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The General

L

p

-Dual Mixed Brightness Integrals

Ping Zhang, Xiaohua Zhang, and Weidong Wang

Abstract—Based on general Lp-mixed brightness integrals

of convex bodies and general Lp-intersection bodies of star

bodies, this paper is going to define the generalLp-dual mixed

brightness integrals. After studying their extremum values and establishing Aleksandrov-Frenchel inequality, cyclic inequality and the Brunn-Minkowski inequality for the general Lp-dual

mixed brightness integrals, we obtain a more general result than the Brunn-Minkowski inequality for the generalLp-dual

mixed brightness integrals.

Index Terms—generalLp-mixed brightness integrals, general Lp-intersection body, general Lp-dual mixed brightness

inte-grals.

I. INTRODUCTION

L

ET Kn denote the set of convex bodies (compact, convex subsets with nonempty interiors) in Euclidean spaceRn. For the set of convex bodies containing the origin in their interiors in Rn, we writeKn

o.Son denotes the set of star bodies (about the origin) in Rn. Let Sn−1 denote the

unite sphere inRn, and letV(K)denote then-dimensional volume of body K. For the standard unit ballB inRn, we writeωn =V(B).

If K ∈ Kn, then its support function, h

K = h(K, .) :

Rn(−∞,), is defined by (see [2])

h(K, x) = max{x·y:y∈K}, x∈Rn,

wherex·y denotes the standard inner product ofxandy. For a compact set K in Rn, which is star shaped with respect to the origin, the radial function,ρK(u) =ρ(K, u), of K is defined by (see [2])

ρK(u) = max{λ≥0 :λu∈K}, u∈Sn−1. (1) If ρK is positive and continuous, then K will be called a star body (about the origin). Two star bodies K and L are said to be dilates (of one another) if ρK(u)/ρL(u) is independent ofu∈Sn−1.

Ifc >0andK∈ Sn

o, thenρ(cK,·) =(K,·).

Let GL(n) denote the group of general (nonsingular) linear transformations, if ϕ GL(n), from (1), we easily have

ρ(ϕK, x) =ρ(K, ϕ−1x), x∈Rn\ {0}, (2)

whereϕ−1 denotes the reverse of transformationϕ.

Manuscript received July 5, 2016; revised November 23, 2016. This work was supported by the National Natural Science Foundations of China(Grant No.11371224),and was supported by the Academic Mainstay Foundation of Hubei Province of China (Grant No.B2016030).

P. Zhang is with the Department of Mathematics, China Three Gorges U-niversity, Yichang, 443002, P. R. China e-mail: (zhangping9978@126.com). X. H. Zhang is with the Department of Mathematics, China Three Gorges University, Yichang, 443002, P. R. China e-mail: (zhangxiao-hua07@163.com).

W. D. Wang is with the Department of Mathematics, China Three Gorges University, Yichang, 443002, P. R. China e-mail: (wangwd722@163.com).

The notion of mixed brightness-integrals of convex bodies was defined by Li(see [8]). After that, Yan and Wang extended mixed brightness-integrals to the general mixed brightness-integrals of convex bodies: For K1,· · ·, Kn

Kn

o,p≥1 andτ∈[1,1], the generalLp-mixed brightness integrals, D()(K1,· · ·, Kn), of K1,· · ·, Kn is defined by

(see [22])

D(pτ)(K1,· · ·, Kn) = 1

n

Sn−1

δp(τ)(K1, u)· · ·δp(τ)(Kn, u)dS(u), where δ()(K, u) = 12hτpK, u) denotes the half general Lp-brightness ofK∈ Kno andΠτpKdenotes the generalLp -projection body ofK∈ Kno. Further, they established some inequalities for the generalLp-mixed brightness integrals(see [22]).

Recently, Wang and Li used the function φτ : R

[0,+)which is given by

φτ(t) =|t| −τ t, τ [1,1] (3)

to define the generalLp-intersection body with parameterτ as follows: ForK ∈ Sn

o, 0 < p <1, and τ [1,1], the generalLp-intersection body,

pK∈ Son,ofKis defined by (see [20])

ρ(IpτK, u)p=i(τ)

K

φτ(u·x)−pdx, u∈Sn−1, (4)

where

i(τ) = (1 +τ)

p(1τ)p

(1 +τ)p+ (1τ)p.

