Reverse Lp -dual Minkowski's inequality

Differential Geometry and its Applications 40 (2015) 243–251

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Differential Geometry and its Applications www.elsevier.com/locate/difgeo

Reverse Lp -dual Minkowski’s inequality Chang-Jian Zhao 1 Department of Mathematics, China Jiliang University, Hangzhou 310018, PR China

a r t i c l e

i n f o

Article history: Received 18 June 2013 Received in revised form 10 February 2015 Available online 19 March 2015 Communicated by B. Ørsted

a b s t r a c t In the paper, we establish an inverse of the well-known Lp -dual Minkowski inequality in convex geometry. As a tool, we also derive a reverse Radon’s integral inequality. © 2015 Elsevier B.V. All rights reserved.

MSC: 52A40 Keywords: Lp -dual mixed volume Lp -dual Minkowski inequality Radon’s inequality Reverse Radon’s integral inequality

1. Preliminaries The setting for this paper is n-dimensional Euclidean space Rn . Let Kn denote the set of convex bodies (compact, convex subsets with non-empty interior) in Rn . We reserve the letter u for unit vectors, and the letter B for the unit ball centered at the origin. The surface of B is S n−1 . We use V (K) for the n-dimensional volume of convex body K. Let h(K, ·) : S n−1 → R, denote the support function of K ∈ Kn ; i.e. for u ∈ S n−1 h(K, u) = max{u · x : x ∈ K}, where u · x denotes the usual inner product u and x in Rn . Let δ denote the Hausdorff metric on Kn , i.e., for K, L ∈ Kn , δ(K, L) = |hK − hL |∞ , where | · |∞ denotes the sup-norm on the space of continuous functions C(S n−1 ). Associated with a compact subset K of Rn , which is star-shaped with respect to the origin, is its radial function ρ(K, ·) : S n−1 → [0, ∞), defined for u ∈ S n−1 , by

1

E-mail addresses: [email protected], [email protected]. Research is supported by National Natural Science Foundation of China (11371334).

http://dx.doi.org/10.1016/j.difgeo.2015.03.002 0926-2245/© 2015 Elsevier B.V. All rights reserved.

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C.-J. Zhao / Differential Geometry and its Applications 40 (2015) 243–251

ρ(K, u) = max{λ ≥ 0 : λu ∈ K}. If ρ(K, ·) is positive and continuous, K will be called a star body. Let S n denote the set of star bodies in Rn . ˜ Let δ˜ denote the radial Hausdorff metric, as follows, if K, L ∈ S n , then δ(K, L) = |ρK − ρL |∞ (see, e.g., [1,2]): For p ≥ 1, the Lp -mixed volume, Vp (K, L), of K, L ∈ Kn , was defined in [3] by n V (K + εL) − V (K) Vp (K, L) = lim . ε→0 p ε

(1.1)

The Lp -mixed volume Vp (K, L) (p ≥ 1), for K, L ∈ Kn , is obtained by Vp (K, L) =



1 n

h(L, u)p dSp (K, u),

(1.2)

S n−1

where Sp (K, ·) denotes the positive Borel measure on S n−1 . The measure Sp (K, ·) is absolutely continuous with respect to S(K, ·), and has Radon–Nikodym derivative dSp (K, ·) = h(K, ·)1−p , dS(K, ·)

(1.3)

where S(K, ·) is a regular Borel measure on S n−1 . The measure Sn−1 (K, ·) is independent of the body K, and is just ordinary Lebesgue measure, S, on S n−1 . Si (B, ·) denotes the i-th surface area measure of the unit ball in Rn . In fact, Si (B, ·) = S for all i. The surface area measure S0 (K, ·) just is S(K, ·) (see [3,4]). For p ≥ 1, the Lp -dual mixed volume, V˜−p (K, L), of K, L ∈ S n , also was defined in [3] by ˆ p ε ◦ L) − V (K) n˜ V (K + V−p (K, L) = lim , ε→0 p ε

(1.4)

ˆ p ε ◦ L is the harmonic p-combination, is defined by where, K + ˆ p ε ◦ L, ·)−p = ρ(K, ·)−p + ερ(L, ·)−p . ρ(K + The Lp -dual mixed volume V˜−p (K, L) (p ≥ 1), for K, L ∈ S n , was obtained by (see [3]) 1 V˜−p (K, L) = n



ρ(K, u)n+p ρ(L, u)−p dS(u).

