Cyclic Brunn–Minkowski inequalities for general width and chord integrals

R E V I E W

Open Access

Cyclic Brunn–Minkowski inequalities for

general width and chord-integrals

Linmei Yu

_{, Yuanyuan Zhang}

_{and Weidong Wang}

*_{Correspondence:}

[email protected]

1_{Department of Mathematics, China}

Three Gorges University, Yichang, China

2_{Three Gorges Mathematical}

Research Center, China Three Gorges University, Yichang, China

Abstract

In this paper, we establish two cyclic Brunn–Minkowski inequalities for the generalith width-integrals and generalith chord-integrals, respectively. Our works bring the cyclic inequality and Brunn–Minkowski inequality together.

MSC: 52A20; 52A40

Keywords: Generalith width-integral; Generalith chord-integral; Cyclic Brunn–Minkowski inequality

1 Introduction and main results

The setting for this paper is the Euclideann-spaceRn. We useSn–1andV(K) to denote the unit sphere and then-dimensional volume of a bodyK, respectively. For the standard unit ballB, we writeV(B) =ωn.

If K is a nonempty compact convex set in Rn_{, then the support function ofK, h} K=

h(K,·) :Rn_{→R, is defined by (see [1])}

h(K,x) =max{x·y:y∈K}

forx∈Rn_{, wherex·yis the standard inner product ofxandy. IfKis a compact convex}

set with nonempty interiors inRn, thenKis called a convex body. LetKndenote the set of convex bodies inRn_.

The radial functionρK=ρ(K,·) :Rn\ {0} →[0,∞) of a compact star-shaped (about the

origin) setK⊂Rn_{is defined by (see [1])}

ρ(K,x) =max{λ≥0 :λx∈K}, x∈Rn\ {0}.

IfρKis continuous, thenKwill be called a star body (about the origin). LetSondenote the

subset of star bodies containing the origin inRn_{. Two star bodiesKandLare dilates (of}

one another) ifρK(u)/ρL(u) is independent ofu∈Sn–1.

The research of width-integrals has a long history. The width-integrals were first con-sidered by Blaschke (see [2]) and were further researched by Hadwiger (see [3]). In 1975, Lutwak [4] gave theith width-integrals as follows:

ForK∈Kn_{and any reali, theith width-integralsB}

i(K) ofKare defined by

Bi(K) =

1 n

Sn–1

b(K,u)n–i_{dS(u). (1.1)}

Hereb(K,u) denotes the half width ofK in the direction u∈Sn–1 _{which is defined by} b(K,u) =1

2h(K,u) + 1

2h(K, –u). If there exists a constantλ> 0 such thatb(K,u) =λb(L,u) for allu∈Sn–1_{, thenK andLare said to have similar width. Further, Lutwak [4]} estab-lished the following Brunn–Minkowski inequality and cyclic inequality for theith width-integrals, respectively.

Theorem 1.A If K,L∈Kn_{and real i<n– 1,then}

Bi(K+L)

1

n–i≤_B

i(K)

1

n–i_+B

i(L)

1

n–i

with equality if and only if K and L have similar width.Here K+L denotes the Minkowski sum of K and L.

Theorem 1.B If K∈Kn_{and reals i,j,k satisfy i<j<k,then}

Bj(K)k–i≤Bi(K)k–jBk(K)j–i

with equality if and only if K is of constant width.

Whereafter, Lutwak [5] showed that the mixed width-integralB(K1, . . . ,Kn) ofK1, . . . , Kn∈Knwas defined by

B(K1, . . . ,Kn) =

1 n

Sn–1b(K1,u)· · ·b(Kn,u)dS(u). (1.2)

In 2016, based on (1.2), Feng [6] introduced the general mixed width-integrals as follows: ForK1, . . . ,Kn∈Knandτ ∈(–1, 1), the general mixed width-integralB(τ)(K1, . . . ,Kn) of

K1, . . . ,Knis given by

B(τ)(K1, . . . ,Kn) =

1 n

Sn–1b

(τ)_(K

1,u)· · ·b(τ)(Kn,u)dS(u), (1.3)

wherebτ_{(K,u) =f}

1(τ)h(K,u) +f2(τ)h(K, –u) and

f1(τ) =

(1 +τ)2

2(1 +τ2₎, f2(τ) =

(1 –τ)2

2(1 +τ2₎. (1.4)

Obviously,

f1(τ) +f2(τ) = 1;

Combined with (1.4), the case ofτ= 0 in (1.3) is just (1.2). If there exists a constantλ> 0 such thatbτ_{(K,u) =λb}τ_{(L,u) for allu∈S}n–1_{, then we say convex bodiesKandLhave} sim-ilar general width.KandLhave joint constant general width means thatb(τ)_(K,u)b(τ)_(L,u) is a constant for allu∈Sn–1.