In this paper, based on the generalLp-intersection bodies and the generalLp-mixed brightness integrals, we are going to define the general Lp-dual mixed brightness integrals of star bodies as follows:

ForK1,· · ·, Kn ∈ Son, 0 < p < 1 and τ [1,1], the generalLp-dual mixed brightness integrals,

p(K1,···, Kn),

ofK1,· · ·, Kn is defined by

Dpτ(K1,· · ·, Kn)

= 1 n

Sn−1

δpτ(K1, u)· · ·δτp(Kn, u)dS(u), (5) whereδτ

p(K, u) =

1 2ρ(I

τ

pK, u)denotes the half generalLp -dual brightness ofK∈ Sn

o in directionu∈Sn−1. Ifτ = 0, we write

p(K1,· · ·, Kn) =Dp(K1,· · ·, Kn) andδp(K, u) = 12ρ(IpK, u), then

Dp(K1,· · ·, Kn) =

1 n

Sn−1

δp(K1, u)· · ·δp(Kn, u)dS(u),

we call Dp(K1,· · ·, Kn) the Lp-dual mixed brightness

integrals ofK1,· · ·, Kn ∈ Son.

IAENG International Journal of Applied Mathematics, 47:2, IJAM_47_2_03

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LetK1=···=Kn−i =KandKn−i+1=···=Kn=L (i= 0,1,· · ·, n), we write

p,i(K, L) =Dpτ(K,· · ·K, L· ··, L).

If iis any real, K, L∈ Sn

o,0< p <1, τ [1,1], then the generalLp-dual mixed brightness integrals,

p,i(K, L), of K andLis defined by

p,i(K, L) = 1 n

Sn−1

δpτ(K, u)n−iδpτ(L, u)idS(u). (6)

Letτ = 0in (6), we write

p,i(K, L) =Dp,i(K, L). Leti= 0in (6), we write

Dp,τ0(K, K) =p(K) = 1 n

Sn−1

δτp(K, u)ndS(u), (7)

which is called the general Lp-dual brightness integrals of K.

Letτ = 0in (7), we write

p(K) =Dp(K); forτ=±1, we write

p(K) =D±p(K).

In this paper, we will establish the following inequalities for the general Lp-dual mixed brightness integrals.

Initinally, we give the extremal values of the general Lp -dual mixed brightness integrals.

Theorem 1.1. IfK∈ Sn

o,0< p <1 [1,1], then

Dp(K)≤Dτp(K)≤Dp±(K), (8)

if K is not origin-symmetric, there is equality in the left inequality if and only if τ = 0 and equality in the right inequality if and only if τ=±1.

Furthermore, we establish the following Fenchel-Aleksandrov type inequality for the general Lp-dual mixed brightness integrals.

Theorem 1.2. If K1,· · ·, Kn∈ Son,0< p <1,τ∈[1,1]

and 1< m≤n, then

Dpτ(K1,· · ·, Kn)m

m

i=1

Dpτ(K1,· · ·, Knm, Kni+1,· · ·, Kni+1) (9)

equality holds if and only ifIτ

pKn−m+1,···, IpτKnare dilates

of each other.

Let τ = 0 in Theorem 1.2, we obtain the following inequality.

Corollary 1.1. If K1,· · ·, Kn ∈ Son, 0 < p <1 and 1 < m≤n, then

Dp(K1,· · ·, Kn)m

m

i=1

Dp(K1,· · ·, Kn−m, Kn−i+1,· · ·, Kn−i+1),

equality holds if and only ifIpKnm+1,···, IpKnare dilates

of each other.

Additionally, we establish the following cyclic inequality for the general Lp-dual mixed brightness integrals.

Theorem 1.3. Iet K, L∈ Sn

o, 0 < p < 1, τ [1,1], if ki

kj >1,then

Dτp,i(K, L)k−jDτp,k(K, L)j−i≥Dp,jτ (K, L)k−i, (10)

equality holds if and only if

pK and IpτL are dilates. If 0< k−i

k−j <1, the inequality (10) is reversed.

Let τ = 0 in Theorem 1.3, we obtain the following inequality.

Corollary 1.2. IetK, L∈ Sn

o,0< p <1, if k−i

kj >1, then

Dp,i(K, L)k−jDp,k(K, L)j−i≥Dp,j(K, L)k−i, (11)

equality holds if and only if IpK and IpL are dilates. If

0< kk−ji <1, the inequality (11) is reversed.

Finally, we obtain the Brunn-Minkowski type inequality for the generalLp-dual mixed brightness integrals as follows: Theorem 1.4. IetK, K′, L∈ Son,0< p <1, τ [1,1],

ifi≤n−p, then

Dp,iτ (Kn−pK

, L)np−i

≤Dp,iτ (K, L)np−i +

p,i(K

, L)np−i, (12)

equality holds if and only if

pK and IpτK

are dilates. If

i≥n−p, the inequality (12) is reversed.

