Computing the best constant in classical Hardy’s inequality

Question 1

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

If $p>1$ , $f(x) \ge 0$ and $F(x)=\int_0^xf(y)\, dy$ then

$\tag{H} \int_0^\infty \left(\frac{F(x)}{x}\right)^p\, dx < C\int_0^\infty (f(x))^p\, dx$

unless $f \equiv 0$ . The best possibile constant is $C=\left(\frac{p}{p-1}\right)^p$ .

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

$f(x)=\begin{cases} 0 & 0\le x <1 \\ x^{-\alpha} & 1\le x \end{cases}$

with $1/p< \alpha < 1$ , then have $\alpha \to 1/p$ . If I do so I get for $C$ the lower bound

$\operatorname{lim sup}_{\alpha \to 1/p}\frac{\alpha p -1}{(1-\alpha)^p}\int_1^\infty (x^{-\alpha}-x^{-1})^p\, dx\le C$

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

$\tag{!!} \left\lvert \int_1^\infty (x^{-\alpha}-x^{-1})^p\, dx - \int_1^\infty x^{-\alpha p }\, dx\right\rvert \to 0\quad \text{as}\ \alpha \to 1/p.$

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

$\int_1^\infty (x^{-2\alpha} -x^{-2\alpha} + 2x^{-\alpha-1}-x^{-2})\, dx \to \int_1^\infty(2x^{-3/2}-x^{-2})\, dx \ne 0.$

Question 2

What you need isn't

$\lim_{\alpha\searrow1/p}\,\left\lvert \int_1^\infty (x^{-\alpha}-x^{-1})^p\mathrm dx - \int_1^\infty x^{-\alpha p }\mathrm dx\right\rvert=0$

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.

Computing the best constant in classical Hardy’s inequality

Answer

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