- Email: [email protected]
A generalization of Hadamard’s inequality for convex functions
Applied Mathematics Letters 21 (2008) 254–257 www.elsevier.com/locate/aml
A generalization of Hadamard’s inequality for convex functions W.H. Yang School of Mathematical Sciences, Fudan University, Shanghai 200433, PR China Received 18 October 2006; received in revised form 19 December 2006; accepted 6 February 2007
Abstract In this note, we give a counterexample to show that Hadamard’s inequality does not hold on a polyhedron in multi-dimensional Euclidean space. Then we give a sufficient condition on the polyhedron for Hadamard’s inequality to hold. Finally, we provide an approach to create a large class of polyhedra on which Hadamard’s inequality is true. c 2007 Elsevier Ltd. All rights reserved.
Keywords: Convex function; Hadamard’s inequality; Jensen’s inequality; Polyhedron; Volume of polyhedron
1. Introduction Let f : I ⊂ R → R be a convex function, where I = [a, b]. The following inequality due to Hermite [4] and Hadamard [2] is well known: Z b f (a) + f (b) a+b 1 f (x)dx ≤ ≤ . (1.1) f 2 b−a a 2 Let v 1 , . . . , v n ∈ Rn be vertices of a polyhedron Ω ⊂ Rn . To generalize (1.1), it is natural to ask whether the following inequality holds for every convex function f defined on Ω : 1 Z v + · · · + vn 1 f (v 1 ) + · · · + f (v n ) f ≤ f (x)dx ≤ . (1.2) n vol(Ω ) Ω n In Section 2, we give an example to show that (1.2) is not true in the general case. In Section 3, we give some conditions under which the generalized Hadamard’s inequality (1.2) holds. We introduce some notation used in this note. For a set S, the convex hull of S is denoted by co(S). A polyhedron Ω ⊂ Rn is defined by Ω := co(v 1 , . . . , v m ), where v 1 , . . . ,Rv m ∈ Rn . The set of all vertices of a polyhedron Ω is denoted by V (Ω ). The volume of Ω is defined as vol(Ω ) := Ω dx. For simplicity, we define the set in := {Ω ⊂ Rn | (1.2) holds for any convex function f defined on Ω }.
E-mail address: [email protected]. c 2007 Elsevier Ltd. All rights reserved. 0893-9659/$ - see front matter doi:10.1016/j.aml.2007.02.024
255
W.H. Yang / Applied Mathematics Letters 21 (2008) 254–257
2. An counterexample We use S 2 to denote the unit sphere in R2 , that is S 2 = {(x, y) | x 2 + y 2 = 1}. Let Pn ⊂ R2 be a sequence of polyhedral sets such that V (Pn ) ⊂ S 2 and Z 1 dx < 3 . 0<π− (2.1) n Pn kn Assume that V (Pn ) = {(ain , bin )}i=1 , where kn is the number of vertices of Pn . We define a polyhedron Ωn ⊂ R3 by !! aknn bknn a1n b1n , . . . , 1, , , Ωn = co θ, 1, , n n n n
where θ = (0, 0, 0) is the origin of R3 . Then V (Ωn ) = {θ, (1, x 2 + y 2 ≤ n12 }. Let Ω¯ n = co(θ, Wn ).
