Hardy's inequality is an inequality in mathematics, named after G. H. Hardy. Its discrete version states that if a 1 , a 2 , a 3 , … {\displaystyle a_{1},a_{2},a_{3},\dots } is a sequence of non-negative real numbers, then for every real number p > 1 one has
∑ n = 1 ∞ ( a 1 + a 2 + ⋯ + a n n ) p ≤ ( p p − 1 ) p ∑ n = 1 ∞ a n p . {\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}.}
If the right-hand side is finite, equality holds if and only if a n = 0 {\displaystyle 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
∫ 0 ∞ ( 1 x ∫ 0 x f ( t ) d t ) p d x ≤ ( p p − 1 ) p ∫ 0 ∞ f ( x ) p d x . {\displaystyle \int _{0}^{\infty }\left({\frac {1}{x}}\int _{0}^{x}f(t)\,dt\right)^{p}\,dx\leq \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.
Statements
General discrete Hardy inequality The general weighted one dimensional version reads as follows: if a n ≥ 0 {\displaystyle a_{n}\geq 0} , λ n > 0 {\displaystyle \lambda _{n}>0} and p > 1 {\displaystyle p>1} ,
∑ n = 1 ∞ λ n ( λ 1 a 1 + ⋯ + λ n a n λ 1 + ⋯ + λ n ) p ≤ ( p p − 1 ) p ∑ n = 1 ∞ λ n a n p . {\displaystyle \sum _{n=1}^{\infty }\lambda _{n}{\Bigl (}{\frac {\lambda _{1}a_{1}+\dotsb +\lambda _{n}a_{n}}{\lambda _{1}+\dotsb +\lambda _{n}}}{\Bigr )}^{p}\leq {\Bigl (}{\frac {p}{p-1}}{\Bigr )}^{p}\sum _{n=1}^{\infty }\lambda _{n}a_{n}^{p}.}
General one-dimensional integral Hardy inequality The general weighted one dimensional version reads as follows:
If α + 1 p < 1 {\displaystyle \alpha +{\tfrac {1}{p}}<1} , then
∫ 0 ∞ ( y α − 1 ∫ 0 y x − α f ( x ) d x ) p d y ≤ 1 ( 1 − α − 1 p ) p ∫ 0 ∞ f ( x ) p d x {\displaystyle \int _{0}^{\infty }{\biggl (}y^{\alpha -1}\int _{0}^{y}x^{-\alpha }f(x)\,dx{\biggr )}^{p}\,dy\leq {\frac {1}{{\bigl (}1-\alpha -{\frac {1}{p}}{\bigr )}^{p}}}\int _{0}^{\infty }f(x)^{p}\,dx}
If α + 1 p > 1 {\displaystyle \alpha +{\tfrac {1}{p}}>1} , then
∫ 0 ∞ ( y α − 1 ∫ y ∞ x − α f ( x ) d x ) p d y ≤ 1 ( α + 1 p − 1 ) p ∫ 0 ∞ f ( x ) p d x . {\displaystyle \int _{0}^{\infty }{\biggl (}y^{\alpha -1}\int _{y}^{\infty }x^{-\alpha }f(x)\,dx{\biggr )}^{p}\,dy\leq {\frac {1}{{\bigl (}\alpha +{\frac {1}{p}}-1{\bigr )}^{p}}}\int _{0}^{\infty }f(x)^{p}\,dx.}
The more general case is weighted integral Hardy inequality
Weighted one-dimensional integral Hardy inequality The statement is following: if 1 < p ≤ q < ∞ {\displaystyle 1<p\leq q<\infty } inequality
( ∫ a b ( ∫ a x f ( t ) d t ) q u ( x ) d x ) 1 / q ≤ C ( ∫ a b f p ( x ) v ( x ) d x ) 1 / p {\displaystyle \left(\int _{a}^{b}{\left(\int _{a}^{x}{f(t)}dt\right)^{q}u(x)}dx\right)^{1/q}\leq C\left(\int _{a}^{b}{f^{p}(x)v(x)}dx\right)^{1/p}}
holds for − ∞ ≤ a < b ≤ ∞ {\displaystyle -\infty \leq a<b\leq \infty } with u , v {\displaystyle u,v} measurable, positive on ( a , b ) {\displaystyle (a,b)} and also f ( x ) ≥ 0 {\displaystyle f(x)\geq 0} any measurable on ( a , b ) {\displaystyle (a,b)} , if and only if
A = sup x ∈ ( a , b ) ( ∫ x b u ( t ) d t ) 1 / q ( ∫ a x v 1 − p ′ ( t ) d t ) 1 / p ′ < ∞ {\displaystyle A=\sup _{x\in (a,b)}{\left(\int _{x}^{b}u(t)dt\right)^{1/q}\left(\int _{a}^{x}v^{1-p'}(t)dt\right)^{1/p'}}<\infty }
And if 1 < q < p < ∞ {\displaystyle 1<q<p<\infty } inequality also holds if and only if
A = ( ∫ a b ( ∫ x b u ( t ) d t ) r / q ( ∫ a x v 1 − p ′ ( t ) d t ) r / q ′ v 1 − p ′ ( x ) d x ) 1 / r < ∞ {\displaystyle A=\left(\int _{a}^{b}{{\left(\int _{x}^{b}u(t)dt\right)^{r/q}\left(\int _{a}^{x}v^{1-p'}(t)dt\right)^{r/q'}}v^{1-p'}(x)dx}\right)^{1/r}<\infty }
where 1 r = 1 q − 1 p , 1 p + 1 p ′ = 1 {\displaystyle {\frac {1}{r}}={\frac {1}{q}}-{\frac {1}{p}},{\frac {1}{p}}+{\frac {1}{p'}}=1} and 1 q + 1 q ′ = 1 {\displaystyle {\frac {1}{q}}+{\frac {1}{q'}}=1}
