Hardy's inequality is an inequality in mathematics , named after G. H. Hardy . It 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.[1] The original formulation was in an integral form slightly different from the above.
General one-dimensional version [ edit ] The general weighted one dimensional version reads as follows:[2] : §329
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.} Multidimensional versions [ edit ] Multidimensional Hardy inequality around a point [ edit ] In the multidimensional case, Hardy's inequality can be extended to L p {\displaystyle L^{p}} -spaces, taking the form [3]
‖ f | x | ‖ L p ( R n ) ≤ p n − p ‖ ∇ f ‖ L p ( R n ) , 2 ≤ n , 1 ≤ p < n , {\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 f ∈ C 0 ∞ ( R n ) {\displaystyle f\in C_{0}^{\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 0 ∞ ( R n ) {\displaystyle f\in C_{0}^{\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 [ edit ] If Ω ⊊ R n {\displaystyle \Omega \subsetneq \mathbb {R} ^{n}} is an nonempty convex open set, then for every f ∈ W 1 , p ( Ω ) {\displaystyle f\in W^{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.[4]
Fractional Hardy inequality [ edit ] 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[5] : Lemma 2
∫ 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 [ edit ] Integral version [ edit ] 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, the last expression equals
∫ 0 1 ( ∫ 0 ∞ f ( x ) p d x ) 1 / p s − 1 / p d s = p p − 1 ( ∫ 0 ∞ f ( x ) p d x ) 1 / p . {\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 [ edit ] Assuming the right-hand side to be finite, we must have a n → 0 {\displaystyle a_{n}\to 0} as n → ∞ {\displaystyle n\to \infty } . Hence, for any positive integer j , there are only finitely many terms bigger than 2 − j {\displaystyle 2^{-j}} . This allows us to construct a decreasing sequence b 1 ≥ b 2 ≥ ⋯ {\displaystyle b_{1}\geq b_{2}\geq \dotsb } containing the same positive terms as the original sequence (but possibly no zero terms). Since a 1 + a 2 + ⋯ + a n ≤ b 1 + b 2 + ⋯ + b n {\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 f ( x ) = b n {\displaystyle f(x)=b_{n}} if n − 1 < x < n {\displaystyle n-1<x<n} and f ( x ) = 0 {\displaystyle f(x)=0} otherwise. Indeed, one has
∫ 0 ∞ f ( x ) p d x = ∑ n = 1 ∞ b n p {\displaystyle \int _{0}^{\infty }f(x)^{p}\,dx=\sum _{n=1}^{\infty }b_{n}^{p}} and, for n − 1 < x < n {\displaystyle n-1<x<n} , there holds
1 x ∫ 0 x f ( t ) d t = b 1 + ⋯ + b n − 1 + ( x − n + 1 ) b n x ≥ b 1 + ⋯ + b n n {\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 ( n − x ) ( b 1 + ⋯ + b n − 1 ) ≥ ( n − 1 ) ( n − x ) b n {\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
∑ n = 1 ∞ ( b 1 + ⋯ + b n n ) p ≤ ∫ 0 ∞ ( 1 x ∫ 0 x f ( t ) d t ) p d x {\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 [ edit ] Let p > 1 {\displaystyle p>1} and let b 1 , … , b n {\displaystyle b_{1},\dots ,b_{n}} be positive real numbers. Set S k = ∑ i = 1 k b i {\displaystyle S_{k}=\sum _{i=1}^{k}b_{i}} . First we prove the inequality
∑ n = 1 N S n p n p ≤ p p − 1 ∑ n = 1 N b n S n p − 1 n p − 1 , {\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 T n = S n n {\displaystyle T_{n}={\frac {S_{n}}{n}}} and let Δ n {\displaystyle \Delta _{n}} be the difference between the n {\displaystyle n} -th terms in the right-hand side and left-hand side of * , that is, Δ n := T n p − p p − 1 b n T n p − 1 {\displaystyle \Delta _{n}:=T_{n}^{p}-{\frac {p}{p-1}}b_{n}T_{n}^{p-1}} . We have:
Δ n = T n p − p p − 1 b n T n p − 1 = T n p − p p − 1 ( n T n − ( n − 1 ) T n − 1 ) T n p − 1 {\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
Δ n = T n p ( 1 − n p p − 1 ) + p ( n − 1 ) p − 1 T n − 1 T n p . {\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:
T n − 1 T n p − 1 ≤ T n − 1 p p + ( p − 1 ) T n p p , {\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:
Δ n ≤ n − 1 p − 1 T n − 1 p − n p − 1 T n p . {\displaystyle \Delta _{n}\leq {\frac {n-1}{p-1}}T_{n-1}^{p}-{\frac {n}{p-1}}T_{n}^{p}.} By telescoping we have:
∑ n = 1 N Δ n ≤ 0 − 1 p − 1 T 1 p + 1 p − 1 T 1 p − 2 p − 1 T 2 p + 2 p − 1 T 2 p − 3 p − 1 T 3 p + ⋯ + N − 1 p − 1 T N − 1 p − N p − 1 T N p = − N p − 1 T N p < 0 , {\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:
∑ n = 1 N S n p n p ≤ p p − 1 ∑ n = 1 N b n S n p − 1 n p − 1 ≤ p p − 1 ( ∑ n = 1 N b n p ) 1 / p ( ∑ n = 1 N S n p n p ) ( p − 1 ) / p {\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:
∑ n = 1 N S n p n p ≤ ( p p − 1 ) p ∑ n = 1 N b n p . {\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 N → ∞ {\displaystyle N\rightarrow \infty } we obtain Hardy's inequality.
See also [ edit ] ^ Hardy, G. H. (1920). "Note on a theorem of Hilbert" . Mathematische Zeitschrift . 6 (3–4): 314–317. doi :10.1007/BF01199965 . S2CID 122571449 . ^ Hardy, G. H.; Littlewood, J.E.; Pólya, G. (1952). Inequalities (Second ed.). Cambridge, UK. {{cite book }}
: CS1 maint: location missing publisher (link ) ^ Ruzhansky, Michael; Suragan, Durvudkhan (2019). Hardy Inequalities on Homogeneous Groups: 100 Years of Hardy Inequalities . Birkhäuser Basel. ISBN 978-3-030-02894-7 . ^ Marcus, Moshe; Mizel, Victor J.; Pinchover, Yehuda (1998). "On the best constant for Hardy's inequality in $\mathbb {R}^n$" . Transactions of the American Mathematical Society . 350 (8): 3237–3255. doi :10.1090/S0002-9947-98-02122-9 . ^ Mironescu, Petru (2018). "The role of the Hardy type inequalities in the theory of function spaces" (PDF) . Revue roumaine de mathématiques pures et appliquées . 63 (4): 447–525. References [ edit ] Hardy, G. H.; Littlewood, J. E.; Pólya, G. (1952). Inequalities (2nd ed.). Cambridge University Press. ISBN 0-521-35880-9 . Masmoudi, Nader (2011), "About the Hardy Inequality", in Dierk Schleicher; Malte Lackmann (eds.), 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 [ edit ]