TakingK1=· · ·=Kn–i=K,Kn–i+1=· · ·=Kn=Bin (1.3) and allowingito be any real, the

generalith width-integralsB(_iτ)(K) ofK∈Kn_{were given by (see [6])}

B(_iτ)(K) = 1 n

Sn–1

b(τ)_(K,u)n–i_{dS(u). (1.5)}

From (1.1), (1.4), and (1.5), we easily see that ifτ= 0, thenB(0)_i (K) =Bi(K).

In 2006, motivated by Lutwak’sith width-integrals and together with the notion of radial function, Li, Yuan, and Leng [7] gave theith chord-integrals as follows: ForK∈S_onandi is any real, theith chord-integralsCi(K) ofKare defined by

Ci(K) =

1 n

Sn–1c(K,u)

n–i_{dS(u). (1.6)}

Here c(K,u) denotes the half chord of K in the direction u and c(K,u) = 1₂ρ(K,u) + 1

2ρ(K, –u). If there exists a constantλ> 0 such thatc(K,u) =λc(L,u) for allu∈S

n–1_{, then} we say thatKandLhave similar chord.

For theith chord-integrals, the authors [7] proved the following Brunn–Minkowski in-equality and cyclic inin-equality.

Theorem 1.C For K,L∈Sn

o and real i=n.If i<n– 1,then

Ci(K+˜L)

1

n–i≤_Ci(K)n1–i_+Ci(L)n1–i_;

if i>n– 1,then

Ci(K+˜L)

1

n–i≥_C

i(K)

1

n–i_+C

i(L)

1

n–i_.

In each inequality,equality holds if and only if K and L are dilates.Here K+˜L denotes the radial sum of K and L.

Theorem 1.D If K∈Sn

o and reals i,j,k satisfy i<j<k,then

Cj(K)k–i≤Ci(K)k–jCk(K)j–i,

with equality if and only if K is of constant chord.

The mixed chord-integrals of star bodies were defined by Lu (see [8]): ForK1, . . . ,Kn∈

Sn

o, the mixed chord-integralsC(K1, . . . ,Kn) ofK1, . . . ,Knare defined by

C(K1, . . . ,Kn) =

1 n

Recently, Feng and Wang [9] gave the general mixed chord-integralsC(τ)_(K

1, . . . ,Kn) of

K1, . . . ,Kn∈Sondefined by

C(τ)_(K

1, . . . ,Kn) =

1 n

Sn–1

c(τ)_(K

1,u)· · ·c(τ)(Kn,u)dS(u), (1.7)

where c(τ)_{(K,u) =f}

1(τ)ρ(K,u) +f2(τ)ρ(K, –u) and functionsf1(τ),f2(τ) satisfy (1.4). By (1.4), letτ = 0 in (1.7), this is just Lu’s mixed chord-integralsC(K1, . . . ,Kn). Star bodiesK

andLare said to have similar general chord mean that there exist constantsλ,μ> 0 such thatλc(τ)_{(K,u) =μc}(τ)_{(L,u) for allu∈S}n–1_{. If the productc}(τ)_(K,u)c(τ)_{(L,u) is constant for} allu∈Sn–1_{, then they are said to have joint constant general chord.}

TakingK1=· · ·=Kn–i=KandKn–i+1=· · ·=Kn=Bin (1.7) and allowingito be any real,

the generalith chord-integralC_i(τ)(K) ofK∈Sn

o was given by (see [9])

C_i(τ)(K) =1 n

Sn–1c

(τ)_(K,u)n–i_{dS(u). (1.8)}

Obviously, (1.4), (1.6), and (1.8) giveC_i(0)(K) =Ci(K).

In this paper, based on Theorems1.A–1.Band Theorems1.C–1.D, we respectively es-tablish two cyclic Brunn–Minkowski inequalities for generalith width-integrals and gen-eralith chord-integrals by using Zhao’s ideas (see [10] and [11]). Our works bring the cyclic inequality and Brunn–Minkowski inequality together. Our main results can be stated as follows.

Theorem 1.1 Let K,L∈Kn_{,τ ∈(–1, 1),and i,j,k all be reals.If j<n– 1and i≤j<k,}

then

B_j(τ)(K+L)nk––ij ≤_B(τ)

i (K)

k–j n–j_B(τ)

k (K)

j–i n–j _+B(τ)

i (L)

k–j n–j_B(τ)

k (L)

j–i

n–j _(1.9)

with equality if and only if K and L have similar general width.If n– 1 <j<n and j≤i<k or j>n and i≤j<k,then inequality(1.9)is reversed.

