Hardy's inequality is an inequality in mathematics, named after G. H. Hardy. It states that if
a1,a2,a3,...
infty | |
\sum | |
n=1 |
\left(
a1+a2+ … +an | |
n |
\right)p\leq\left(
p | |
p-1 |
\right
infty | |
) | |
n=1 |
p. | |
a | |
n |
an=0
An integral version of Hardy's inequality states the following: if f is a measurable function with non-negative values, then
infty | |
\int | |
0 |
\left(
1 | |
x |
x | |
\int | |
0 |
f(t)dt\right)pdx\le\left(
p | |
p-1 |
\right
infty | |
) | |
0 |
f(x)pdx.
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.
The general weighted one dimensional version reads as follows:[2]
\alpha+\tfrac{1}{p}<1
infty | |
\int | |
0 |
l(y\alpha
y | |
\int | |
0 |
x-\alphaf(x)dxr)pdy\le
1 | |||||
|
infty | |
\int | |
0 |
f(x)pdx
\alpha+\tfrac{1}{p}>1
infty | |
\int | |
0 |
l(y\alpha
infty | |
\int | |
y |
x-\alphaf(x)dxr)pdy\le
1 | |||||
|
infty | |
\int | |
0 |
f(x)pdx.
In the multidimensional case, Hardy's inequality can be extended to
Lp
\left\| | f |
|x| |
\right\| | |
Lp(Rn) |
\le
p | |
n-p |
\|\nabla
f\| | |
Lp(Rn) |
,2\len,1\lep<n,
f\in
infty | |
C | |
0 |
(Rn)
p | |
n-p |
W1,(Rn)
Similarly, if
p>n\ge2
f\in
infty | |
C | |
0 |
(Rn)
(1-
n | |
p |
)p
\int | |
Rn |
\vertf(x)-f(0)\vertp | |
|x|p |
dx\le
\int | |
Rn |
\vert\nablaf\vertp.
If
\Omega\subsetneqRn
f\inW1,(\Omega)
(1-
1 | |
p |
p\int | |
) | |
\Omega |
\vertf(x)\vertp | |
\operatorname{dist |
(x,\partial\Omega)p}dx \le\int\Omega\vert\nablaf\vertp,
If
1\lep<infty
0<λ<infty
λ\ne1
C
f:(0,infty)\toR
infty | |
\int | |
0 |
\vertf(x)\vertp/xλdx<infty
infty | |
\int | |
0 |
\vertf(x)\vertp | |
xλ |
dx\leC
infty | |
\int | |
0 |
infty | |
\int | |
0 |
\vertf(x)-f(y)\vertp | |
\vertx-y\vert1+λ |
dxdy.
A change of variables gives
| ||||
\left(\int | ||||
0 |
x | |
\int | |
0 |
f(t)dt\right)p dx\right)1/p
1 | |
=\left(\int | |
0 |
f(sx)ds\right)pdx\right)1/p,
infty | |
\int | |
0 |
f(sx)pdx\right)1/pds
infty | |
\int | |
0 |
f(x)pdx\right)1/ps-1/pds=
p | |
p-1 |
infty | |
\left(\int | |
0 |
f(x)pdx\right)1/p.
Assuming the right-hand side to be finite, we must have
an\to0
n\toinfty
2-j
b1\geb2\ge...b
a1+a2+...b+an\leb1+b2+...b+bn
f(x)=bn
n-1<x<n
f(x)=0
infty | |
\int | |
0 |
infty | |
f(x) | |
n=1 |
p | |
b | |
n |
n-1<x<n
1 | |
x |
x | ||
\int | f(t)dt= | |
0 |
b1+...+bn-1+(x-n+1)bn | |
x |
\ge
b1+...+bn | |
n |
(n-x)(b1+...+bn-1)\ge(n-1)(n-x)bn
| ||||
\sum | ||||
n=1 |
| ||||
\right) | ||||
0 |
x | |
\int | |
0 |
f(t)dt\right)pdx
Let
p>1
b1,...,bn
Sk=
k | |
\sum | |
i=1 |
bi
Let
Tn=
Sn | |
n |
\Deltan
n
\Deltan:=
p | |
T | |
n |
-
p | |
p-1 |
bn
p-1 | |
T | |
n |
\Deltan=
p | |
T | |
n |
-
p | |
p-1 |
bn
p-1 | |
T | |
n |
=
p | |
T | |
n |
-
p | |
p-1 |
(nTn-(n-1)Tn-1)
p-1 | |
T | |
n |
or
\Deltan=
p | |
T | |
n |
\left(1-
np | |
p-1 |
\right)+
p(n-1) | |
p-1 |
Tn-1
p | |
T | |
n |
.
According to Young's inequality we have:
Tn-1
p-1 | |
T | |
n |
\leq
| |||||||
p |
+(p-1)
| |||||||
p |
,
from which it follows that:
\Deltan\leq
n-1 | |
p-1 |
p | |
T | |
n-1 |
-
n | |
p-1 |
p | |
T | |
n |
.
By telescoping we have:
N | |
\begin{align} \sum | |
n=1 |
\Deltan&\leq0-
1 | |
p-1 |
p | |
T | |
1 |
+
1 | |
p-1 |
p | |
T | |
1 |
-
2 | |
p-1 |
p | |
T | |
2 |
+
2 | |
p-1 |
p | |
T | |
2 |
-
3 | |
p-1 |
p | |
T | |
3 |
+...b+
N-1 | |
p-1 |
p | |
T | |
N-1 |
-
N | |
p-1 |
p | |
T | |
N |
\\ &=-
N | |
p-1 |
p | |
T | |
N |
<0, \end{align}
proving . Applying Hölder's inequality to the right-hand side of we have:
N | |
\sum | |
n=1 |
| |||||||
np |
\leq
p | |
p-1 |
N | |
\sum | |
n=1 |
| |||||||||||||
np-1 |
\leq
p | |
p-1 |
\left(
N | |
\sum | |
n=1 |
p | |
b | |
n |
\right)1/p\left(
N | |
\sum | |
n=1 |
| |||||||
np |
\right)(p-1)/p
from which we immediately obtain:
N | |
\sum | |
n=1 |
| |||||||
np |
\leq\left(
p | |
p-1 |
\right)p
N | |
\sum | |
n=1 |
p | |
b | |
n |
.
Letting
N → infty