Hardy–Littlewood inequality

In mathematical analysis, the Hardy–Littlewood inequality, named after G. H. Hardy and John Edensor Littlewood, states that if f and g are nonnegative measurable real functions vanishing at infinity that are defined on n-dimensional Euclidean space Rⁿ then

${\displaystyle \int _{\mathbb {R} ^{n}}f(x)g(x)\,dx\leq \int _{\mathbb {R} ^{n}}f^{*}(x)g^{*}(x)\,dx}$

where f^* and g^* are the symmetric decreasing rearrangements of f(x) and g(x), respectively.^[1][2]

Proof[]

From layer cake representation we have:^[1][2]

${\displaystyle f(x)=\int _{0}^{\infty }\chi _{f(x)>r}\,dr}$

${\displaystyle g(x)=\int _{0}^{\infty }\chi _{g(x)>s}\,ds}$

where ${\displaystyle \chi _{f(x)>r}}$ denotes the indicator function of the subset E _f given by

${\displaystyle E_{f}=\left\{x\in X:f(x)>r\right\}}$

Analogously, ${\displaystyle \chi _{g(x)>s}}$ denotes the indicator function of the subset E _g given by

${\displaystyle E_{g}=\left\{x\in X:g(x)>s\right\}}$

${\displaystyle \int _{\mathbb {R} ^{n}}f(x)g(x)\,dx=\displaystyle \int _{\mathbb {R} ^{n}}\int _{0}^{\infty }\int _{0}^{\infty }\chi _{f(x)>r}\chi _{g(x)>s}\,dr\,ds\,dx}$

${\displaystyle =\int _{0}^{\infty }\int _{0}^{\infty }\int _{\mathbb {R} ^{n}}\chi _{f(x)>r\cap g(x)>s}\,dx\,dr\,ds}$

${\displaystyle =\int _{0}^{\infty }\int _{0}^{\infty }\mu \left(\left\{f(x)>r\right\}\cap \left\{g(x)>s\right\}\right)\,dr\,ds}$

${\displaystyle \leq \int _{0}^{\infty }\int _{0}^{\infty }\min \left(\mu \left(f(x)>r\right);\mu \left(g(x)>s\right)\right)\,dr\,ds}$

${\displaystyle =\int _{0}^{\infty }\int _{0}^{\infty }\min \left(\mu \left(f^{*}(x)>r\right);\mu \left(g^{*}(x)>s\right)\right)\,dr\,ds}$

${\displaystyle =\int _{0}^{\infty }\int _{0}^{\infty }\mu \left(\left\{f^{\ast }(x)>r\right\}\cap \left\{g^{\ast }(x)>s\right\}\right)\,dr\,ds}$

${\displaystyle =\int _{\mathbb {R} ^{n}}f^{*}(x)g^{*}(x)\,dx}$

An application[]

Let random variable ${\displaystyle X}$ is Normally distributed with mean ${\displaystyle \mu }$ and finite non-zero variance ${\displaystyle \sigma ^{2}}$, then using the Hardy–Littlewood inequality, it can be proved that for ${\displaystyle 0<\delta <1}$ the ${\displaystyle \delta ^{\text{th}}}$ reciprocal moment for the absolute value of ${\displaystyle X}$ is

${\displaystyle {\begin{aligned}\operatorname {E} \left[{\frac {1}{\vert X\vert ^{\delta }}}\right]&\leq 2^{\frac {(1-\delta )}{2}}{\frac {\Gamma \left({\frac {1-\delta }{2}}\right)}{\sigma ^{\delta }{\sqrt {2\pi }}}}{\text{ irrespective of the value of }}\mu \in \mathbb {R} .\end{aligned}}}$^[3]

The technique that is used to obtain the above property of the Normal distribution can be utilized for other unimodal distributions.

See also[]

• Rearrangement inequality
• Chebyshev's sum inequality
• Lorentz space

References[]