Hardy–Littlewood inequality

The layer cake representation[1][2] allows us to write the general functions ${\displaystyle f}$ and ${\displaystyle g}$ in the form

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

where ${\displaystyle r\mapsto \chi _{f(x)>r}}$ equals ${\displaystyle 1}$ for ${\displaystyle r<f(x)}$ and ${\displaystyle 0}$ otherwise. Analogously, ${\displaystyle s\mapsto \chi _{g(x)>s}}$ equals ${\displaystyle 1}$ for ${\displaystyle s<g(x)}$ and ${\displaystyle 0}$ otherwise.

Now the proof can be obtained by first using Fubini's theorem to interchange the order of integration. When integrating with respect to ${\displaystyle x\in \mathbb {R} ^{n}}$ the conditions ${\displaystyle f(x)>r}$ and ${\displaystyle g(x)>s}$ the indicator functions ${\displaystyle x\mapsto \chi _{E_{f}(r)}(x)}$ and ${\displaystyle x\mapsto \chi _{E_{g}(s)}(x)}$ appear with the superlevel sets ${\displaystyle E_{f}(r)}$ and ${\displaystyle E_{g}(s)}$ as introduced above:

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

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

Denoting by ${\displaystyle \mu }$ the ${\displaystyle n}$-dimensional Lebesgue measure we continue by estimating the volume of the intersection by the minimum of the volumes of the two sets. Then, we can use the equality of the volumes of the superlevel sets for the rearrangements:

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

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

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

Now, we use that the superlevel sets ${\displaystyle E_{f^{*}}(r)}$ and ${\displaystyle E_{g^{*}}(s)}$ are balls in ${\displaystyle \mathbb {R} ^{n}}$ centered at ${\displaystyle x=0}$, which implies that ${\displaystyle E_{f^{*}}(r)\,\cap \,E_{g^{*}}(s)}$ is exactly the smaller one of the two balls:

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

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

The last identity follows by reversing the initial five steps that even work for general functions. This finishes the proof.

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