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


 * $$\int_0^\infty \left (\frac{1}{x}\int_0^x f(t)\, dt\right)^p\, dx\le\left (\frac{p}{p-1}\right )^p\int_0^\infty f(x)^p\, dx.$$

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 \le

\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 \le

\frac{1}{\bigl(\alpha + \frac{1}{p} - 1\bigr)^p} \int_0^\infty f(x)^p\, dx. $$

Multidimensional Hardy inequality around a point
In the multidimensional case, Hardy's inequality can be extended to $$L^{p}$$-spaces, taking the form


 * $$\left\|\frac{f}{|x|}\right\|_{L^{p}(\mathbb{R}^{n})}\le \frac{p}{n-p}\|\nabla f\|_{L^{p}(\mathbb{R}^{n})}, 2\le n, 1\le p n \ge 2$$, 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 \le \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 \le \int_{\Omega}\vert \nabla f \vert^p, $$ and the constant cannot be improved.

Fractional Hardy inequality
If $$1 \le p < \infty$$ and $$0 < \lambda < \infty$$, $$\lambda \ne 1$$, there exists a constant $$C$$ such that for every $$f : (0, \infty) \to \mathbb{R}$$ 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 \le C \int_0^\infty \int_0^\infty \frac{\vert f (x) - f (y)\vert^p}{\vert x - y\vert^{1+\lambda}} \,dx \, dy. $$

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 $$b_1\ge b_2\ge\dotsb$$ containing the same positive terms as the original sequence (but possibly no zero terms). Since $$a_1+a_2+\dotsb +a_n\le b_1+b_2+\dotsb +b_n$$ 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 $$n-1 1$$ 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

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:


 * $$\Delta_n = T_n^p - \frac{p}{p-1} b_n T_n^{p-1} = T_n^p - \frac{p}{p-1} (n T_n - (n-1) T_{n-1}) T_n^{p-1}$$

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{align}

\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{align} $$

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.