aknn a1n b1n n , n ), . . . , (1, n
,
bknn n
)}. Define Wn := {(1, x, y) |
Let f : R3 → R be defined by f (x, y, z) = 1 − x. From (2.1), it follows that Z Z 1 1 0< f (x) − f (x)dx < . 0 < vol(Ω¯ n ) − vol(Ωn ) < 5 , 5 ¯ 3n 12n Ωn Ωn
(2.2)
By the definition of f , 1 lim n→∞ 1 + kn
f (θ ) +
kn X
! f (1, ain , ain ) = lim
n→∞
i=1
1 = 0. 1 + kn
(2.3)
Notice that 1
Z
vol(Ω¯ n ) Ω¯ n
f (x)dx =
3n 2 π
1
Z 0
π x2 1 (1 − x)dx = . 4 n2
By (2.2)–(2.4), we must have 1 1 + kn
f (θ ) +
kn X
! f (1, ain , ain )
i=1
<
Z 1 f (x)dx vol(Ωn ) Ωn
for n large enough. So (1.2) does not hold for f defined on the polyhedron Ωn for some n that is large enough. 3. Main results Let xi ∈ Rn , i = 1, . . . , n. Set f¯k,n = f¯k,n (x1 , . . . , xn ) X 1 1 = f (xi + · · · + xik ) . k 1 n + k − 1 1≤i ≤···≤i ≤n k 1 k In Pe˘cari´c and Svrtan [5, Theorem 1], the following infinite refinement of the Jensen inequality is proved. Lemma 3.1. f
n 1X xi n i=1
!
n 1X ≤ · · · ≤ f¯k+1,n ≤ f¯k,n ≤ · · · ≤ f¯1,n = f (xi ). n i=1
(2.4)
256
W.H. Yang / Applied Mathematics Letters 21 (2008) 254–257
Lemma 3.2. For a < b, let ηn = then
b−a n
n P
and let ai = a + iηn , i = 0, . . . , n. If f : [a, b] → R is a convex function,
f (a + iηn )
n+1 P
Rb f (a + iηn+1 ) f (x)dx f (a) + f (b) i=0 i=0 ≥ ··· ≥ ≥ ≥ ··· ≥ a . 2 n+1 n+2 b−a 1 Pn ¯ Proof. Since n+1 i=0 f (a + iηn ) is just f n,2 (a, b) and Z b n 1 X 1 lim f (x)dx, f (a + iηn ) = n→∞ n + 1 b−a a i=0 (3.1) follows from Lemma 3.1 immediately.
(3.1)
Theorem 3.1. If Ω ∈ in , then Ω 0 = Ω × [a, b] ∈ in+1 for any real numbers a < b. Proof. Let V (Ω ) = {v 1 , . . . , v m } be the vertices of Ω . Then v i × {a}, v i × {b}, i = 1, . . . , m, are the vertices of Ω 0 . 0 Let ηn = b−a n . Define ai := a + iηn , i = 0, . . . , n. For each convex function f defined on Ω , we have Z n Z X 1 1 f (x)dx = lim f (y)dy · ηn vol(Ω 0 ) Ω 0 (b − a)vol(Ω ) n→∞ i=0 Ω ×{ai } m P j × {a }) f (v i n b−a X 1 j=1 · lim (3.2) ≤ n b − a n→∞ i=0 m !
= lim
m X n 1 X f (v j × {ai }) m j=1 i=0 n+1
≤ lim
m 1 X f (v j × {a}) + f (v j × {b}) m j=1 2
n→∞
n→∞
P w∈V (Ω 0 )
!
f (w)
, (3.3) 2m where (3.2) holds thanks to Ω ∈ in and (3.3) follows from (3.1). The first inequality of (1.2) for Ω 0 can be proved similarly. =
n [a , b ] in Rn belongs to i . In [6] (or see [3]), Zhang showed that a By Theorem 3.1 and (1.1), a box Ω = Πi=1 i i n simplex Σ ∈ in . Then we can also construct more polyhedral sets belong to in with the help of Theorem 3.1. In the remainder, we show the invariance of in under nonsingular linear transformations.
