How prove this integral inequality$\left(\int_{0}^{+\infty}\left(\frac{1}{x}\int_{0}^{x}|f(t)|dt\right)^pdx\right)^{\frac{1}{p}}$?

Question

let $p>1$, and $f$ is continuous and $\displaystyle\int_{0}^{+\infty}|f(t)|^p|dt$ is convergence,show that

$$\left(\int_{0}^{+\infty}\left(\dfrac{1}{x}\int_{0}^{x}|f(t)|dt\right)^pdx\right)^{\frac{1}{p}}\le\dfrac{p}{p-1}\left(\int_{0}^{+\infty}|f(t)|^pdt\right)^{\frac{1}{p}}$$

Answer 1

This is the Hardy's inequality, you can treat it as the convolution of $|f(x)|x^{1/p}$ with the function $x^{1/p-1}\chi_{[1,\infty)}(x)$ on the multiplicative group $(\mathbb{R}^+,\frac{dt}{t})$ and use the Minkowski's inequality:

$\|g*f\|_{L^p(G)}\leq\|g\|_{L^1(G)}\|f\|_{L^p(G)}$