Multidimensional Hardy inequality around a point

In the multidimensional case, Hardy's inequality can be extended to ${\displaystyle L^{p}}$-spaces, taking the form ^[3]

${\displaystyle \left\|{\frac {f}{|x|}}\right\|_{L^{p}(\mathbb {R} ^{n})}\leq {\frac {p}{n-p}}\|\nabla f\|_{L^{p}(\mathbb {R} ^{n})},2\leq n,1\leq p<n,}$

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

Similarly, if ${\displaystyle p>n\geq 2}$, then one has for every ${\displaystyle f\in C_{0}^{\infty }(\mathbb {R} ^{n})}$

${\displaystyle {\Big (}1-{\frac {n}{p}}{\Big )}^{p}\int _{\mathbb {R} ^{n}}{\frac {\vert f(x)-f(0)\vert ^{p}}{|x|^{p}}}dx\leq \int _{\mathbb {R} ^{n}}\vert \nabla f\vert ^{p}.}$

Multidimensional Hardy inequality near the boundary

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

${\displaystyle {\Big (}1-{\frac {1}{p}}{\Big )}^{p}\int _{\Omega }{\frac {\vert f(x)\vert ^{p}}{\operatorname {dist} (x,\partial \Omega )^{p}}}\,dx\leq \int _{\Omega }\vert \nabla f\vert ^{p},}$

and the constant cannot be improved.^[4]

Integral version

A change of variables gives

${\displaystyle \left(\int _{0}^{\infty }\left({\frac {1}{x}}\int _{0}^{x}f(t)\,dt\right)^{p}\ dx\right)^{1/p}=\left(\int _{0}^{\infty }\left(\int _{0}^{1}f(sx)\,ds\right)^{p}\,dx\right)^{1/p},}$

which is less or equal than ${\displaystyle \int _{0}^{1}\left(\int _{0}^{\infty }f(sx)^{p}\,dx\right)^{1/p}\,ds}$ by Minkowski's integral inequality. Finally, by another change of variables, the last expression equals

${\displaystyle \int _{0}^{1}\left(\int _{0}^{\infty }f(x)^{p}\,dx\right)^{1/p}s^{-1/p}\,ds={\frac {p}{p-1}}\left(\int _{0}^{\infty }f(x)^{p}\,dx\right)^{1/p}.}$

Discrete version: from the continuous version

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

${\displaystyle \int _{0}^{\infty }f(x)^{p}\,dx=\sum _{n=1}^{\infty }b_{n}^{p}}$

and, for ${\displaystyle n-1<x<n}$, there holds

${\displaystyle {\frac {1}{x}}\int _{0}^{x}f(t)\,dt={\frac {b_{1}+\dots +b_{n-1}+(x-n+1)b_{n}}{x}}\geq {\frac {b_{1}+\dots +b_{n}}{n}}}$

(the last inequality is equivalent to ${\displaystyle (n-x)(b_{1}+\dots +b_{n-1})\geq (n-1)(n-x)b_{n}}$, which is true as the new sequence is decreasing) and thus

${\displaystyle \sum _{n=1}^{\infty }\left({\frac {b_{1}+\dots +b_{n}}{n}}\right)^{p}\leq \int _{0}^{\infty }\left({\frac {1}{x}}\int _{0}^{x}f(t)\,dt\right)^{p}\,dx}$.

Discrete version: Direct proof

Let ${\displaystyle p>1}$ and let ${\displaystyle b_{1},\dots ,b_{n}}$ be positive real numbers. Set ${\displaystyle S_{k}=\sum _{i=1}^{k}b_{i}}$. First we prove the inequality

${\displaystyle \sum _{n=1}^{N}{\frac {S_{n}^{p}}{n^{p}}}\leq {\frac {p}{p-1}}\sum _{n=1}^{N}{\frac {b_{n}S_{n}^{p-1}}{n^{p-1}}},}$

(*)

Let ${\displaystyle T_{n}={\frac {S_{n}}{n}}}$ and let ${\displaystyle \Delta _{n}}$ be the difference between the ${\displaystyle n}$-th terms in the right-hand side and left-hand side of *, that is, ${\displaystyle \Delta _{n}:=T_{n}^{p}-{\frac {p}{p-1}}b_{n}T_{n}^{p-1}}$. We have:

${\displaystyle \Delta _{n}=T_{n}^{p}-{\frac {p}{p-1}}b_{n}T_{n}^{p-1}=T_{n}^{p}-{\frac {p}{p-1}}(nT_{n}-(n-1)T_{n-1})T_{n}^{p-1}}$

or

${\displaystyle \Delta _{n}=T_{n}^{p}\left(1-{\frac {np}{p-1}}\right)+{\frac {p(n-1)}{p-1}}T_{n-1}T_{n}^{p}.}$

According to Young's inequality we have:

${\displaystyle T_{n-1}T_{n}^{p-1}\leq {\frac {T_{n-1}^{p}}{p}}+(p-1){\frac {T_{n}^{p}}{p}},}$

from which it follows that:

${\displaystyle \Delta _{n}\leq {\frac {n-1}{p-1}}T_{n-1}^{p}-{\frac {n}{p-1}}T_{n}^{p}.}$

By telescoping we have:

${\displaystyle {\begin{aligned}\sum _{n=1}^{N}\Delta _{n}&\leq 0-{\frac {1}{p-1}}T_{1}^{p}+{\frac {1}{p-1}}T_{1}^{p}-{\frac {2}{p-1}}T_{2}^{p}+{\frac {2}{p-1}}T_{2}^{p}-{\frac {3}{p-1}}T_{3}^{p}+\dotsb +{\frac {N-1}{p-1}}T_{N-1}^{p}-{\frac {N}{p-1}}T_{N}^{p}\\&=-{\frac {N}{p-1}}T_{N}^{p}<0,\end{aligned}}}$

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

${\displaystyle \sum _{n=1}^{N}{\frac {S_{n}^{p}}{n^{p}}}\leq {\frac {p}{p-1}}\sum _{n=1}^{N}{\frac {b_{n}S_{n}^{p-1}}{n^{p-1}}}\leq {\frac {p}{p-1}}\left(\sum _{n=1}^{N}b_{n}^{p}\right)^{1/p}\left(\sum _{n=1}^{N}{\frac {S_{n}^{p}}{n^{p}}}\right)^{(p-1)/p}}$

from which we immediately obtain:

${\displaystyle \sum _{n=1}^{N}{\frac {S_{n}^{p}}{n^{p}}}\leq \left({\frac {p}{p-1}}\right)^{p}\sum _{n=1}^{N}b_{n}^{p}.}$

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