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How can I prove the following inequality:

If $r>p$, then \begin{equation} \left\|X\right\|_{p}\le\left(\frac{r}{r-p}\right)^{1/p}\left\|X\right\| _{r,\infty}, \end{equation} where \begin{equation} \left\|X\right\|_p=\left(\operatorname{E}\left|X\right|^p\right)^{1/p}\qquad\text{and}\qquad\left\|X\right\| _{r,\infty}=\left(\sup _{t>0}\ t^r\Pr\left(\left|X\right|>t\right) \right)^{1/r}. \end{equation}

I found this inequality on page 10 of the book by Ledoux and Talagrand.

Thank you for your answers!

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1 Answer 1

up vote 3 down vote accepted

The RHS is the upper bound of $\displaystyle\|X\|_p^p=\mathrm E(|X|^p)=p\int_0^{+\infty}x^{p-1}\,\mathrm P(|X|\geqslant x)\,\mathrm dx $ that one gets using the upper bound $ \mathrm P(|X|\geqslant x)\leqslant\min\{1,\|X\|_{r,\infty}^rx^{-r}\} $ for every $x\gt0$.

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