Hardy's inequality

Template:Short description Hardy's inequality is an inequality in mathematics, named after G. H. Hardy.

Its discrete version 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

${\int }_0^{\mathrm{\infty }}{(\frac{1}{x}{\int }_0^{x}f(t)dt)}^{p}dx\le {\left(\frac{p}{p-1}\right)}^p{\int }_0^{\mathrm{\infty }}f(x{)}^{p}dx.$

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.

Statements

General discrete Hardy inequality

The general weighted one dimensional version reads as follows:Template:Rp if $a_n\ge 0$, ${\lambda }_n>0$ and $p>1$,

$\sum _{n=1}^{\mathrm{\infty }}{\lambda }_n(\frac{{\lambda }_1{a}_1+\cdots +{\lambda }_n{a}_n}{{\lambda }_1+\cdots +{\lambda }_n}{)}^p\le (\frac{p}{p-1}{)}^p\sum _{n=1}^{\mathrm{\infty }}{\lambda }_n{a}_n^p.$

General one-dimensional integral Hardy inequality

The general weighted one dimensional version reads as follows:Template:Rp

• If $\alpha +\frac{1}{p}<1$, then

${\int }_0^{\mathrm{\infty }}(y^{\alpha -1}{\int }_0^y{x}^{-\alpha }f(x)dx{)}^{p}dy\le \frac{1}{(1-\alpha -\frac{1}{p}{)}^p}{\int }_0^{\mathrm{\infty }}f(x{)}^{p}dx$

• If $\alpha +\frac{1}{p}>1$, then

${\int }_0^{\mathrm{\infty }}(y^{\alpha -1}{\int }_y^{\mathrm{\infty }}{x}^{-\alpha }f(x)dx{)}^{p}dy\le \frac{1}{(\alpha +\frac{1}{p}-1{)}^p}{\int }_0^{\mathrm{\infty }}f(x{)}^{p}dx.$

The more general case is weighted integral Hardy inequality

Weighted one-dimensional integral Hardy inequality

The statement is following: if $1<p\le q<\mathrm{\infty }$ inequality

${\left({\int }_a^b{\left({\int }_a^{x}f(t)dt\right)}^{q}u(x)dx\right)}^{1/q}\le C{\left({\int }_a^b{f}^p(x)v(x)dx\right)}^{1/p}$

holds for $-\mathrm{\infty }\le a<b\le \mathrm{\infty }$ with $u,v$ measurable, positive on $(a,b)$ and also $f(x)\ge 0$ any measurable on $(a,b)$, if and only if

$A=\underset{x\in (a,b)}{sup}{({\int }_x^{b}u(t)dt)}^{1/q}{({\int }_a^x{v}^{1-p^{\prime }}(t)dt)}^{1/p^{\prime }}<\mathrm{\infty }$

And if $1<q<p<\mathrm{\infty }$ inequality also holds if and only if

$A={\left({\int }_a^b{({\int }_x^{b}u(t)dt)}^{r/q}{({\int }_a^x{v}^{1-p^{\prime }}(t)dt)}^{r/q^{\prime }}{v}^{1-p^{\prime }}(x)dx\right)}^{1/r}<\mathrm{\infty }$

where $\frac{1}{r}=\frac{1}{q}-\frac{1}{p},\frac{1}{p}+\frac{1}{p^{\prime }}=1$ and $\frac{1}{q}+\frac{1}{q^{\prime }}=1$

Multidimensional Hardy inequalities with gradient

Multidimensional Hardy inequality around a point

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

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

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

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

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

Multidimensional Hardy inequality near the boundary

If $\mathrm{\Omega }⊊{ℝ}^n$ is an nonempty convex open set, then for $1<p<\mathrm{\infty }$ and every $f\in W_0^{1,p}(\mathrm{\Omega })$,

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

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 $f:(0,\mathrm{\infty })\to ℝ$ satisfying ${\int }_0^{\mathrm{\infty }}|f(x){|}^p/x^{\lambda }dx<\mathrm{\infty }$, one hasTemplate:Rp

${\int }_0^{\mathrm{\infty }}\frac{|f(x){|}^p}{x^{\lambda }}dx\le C{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{|f(x)-f(y){|}^p}{|x-y{|}^{1+\lambda }}dxdy.$

Proof of the inequality

Integral version (integration by parts and Hölder)

Hardy’s original proofTemplate:Rp begins with an integration by parts to get

$\begin{array}{cc}{\int }_0^{\mathrm{\infty }}{(\frac{1}{x}{\int }_0^{x}f(t)dt)}^{p}dx& ={\int }_0^{\mathrm{\infty }}{({\int }_0^{x}f(t)dt)}^p\frac{1}{x^p}dx\\ & =\frac{p}{p-1}{\int }_0^{\mathrm{\infty }}{({\int }_0^{x}f(t)dt)}^{p-1}\frac{f(x)}{x^{p-1}}dx\\ & =\frac{p}{p-1}{\int }_0^{\mathrm{\infty }}{(\frac{1}{x}{\int }_0^{x}f(t)dt)}^{p-1}f(x)dx\end{array}$

Then, by Hölder's inequality,

${\int }_0^{\mathrm{\infty }}{(\frac{1}{x}{\int }_0^{x}f(t)dt)}^{p}dx\le \frac{p}{p-1}{\left({\int }_0^{\mathrm{\infty }}{(\frac{1}{x}{\int }_0^{x}f(t)dt)}^{p}dx\right)}^{1-\frac{1}{p}}{({\int }_0^{\mathrm{\infty }}f(x{)}^{p}dx)}^{\frac{1}{p}},$

and the conclusion follows.

Integral version (scaling and Minkowski)

A change of variables gives

${\left({\int }_0^{\mathrm{\infty }}{(\frac{1}{x}{\int }_0^{x}f(t)dt)}^{p}dx\right)}^{1/p}={\left({\int }_0^{\mathrm{\infty }}{({\int }_0^{1}f(sx)ds)}^{p}dx\right)}^{1/p},$

which is less or equal than ${\int }_0^1{({\int }_0^{\mathrm{\infty }}f(sx{)}^{p}dx)}^{1/p}ds$ by Minkowski's integral inequality. Finally, by another change of variables, the last expression equals

${\int }_0^1{({\int }_0^{\mathrm{\infty }}f(x{)}^{p}dx)}^{1/p}{s}^{-1/p}ds=\frac{p}{p-1}{({\int }_0^{\mathrm{\infty }}f(x{)}^{p}dx)}^{1/p}.$

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 Template:Math, 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 Template:Math, 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

${\int }_0^{\mathrm{\infty }}f(x{)}^{p}dx=\sum _{n=1}^{\mathrm{\infty }}{b}_n^p$

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

$\frac{1}{x}{\int }_0^{x}f(t)dt=\frac{b_1+\dots +b_{n-1}+(x-n+1)b_n}{x}\ge \frac{b_1+\dots +b_n}{n}$

(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

$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{b_1+\dots +b_n}{n}\right)}^p\le {\int }_0^{\mathrm{\infty }}{(\frac{1}{x}{\int }_0^{x}f(t)dt)}^{p}dx$.

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 Template:NumBlk

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 Template:EquationNote, 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:

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

proving Template:EquationNote. Applying Hölder's inequality to the right-hand side of Template:EquationNote 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

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External links

• Template:Springer