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

$\sum _{n=1}^{\infty }\left({\frac {a_{1}+a_{2}+\cdots +a_{n}}{n}}\right)^{p}\leq \left({\frac {p}{p-1}}\right)^{p}\sum _{n=1}^{\infty }a_{n}^{p}.$

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

$\int _{0}^{\infty }\left({\frac {1}{x}}\int _{0}^{x}f(t)\,dt\right)^{p}\,dx\leq \left({\frac {p}{p-1}}\right)^{p}\int _{0}^{\infty }f(x)^{p}\,dx.$

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.

General one-dimensional version

The general weighted one dimensional version reads as follows:

• If $\alpha +{\tfrac {1}{p}}<1$, then

$\int _{0}^{\infty }{\biggl (}y^{\alpha -1}\int _{0}^{y}x^{-\alpha }f(x)\,dx{\biggr )}^{p}\,dy\leq {\frac {1}{{\bigl (}1-\alpha -{\frac {1}{p}}{\bigr )}^{p}}}\int _{0}^{\infty }f(x)^{p}\,dx$

• If $\alpha +{\tfrac {1}{p}}>1$, then

$\int _{0}^{\infty }{\biggl (}y^{\alpha -1}\int _{y}^{\infty }x^{-\alpha }f(x)\,dx{\biggr )}^{p}\,dy\leq {\frac {1}{{\bigl (}\alpha +{\frac {1}{p}}-1{\bigr )}^{p}}}\int _{0}^{\infty }f(x)^{p}\,dx.$

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

$\left\|{\frac {f}{|x|}}\right\|_{L^{p}(\mathbb {R} ^{n})}\leq {\frac {p}{n-p}}\|\nabla f\|_{L^{p}(\mathbb {R} ^{n})},2\leq n,1\leq p<n,$

where $f\in C_{0}^{\infty }(\mathbb {R} ^{n})$, 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 $p>n\geq 2$, then one has for every $f\in C_{0}^{\infty }(\mathbb {R} ^{n})$

${\Big (}1-{\frac {n}{p}}{\Big )}^{p}\int _{\mathbb {R} ^{n}}{\frac {\vert f(x)-f(0)\vert ^{p}}{|x|^{p}}}dx\leq \int _{\mathbb {R} ^{n}}\vert \nabla f\vert ^{p}.$

Multidimensional Hardy inequality near the boundary

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

${\Big (}1-{\frac {1}{p}}{\Big )}^{p}\int _{\Omega }{\frac {\vert f(x)\vert ^{p}}{\operatorname {dist} (x,\partial \Omega )^{p}}}\,dx\leq \int _{\Omega }\vert \nabla f\vert ^{p},$

and the constant cannot be improved.

Fractional Hardy inequality

If $1\leq p<\infty$ and $0<\lambda <\infty$, $\lambda \neq 1$, there exists a constant $C$ such that for every $f:(0,\infty )\to \mathbb {R}$ satisfying $\int _{0}^{\infty }\vert f(x)\vert ^{p}/x^{\lambda }\,dx<\infty$, one has

$\int _{0}^{\infty }{\frac {\vert f(x)\vert ^{p}}{x^{\lambda }}}\,dx\leq C\int _{0}^{\infty }\int _{0}^{\infty }{\frac {\vert f(x)-f(y)\vert ^{p}}{\vert x-y\vert ^{1+\lambda }}}\,dx\,dy.$

Proof of the inequality

Integral version

A change of variables gives

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

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

$\int _{0}^{1}\left(\int _{0}^{\infty }f(x)^{p}\,dx\right)^{1/p}s^{-1/p}\,ds={\frac {p}{p-1}}\left(\int _{0}^{\infty }f(x)^{p}\,dx\right)^{1/p}.$

