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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

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

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Addendum: The proof is complete if it is shown that $\operatorname{lim sup}_{\alpha \to 1/p} (\ldots) \ge \left( \frac{p}{p-1}\right)^p$, which is what we get if we formally neglect that $x^{-1}$. However I can't find a rigorous way to do that. – Giuseppe Negro Apr 1 '12 at 14:24
This is also an exercise in Rudin's book Real and Complex. – AD. Apr 3 '12 at 21:11
up vote 16 down vote accepted

What you need isn't

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


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

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 $1$, and 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.

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In moving from the fourth to the fifth line in the big eqnarray, how do you get $$\int_b^\infty (x^{-\alpha})^p\mathrm dx = b^{1-\alpha p}\int_1^\infty (x^{-\alpha})^p\mathrm dx?$$ – Antonio Vargas Apr 3 '12 at 18:15
@Antonio: Substitute $u=x/b$. – joriki Apr 3 '12 at 18:22
Ah, of course. Very nice, by the way. – Antonio Vargas Apr 3 '12 at 18:26
+1 Nice estimate. – AD. Apr 3 '12 at 20:59
This is a master's solution and I learned a lot in working it out! Thank you very much. – Giuseppe Negro Apr 4 '12 at 10:07

We have the operator $T: L^p(\mathbb{R}^+) \to L^p(\mathbb{R}^+)$ with $p \in (1, \infty)$, defined by$$(Tf)(x) := {1\over x} \int_0^x f(t)\,dt.$$Calculate $\|T\|$.

For the operator $T$ defined above, the operator norm is $p/(p - 1)$. We will also note that this is also a bounded operator for $p = \infty$, but not for $p = 1$.

Assume $1 < p < \infty$, and let $q$ be the dual exponent, $1/p + 1/q = 1$. By the theorem often referred to as "converse Hölder,"$$\|Tf\|_p = \sup_{\|g\|_q = 1}\left|\int_0^\infty (Tf)(x)g(x)\,dx\right|.$$So, assume that$\|g\|_q = 1$,\begin{align*} \left| \int_0^\infty (Tf)(x)g(x)\,dx\right| & \le \int_0^\infty |Tf(x)||g(x)|\,dx \le \int_0^\infty \int_0^x {1\over x}|f(t)||g(x)|\,dt\,dx \\ & = \int_0^\infty \int_0^1 |f(ux)||g(x)|\,du\,dx = \int_0^1 \int_0^\infty |f(ux)||g(x)|\,dx\,du \\ & \le \int_0^1 \left(\int_0^\infty |f(ux)|^pdx\right)^{1\over p} \left(\int_0^\infty |g(x)|^q dx\right)^{1\over q}du \\ & = \int_0^1 u^{-{1\over p}}\|f\|_p \|g\|_q du = {p\over{p - 1}}\|f\|_p.\end{align*}So that gives us that the operator norm is at most $p/(p - 1)$. To show that this is tight, let $f(x) = 1$ on $(0, 1]$ and zero otherwise. We have $\|f\|_p = 1$ for all $p$. We can then compute that$$(Tf)(x) = \begin{cases} 1 & 0 < x \le 1 \\ {1\over x} & x > 1\end{cases}$$and by direct computation, $\|Tf\|_p = p/(p - 1)$ for $p > 1$.

The same example also shows that we can have $f \in L^1$ but $Tf \notin L^1$, so we must restrict to $p > 1$. However, it is straightforward to show that $T$ is bounded from $L^\infty \to L^\infty$ with norm $1$. Note that the range of the operator in that case is contained within the bounded continuous functions on $(0, \infty)$.

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Thank you for your interest! I see a problem here. The book's statement is that the inequality is strict (except for the trivial case $f=0$). You claim that we have equality for $f(x)=\mathbf{1}_{0<x<1}$. The two statements contradict each other so one of them is wrong. – Giuseppe Negro Jun 14 at 8:56
I think I have found an error. I suspect that it is not true that $\| Tf\|_p=\frac{p}{p-1}$. Instead, it is correct that $\| Tf\|_p = \left(\frac{p}{p-1}\right)^{\frac1p}$. – Giuseppe Negro Jun 14 at 9:00

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