THE ORLICZ BRUNN-MINKOWSKI INEQUALITY

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1 THE ORLICZ BRUNN-MINKOWSKI INEQUALITY DONGMENG XI, HAILIN JIN, AND GANGSONG LENG Abstract. The Orlicz Bru-Mikowski theory origiated with the work of Lutwak, Yag, ad Zhag i 200. I this paper, we first itroduce the Orlicz additio of covex bodies cotaiig the origi i their iteriors, ad the exted the L p Bru-Mikowski iequality to the Orlicz Bru-Mikowski iequality. Furthermore, we exted the L p Mikowski mixed volume iequality to the Orlicz mixed volume iequality by usig the Orlicz Bru-Mikowski iequality.. Itroductio The classical Bru-Mikowski iequality was ispired by questios aroud the isoperimetric problem. May other cosequeces i covex geometry make it a corerstoe of the Bru-Mikowski theory, which provides a beautiful ad powerful apparatus for coquerig all sorts of geometrical problems ivolvig metric quatities such as volume, surface area, ad mea width. The classical Bru-Mikowski iequality see [3] states that for covex bodies K, L i Euclidea -space R, the volume of the bodies ad of their Mikowski sum K + L = {x + y : x Kad y L} are related by V K + L V K + V L, Mathematics Subject Classificatio. 52A20, 52A40. Key words ad phrases. Covex body, Orlicz additio, Orlicz Bru-Mikowski iequality, Orlicz mixed volume. The authors would like to ackowledge the support from the Natioal Natural Sciece Foudatio of Chia 09728, the Natioal Natural Sciece Foudatio of Chia 27244, ad the Shaghai Leadig Academic Disciplie Project S3004.

2 2 D.M. XI, H.L. JIN, AND G.S. LENG with equality if ad oly if K ad L are homothetic. I his survey, Garder [3] summarized the history of the Bru- Mikowski iequality ad some applicatios i other fields such as: probability ad statistics, iformatio theory, physics, elliptic partial differetial equatios, combiatorics, iteractig gases, shapes of crystals ad algebraic geometry. I the early 960s, Firey [2] defied for each p, what have become kow as Mikowski-Firey L p -additios or simply L p -additios of covex bodies. For the L p -additios, Firey [2] also established the L p Bru-Mikowski iequality a iequality that is also kow as the Bru-Mikowski-Firey iequality, see [35]. If p >, ad K, L R are covex bodies cotaiig the origi i their iteriors, the V K + p L p V K p + V L p,.2 with equality if ad oly if K ad L are dilates. The mixed volume V K, L of covex bodies K, L is defied by V K, L := V K + ϵl V K = h L uds K u, S.3 where S K is the surface area measure of K. The Mikowski mixed volume iequality for covex bodies K, L states that V K, L V K V L,.4 with equality if ad oly if K ad L are homothetic. For p >, the L p mixed volume of covex bodies K, L cotaiig the origi i their iteriors is defied by Lutwak [35] as V p K, L := p V K + p ϵ L V K. Lutwak [35] showed that the L p mixed volume has the followig itegral represetatio: V p K, L = S hl u phk uds K u..5 h K u