Lemma 3.3 ([1, Lemma 8.1]). Let T x = Ax + b be an affine map, where A is a positive definite n × n matrix and b ∈ Rn . Then vol(T (Ω )) = |det(A)|vol(Ω ). Proposition 3.1. Let T x = Ax + b be an affine map, where A is a positive definite n × n matrix and b ∈ Rn . Define f¯(y) = f (T y). Then Z Z 1 1 f (x)dx = f¯(y)dy. vol(Ω ) Ω vol(T −1 (Ω )) T −1 (Ω ) Proof. By Lemma 3.3, we have vol(Ω ) = |det(A)|vol(T −1 (Ω )). Let x = T y. Then Z Z 1 1 f (x)dx = f¯(y)|det(J (y))|dy vol(Ω ) Ω vol(Ω ) T −1 (Ω )
W.H. Yang / Applied Mathematics Letters 21 (2008) 254–257
257
Z
|det(A)| f¯(y)dy vol(Ω ) T −1 (Ω ) Z 1 = f¯(y)dy vol(T −1 (Ω )) T −1 (Ω ) =
where J (y) is the Jacobian matrix associated with T (·).
Theorem 3.2. Let T be as in Proposition 3.1. Then in = T (in ). Proof. Let V (Ω ) = {v 1 , . . . , v m } for some m ∈ N. Then, V (T (Ω )) = {T v 1 , . . . , T v m }. Notice that f is a convex function if and only if f¯ is. By Proposition 3.1, it is easy to see that Ω ∈ in if and only if T (Ω ) ∈ in . References [1] D. Bertsimas, J.N. Tsitsiklis, Introduction to Linear Optimization, Athena Scientific, Belmont, Mass, 1997. [2] J. Hadamard, Etude sur les propri´et´ees des fonctions enti´eres et en particulier d’une fonction consid´er´ee par Riemann, J. Math. Pure Appl. 58 (1883) 171–215. [3] J. Kuang, Applied Inequalities, 3rd ed., Shandong Science and Technology Press, Jinan, 2004. [4] D.S. Mitrinovi´c, I.B. Lackovi´c, Hermite and convexity, Aequationes Math. 28 (1985) 225–232. [5] J. Pe˘cari´c, D. Svrtan, New refinements of the Jensen inequalities based on samples with repetitions, J. Math. Anal. Appl. 222 (1998) 365–373. [6] Y. Zhang, Hadamard inequalities for convex functions, J. Math. 7 (1987) 385–386 (in Chinese).
A generalization of Hadamard’s inequality for convex functions W.H. Yang School of Mathematical Sciences, Fudan University, Shanghai 200433, PR China Received 18 October 2006; received in revised form 19 December 2006; accepted 6 February 2007
Abstract In this note, we give a counterexample to show that Hadamard’s inequality does not hold on a polyhedron in multi-dimensional Euclidean space. Then we give a sufficient condition on the polyhedron for Hadamard’s inequality to hold. Finally, we provide an approach to create a large class of polyhedra on which Hadamard’s inequality is true. c 2007 Elsevier Ltd. All rights reserved.
Keywords: Convex function; Hadamard’s inequality; Jensen’s inequality; Polyhedron; Volume of polyhedron
1. Introduction Let f : I ⊂ R → R be a convex function, where I = [a, b]. The following inequality due to Hermite [4] and Hadamard [2] is well known: Z b f (a) + f (b) a+b 1 f (x)dx ≤ ≤ . (1.1) f 2 b−a a 2 Let v 1 , . . . , v n ∈ Rn be vertices of a polyhedron Ω ⊂ Rn . To generalize (1.1), it is natural to ask whether the following inequality holds for every convex function f defined on Ω : 1 Z v + · · · + vn 1 f (v 1 ) + · · · + f (v n ) f ≤ f (x)dx ≤ . (1.2) n vol(Ω ) Ω n In Section 2, we give an example to show that (1.2) is not true in the general case. In Section 3, we give some conditions under which the generalized Hadamard’s inequality (1.2) holds. We introduce some notation used in this note. For a set S, the convex hull of S is denoted by co(S). A polyhedron Ω ⊂ Rn is defined by Ω := co(v 1 , . . . , v m ), where v 1 , . . . ,Rv m ∈ Rn . The set of all vertices of a polyhedron Ω is denoted by V (Ω ). The volume of Ω is defined as vol(Ω ) := Ω dx. For simplicity, we define the set in := {Ω ⊂ Rn | (1.2) holds for any convex function f defined on Ω }.
