Hardy's inequality

Hardy's inequality is an inequality in mathematics, named after G. H. Hardy. It states that if $a_1,a_2,a_3,\dots$ is a sequence of non-negative real numbers, then for every real number p > 1 one has

$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{a_1+a_2+\cdots +a_n}{n}\right)}^p\le {\left(\frac{p}{p-1}\right)}^p\sum _{n=1}^{\mathrm{\infty }}{a}_n^p.$

If the right-hand side is finite, equality holds if and only if $a_n=0$ for all n.

An integral version of Hardy's inequality states the following: if f is a measurable function with non-negative values, then

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{(\frac{1}{x}\underset{0}{\overset{x}{\int }}f(t)dt)}^{p}dx\le {\left(\frac{p}{p-1}\right)}^p\underset{0}{\overset{\mathrm{\infty }}{\int }}f(x{)}^{p}dx.$

If the right-hand side is finite, equality holds if and only if f(x) = 0 almost everywhere.

Hardy's inequality was first published and proved (at least the discrete version with a worse constant) in 1920 in a note by Hardy. The original formulation was in an integral form slightly different from the above.

General one-dimensional version

The general weighted one dimensional version reads as follows:

• If $\alpha +\frac{1}{p}<1$, then

$\underset{0}{\overset{\mathrm{\infty }}{\int }}(y^{\alpha -1}\underset{0}{\overset{y}{\int }}{x}^{-\alpha }f(x)dx{)}^{p}dy\le \frac{1}{(1-\alpha -\frac{1}{p}{)}^p}\underset{0}{\overset{\mathrm{\infty }}{\int }}f(x{)}^{p}dx$

• If $\alpha +\frac{1}{p}>1$, then

$\underset{0}{\overset{\mathrm{\infty }}{\int }}(y^{\alpha -1}\underset{y}{\overset{\mathrm{\infty }}{\int }}{x}^{-\alpha }f(x)dx{)}^{p}dy\le \frac{1}{(\alpha +\frac{1}{p}-1{)}^p}\underset{0}{\overset{\mathrm{\infty }}{\int }}f(x{)}^{p}dx.$

Multidimensional versions

Multidimensional Hardy inequality around a point

In the multidimensional case, Hardy's inequality can be extended to $L^p$-spaces, taking the form

${\Vert \frac{f}{|x|}\Vert }_{L^p({\mathbb{ℝ}}^n)}\le \frac{p}{n-p}\Vert \mathrm{\nabla }f{\Vert }_{L^p({\mathbb{ℝ}}^n)},2\le n,1\le p<n,$

where $f\in C_0^{\mathrm{\infty }}({\mathbb{ℝ}}^n)$, and where the constant $\frac{p}{n-p}$ is known to be sharp; by density it extends then to the Sobolev space $W^{1,p}({\mathbb{ℝ}}^n)$.

Similarly, if $p>n\ge 2$, then one has for every $f\in C_0^{\mathrm{\infty }}({\mathbb{ℝ}}^n)$

$(1-\frac{n}{p}{)}^p\underset{{\mathbb{ℝ}}^n}{\int }\frac{|f(x)-f(0){|}^p}{|x{|}^p}dx\le \underset{{\mathbb{ℝ}}^n}{\int }|\mathrm{\nabla }f{|}^p.$

Multidimensional Hardy inequality near the boundary

If $\mathrm{\Omega }⊊{\mathbb{ℝ}}^n$ is an nonempty convex open set, then for every $f\in W^{1,p}(\mathrm{\Omega })$,

$(1-\frac{1}{p}{)}^p\underset{\mathrm{\Omega }}{\int }\frac{|f(x){|}^p}{\mathrm{dist}(x,\partial \mathrm{\Omega }{)}^p}dx\le \underset{\mathrm{\Omega }}{\int }|\mathrm{\nabla }f{|}^p,$

and the constant cannot be improved.

Fractional Hardy inequality

If $1\le p<\mathrm{\infty }$ and $0<\lambda <\mathrm{\infty }$, $\lambda \ne 1$, there exists a constant $C$ such that for every $f:(0,\mathrm{\infty })\to \mathbb{ℝ}$ satisfying $\underset{0}{\overset{\mathrm{\infty }}{\int }}|f(x){|}^p/x^{\lambda }dx<\mathrm{\infty }$, one has

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{|f(x){|}^p}{x^{\lambda }}dx\le C\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{|f(x)-f(y){|}^p}{|x-y{|}^{1+\lambda }}dxdy.$

Proof of the inequality

Integral version

A change of variables gives

${\left(\underset{0}{\overset{\mathrm{\infty }}{\int }}{(\frac{1}{x}\underset{0}{\overset{x}{\int }}f(t)dt)}^{p}dx\right)}^{1/p}={\left(\underset{0}{\overset{\mathrm{\infty }}{\int }}{(\underset{0}{\overset{1}{\int }}f(sx)ds)}^{p}dx\right)}^{1/p},$

which is less or equal than $\underset{0}{\overset{1}{\int }}{(\underset{0}{\overset{\mathrm{\infty }}{\int }}f(sx{)}^{p}dx)}^{1/p}ds$ by Minkowski's integral inequality. Finally, by another change of variables, the last expression equals

$\underset{0}{\overset{1}{\int }}{(\underset{0}{\overset{\mathrm{\infty }}{\int }}f(x{)}^{p}dx)}^{1/p}{s}^{-1/p}ds=\frac{p}{p-1}{(\underset{0}{\overset{\mathrm{\infty }}{\int }}f(x{)}^{p}dx)}^{1/p}.$