Discrete version: from the continuous version

Assuming the right-hand side to be finite, we must have $a_{n}\to 0$ as $n\to \infty$. Hence, for any positive integer j, there are only finitely many terms bigger than $2^{-j}$. This allows us to construct a decreasing sequence $b_{1}\geq b_{2}\geq \dotsb$ containing the same positive terms as the original sequence (but possibly no zero terms). Since $a_{1}+a_{2}+\dotsb +a_{n}\leq b_{1}+b_{2}+\dotsb +b_{n}$ 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

$\int _{0}^{\infty }f(x)^{p}\,dx=\sum _{n=1}^{\infty }b_{n}^{p}$

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

${\frac {1}{x}}\int _{0}^{x}f(t)\,dt={\frac {b_{1}+\dots +b_{n-1}+(x-n+1)b_{n}}{x}}\geq {\frac {b_{1}+\dots +b_{n}}{n}}$

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

$\sum _{n=1}^{\infty }\left({\frac {b_{1}+\dots +b_{n}}{n}}\right)^{p}\leq \int _{0}^{\infty }\left({\frac {1}{x}}\int _{0}^{x}f(t)\,dt\right)^{p}\,dx$.

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

$\sum _{n=1}^{N}{\frac {S_{n}^{p}}{n^{p}}}\leq {\frac {p}{p-1}}\sum _{n=1}^{N}{\frac {b_{n}S_{n}^{p-1}}{n^{p-1}}},$

(*)

Let $T_{n}={\frac {S_{n}}{n}}$ and let $\Delta _{n}$ be the difference between the $n$-th terms in the right-hand side and left-hand side of *, that is, $\Delta _{n}:=T_{n}^{p}-{\frac {p}{p-1}}b_{n}T_{n}^{p-1}$. We have:

$\Delta _{n}=T_{n}^{p}-{\frac {p}{p-1}}b_{n}T_{n}^{p-1}=T_{n}^{p}-{\frac {p}{p-1}}(nT_{n}-(n-1)T_{n-1})T_{n}^{p-1}$

or

$\Delta _{n}=T_{n}^{p}\left(1-{\frac {np}{p-1}}\right)+{\frac {p(n-1)}{p-1}}T_{n-1}T_{n}^{p}.$

According to Young's inequality we have:

$T_{n-1}T_{n}^{p-1}\leq {\frac {T_{n-1}^{p}}{p}}+(p-1){\frac {T_{n}^{p}}{p}},$

from which it follows that:

$\Delta _{n}\leq {\frac {n-1}{p-1}}T_{n-1}^{p}-{\frac {n}{p-1}}T_{n}^{p}.$

By telescoping we have:

${\begin{aligned}\sum _{n=1}^{N}\Delta _{n}&\leq 0-{\frac {1}{p-1}}T_{1}^{p}+{\frac {1}{p-1}}T_{1}^{p}-{\frac {2}{p-1}}T_{2}^{p}+{\frac {2}{p-1}}T_{2}^{p}-{\frac {3}{p-1}}T_{3}^{p}+\dotsb +{\frac {N-1}{p-1}}T_{N-1}^{p}-{\frac {N}{p-1}}T_{N}^{p}\\&=-{\frac {N}{p-1}}T_{N}^{p}<0,\end{aligned}}$

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

$\sum _{n=1}^{N}{\frac {S_{n}^{p}}{n^{p}}}\leq {\frac {p}{p-1}}\sum _{n=1}^{N}{\frac {b_{n}S_{n}^{p-1}}{n^{p-1}}}\leq {\frac {p}{p-1}}\left(\sum _{n=1}^{N}b_{n}^{p}\right)^{1/p}\left(\sum _{n=1}^{N}{\frac {S_{n}^{p}}{n^{p}}}\right)^{(p-1)/p}$

from which we immediately obtain:

$\sum _{n=1}^{N}{\frac {S_{n}^{p}}{n^{p}}}\leq \left({\frac {p}{p-1}}\right)^{p}\sum _{n=1}^{N}b_{n}^{p}.$

Letting $N\rightarrow \infty$ 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 (eds.), 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

• "Hardy inequality", Encyclopedia of Mathematics, EMS Press, 2001 [1994]