(1.5)

S n−1

2. Statement of main results In convex geometry, the well-known Lp -Minkowski inequality for p-mixed volume Vp is following (see [3]): Theorem A. If K1 , K2 are convex bodies in Rn and 1 ≤ p < ∞, then Vp (K1 , K2 ) ≥ V (K1 )

n−p n

p

V (K2 ) n ,

with equality if and only if K1 and K2 are homothetic. As dual case, Lp -dual Minkowski inequality for p-dual mixed volume V˜−p is following (see [3]):

C.-J. Zhao / Differential Geometry and its Applications 40 (2015) 243–251

245

Theorem B. If K1 , K2 are star bodies in Rn and 1 ≤ p < ∞, then n+p p V˜−p (K1 , K2 ) ≥ V (K1 ) n V (K2 )− n ,

with equality if and only if K1 and K2 are dilates. The main aim of this paper is to consider the following reverse problem. Question. For which values of  and every pair of star bodies K1 and K2 in Rn and 1 ≤ p < ∞, is it true that n+p p V˜−p (K1 , K2 ) ≤  · V (K1 ) n V (K2 )− n ?

We prove the following reverse Lp -dual Minkowski inequality for the p-dual mixed volume V˜−p . Theorem 2.1 (Reverse Lp -dual Minkowski inequality). If K1 , K2 are convex bodies in Rn and 1 ≤ p < ∞, then n+p p V˜−p (K1 , K2 ) ≤  · V (K1 ) n V (K2 )− n ,

(2.1)

with equality if and only if K1 and K2 are dilates, where  =





max C n+p , n+p n

p

   n+p 2np n r1n r2n R1n R2n r1n r2n+p n+p n+p , , , , C 2np 2np n , p Rn Rn ω rn rn ω r1n m2n+p ωn 1 2 n R1n R2n+p ωn 1 2 n 2np

R1n R2n+p

Cα,β (ξ, η) =

ξ/α + η/β , ξ 1/α η 1/β

(2.2)

(2.3)

and α1 + β1 = 1, α > 1, and rj denotes the radius of the largest ball contained in Kj , and Rj denotes the radius of the smallest ball containing Kj (j = 1, 2), and ωn is the volume of the unit n-ball. The well-known inequality due to Radon can be stated as follows (see [5], p. 61 or [6]). Theorem C (Radon’s inequality). If {ai } and {bi } are positive real sequences and p > 1 or p < 0, then n n p  ( i=1 ai ) api ≥ n p−1 , bp−1 ( i=1 bi ) i=1 i with equality if and only if {ai } and {bi } are proportional. This inequality is reversed if 0 < p < 1. An integral form of (1.1) easy follows. Theorem D. If f (x) and g(x) are positive continuous functions on [a, b] and p > 1 or p < 0, then b a

p f (x)dx a f (x) dx ≥

p−1 , g(x)p−1 b g(x)dx a p



b

with equality if and only if f (x) and g(x) are proportional. This inequality is reversed if 0 < p < 1.

C.-J. Zhao / Differential Geometry and its Applications 40 (2015) 243–251

246

Radon’s inequality was studied extensively and numerous variants, generalizations, and extensions appeared in the literatures (see e.g. [7–13]) and the references cited therein. Research on reverse Radon’s inequality is rare. To prove the main result, as tool, we derived the following reverse Radon’s integral inequality. Theorem 2.2. Let f (x) and g(x) be positive continuous functions on [a, b]. If 0 < m1 ≤ f (x) ≤ M1 , 0 < m2 ≤ g(x) ≤ M2 and m > 0, then b a

f



b

m+1

m+1 f (x)dx

a (x) m , dx ≤ k ·

b g m (x) g(x)dx a

(2.4)

with equality if and only if f (x) and g(x) are proportional, where  k=

 m1 m 2 , max Cm+1, m+1 , 2m m m1 m2m+1 (b − a) M1 M2 (b − a) m+1  2m m1 m2m+1 M 1 M2 , , Cm+1, m+1 2m m M M m+1 (b − a) m1 m2 (b − a) 



1

2m

M1 M2m+1

(2.5)

2

and Cm+1, m+1 (ξ, η) is as in (2.3). m

3. Proof In this section, we start with three auxiliary results (Lemmas 3.1, 3.2 and 3.3), which will be the base of our further study. Our main result is given in the following theorem (Theorem 3.4). Lemma 3.1. (See [14].) If 0 < m1 ≤ a ≤ M1 , 0 < m2 ≤ b ≤ M2 ,