Theorem 1.2 Let K,L∈Sn

o,τ∈(–1, 1),and i,j,k all be reals.If j<n– 1and i≤j<k,then

C_j(τ)(K+˜L)nk––ij ≤_C(τ)

i (K)

k–j n–j_C(τ)

k (K)

j–i n–j _+C(τ)

i (L)

k–j n–j_C(τ)

k (L)

j–i

n–j _(1.10)

with equality if and only if K and L have similar general chord.If n– 1 <j<n and j≤i<k or j>n and i≤j<k,then inequality(1.10)is reversed.

Remark 1.1 Let i=jin Theorem 1.1 and Theorem 1.2, we may obtain the following Brunn–Minkowski inequalities for general ith width-integrals and general ith chord-integrals, respectively.

Corollary 1.1 Let K,L∈Kn_{,real i=n,andτ∈(–1, 1).If i<n– 1,then}

B(_iτ)(K+L)n1–i≤_B(τ)

i (K)

1

n–i_+B(τ)

i (L)

1

n–i _(1.11)

Corollary 1.2 Let K,L∈Sn

o,real i=n,andτ∈(–1, 1).If i<n– 1,then

C_i(τ)(K+˜L)n1–i≤_C(τ)

i (K)

1

n–i_+C(τ)

i (L)

1

n–i

with equality holds if and only if K and L have similar general chord.If i>n– 1,then the above inequality is reversed.

Obviously, ifτ= 0, then inequality (1.11) yields Theorem1.A, Corollary1.2gives The-orem1.C, respectively.

Remark1.2 IfKandLare nonempty compact convex sets, then Theorem1.1also is true. From this, takeL={o}in Theorem1.1. SinceK+{o}=Kand notice thatBτ

i({o}) = 0, thus

by inequality (1.9) we have the following.

Corollary 1.3 If K∈Kn_{,τ∈(–1, 1),and i<j<k,then}

B(_jτ)(K)k–i≤B(_iτ)(K)k–jB(_kτ)(K)j–i

with equality if and only if K is of constant width.

Because of{o} ∈S_on, hence letL={o}in Theorem1.2, we may obtain the following. Corollary 1.4 If K∈Sn

o,τ∈(–1, 1),and i<j<k,then

C_j(τ)(K)k–i≤C(_iτ)(K)k–jC(_kτ)(K)j–i

with equality if and only if K is of constant chord.

Ifτ = 0, then Corollary1.3and Corollary1.4respectively give Theorem1.Band Theo-rem1.D.

Our works belong to the asymmetric Brunn–Minkowski theory, which has its starting point in the theory of valuations in connection with isoperimetric and analytic inequali-ties. As an important research object in convex geometry, asymmetric Brunn–Minkowski theory has gotten rich development, readers can refer to [12–19].

2 Preliminaries

For nonempty compact convex bodiesKandL,λ,μ≥0 (not both zero), the Minkowski combinationλK+μLofKandLis defined by (see [1])

h(λK+μL,·) =λh(K,·) +μh(L,·), (2.1)

where “+” and “λK” are called Minkowski addition and Minkowski scalar multiplication, respectively. Whenλ=μ= 1, theK+Lis called Minkowski sum.

ForK,L∈Sn

o,λ,μ≥0 (not both zero), the radial combinationλ◦K+˜μ◦LofK,Lis

given by (see [1,20])

where “+” and “˜ λ◦K” are radial addition and radial scalar multiplication, respectively. Whenλ=μ= 1, theK+˜Lis radial sum ofKandL.

3 Proofs of theorems

In this part, we give the proofs of Theorem1.1and Theorem1.2. First, we give the follow-ing lemmas.

Lemma 3.1 Let f ∈Lp_(E),g∈Lq_{(E),real number p,q= 0,and} 1

p+

1

q= 1,if p> 1,then

E

f(x)pdx 1

p

E

g(x)qdx 1

q

≥

E

f(x)g(x)dx (3.1)

with equality if and only if there exists constants c1 and c2 such that c1|f(x)|p=c2|g(x)|q. The inequality is reversed if p< 0or0 <p< 1.Here Lp(E)denotes all function sets defined on a measurable set E in Lpspaces.

Lemma 3.2 Let f,g∈Lp_{(E),if real number p= 0and p> 1,then}

E

f(x)pdx 1

p

+

E

g(x)pdx 1

p

≥

E

f(x) +g(x)p 1

p

dx (3.2)

with equality if and only if there exists constants c1 and c2 such that c1|f(x)|p=c2|g(x)|q. The inequality is reversed if p< 0or0 <p< 1.

