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Let $0<b<\infty$ and $1\le p<\infty$. Show that

$$\bigg(\int_{0}^{\infty}\bigg(\int_{x}^{\infty}|f(t)|dt\bigg)^{p}x^{b-1}dx\bigg)^{\frac{1}{p}}\le\frac{p}{b}\bigg(\int_{0}^{\infty}|f(t)|^{p}t^{p+b-1}dt\bigg)^{\frac{1}{p}}$$

Here is me attempt :

\begin{align} \bigg(\int_{0}^{\infty}\bigg(\int_{x}^{\infty} |f(t)|dt\bigg)^{p}x^{b-1}dx\bigg)^{\frac{1}{p}}&=\bigg(\int_{0}^{\infty}\bigg(\int_{x}^{\infty} x^{\frac{b-1}{p}}|f(t)|dt\bigg)^{p}dx\bigg)^{\frac{1}{p}}\\ &=\bigg(\int_{0}^{\infty}\bigg(\int_{1}^{\infty} x^{\frac{b-1}{p}+1}~|f(xt)|dt\bigg)^{p}dx\bigg)^{\frac{1}{p}}\\ &\le \int_{1}^{\infty}\bigg(\int_{0}^{\infty} x^{p+b-1}|f(xt)|^{p}dx\bigg)^{\frac{1}{p}}dt\\ &=\int_{1}^{\infty}\bigg(\int_{0}^{\infty} (xt)^{p+b-1}\frac{1}{t^{p+b-1}}|f(xt)|^{p}dx\bigg)^{\frac{1}{p}}dt\\ &=\int_{1}^{\infty}\bigg(\int_{0}^{\infty} \eta^{p+b-1}\frac{1}{t^{p+b-1}}|f(\eta)|^{p}~\frac{d\eta}{t}\bigg)^{\frac{1}{p}}dt\\ &=\int_{1}^{\infty}\bigg(\int_{0}^{\infty} \eta^{p+b-1}\frac{1}{t^{p+b}}|f(\eta)|^{p}~d\eta\bigg)^{\frac{1}{p}}dt\\ &=\int_{1}^{\infty}\frac{1}{t^{1+\frac{b}{p}}}~dt\bigg(\int_{0}^{\infty} \eta^{p+b-1}|f(\eta)|^{p}~d\eta\bigg)^{\frac{1}{p}}\\ &=\frac{p}{b}\bigg(\int_{0}^{\infty} \eta^{p+b-1}|f(\eta)|^{p}~d\eta\bigg)^{\frac{1}{p}}\\ \end{align}

,where $\eta=xt$ and hence $dx=\frac{d\eta}{t}$ and the first inequality is according to the integral form of Minkowski's inequality.

If you have the time , please checking my proof for validity. Any valuable suggestion and advice will be appreciated.

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This is the so called Hardy inequality and your answer seems perfectly done to me you find another way to prove it here page 55. As well it should stressed out that your version is more general than the following version of Hardy inequality here: Hardy's Inequality for Integrals and it is a third part version of this one from which it sufficed to set $$ f(t) = \frac{g(1/x)}{x^2}$$ to deduce your version.

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  • $\begingroup$ It is the exercise of my text book :Classical Fourier Analysis of Loukas Grafakos. $\endgroup$ – user1992 Jan 1 '18 at 10:51
  • $\begingroup$ @user1992 I am a fan of that book too I see which exercise you mean they gave some hint right? $\endgroup$ – Guy Fsone Jan 1 '18 at 10:53
  • $\begingroup$ Yes,but I think just use change of variable and Minkowski will work so, I try to do . $\endgroup$ – user1992 Jan 1 '18 at 10:55
  • $\begingroup$ @user1992 see the update $\endgroup$ – Guy Fsone Jan 1 '18 at 11:00
  • $\begingroup$ You are right . $\endgroup$ – user1992 Jan 1 '18 at 11:07

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