1 α

+

1 β

= 1 and α > 1, then

max{Cα,β (M1 , m2 ), Cα,β (m1 , M2 )} · a1/α b1/β ≥

b a + , α β

(3.1)

with equality if and only if either (a, b) = (m1 , M2 ) or (a, b) = (M1 , m2 ), where Cα,β (ξ, η) is as in (2.3). Lemma 3.2. If f (x) and g(x) are non-negative continuous functions on [a, b], and f α (x) and g β (x) be integrable on [a, b]. If 0 < m1 ≤ f (x) ≤ M1 , 0 < m2 ≤ g(x) ≤ M2 , α1 + β1 = 1, and α > 1, then ⎞1/α ⎛ b ⎞1/β ⎛ b   b ⎝ f α (x)dx⎠ ⎝ g β (x)dx⎠ ≤ cα,β · f (x)g(x)dx, a

a

(3.2)

a

with equality if and only if f α (x) and g β (x) are proportional, where  cα,β = max Cα,β and Cα,β is as in (2.3).



mβ2 M1α , β mα 1 (b − a) M2 (b − a)



 , Cα,β

M2β mα 1 , M1α (b − a) mβ2 (b − a)

 ,

C.-J. Zhao / Differential Geometry and its Applications 40 (2015) 243–251

247

In this article, Lemma 3.2 is an important and useful result, which has been shown in [15]. For completeness, we listed below: Proof. If we set successively f α (x) , X= a= X

b f α (x)dx, a

b=

g β (x) , Y = Y

b g β (x)dx. a

Notice that M1α mα 1 ≤ a ≤ , M1α (b − a) mα 1 (b − a)

(3.3)

and mβ2 M2β (b − a)

≤b≤

M2β mβ2 (b − a)

.

(3.4)

By using Lemma 3.1, we have  max Cα,β ≥



mβ2 M1α , β mα 1 (b − a) M2 (b − a)



 , Cα,β

M2β mα 1 , M1α (b − a) mβ2 (b − a)

 ·

1 g β (x) 1 f α (x) + , α X β Y

f (x)g(x) X 1/α Y 1/β (3.5)

with equality if and only if either  (a, b) =

Mβ mα 1 , β 2 α M1 (b − a) m2 (b − a)



or  (a, b) =

mβ2 M1α , β mα 1 (b − a) M2 (b − a)

 .

Therefore cα,β ·

b

b

a

a

f (x)g(x)dx 1 ≥ α X 1/α Y 1/β

f α (x)dx 1 + X β

b a

g β (x)dx = 1. Y

(3.6)

The inequality (3.2) in Lemma 3.2 easy follows. In the following, we discuss the equality condition of (3.6). In view of the equality conditions of Lemma 3.1 and (3.5), the equality in (3.6) holds if and only if 

b a

f α (x) f α (x)dx

, b a

g β (x) g β (x)dx



 =

M2β mα 1 , M1α (b − a) mβ2 (b − a)



C.-J. Zhao / Differential Geometry and its Applications 40 (2015) 243–251

248

or 

b a

f α (x) f α (x)dx

, b a



g β (x) g β (x)dx

 =

mβ2 M1α , β mα 1 (b − a) M2 (b − a)

 .

Hence f α (x) = μg β (x), where,

μ=

β mα 1 m2

M2β M1α

b a

f α (x)dx

a

g β (x)dx

b

or μ=

M1α M2β mβ2 mα 1

b a

f α (x)dx

a

g β (x)dx

b

is a constant. This follows the equality of (3.6) holds if and only if f α (x) and g β (x) are proportional. This proof is completed. 2 Lemma 3.3 (Reverse Radon’s integral inequality). Let f (x) and g(x) be positive continuous functions on [a, b]. If 0 < m1 ≤ f (x) ≤ M1 , 0 < m2 ≤ g(x) ≤ M2 and m > 0, then b a

m+1

b f (x)dx a f m+1 (x) m , dx ≤ k ·

b g m (x) g(x)dx a

(3.7)

with equality if and only if f (x) and g(x) are proportional, where  k=

 m1 m 2 , max Cm+1, m+1 , 2m m m1 m2m+1 (b − a) M1 M2 (b − a) m+1  2m m1 m2m+1 M 1 M2 , , Cm+1, m+1 2m m M M m+1 (b − a) m1 m2 (b − a) 