Actually, we prove a more general result than Theorem 1.4 in Section III.

Our work belongs to a new and rapidly evolving asymmet-ricLpdual Brunn-Minkowski theory that has its own origin in the work of Ludwig, Haberl and Schuster (see [3], [4], [5], [6], [10], [11]). For the further researches of asymmetric LpBrunn-Minkowski theory, we can refer to papers [1], [7], [14], [15], [16], [17], [18], [19], [20], [21].

II. PRELIMINARIES

A. Dual mixed volumes

In 1975, Lutwak (see [9]) gave the notion of dual mixed volumes as follows: For K1, K2,· · ·, Kn ∈ Son, the dual mixed volume, Ve(K1, K2,· · ·, Kn), of K1, K2,· · ·, Kn is defined by

e

V(K1,· · ·, Kn) = 1

n

Sn−1

ρ(K1, u)· · ·ρ(Kn, u)dS(u). (13)

If K1 = · · · = Kn−i = K, Kn−i+1 = · · · =Kn = L in (13), we write Vei(K, L) = Ve(K, n−i;L, i), where K appearsn−itimes andLappearsitimes. Letibe any real, we have

e

Vi(K, L) = 1 n

Sn−1

ρ(K, u)n−iρ(L, u)idS(u). (14)

Leti= 0 in (14), then

e

V0(K, L) =V(K) =

1 n

Sn−1

ρ(K, u)ndS(u). (15)

B. SomeLp-combinations

ForK, L ∈ Son, p > 0 and λ, µ 0(not both zero), the Lp-radial linear combination,λ◦Kpµ◦L∈ Son,ofK and Lis defined by(see [12])

ρ(λ◦Kpµ◦L,·)p =λρ(K,·)p+µρ(L,·)p. (16)

Forϕ∈GL(n), K, L∈ Son, p >0andλ, µ≥0(not both zero), from (2) and (16), we have

ϕ(λ◦Kpµ◦L,·) =λ◦ϕKpµ◦ϕL. (17)

ForK, L∈ Sn

o,0< p <1,τ [1,1], from (4),(16) and a transformation to polar coordinate, we obtain

ρ(Ipτ(Kn−pL))p=ρ(IpτK,·) p

+ρ(IpτL,·) p

, (18)

IAENG International Journal of Applied Mathematics, 47:2, IJAM_47_2_03

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i.e.,

Ipτ(Kn−pL) =IpτKpIpτL. (19)

From (17) and (19), we get

Ipτ(ϕ(KnpL)) =IpτϕKpI τ

pϕL. (20)

III. PROOFS OFTHEOREMS

In this section, firstly we shall prove Theorems 1.1-1.3 , then we prove a more general result than Theorem 1.4, i.e., a quotient form of the Brunn-Minkowski type inequality for the general Lp-dual mixed brightness integrals.

In order to prove Theorem 1.1, we need the following inequality (see [20]).

Lemma 3.1. IfK∈ Sn

o, 0< p <1, τ [1,1], then

V(IpK)≤V(IpτK)≤V(Ip±K). (21)

If K is not origin-symmetric, there is equality in the left inequality if and only if τ = 0 and equality in the right inequality if and only if τ=±1.

Proof of Theorem 1.1. IfK, L∈ Sn

o, from (6), then

Dp,iτ (K, L) = 1 n

Sn−1

δpτ(K, u)n−iδpτ(L, u)idS(u)

= 1 2n

1 n

Sn−1

ρ(IpτK, u)n−iρ(IpτL, u)idS(u)

= 1 2nVei(I

τ pK, I

τ

pL). (22)

Leti= 0in (22), and from (15), we have

p,0(K, L) =p(K) = 1 2nV(I

τ pK).

According to (21), we get

1

2nV(IpK) 1 2nV(I

τ pK)

1 2nV(I

±

pK).

i.e.,

Dp(K)≤Dτp(K)≤Dp±(K). (23)

According to (21), we know that if K is not origin-symmetric, there is equality in the left inequality if and only if τ = 0 and equality in the right inequality if and only if τ =±1 in (23). And (23) is just the inequality (8).

The proof of Theorem 1.2 requires the following extension of the Ho¨lder inequality (see [8] [13]).

