Computing the best constant in classical Hardy’s inequality

Classical Hardy’s inequality (cfr. Hardy-Littlewood-Polya Inequalities, Theorem 327)

If p>1, f(x)0 and F(x)=x0f(y)dy then

0(F(x)x)pdx<C0(f(x))pdx

unless f0. The best possibile constant is C=(pp1)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)={00x<1xα1x

with 1/p<α<1, then have α1/p. If I do so I get for C the lower bound

lim supα1/pαp1(1α)p1(xαx1)pdxC

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(xαx1)pdx1xαpdx|0as α1/p.

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

1(x2αx2α+2xα1x2)dx1(2x3/2x2)dx0.

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

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