E-mail address: [email protected]. c 2007 Elsevier Ltd. All rights reserved. 0893-9659/$ - see front matter doi:10.1016/j.aml.2007.02.024
255
W.H. Yang / Applied Mathematics Letters 21 (2008) 254–257
2. An counterexample We use S 2 to denote the unit sphere in R2 , that is S 2 = {(x, y) | x 2 + y 2 = 1}. Let Pn ⊂ R2 be a sequence of polyhedral sets such that V (Pn ) ⊂ S 2 and Z 1 dx < 3 . 0<π− (2.1) n Pn kn Assume that V (Pn ) = {(ain , bin )}i=1 , where kn is the number of vertices of Pn . We define a polyhedron Ωn ⊂ R3 by !! aknn bknn a1n b1n , . . . , 1, , , Ωn = co θ, 1, , n n n n
where θ = (0, 0, 0) is the origin of R3 . Then V (Ωn ) = {θ, (1, x 2 + y 2 ≤ n12 }. Let Ω¯ n = co(θ, Wn ).
aknn a1n b1n n , n ), . . . , (1, n
,
bknn n
)}. Define Wn := {(1, x, y) |
Let f : R3 → R be defined by f (x, y, z) = 1 − x. From (2.1), it follows that Z Z 1 1 0< f (x) − f (x)dx < . 0 < vol(Ω¯ n ) − vol(Ωn ) < 5 , 5 ¯ 3n 12n Ωn Ωn
(2.2)
By the definition of f , 1 lim n→∞ 1 + kn
f (θ ) +
kn X
! f (1, ain , ain ) = lim
n→∞
i=1
1 = 0. 1 + kn
(2.3)
Notice that 1
Z
vol(Ω¯ n ) Ω¯ n
f (x)dx =
3n 2 π
1
Z 0
π x2 1 (1 − x)dx = . 4 n2
By (2.2)–(2.4), we must have 1 1 + kn
f (θ ) +
kn X
! f (1, ain , ain )
i=1
<
Z 1 f (x)dx vol(Ωn ) Ωn
for n large enough. So (1.2) does not hold for f defined on the polyhedron Ωn for some n that is large enough. 3. Main results Let xi ∈ Rn , i = 1, . . . , n. Set f¯k,n = f¯k,n (x1 , . . . , xn ) X 1 1 = f (xi + · · · + xik ) . k 1 n + k − 1 1≤i ≤···≤i ≤n k 1 k In Pe˘cari´c and Svrtan [5, Theorem 1], the following infinite refinement of the Jensen inequality is proved. Lemma 3.1. f
n 1X xi n i=1
!
n 1X ≤ · · · ≤ f¯k+1,n ≤ f¯k,n ≤ · · · ≤ f¯1,n = f (xi ). n i=1
(2.4)
256
W.H. Yang / Applied Mathematics Letters 21 (2008) 254–257
Lemma 3.2. For a < b, let ηn = then
b−a n
n P
and let ai = a + iηn , i = 0, . . . , n. If f : [a, b] → R is a convex function,
f (a + iηn )
n+1 P
Rb f (a + iηn+1 ) f (x)dx f (a) + f (b) i=0 i=0 ≥ ··· ≥ ≥ ≥ ··· ≥ a . 2 n+1 n+2 b−a 1 Pn ¯ Proof. Since n+1 i=0 f (a + iηn ) is just f n,2 (a, b) and Z b n 1 X 1 lim f (x)dx, f (a + iηn ) = n→∞ n + 1 b−a a i=0 (3.1) follows from Lemma 3.1 immediately.