Lemma 3.2. Iff0, f1,· · ·, fm are (strictly) positive

contin-uous functions defined onSn−1 andλ1,· · ·, λmare positive

constants the sum of whose reciprocals is unity, then

Sn−1

f0(u)f1(u)· · ·fm(u)dS(u)

m

i=1

( ∫

Sn−1

f0(u)fiλi(u)dS(u)

)1

λi

, (24)

with equality if and only if there exist positive constantsα1,··

·, αm such that α1f1λ1(u) = · · · =αmfmλm(u) for all u∈ Sn−1.

Proof of Theorem 1.2. If K1,· · ·, Kn ∈ Son,0< p < 1, τ [1,1],1< m≤n,and letλi=m(1≤i≤m),and

f0(u) =δpτ(K1, u)· · ·δpτ(Kn−m, u), (f0= 1if m=n),

fi(u) =δτp(Kn−i+1, u), (1≤i≤m).

According to (24), we have

Sn−1

δτp(K1, u)· · ·δpτ(Kn−m, u)· · ·δ τ

p(Kn, u)dS(u)

m

i=1

( ∫

Sn−1

δτp(K1, u)× · · ·

×δpτ(Kn−m, u)δpτ(Kn−i+1, u)mdS(u)

)1

m

.

So

( ∫

Sn−1

δτp(K1, u)· · ·δτp(Kn−m, u)· · ·δ τ

p(Kn, u)dS(u)

)m

m

i=1

Sn−1

δpτ(K1, u)···δpτ(Kn−m, u)δτp(Kn−i+1, u)mdS(u).

i.e.,

p(K1,· · ·, Kn)m

m

i=1

Dpτ(K1,· · ·, Kn−m, Kn−i+1,· · ·, Kn−i+1). (25)

The equality condition in (25) can be got from the equality condition in inequality (24) if and only if there exist positive constantsα1,· · ·, αmsuch that

α1δpτ(Kn−m+1, u)m=α2δpτ(Kn−m+2, u)m

=· · ·=αmδpτ(Kn, u)m

for allu Sn−1. So equality holds in (25) if and only if

pKn−m+1,· · ·, IpτKnare dilates of each other. And (25) is just the inequality (9).

Proof of Theorem 1.3. Let K, L ∈ Sn

o, 0 < p < 1, τ [1,1], if kkji >1, according to (6), and the H¨older inequality, we have

Dp,iτ (K, L)kk−−jiDτ

p,k(K, L)

j−i k−i

=

(

1 n

Sn−1

δpτ(K, u)n−iδpτ(L, u)idS(u)

)k−j k−i

×

(

1 n

Sn−1

δpτ(K, u)n−kδτp(L, u)kdS(u)

)j−i k−i

=

(

1 n

Sn−1

[δτp(K, u)

(ni)kkji

δτp(L, u) ikkji

]kk−−ijdS(u) )k−j

k−i

×

(

1 n

Sn−1

[δτp(K, u)(n−k)jk−−iiδτ

p(L, u)

kkj−ii]jkiidS(u)

)j−i k−i

1

n

Sn−1

δpτ(K, u)(n−i)kk−−jiδτ

p(K, u)

(n−k)kj−ii

δpτ(L, u)kjk−−iiδτ

p(L, u) ik−j

k−idS(u)

= 1 n

Sn−1

δpτ(K, u)n−jδτp(L, u)jdS(u) =Dp,jτ (K, L).

i.e.,

p,i(K, L)kk−−jiDτ

p,k(K, L)

j−i k−i ≥Dτ

p,j(K, L). (26)

The equality condition in (26) can be got from the equality condition in the H¨older inequalitiy if and only if IpτK and

IAENG International Journal of Applied Mathematics, 47:2, IJAM_47_2_03

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pL are dilates. Similarly, if0 < ki

k−j <1, we can obtain the reverse form of (26). And (26) is just the inequality (10). Now, we give a more general result than Theorem 1.4 as follows:

Theorem 3.1. ForK, L, K′ ∈ Son,ϕ∈GL(n),0< p <1,

τ [1,1], if0≤n−j≤p≤n−i, then

(

Dp,iτ (ϕ(Kn−pK

), L)

p,j(ϕ(KnpK′), L)

) p

j−i

(Dτ

p,i(ϕK, L)

p,j(ϕK, L)

) p

j−i

+

(Dτ p,i(ϕK

, L)

p,j(ϕK

, L)

) p j−i

, (27)

with equality holds in (27) if and only ifIτ

pϕK andIpτϕK

are dilates. Ifn−j 0< n−i≤p,the inequality (27) is reversed.

The proof of Theorem 3.1 requires the following Dresher’s inequality(see [23]).