Proof of Theorem1.1 Ifj<n– 1, i.e.,n–j> 1, then from (1.5), (2.1), and (3.2), we have

B(_jτ)(K+L)n1–j ₌

1 n

Sn–1b

(τ)_(K+L,u)n–j_dS(u)

1

n–j

=

1 n

Sn–1

f1(τ)h(K+L,u) +f2(τ)h(K+L, –u) n–jdS(u) 1

n–j

=

1 n

Sn–1

f1(τ)h(K,u) +f1(τ)h(L,u)

+f2(τ)h(K, –u) +f2(τ)h(L, –u) n–jdS(u) 1

n–j

=

1 n

Sn–1

b(τ)_{(K,u) +b}(τ)_(L,u) n–j

dS(u) 1

n–j

=

1 n

Sn–1

b(τ)(K,u)

(k–j)(n–i)

(k–i)(n–j)_b(τ)_(K,u) (j–i)(n–k) (k–i)(n–j)

+b(τ)_(L,u)((kk––ij)()(nn––ij))_b(τ)_(L,u) (j–i)(n–k)

(k–i)(n–j) n–j_dS(u)

1

n–j

≤

1 n

Sn–1b

(τ)_(K,u)(k–_kj)(_–n_i–i)_b(τ)_(K,u)(j–i_k)(_–n_i–k)_dS(u) 1

n–j

+

1 n

Sn–1

b(τ)_(L,u)(k–_kj)(_–n_i–i)_b(τ)_(L,u)(j–i_k)(_–n_i–k)_dS(u) 1

n–j

In the inequality on the right above, notice thati<j<kmeans k_k–_–i_j > 1, thus by (3.1) we obtain

1 n

Sn–1b

(τ)_(K,u)(k–_kj)(_–n_i–i)_b(τ)_(K,u)(j–i_k)(_–n_i–k)_dS(u)

≤

1 n

Sn–1

b(τ)(K,u)(k–kj)(–ni–i) k–i k–j_dS(u)

k–j k–i

×

1 n

Sn–1

b(τ)(K,u)(j–ik)(–ni–k) k–i

j–i_dS(u)

j–i k–i

=B(_iτ)(K)kk––ji_B(τ)

k (K)

j–i

k–i_{. (3.4)}

From this, forj<n– 1, we can get the following inequality by (3.4):

1 n

Sn–1b

(τ)_(K,u)(k–_kj)(_–n_i–i)_b(τ)_(K,u)(j–i_k)(_–n_i–k)_dS(u) 1

n–j

≤B(_iτ)(K)

k–j

(k–i)(n–j)_B(τ)

k (K)

j–i

(k–i)(n–j)_{. (3.5)}

Similarly,

1 n

Sn–1b

(τ)_(L,u)(k–_kj)(_–n_i–i)_b(τ)_(L,u)(j–i_k)(_–n_i–k)_dS(u) 1

n–j

≤B(_iτ)(L)

k–j

(k–i)(n–j)_B(τ)

k (L)

j–i

(k–i)(n–j)_{. (3.6)}

Hence, by (3.3), (3.5), and (3.6) we have

B(_jτ)(K+L)n1–j ≤_B(τ)

i (K)

k–j

(k–i)(n–j)_B(τ)

k (K)

j–i

(k–i)(n–j) _+B(τ)

i (L)

k–j

(k–i)(n–j)_B(τ)

k (L)

j–i

(k–i)(n–j)_.

This yields inequality (1.9).

Ifn– 1 <j<n, then 0 <n–j< 1, this gives (3.3) is reversed. Butj<i<kmeans 0 <k_k–_–i_j< 1, thus (3.4) is reversed. Hence inequality (3.5) is reversed. Similarly, inequality (3.6) is also reversed. From this, we know that inequality (1.9) is reversed.

Ifj>n, thenn–j< 0 implies that (3.3) is reversed. But byn–j< 0 and (3.4), we see inequality (3.5) is reversed (inequality (3.6) is also reversed). These yield that inequality (1.9) is reversed.

Fori=j, by (3.3) (or its reverse) we easily get inequality (1.9) (or its reverse).

According to the equality conditions of inequalities (3.1) and (3.2), we see that equality holds in (1.9) (or its reverse) if and only ifKandLhave similar general width.

