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

:$\sum_^\infty \left (\frac\right )^p\leq\left (\frac\right )^p\sum_^\infty a_n^p.$

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

:$\int_0^\infty \left (\frac\int_0^x f(t)\, dt\right)^p\, dx\le\left (\frac\right )^p\int_0^\infty f(x)^p\, dx.$

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 $\alpha + \tfrac < 1$, then
:$\int_0^\infty \biggl(y^ \int_0^y x^ f(x)\,dx \biggr)^p \,dy \le \frac \int_0^\infty f(x)^p\, dx$
* If $\alpha + \tfrac > 1$, then
:$\int_0^\infty \biggl(y^ \int_y^\infty x^ f(x)\,dx \biggr)^p\,dy \le \frac \int_0^\infty f(x)^p\, dx.$

Multidimensional version

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

:\left\, \frac\right\, _\le \frac\, \nabla f\, _, 2\le n, 1\le p
where $f\in C_^(R^)$, and where the constant $\frac$ is known to be sharp.

Fractional Hardy inequality

If $1 \le p < \infty$ and $0 < \lambda < \infty$, $\lambda \ne 1$, there exists a constant $C$ such that for every $f : (0, \infty) \to \mathbb$ satisfying $\int_0^\infty \vert f (x)\vert^p/x^ \,dx < \infty$, one has
:$\int_0^\infty \frac \,dx \le C \int_0^\infty \int_0^\infty \frac \,dx \, dy.$

Proof of the inequality

Integral version

A change of variables gives
$\left(\int_0^\infty\left(\frac\int_0^x f(t)\,dt\right)^p\ dx\right)^=\left(\int_0^\infty\left(\int_0^1 f(sx)\,ds\right)^p\,dx\right)^$,
which is less or equal than $\int_0^1\left(\int_0^\infty f(sx)^p\,dx\right)^\,ds$ by Minkowski's integral inequality. Finally, by another change of variables, the last expression equals
$\int_0^1\left(\int_0^\infty f(x)^p\,dx\right)^s^\,ds=\frac\left(\int_0^\infty f(x)^p\,dx\right)^$.

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 $2^$. 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 and $f(x)=0$ otherwise. Indeed, one has
$\int_0^\infty f(x)^p\,dx=\sum_^\infty b_n^p$
and, for n-1, there holds
$\frac\int_0^x f(t)\,dt=\frac \ge \frac$
(the last inequality is equivalent to $(n-x)(b_1+\dots+b_)\ge (n-1)(n-x)b_n$, which is true as the new sequence is decreasing) and thus
$\sum_^\infty\left(\frac\right)^p\le\int_0^\infty\left(\frac\int_0^x f(t)\,dt\right)^p\,dx$.

Discrete version: Direct proof

Let $p > 1$ and let $b_1 , \dots , b_n$ be positive real numbers. Set $S_k = \sum_^k b_i$ First we prove the inequality
$\sum_^N \frac \leq \frac \sum_^N \frac \quad (*)$,

Let $T_n = \frac$ and let $\Delta_n$ be the difference between the $n$-th terms in the RHS and LHS of $(*)$, that is, $\Delta_n := T_n^p - \frac b_n T_n^$. We have:

:$\Delta_n = T_n^p - \frac b_n T_n^ = T_n^p - \frac (n T_n - (n-1) T_) T_n^$

or

:$\Delta_n = T_n^p \left( 1 - \frac \right) + \frac T_ T_n^p .$

According to Young's inequality we have:

:$T_ T_n^ \leq \frac + (p-1) \frac ,$

from which it follows that:

:$\Delta_n \leq \frac T_^p - \frac T_n^p .$

By telescoping we have:

:$\begin \sum_^N \Delta_n &\leq 0 - \frac T_1^p \\ &+ \frac T_1^p - \frac T_2^p \\ &+ \frac T_2^p - \frac T_3^p \\ & \vdots \\ &+ \frac T_^p - \frac T_N^p \\ &= - \frac T_N^p < 0 , \end$

proving $(*)$. By applying Hölder's inequality to the RHS of $(*)$ we have:

:$\sum_^N \frac \leq \frac \sum_^N \frac \leq \frac \left( \sum_^N b_n^p \right)^ \left( \sum_^N \frac \right)^$

from which we immediately obtain:

:$\sum_^N \frac \leq \left( \frac \right)^p \sum_^N b_n^p .$

Letting $N \rightarrow \infty$ we obtain Hardy's inequality.

See also

* Carleman's inequality

Notes

References

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External links

* {{springer, title=Hardy inequality, id=p/h046340

Inequalities
Theorems in real analysis