Hardy's inequality is an inequality in mathematics, named after G. H. Hardy. It states that if
is a sequence of non-negative real numbers, then for every real number p > 1 one has
![{\displaystyle \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}.}](//wikimedia.org/api/rest_v1/media/math/render/svg/4c957db6870f68b9e28f81a8811b43017648902f)
If the right-hand side is finite, equality holds if and only if
for all n.
An integral version of Hardy's inequality states the following: if f is a measurable function with non-negative values, then
![{\displaystyle \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.}](//wikimedia.org/api/rest_v1/media/math/render/svg/524add9177d9df0abd08b3397ed1bb065485be8b)
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.[1] The original formulation was in an integral form slightly different from the above.
Multidimensional Hardy inequality around a point
In the multidimensional case, Hardy's inequality can be extended to
-spaces, taking the form [3]
![{\displaystyle \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,}](//wikimedia.org/api/rest_v1/media/math/render/svg/fb8386c2de43f9b3d2c0dc29639271043d063443)
where
, and where the constant
is known to be sharp; by density it extends then to the Sobolev space
.
Similarly, if
, then one has for every ![{\displaystyle f\in C_{0}^{\infty }(\mathbb {R} ^{n})}](//wikimedia.org/api/rest_v1/media/math/render/svg/db3b30ba7cd7193df0852f878c8fe3d031417898)
![{\displaystyle {\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}.}](//wikimedia.org/api/rest_v1/media/math/render/svg/b065ae7c3f952c02847822e1928cf486235bbb0f)
Multidimensional Hardy inequality near the boundary
If
is an nonempty convex open set, then for every
,
![{\displaystyle {\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},}](//wikimedia.org/api/rest_v1/media/math/render/svg/62edef0cfc32ef24aabba6b563d20acd837778fe)
and the constant cannot be improved.[4]
Integral version
A change of variables gives
![{\displaystyle \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},}](//wikimedia.org/api/rest_v1/media/math/render/svg/b596a59281fb2f87ff9750afab3b474ce643d044)
which is less or equal than
by Minkowski's integral inequality.
Finally, by another change of variables, the last expression equals
![{\displaystyle \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}.}](//wikimedia.org/api/rest_v1/media/math/render/svg/0c144c23e1f686b951396881be2e06cbd67980b6)
Discrete version: from the continuous version
Assuming the right-hand side to be finite, we must have
as
. Hence, for any positive integer j, there are only finitely many terms bigger than
.
This allows us to construct a decreasing sequence
containing the same positive terms as the original sequence (but possibly no zero terms). Since
for every n, it suffices to show the inequality for the new sequence. This follows directly from the integral form, defining
if
and
otherwise. Indeed, one has
![{\displaystyle \int _{0}^{\infty }f(x)^{p}\,dx=\sum _{n=1}^{\infty }b_{n}^{p}}](//wikimedia.org/api/rest_v1/media/math/render/svg/6aa5f3f6cdd14f3b3a3a841c7b41187281df1c36)
and, for
, there holds
![{\displaystyle {\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}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/b8807271a696c1e570e6d844c15b8603db99f816)
(the last inequality is equivalent to
, which is true as the new sequence is decreasing) and thus
.
Discrete version: Direct proof
Let
and let
be positive real numbers. Set
.
First we prove the inequality
![{\displaystyle \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}}},}](//wikimedia.org/api/rest_v1/media/math/render/svg/47c12077552b381a96c4a25c5b464834c0868a6a) | | (*) |
Let
and let
be the difference between the
-th terms in the right-hand side and left-hand side of *, that is,
. We have:
![{\displaystyle \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}}](//wikimedia.org/api/rest_v1/media/math/render/svg/f8762d62754f441abe01ff9f002aca50f8b6377c)
or
![{\displaystyle \Delta _{n}=T_{n}^{p}\left(1-{\frac {np}{p-1}}\right)+{\frac {p(n-1)}{p-1}}T_{n-1}T_{n}^{p}.}](//wikimedia.org/api/rest_v1/media/math/render/svg/4487308a335a892e05889ec2db0e150f0f96cd3b)
According to Young's inequality we have:
![{\displaystyle T_{n-1}T_{n}^{p-1}\leq {\frac {T_{n-1}^{p}}{p}}+(p-1){\frac {T_{n}^{p}}{p}},}](//wikimedia.org/api/rest_v1/media/math/render/svg/bfaf907e88ddd5f6c602f05cb90008469984293d)
from which it follows that:
![{\displaystyle \Delta _{n}\leq {\frac {n-1}{p-1}}T_{n-1}^{p}-{\frac {n}{p-1}}T_{n}^{p}.}](//wikimedia.org/api/rest_v1/media/math/render/svg/10d4581a59949f1d4d824c38a289eb758bf90379)
By telescoping we have:
![{\displaystyle {\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}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/769d67f08fe3de74c94c262f4bb117c7abf24ffa)
proving *.
Applying Hölder's inequality to the right-hand side of * we have:
![{\displaystyle \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}}](//wikimedia.org/api/rest_v1/media/math/render/svg/fd105108ae22d92a49bc40aaf8734ccd44de4080)
from which we immediately obtain:
![{\displaystyle \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}.}](//wikimedia.org/api/rest_v1/media/math/render/svg/3700a8f9f798cd721a991eab8ee714c5c3ff583a)
Letting
we obtain Hardy's inequality.
Hardy, G. H.; Littlewood, J.E.; Pólya, G. (1952). Inequalities (Second ed.). Cambridge, UK.{{cite book}}
: CS1 maint: location missing publisher (link)
- Hardy, G. H.; Littlewood, J. E.; Pólya, G. (1952). Inequalities (2nd ed.). Cambridge University Press. ISBN 0-521-35880-9.
- 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.