1

2m

M1 M2m+1

2

and Cm+1, m+1 (ξ, η) is defined in (2.3). m

Proof. Let α = m + 1, β = (m + 1)/m and replacing f (x) and g(x) by u(x) and v(x) in (3.2), respectively, and in view of 0 ≤ m1 ≤ u(x) ≤ M1 and 0 ≤ m2 ≤ v(x) ≤ M2 , we obtain ⎞1/(m+1) ⎛ b ⎞m/(m+1) ⎛ b   b ⎝ u(x)m+1 dx⎠ ⎝ v(x)(m+1)/m dx⎠ ≤ c · u(x)v(x)dx, a

a

(3.8)

a

with equality if and only if u(x)m+1 and v(x)(m+1)/m are proportional, where 



c = max Cm+1, m+1 m

m+1

m m M1m+1 , m+12 m+1 m1 (b − a) M m (b − a) 2

and Cm+1, m+1 (ξ, η) is as in (2.3). m



 , Cm+1, m+1 m

m+1

M m mm+1 1 , m+12 m+1 M1 (b − a) m m (b − a) 2

 ,

C.-J. Zhao / Differential Geometry and its Applications 40 (2015) 243–251

249

Notice that 

M1 m2

1/(m+1)

 ≥

f (x) g(x)

1/(m+1)

 ≥

m1 M2

1/(m+1) ,

and m/(m+1)

M1

1/(m+1)

M2

m/(m+1)

≥ f m/(m+1) (x)g 1/(m+1) (x) ≥ m1

1/(m+1)

m2

.

Taking for  u(x) =

f (x) g(x)

1/(m+1) , v(x) = f m/(m+1) (x)g 1/(m+1) (x)

in (3.8), we obtain b d·

⎞1/(m+1) ⎛ b ⎛ b ⎞m/(m+1)   f (x) ⎝ f (x)g 1/m (x)dx⎠ dx⎠ f (x)dx ≥ ⎝ , g(x)

a

a

a

with equality if and only if f (x)/g(x) and f (x)g(x)1/m are proportional, where 



d = max Cm+1, m+1 m

1/m

m1 m2 M1 M2 , m1 m2 (b − a) M1 M 1/m (b − a)



 , Cm+1, m+1 m

2

1/m

M 1 M2 m 1 m2 , M1 M2 (b − a) m1 m1/m (b − a)

 .

2

Hence b

⎛ f (x) dx ≤ ⎝d · g(x)

a

b

⎞m+1 ⎛ b ⎞−m  ⎝ f (x)g 1/m (x)dx⎠ f (x)dx⎠ ,

a

(3.9)

a

with equality if and only if f (x)/g(x) and f (x)g(x)1/m are proportional. On the other hand, notice that M1 m/(m+1) m2

≥

f (x) g m/(m+1) (x)

≥

m1 , m/(m+1) M2

and m/(m+1)

M2

m/(m+1)

≥ g m/(m+1) (x) ≥ m2

.

Replacing f (x) and g(x) by u(x) and v(x) in (3.9), respectively, and let u(x) = f (x) and v(x) = we have b a

⎛ b ⎞m+1 ⎛ b ⎞−m   f (x) ⎝ g(x)dx⎠ dx ≤ k · ⎝ f (x)dx⎠ , g m (x) m+1

a

a

with equality if and only if f (x) and g(x) are proportional, where,

g(x) f (x)

m ,

C.-J. Zhao / Differential Geometry and its Applications 40 (2015) 243–251

250

 k=

 m1 m 2 , max Cm+1, m+1 , 2m m m1 m2m+1 (b − a) M1 M2 (b − a) m+1  2m M 1 M2 m1 m2m+1 Cm+1, m+1 , . 2m m M M m+1 (b − a) m1 m2 (b − a) 



1

2m

M1 M2m+1

2

This proof is completed. 2 In the following, we establish an inverse of the well-known Lp -dual Minkowski inequality. Theorem 3.4 (Reverse Lp -dual Minkowski inequality). If K1 , K2 are convex bodies in Rn and 1 ≤ p < ∞, then n+p p V˜−p (K1 , K2 ) ≤  · V (K1 ) n V (K2 )− n ,

(3.10)

with equality if and only if K1 and K2 are dilates, where  is as in (2.2). Proof. From the hypotheses, for K1 ∈ Kn and any u ∈ S n−1 , we have ρ(K1 , u) ≥ r1 , and h(K1 , u) ≤ R1 . Moreover, for K1 ∈ Kn and u ∈ S n−1 , we have ρ(K1 , u) ≤ h(K1 , u). Hence, for K1 ∈ Kn and u ∈ S n−1 r1n ≤ ρ(K1 , u)n ≤ R1n .