Proof of Theorem1.2 Forj<n– 1 andi≤j<k, sincen–j> 1, thus by (1.8), (2.2), and (3.2) we get

C_j(τ)(K+˜L)n1–j ₌

1 n

Sn–1c

(τ)_(K+˜L,u)n–j_dS(u)

1

n–j

=

1 n

Sn–1

f1(τ)ρ(K+˜L,u) +f2(τ)ρ(K+˜L, –u)

n–j

dS(u) 1

=

1 n

Sn–1

f1(τ)ρ(K,u) +f1(τ)ρ(L,u)

+f2(τ)ρ(K, –u) +f2(τ)ρ(L, –u)

n–j

dS(u) 1

n–j

=

1 n

Sn–1

c(τ)(K,u) +c(τ)(L,u)n–jdS(u) 1

n–j

=

1 n

Sn–1

c(τ)(K,u)

(k–j)(n–i)

(k–i)(n–j)_c(τ)_(K,u) (j–i)(n–k) (k–i)(n–j)

+c(τ)(L,u)

(k–j)(n–i)

(k–i)(n–j)_c(τ)_(L,u) (j–i)(n–k)

(k–i)(n–j) n–j_dS(u)

1

n–j

≤

1 n

Sn–1

c(τ)_(K,u)(k–_kj)(_–n_i–i)_c(τ)_(K,u)(j–i_k)(_–n_i–k)_dS(u) 1

n–j

+

1 n

Sn–1c

(τ)_(L,u)(k–_kj)(_–n_i–i)_c(τ)_(L,u)(j–i_k)(_–n_i–k)_dS(u) 1

n–j

. (3.7)

On the right-hand side of the above inequality, wheni<j<k, i.e., k_k–_–i_j> 1, by (3.1) we have

1 n

Sn–1c

(τ)_(K,u)(k–_kj)(_–n_i–i)_c(τ)_(K,u)(j–i_k)(_–n_i–k)_dS(u)

≤

1 n

Sn–1

c(τ)_(K,u)(k–j_k)(_–n_i–i) kk––ij_dS(u)

k–j k–i

×

1 n

Sn–1

c(τ)(K,u)(j–ik)(–ni–k) k–i j–i_dS(u)

j–i k–i

=C_i(τ)(K)kk––ji_C(τ)

k (K)

j–i

k–i_{. (3.8)}

Hence, forj<n– 1,

1 n

Sn–1c

(τ)_(K,u)(k–_kj)(_–n_i–i)_c(τ)_(K,u)(j–i_k)(_–n_i–k)_dS(u) 1

n–j

≤C(_iτ)(K)

k–j

(k–i)(n–j)_C(τ)

k (K)

j–i

(k–i)(n–j)_{. (3.9)}

Similarly,

1 n

Sn–1

c(τ)_(L,u)(k–_kj)(_–n_i–i)_c(τ)_(L,u)(j–i_k)(_–n_i–k)_dS(u) 1

n–j

≤C(_iτ)(L)

k–j

(k–i)(n–j)_C(τ)

k (L)

j–i

(k–i)(n–j)_{. (3.10)}

According to (3.7), (3.9), and (3.10), we see that

C_j(τ)(K+˜L)n1–j ≤_C(τ)

i (K)

k–j

(k–i)(n–j)_C(τ)

k (K)

j–i

(k–i)(n–j)_+C(τ)

i (L)

k–j

(k–i)(n–j)_C(τ)

k (L)

j–i

(k–i)(n–j)_.

Forn– 1 <j<nandj≤i<k, we easily obtain that (3.7) and (3.8) are reversed, thus inequalities (3.9) and (3.10) both are reversed. So, we can get that inequality (1.10) is re-versed.

Forj>nandi≤j<k, we know that inequality (3.7) is reversed, notice that inequality (3.8) still holds, thus by (3.8) andn–j< 0, inequality (3.9) is reversed. Similarly, inequality (3.10) is also reversed. Therefore, inequality (1.10) is reversed.

Wheni=j, by (3.7) (or its reverse) we easily get inequality (1.10) (or its reverse). The equality conditions of Lemma3.1and Lemma3.2show that equality holds in (1.10) (or its reverse) if and only ifKandLhave similar general chord.

Acknowledgements

The authors want to express earnest thankfulness for the referees who provided extremely precious and helpful comments and suggestions. This made the article more accurate and readable.

Funding

Research is supported in part by the Natural Science Foundation of China (Grant No. 11371224) and the Innovation Foundation of Graduate Student of China Three Gorges University (Grant No. 2019SSPY147).

Competing interests

The authors state that they have no competing interests.

Authors’ contributions

The authors were devoted equally to the writing of this article. The authors read and approved the final manuscript.

Publisher’s Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

Received: 17 March 2019 Accepted: 3 July 2019

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