Multidimensional Hardy inequalities with gradient
Multidimensional Hardy inequality around a point In the multidimensional case, Hardy's inequality can be extended to L p {\displaystyle L^{p}} -spaces, taking the form
‖ f | ⋅ | ‖ L p ( R n ) ≤ p n − p ‖ ∇ f ‖ L p ( R n ) , 2 ≤ n , 1 ≤ p < n , {\displaystyle \left\|{\frac {f}{|\cdot |}}\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 f ∈ C c ∞ ( R n ) {\displaystyle f\in C_{c}^{\infty }(\mathbb {R} ^{n})} , and where the constant p n − p {\displaystyle {\frac {p}{n-p}}} is known to be sharp; by density it extends then to the Sobolev space W 1 , p ( R n ) {\displaystyle W^{1,p}(\mathbb {R} ^{n})} . Similarly, if p > n ≥ 2 {\displaystyle p>n\geq 2} , then one has for every f ∈ C c ∞ ( R n ) {\displaystyle f\in C_{c}^{\infty }(\mathbb {R} ^{n})}
( 1 − n p ) p ∫ R n | f ( x ) − f ( 0 ) | p | x | p d x ≤ ∫ R n | ∇ f | p . {\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 Ω ⊊ R n {\displaystyle \Omega \subsetneq \mathbb {R} ^{n}} is an nonempty convex open set, then for 1 < p < ∞ {\displaystyle 1<p<\infty } and every f ∈ W 0 1 , p ( Ω ) {\displaystyle f\in W_{0}^{1,p}(\Omega )} ,
( 1 − 1 p ) p ∫ Ω | f ( x ) | p dist ( x , ∂ Ω ) p d x ≤ ∫ Ω | ∇ f | p , {\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.
Fractional Hardy inequality If 1 ≤ p < ∞ {\displaystyle 1\leq p<\infty } and 0 < λ < ∞ {\displaystyle 0<\lambda <\infty } , λ ≠ 1 {\displaystyle \lambda \neq 1} , there exists a constant C {\displaystyle C} such that for every f : ( 0 , ∞ ) → R {\displaystyle f:(0,\infty )\to \mathbb {R} } satisfying ∫ 0 ∞ | f ( x ) | p / x λ d x < ∞ {\displaystyle \int _{0}^{\infty }\vert f(x)\vert ^{p}/x^{\lambda }\,dx<\infty } , one has
∫ 0 ∞ | f ( x ) | p x λ d x ≤ C ∫ 0 ∞ ∫ 0 ∞ | f ( x ) − f ( y ) | p | x − y | 1 + λ d x d y . {\displaystyle \int _{0}^{\infty }{\frac {\vert f(x)\vert ^{p}}{x^{\lambda }}}\,dx\leq C\int _{0}^{\infty }\int _{0}^{\infty }{\frac {\vert f(x)-f(y)\vert ^{p}}{\vert x-y\vert ^{1+\lambda }}}\,dx\,dy.}
Proof of the inequality
Integral version (integration by parts and Hölder) Hardy’s original proof begins with an integration by parts to get
∫ 0 ∞ ( 1 x ∫ 0 x f ( t ) d t ) p d x = ∫ 0 ∞ ( ∫ 0 x f ( t ) d t ) p 1 x p d x = p p − 1 ∫ 0 ∞ ( ∫ 0 x f ( t ) d t ) p − 1 f ( x ) x p − 1 d x = p p − 1 ∫ 0 ∞ ( 1 x ∫ 0 x f ( t ) d t ) p − 1 f ( x ) d x {\displaystyle {\begin{aligned}\int _{0}^{\infty }\left({\frac {1}{x}}\int _{0}^{x}f(t)\,dt\right)^{p}dx&=\int _{0}^{\infty }\left(\int _{0}^{x}f(t)\,dt\right)^{p}{\frac {1}{x^{p}}}dx\\[0.2em]&={\frac {p}{p-1}}\int _{0}^{\infty }\left(\int _{0}^{x}f(t)\,dt\right)^{p-1}{\frac {f(x)}{x^{p-1}}}dx\\[0.2em]&={\frac {p}{p-1}}\int _{0}^{\infty }\left({\frac {1}{x}}\int _{0}^{x}f(t)\,dt\right)^{p-1}f(x)dx\end{aligned}}}
Then, by Hölder's inequality,
∫ 0 ∞ ( 1 x ∫ 0 x f ( t ) d t ) p d x ≤ p p − 1 ( ∫ 0 ∞ ( 1 x ∫ 0 x f ( t ) d t ) p d x ) 1 − 1 p ( ∫ 0 ∞ f ( x ) p d x ) 1 p , {\displaystyle \int _{0}^{\infty }\left({\frac {1}{x}}\int _{0}^{x}f(t)\,dt\right)^{p}dx\leq {\frac {p}{p-1}}\left(\int _{0}^{\infty }\left({\frac {1}{x}}\int _{0}^{x}f(t)\,dt\right)^{p}dx\right)^{1-{\frac {1}{p}}}\left(\int _{0}^{\infty }f(x)^{p}\,dx\right)^{\frac {1}{p}},}
and the conclusion follows.
Integral version (scaling and Minkowski) A change of variables gives
( ∫ 0 ∞ ( 1 x ∫ 0 x f ( t ) d t ) p d x ) 1 / p = ( ∫ 0 ∞ ( ∫ 0 1 f ( s x ) d s ) p d x ) 1 / p , {\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 ∫ 0 1 ( ∫ 0 ∞ f ( s x ) p d x ) 1 / p d s {\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