# Computing the best constant in classical Hardy’s inequality

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

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

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

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

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

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

What you need isn't

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

but

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

which is indeed the case, since as $$\alpha\searrow1/p\alpha\searrow1/p$$, the integrals are more and more dominated by regions where $$x^{-1}\ll x^{-\alpha}x^{-1}\ll x^{-\alpha}$$. For arbitrary $$b\gt1b\gt1$$ and $$1/p\lt\alpha\lt11/p\lt\alpha\lt1$$, we have

$$\begin{eqnarray} \int_1^\infty (x^{-\alpha})^p\mathrm dx &\gt& \int_1^\infty (x^{-\alpha}-x^{-1})^p\mathrm dx \\ &\gt& \int_b^\infty (x^{-\alpha}-x^{-1})^p\mathrm dx \\ &\gt& \int_b^\infty (x^{-\alpha}-b^{\alpha-1}x^{-\alpha})^p\mathrm dx \\ &=& (1-b^{\alpha-1})^p\int_b^\infty (x^{-\alpha})^p\mathrm dx \\ &=& (1-b^{\alpha-1})^pb^{1-\alpha p}\int_1^\infty (x^{-\alpha})^p\mathrm dx \\ &=& (b^{1/p-\alpha}-b^{1/p-1})^p\int_1^\infty (x^{-\alpha})^p\mathrm dx\;. \end{eqnarray} \begin{eqnarray} \int_1^\infty (x^{-\alpha})^p\mathrm dx &\gt& \int_1^\infty (x^{-\alpha}-x^{-1})^p\mathrm dx \\ &\gt& \int_b^\infty (x^{-\alpha}-x^{-1})^p\mathrm dx \\ &\gt& \int_b^\infty (x^{-\alpha}-b^{\alpha-1}x^{-\alpha})^p\mathrm dx \\ &=& (1-b^{\alpha-1})^p\int_b^\infty (x^{-\alpha})^p\mathrm dx \\ &=& (1-b^{\alpha-1})^pb^{1-\alpha p}\int_1^\infty (x^{-\alpha})^p\mathrm dx \\ &=& (b^{1/p-\alpha}-b^{1/p-1})^p\int_1^\infty (x^{-\alpha})^p\mathrm dx\;. \end{eqnarray}$$

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

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

Since $$\beta\to0\beta\to0$$ as $$\alpha\searrow1/p\alpha\searrow1/p$$, the factor on the right goes to $$11$$; thus,

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

as required.