Hilbert's Inequality

Solution 1:

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.