Hardy's inequality

Hardy's inequality is an inequality in mathematics, named after G. H. Hardy. It states that if [math]\displaystyle{ a_1, a_2, a_3, \dots }[/math] is a sequence of non-negative real numbers, then for every real number p > 1 one has

[math]\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. }[/math]

If the right-hand side is finite, equality holds if and only if [math]\displaystyle{ a_n = 0 }[/math] for all n.

An integral version of Hardy's inequality states the following: if f is a measurable function with non-negative values, then

[math]\displaystyle{ \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. }[/math]

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 [math]\displaystyle{ \alpha + \tfrac{1}{p} \lt 1 }[/math], then

[math]\displaystyle{ \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 }[/math]

• If [math]\displaystyle{ \alpha + \tfrac{1}{p} \gt 1 }[/math], then

[math]\displaystyle{ \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. }[/math]

Multidimensional versions

Multidimensional Hardy inequality around a point

In the multidimensional case, Hardy's inequality can be extended to [math]\displaystyle{ L^{p} }[/math]-spaces, taking the form

[math]\displaystyle{ \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\lt n, }[/math]

where [math]\displaystyle{ f\in C_{0}^{\infty}(\mathbb{R}^{n}) }[/math], and where the constant [math]\displaystyle{ \frac{p}{n-p} }[/math] is known to be sharp; by density it extends then to the Sobolev space [math]\displaystyle{ W^{1, p} (\mathbb{R}^n) }[/math].

Similarly, if [math]\displaystyle{ p \gt n \ge 2 }[/math], then one has for every [math]\displaystyle{ f\in C_{0}^{\infty}(\mathbb{R}^{n}) }[/math]

[math]\displaystyle{ \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. }[/math]

Multidimensional Hardy inequality near the boundary

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

[math]\displaystyle{ \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, }[/math]

and the constant cannot be improved.

Fractional Hardy inequality

If [math]\displaystyle{ 1 \le p \lt \infty }[/math] and [math]\displaystyle{ 0 \lt \lambda \lt \infty }[/math], [math]\displaystyle{ \lambda \ne 1 }[/math], there exists a constant [math]\displaystyle{ C }[/math] such that for every [math]\displaystyle{ f : (0, \infty) \to \mathbb{R} }[/math] satisfying [math]\displaystyle{ \int_0^\infty \vert f (x)\vert^p/x^{\lambda} \,dx \lt \infty }[/math], one has

[math]\displaystyle{ \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. }[/math]

Proof of the inequality

Integral version

A change of variables gives

[math]\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}, }[/math]

which is less or equal than [math]\displaystyle{ \int_0^1\left(\int_0^\infty f(sx)^p\,dx\right)^{1/p}\,ds }[/math] by Minkowski's integral inequality. Finally, by another change of variables, the last expression equals

[math]\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}. }[/math]

Discrete version: from the continuous version

Assuming the right-hand side to be finite, we must have [math]\displaystyle{ a_n\to 0 }[/math] as [math]\displaystyle{ n\to\infty }[/math]. Hence, for any positive integer j, there are only finitely many terms bigger than [math]\displaystyle{ 2^{-j} }[/math]. This allows us to construct a decreasing sequence [math]\displaystyle{ b_1\ge b_2\ge\dotsb }[/math] containing the same positive terms as the original sequence (but possibly no zero terms). Since [math]\displaystyle{ a_1+a_2+\dotsb +a_n\le b_1+b_2+\dotsb +b_n }[/math] for every n, it suffices to show the inequality for the new sequence. This follows directly from the integral form, defining [math]\displaystyle{ f(x)=b_n }[/math] if [math]\displaystyle{ n-1\lt x\lt n }[/math] and [math]\displaystyle{ f(x)=0 }[/math] otherwise. Indeed, one has

[math]\displaystyle{ \int_0^\infty f(x)^p\,dx=\sum_{n=1}^\infty b_n^p }[/math]

and, for [math]\displaystyle{ n-1\lt x\lt n }[/math], there holds

[math]\displaystyle{ \frac{1}{x}\int_0^x f(t)\,dt=\frac{b_1+\dots+b_{n-1}+(x-n+1)b_n}{x} \ge \frac{b_1+\dots+b_n}{n} }[/math]

(the last inequality is equivalent to [math]\displaystyle{ (n-x)(b_1+\dots+b_{n-1})\ge (n-1)(n-x)b_n }[/math], which is true as the new sequence is decreasing) and thus

[math]\displaystyle{ \sum_{n=1}^\infty\left(\frac{b_1+\dots+b_n}{n}\right)^p\le\int_0^\infty\left(\frac{1}{x}\int_0^x f(t)\,dt\right)^p\,dx }[/math].

Discrete version: Direct proof

Let [math]\displaystyle{ p \gt 1 }[/math] and let [math]\displaystyle{ b_1 , \dots , b_n }[/math] be positive real numbers. Set [math]\displaystyle{ S_k = \sum_{i=1}^k b_i }[/math]. First we prove the inequality

[math]\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}}, }[/math]

(*)

Let [math]\displaystyle{ T_n = \frac{S_n}{n} }[/math] and let [math]\displaystyle{ \Delta_n }[/math] be the difference between the [math]\displaystyle{ n }[/math]-th terms in the right-hand side and left-hand side of *, that is, [math]\displaystyle{ \Delta_n := T_n^p - \frac{p}{p-1} b_n T_n^{p-1} }[/math]. We have:

[math]\displaystyle{ \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} }[/math]

or

[math]\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 . }[/math]

According to Young's inequality we have:

[math]\displaystyle{ T_{n-1} T_n^{p-1} \leq \frac{T_{n-1}^p}{p} + (p-1) \frac{T_n^p}{p} , }[/math]

from which it follows that:

[math]\displaystyle{ \Delta_n \leq \frac{n-1}{p-1} T_{n-1}^p - \frac{n}{p-1} T_n^p . }[/math]

By telescoping we have:

[math]\displaystyle{ \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 \lt 0 , \end{align} }[/math]

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

[math]\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} }[/math]

from which we immediately obtain:

[math]\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 . }[/math]

Letting [math]\displaystyle{ N \rightarrow \infty }[/math] 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, 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

• Hazewinkel, Michiel, ed. (2001), "Hardy inequality", Encyclopedia of Mathematics, Springer Science+Business Media B.V. / Kluwer Academic Publishers, ISBN 978-1-55608-010-4, https://www.encyclopediaofmath.org/index.php?title=p/h046340