt\\ &=\int\limits_0^\infty\frac{1}{1+t}\left(\int\limits_0^\infty|f(tx)|^p dx\right)^{1/p}dt\\ &=\int\limits_0^\infty\frac{1}{1+t}\left(\int\limits_0^\infty|f(x)|^p \frac{dx}{t}\right)^{1/p}dt\\ &=\int\limits_0^\infty\frac{1}{(1+t)t^{1/p}}\left(\int\limits_0^\infty|f(x)|^p dx\right)^{1/p}dt\\ &=\int\limits_0^\infty\frac{ t^{-1/p}dt}{1+t}\Vert f\Vert_p dt\\ &=C_p\Vert f\Vert_p \end{align} $$ Q.E.D.
Norbert
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This might be a late comment but does Minkowski's inequality works only when the function $F(x, t)$ (here is $\frac{f(t x)}{1 + t}$) is integrable in the product measure space? – Sanae Dec 01 '20 at 23:25
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It is indeed integrable. It follows from the inequality I've proved. – Norbert Dec 02 '20 at 10:23
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Got it now. Thank you. – Sanae Dec 02 '20 at 21:18