Let $1<p<\infty,$show that $$\bigg(\int_{0}^{\infty}\bigg(\frac{1}{x}\int_{0}^{x}|f(t)|dt\bigg)^{p}dx\bigg)^{\frac{1}{p}}\le\frac{p}{p-1}\lVert f\rVert_{L^{p}(0,\infty)}$$
Here is my attempt :
\begin{align} \bigg(\int_{0}^{\infty}\bigg(\frac{1}{x}\int_{0}^{x}|f(t)|dt\bigg)^{p}dx\bigg)^{\frac{1}{p}}&\le\int_{0}^{\infty}\bigg(\int_{t}^{\infty}\frac{|f(t)|^{p}}{x^{p}}~dx\bigg)^{\frac{1}{p}}dt\\ &\le\int_{0}^{\infty}\bigg(|f(t)|^{p}\int_{t}^{\infty}\frac{1}{x^{p}}~dx\bigg)^{\frac{1}{p}}dt\\ &=\int_{0}^{\infty}\bigg(|f(t)|^{p}\frac{t^{1-p}}{p-1}\bigg)^{\frac{1}{p}}dt \end{align}
,where the first I used minkowski inequality of integral form.
Then I stuck with it . But I got a hint to apply the convolution with some function .
Is there anybody that can show more details ? Thanks for considering my request .