Discrete version: from the continuous version

Assuming the right-hand side to be finite, we must have $a_n\to 0$ as $n\to \mathrm{\infty }$. Hence, for any positive integer j, there are only finitely many terms bigger than $2^{-j}$. This allows us to construct a decreasing sequence $b_1\ge b_2\ge \cdots$ containing the same positive terms as the original sequence (but possibly no zero terms). Since $a_1+a_2+\cdots +a_n\le b_1+b_2+\cdots +b_n$ for every n, it suffices to show the inequality for the new sequence. This follows directly from the integral form, defining $f(x)=b_n$ if $n-1<x<n$ and $f(x)=0$ otherwise. Indeed, one has

$\underset{0}{\overset{\mathrm{\infty }}{\int }}f(x{)}^{p}dx=\sum _{n=1}^{\mathrm{\infty }}{b}_n^p$

and, for $n-1<x<n$, there holds

$\frac{1}{x}\underset{0}{\overset{x}{\int }}f(t)dt=\frac{b_1+\dots +b_{n-1}+(x-n+1)b_n}{x}\ge \frac{b_1+\dots +b_n}{n}$

(the last inequality is equivalent to $(n-x)(b_1+\dots +b_{n-1})\ge (n-1)(n-x)b_n$, which is true as the new sequence is decreasing) and thus

$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{b_1+\dots +b_n}{n}\right)}^p\le \underset{0}{\overset{\mathrm{\infty }}{\int }}{(\frac{1}{x}\underset{0}{\overset{x}{\int }}f(t)dt)}^{p}dx$.

Discrete version: Direct proof

Let $p>1$ and let $b_1,\dots ,b_n$ be positive real numbers. Set $S_k=\sum _{i=1}^k{b}_i$. First we prove the inequality

$\sum _{n=1}^N\frac{S_n^p}{n^p}\le \frac{p}{p-1}\sum _{n=1}^N\frac{b_n{S}_n^{p-1}}{n^{p-1}},$

(*)

Let $T_n=\frac{S_n}{n}$ and let ${\mathrm{\Delta }}_n$ be the difference between the $n$-th terms in the right-hand side and left-hand side of *, that is, ${\mathrm{\Delta }}_n:=T_n^p-\frac{p}{p-1}{b}_n{T}_n^{p-1}$. We have:

${\mathrm{\Delta }}_n=T_n^p-\frac{p}{p-1}{b}_n{T}_n^{p-1}=T_n^p-\frac{p}{p-1}(n{T}_n-(n-1)T_{n-1})T_n^{p-1}$

or

${\mathrm{\Delta }}_n=T_n^p(1-\frac{np}{p-1})+\frac{p(n-1)}{p-1}{T}_{n-1}{T}_n^p.$

According to Young's inequality we have:

$T_{n-1}{T}_n^{p-1}\le \frac{T_{n-1}^p}{p}+(p-1)\frac{T_n^p}{p},$

from which it follows that:

${\mathrm{\Delta }}_n\le \frac{n-1}{p-1}{T}_{n-1}^p-\frac{n}{p-1}{T}_n^p.$

By telescoping we have:

$\begin{array}{rl}\sum _{n=1}^N{\mathrm{\Delta }}_n& \le 0-\frac{1}{p-1}{T}_1^p+\frac{1}{p-1}{T}_1^p-\frac{2}{p-1}{T}_2^p+\frac{2}{p-1}{T}_2^p-\frac{3}{p-1}{T}_3^p+\cdots +\frac{N-1}{p-1}{T}_{N-1}^p-\frac{N}{p-1}{T}_N^p\\ & =-\frac{N}{p-1}{T}_N^p<0,\end{array}$

Applying Hölder's inequality to the right-hand side of * we have:

$\sum _{n=1}^N\frac{S_n^p}{n^p}\le \frac{p}{p-1}\sum _{n=1}^N\frac{b_n{S}_n^{p-1}}{n^{p-1}}\le \frac{p}{p-1}{\left(\sum _{n=1}^N{b}_n^p\right)}^{1/p}{\left(\sum _{n=1}^N\frac{S_n^p}{n^p}\right)}^{(p-1)/p}$

from which we immediately obtain:

$\sum _{n=1}^N\frac{S_n^p}{n^p}\le {\left(\frac{p}{p-1}\right)}^p\sum _{n=1}^N{b}_n^p.$

Letting $N\to \mathrm{\infty }$ we obtain Hardy's inequality.

See also

• Carleman's inequality

References

• Hardy, G. H.; Littlewood, J. E.; Pólya, G. (1952). Inequalities (2nd ed.). Cambridge University Press. ISBN 0-521-35880-9. 

• Kufner, Alois; Persson, Lars-Erik (2003). Weighted inequalities of Hardy type. World Scientific Publishing. ISBN 981-238-195-3. 

• Masmoudi, Nader (2011), "About the Hardy Inequality", in Dierk Schleicher; Malte Lackmann, An Invitation to Mathematics, Springer Berlin Heidelberg, ISBN 978-3-642-19533-4 .
• Ruzhansky, Michael; Suragan, Durvudkhan (2019). Hardy Inequalities on Homogeneous Groups: 100 Years of Hardy Inequalities. Birkhäuser Basel. ISBN 978-3-030-02895-4. 

External links

• Hazewinkel, Michiel, ed. (2001), "Hardy inequality", Encyclopedia of Mathematics, Springer Science+Business Media B.V. / Kluwer Academic Publishers, ISBN 978-1-55608-010-4, https://www.encyclopediaofmath.org/index.php?title=p/h046340