(3.11)

Similarly, for K2 ∈ Kn and u ∈ S n−1 , we have r2n ≤ ρ(K2 , u)n ≤ R2n .

(3.12)

On the other hand, from (1.5), (3.11), (3.12) and in view of the reverse Radon’s integral inequality, we obtain  1 V˜−p (K1 , K2 ) = ρ(K1 , u)n+p ρ(K2 , u)−p dS(u) n S n−1

=



1 n

p

p

[ρ(K1 , u)n ] n +1 [ρ(K2 , u)n ]− n dS(u)

S n−1







   n+p 2np n r1n r2n r1n r2n+p R1n R2n n+p n+p ≤ max C n+p , n+p , , C , 2np 2np n p n , p Rn Rn ω rn rn ω r1n m2n+p ωn 1 2 n R1n R2n+p ωn 1 2 n ⎛ ⎞ n+p ⎞− np ⎛ n   ⎝ ×⎝ ρ(K1 , u)n dS(u)⎠ ρ(K2 , u)n dS(u)⎠ 2np

R1n R2n+p

S n−1

=  · V (K1 )

S n−1 n+p n

p −n

V (K2 )

,

(3.13)

C.-J. Zhao / Differential Geometry and its Applications 40 (2015) 243–251

251

where  =





max C n+p , n+p n

p

   n+p 2np n r1n r2n r1n r2n+p R1n R2n n+p n+p , , C , . 2np 2np n , p Rn Rn ω rn rn ω r1n m2n+p ωn 1 2 n R1n R2n+p ωn 1 2 n 2np

R1n R2n+p

From the equality condition of the reverse Radon’s integral inequality, we get that the equality of (3.13) holds if and only if ρ(K1 , u)n and ρ(K2 , u)n are proportional, it follows the equality holds if and only if K1 and K2 are dilates. This proof is completed. 2 Acknowledgements The author expresses his grateful thanks to the referee for his suggestions and comments. References [1] R.J. Gardner, Geometric Tomography, Cambridge Univ. Press, New York, 1996. [2] R. Schneider, Convex Bodies: The Brunn–Minkowski Theory, Cambridge Univ. Press, 1993. [3] E. Lutwak, The Brunn–Minkowski–Firey theory I: mixed volumes and the Minkowski problem, J. Differ. Geom. 38 (1993) 131–150. [4] E. Lutwak, Inequalities for mixed intersection bodies, Trans. Am. Math. Soc. 339 (1993) 901–916. [5] G.H. Hardy, J.E. Littlewood, G. Pölya, Inequalities, Cambridge Univ. Press, Cambridge, U.K., 1934. [6] J. Radon, Uber die absolut additiven Mengenfunktionen, Wiener Sitzungsber. 122 (1913) 1295–1438. [7] D.M. Batinetu-Giurgiu, D. Marghidanu, O.T. Pop, A new generalization of Radon’s inequalities and applications, Creative Math. Inform. 20 (2011) 62–73. [8] L. Ciurdariu, Integral inequalities, J. Sci. Arts 4 (2011) 369–376. [9] F. Furuichi, N. Minculete, F.C. Mitroi, Some inequalities on generalized entropies, J. Inequal. Appl. 2012 (2012) 226. [10] D. Marghidanu, Generalizations and refinements for Bergstrom and Radon’s inequalities, J. Sci. Arts 8 (2008) 57–62. [11] C. Mortici, A new refinement of the Radon inequality, Math. Commun. 16 (2011) 319–324. [12] K. Wen, New generalization of Radon inequality and the application to the generalization of circular inequality, J. Bijie Univ. 25 (2007) 7–11. [13] S. Wu, A class of new Radon type inequalities and their applications, Math. Pract. Theory 36 (2006) 219–224. [14] Y. Zhang, On inverse of the Hölder inequality, J. Math. Anal. Appl. 161 (1991) 566–575. [15] C. Zhao, On reverse Minkowski-type inequalities, Mediterr. J. Math. (2015), http://dx.doi.org/10.1007/s00009-014-0475-1, in press.