Classical Hardy’s inequality(cfr. Hardy-Littlewood-PolyaInequalities, Theorem 327)If p>1, f(x)≥0 and F(x)=∫x0f(y)dy then

∫∞0(F(x)x)pdx<C∫∞0(f(x))pdx

unless f≡0.

The best possibile constant is C=(pp−1)p.I would like to prove the statement in italic regarding the best constant. As already noted by Will Jagy here, the book suggests stress-testing the inequality with

f(x)={00≤x<1x−α1≤x

with 1/p<α<1, then have α→1/p. If I do so I get for C the lower bound

lim supα→1/pαp−1(1−α)p∫∞1(x−α−x−1)pdx≤C

but now I find myself in trouble in computing that lim sup. Can someone lend me a hand, please?

UPDATE: A first attempt, based on an idea by Davide Giraudo, unfortunately failed. Davide pointed out that the claim would easily follow from

|∫∞1(x−α−x−1)pdx−∫∞1x−αpdx|→0as α→1/p.

But this is false in general: for example if p=2 we get

∫∞1(x−2α−x−2α+2x−α−1−x−2)dx→∫∞1(2x−3/2−x−2)dx≠0.

**Answer**

What you need isn't

lim

but

\lim_{\alpha\searrow1/p}\frac{\int_1^\infty (x^{-\alpha}-x^{-1})^p\mathrm dx}{\int_1^\infty x^{-\alpha p }\mathrm dx}=1\;,

which is indeed the case, since as \alpha\searrow1/p, the integrals are more and more dominated by regions where x^{-1}\ll x^{-\alpha}. For arbitrary b\gt1 and 1/p\lt\alpha\lt1, we have

\begin{eqnarray}

\int_1^\infty (x^{-\alpha})^p\mathrm dx

&\gt&

\int_1^\infty (x^{-\alpha}-x^{-1})^p\mathrm dx

\\

&\gt&

\int_b^\infty (x^{-\alpha}-x^{-1})^p\mathrm dx

\\

&\gt&

\int_b^\infty (x^{-\alpha}-b^{\alpha-1}x^{-\alpha})^p\mathrm dx

\\

&=&

(1-b^{\alpha-1})^p\int_b^\infty (x^{-\alpha})^p\mathrm dx

\\

&=&

(1-b^{\alpha-1})^pb^{1-\alpha p}\int_1^\infty (x^{-\alpha})^p\mathrm dx

\\

&=&

(b^{1/p-\alpha}-b^{1/p-1})^p\int_1^\infty (x^{-\alpha})^p\mathrm dx\;.

\end{eqnarray}

Then choosing b=2^{1/\beta} with \beta=\sqrt{\alpha-1/p} yields

\int_1^\infty (x^{-\alpha})^p\mathrm dx

\gt

\int_1^\infty (x^{-\alpha}-x^{-1})^p\mathrm dx

\gt

(2^{-\beta}-2^{(1/p-1)/\beta})^p\int_1^\infty (x^{-\alpha})^p\mathrm dx\;.

Since \beta\to0 as \alpha\searrow1/p, the factor on the right goes to 1; thus,

\int_1^\infty (x^{-\alpha}-x^{-1})^p\mathrm dx\sim\int_1^\infty x^{-\alpha p }\mathrm dx\quad\text{as}\quad\alpha\searrow1/p,

as required.

**Attribution***Source : Link , Question Author : Giuseppe Negro , Answer Author : Oliver Diaz*