Lemma 3.3. Let functionsf1, f2, g1, g20,Eis a bounded

measurable subset inRn. If p1r0, then

(∫

E(f1+f2) pdx

E(g1+g2) rdx

) 1

p−r

(∫ Ef p 1dxEg r 1dx ) 1

p−r

+ (∫ Ef p 2dxEg r 2dx ) 1

p−r

, (28)

equality holds if and only if f1 f2 =

g1

g2 . If1≥p >0> r, the inequality (28) is reversed.

Proof of Theorem 3.1. ForK, K′, L∈ Sn

o, ϕ∈GL(n), 0< p <1,τ∈[1,1], if0≤n−j≤p≤n−i, according to (16),(20) and (28), we have

(Dτ

p,i(ϕ(Kn−pK

), L)

p,j(ϕ(Kn−pK′), L)

) p j−i

=

( 1

n

Sn−1δ τ

p(ϕ(Kn−pK

), u)n−iδpτ(L, u)idS(u)

1

n

Sn−1δτp(ϕ(Kn−pK′), u)n−jδpτ(L, u)jdS(u)

) p

j−i

=

( 1

n

Sn−1ρ(I τ

p(ϕ(KnpK

)), u)n−iρ(Iτ

pL, u)idS(u)

1

n

Sn−1ρ(Ipτ(ϕ(Kn−pK′)), u)n−jρ(IpτL, u)jdS(u)

) p

j−i

( 1

n

Sn−1ρ(I τ

pϕK, u)n−iρ(IpτL, u)idS(u)

1

n

Sn−1ρ(IpτϕK, u)n−jρ(IpτL, u)jdS(u)

) p j−i

+

(1

n

Sn−1ρ(I τ pϕK

, u)n−iρ(Iτ

pL, u)idS(u)

1

n

Sn−1ρ(IpτϕK

, u)n−jρ(Iτ

pL, u)jdS(u)

) p

j−i

=

(Dτ

p,i(ϕK, L)

p,j(ϕK, L)

) p

j−i

+

(Dτ p,i(ϕK

, L)

p,j(ϕK

, L)

) p j−i

.

The equality condition in (27) can be got from the equality condition in (28) if and only if

pϕKandIpτϕK

are dilates. Ifn−j≤0< n−i≤p, similarly, we can prove that the reverse of the inequality (27) is true.

Ifϕis identic, then we get the following inequality. Theorem 3.2. For K, L, K′ ∈ Son, 0 < p < 1, if 0 n−j ≤p≤n−i, then

(Dτ

p,i(Kn−pK

, L)

p,j(KnpK′, L)

) p

j−i

(Dτ p,i(K, L)

p,j(K, L)

) p

j−i

+

(Dτ p,i(K

, L)

p,j(K

, L)

) p j−i

, (29)

with equality holds in (29) if and only ifIτ

pK andIpτK

are dilates. If n−j 0 < n−i p, the inequality (29) is reversed.

Proof of Theorem 1.4. Let j =n in the inequality (29) and notice that

p,n(M, L) =Dτp(L)by (6), fori≤n−p and anyL∈ Sn

o, we get

p,i(Kn−pK

, L)np−i

≤Dp,iτ (K, L)np−i +Dτ

p,i(K

, L)np−i, (30)

which is just the inequality (12). From the equality condition of (29), we see that equality holds in (30) if and only ifIpτK and

pK

are dilates.

Similarly, letj =nin the reverse of the inequality (29), and for i n−p, we can obtain that the reverse of the inequality (30) is true.

ACKNOWLEDGMENT

The author is indebted to the editors and the anonymous referees for many valuable suggestions and comments.

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-intersection bodies,”Acta Mathematica Sinica, English Series, vol. 31, no. 5, pp. 777-786, May. 2015.

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(5)

[20] W. D. Wang and Y. N. Li, “GeneralLp-intersection bodies,”Taiwanese

Journal of Mathematics, vol. 19, no. 4, pp. 1247-1259, Aug. 2015. [21] M. Weberndorfer, “Shadow systems of asymmetric Lp zonotopes,”

Advances in Mathematics, vol. 240, pp. 613-635, Jun. 2013. [22] L. Yan and W. D. Wang, “The GeneralLp-mixed brightness integrals,”

Journal of Inequalities and Applications, vol. 2015, no. 190, pp.1-11, Jun. 2015.

[23] P. Zhang and W. D. Wang, and X. H. Zhang. “On dual mixed quermassintegral quotient,”Journal of Inequalities and Applications, vol. 2015, no. 340, pp.1-9, Oct. 2015.

IAENG International Journal of Applied Mathematics, 47:2, IJAM_47_2_03

References

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