(3.1)
Theorem 3.1. If Ω ∈ in , then Ω 0 = Ω × [a, b] ∈ in+1 for any real numbers a < b. Proof. Let V (Ω ) = {v 1 , . . . , v m } be the vertices of Ω . Then v i × {a}, v i × {b}, i = 1, . . . , m, are the vertices of Ω 0 . 0 Let ηn = b−a n . Define ai := a + iηn , i = 0, . . . , n. For each convex function f defined on Ω , we have Z n Z X 1 1 f (x)dx = lim f (y)dy · ηn vol(Ω 0 ) Ω 0 (b − a)vol(Ω ) n→∞ i=0 Ω ×{ai } m P j × {a }) f (v i n b−a X 1 j=1 · lim (3.2) ≤ n b − a n→∞ i=0 m !
= lim
m X n 1 X f (v j × {ai }) m j=1 i=0 n+1
≤ lim
m 1 X f (v j × {a}) + f (v j × {b}) m j=1 2
n→∞
n→∞
P w∈V (Ω 0 )
!
f (w)
, (3.3) 2m where (3.2) holds thanks to Ω ∈ in and (3.3) follows from (3.1). The first inequality of (1.2) for Ω 0 can be proved similarly. =
n [a , b ] in Rn belongs to i . In [6] (or see [3]), Zhang showed that a By Theorem 3.1 and (1.1), a box Ω = Πi=1 i i n simplex Σ ∈ in . Then we can also construct more polyhedral sets belong to in with the help of Theorem 3.1. In the remainder, we show the invariance of in under nonsingular linear transformations.
Lemma 3.3 ([1, Lemma 8.1]). Let T x = Ax + b be an affine map, where A is a positive definite n × n matrix and b ∈ Rn . Then vol(T (Ω )) = |det(A)|vol(Ω ). Proposition 3.1. Let T x = Ax + b be an affine map, where A is a positive definite n × n matrix and b ∈ Rn . Define f¯(y) = f (T y). Then Z Z 1 1 f (x)dx = f¯(y)dy. vol(Ω ) Ω vol(T −1 (Ω )) T −1 (Ω ) Proof. By Lemma 3.3, we have vol(Ω ) = |det(A)|vol(T −1 (Ω )). Let x = T y. Then Z Z 1 1 f (x)dx = f¯(y)|det(J (y))|dy vol(Ω ) Ω vol(Ω ) T −1 (Ω )
W.H. Yang / Applied Mathematics Letters 21 (2008) 254–257
257
Z
|det(A)| f¯(y)dy vol(Ω ) T −1 (Ω ) Z 1 = f¯(y)dy vol(T −1 (Ω )) T −1 (Ω ) =
where J (y) is the Jacobian matrix associated with T (·).
Theorem 3.2. Let T be as in Proposition 3.1. Then in = T (in ). Proof. Let V (Ω ) = {v 1 , . . . , v m } for some m ∈ N. Then, V (T (Ω )) = {T v 1 , . . . , T v m }. Notice that f is a convex function if and only if f¯ is. By Proposition 3.1, it is easy to see that Ω ∈ in if and only if T (Ω ) ∈ in . References [1] D. Bertsimas, J.N. Tsitsiklis, Introduction to Linear Optimization, Athena Scientific, Belmont, Mass, 1997. [2] J. Hadamard, Etude sur les propri´et´ees des fonctions enti´eres et en particulier d’une fonction consid´er´ee par Riemann, J. Math. Pure Appl. 58 (1883) 171–215. [3] J. Kuang, Applied Inequalities, 3rd ed., Shandong Science and Technology Press, Jinan, 2004. [4] D.S. Mitrinovi´c, I.B. Lackovi´c, Hermite and convexity, Aequationes Math. 28 (1985) 225–232. [5] J. Pe˘cari´c, D. Svrtan, New refinements of the Jensen inequalities based on samples with repetitions, J. Math. Anal. Appl. 222 (1998) 365–373. [6] Y. Zhang, Hadamard inequalities for convex functions, J. Math. 7 (1987) 385–386 (in Chinese).