3 THE ORLICZ BRUNN-MINKOWSKI INEQUALITY 3 Lutwak s L p Mikowski mixed volume iequality [35] states V p K, L V K p p V L,.6 with equality if ad oly if K ad L are dilates. I the mid 990s, it was show i [35] ad [36] that a study of the volume of Mikowski-Firey L p -additios leads to a L p Bru- Mikowski theory. This theory has expaded rapidly see, e.g., [,2,4 0, 7 9, 2 23, 27, 29 32, 34 45, 48 5, 53 55, 58]. The Orlicz Bru-Mikowski theory origiated with the work of Lutwak, Yag, ad Zhag i 200. Precisely, Lutwak, Yag, ad Zhag [46, 47] itroduced Orlicz projectio bodies ad Orlicz cetroid bodies, ad they successively established the fudametal affie iequalities for these bodies. Haberl, Lutwak, Yag, ad Zhag [20] dealt with the eve Orlicz Mikowski problem. For related work, see also [2, 22, 32, 33]. Ludwig ad Reitzer [33] itroduced what soo came to be see as Orlicz affie surface area ad Ludwig [32] established its fudametal affie iequalities. For the developmet of the Orlicz Bru-Mikowski theory, see [24, 56, 60]. It seems atural, ow, to defie the Orlicz additio ad to give the Orlicz Bru-Mikowski iequality. We cosider the Orlicz additio, which is a extesio of L p -additio. Let C be the class of covex, strictly icreasig fuctios ϕ : [0, [0, + satisfyig ϕ0 = 0. It is ot hard to coclude from [52, p ] that ϕ C is cotiuous o [0, +, ad the left derivative ϕ l ad right derivative ϕ r exist. Furthermore, ϕ l is left-cotiuous o 0, +, ϕ r is right-cotiuous o [0, +, ad ϕ l ad ϕ r are positive o 0, +. Defiitio. Let ϕ C, ad let K, L R be covex bodies cotaiig the origi i their iteriors. We defie the Orlicz sum K + ϕ L by h K+ϕ Lu = if{τ > 0 : ϕ h Ku τ + ϕ h Lu }. τ

4 4 D.M. XI, H.L. JIN, AND G.S. LENG If ϕt = t p, p, the K + ϕ L = K + p L. Theorem is what we are callig the Orlicz Bru-Mikowski iequality. Theorem. Let ϕ C, ad let K, L R be covex bodies cotaiig the origi i their iteriors. The, we have V K ϕ V K + ϕ L V L + ϕ V K + ϕ L..7 Equality holds if K ad L are dilates. Whe ϕ is strictly covex, equality holds if ad oly if K ad L are dilates. Next we give the defiitio of Orlicz combiatio. Defiitio 2. Let ϕ C, ad let K, L R be covex bodies cotaiig the origi i their iteriors. Suppose α > 0 ad β 0. We defie the Orlicz combiatio M ϕ α, β; K, L or the Orlicz mea of covex bodies by h Mϕ α,β;k,lu = if{τ > 0 : αϕ h Ku τ + βϕ h Lu }..8 τ Sice the fuctio z αϕ h Ku + βϕ h Lu is strictly decreasig, we z z have h Mϕ α,β;k,lu = τ u, if ad oly if αϕ h Ku + βϕ h Lu =. τ u τ u.9 It is obvious that M ϕ, ; K, L = K + ϕ L. I Sectio 2, we will show that h Mϕ α,β;k,l is ideed a support fuctio of a covex body which cotais the origi i its iterior. Whe ϕt = t p p, the covex body M ϕ α, β; K, L is precisely the Firey combiatio see [2,35] α K + p β L. However, for geeral ϕ C, the could ot be defied, which meas we caot write α K + ϕ β L istead of M ϕ α, β; K, L.

5 THE ORLICZ BRUNN-MINKOWSKI INEQUALITY 5 Defiitio 3. Let ϕ C satisfy ϕ =, ad let K, L R be covex bodies cotaiig the origi i their iteriors. The Orlicz mixed volume is defied by V ϕ K, L = ϕ l V M ϕ, ϵ; K, L V K. The followig theorem shows that the it i Defiitio 3 exists ad has a itegral represetatio, which is a extesio of.5. Theorem 2. Let ϕ C satisfy ϕ =, ad let K, L R be covex bodies cotaiig the origi i their iteriors. The, we have V ϕ K, L = ϕ l V M ϕ, ϵ; K, L V K = hl u h K uds K u..0 h K u S ϕ The followig is the Orlicz mixed volume iequality. Theorem 3. Let ϕ C satisfy ϕ =, ad let K, L R be covex bodies cotaiig the origi i their iteriors. The, V L V ϕ K, L V Kϕ.. V K Equality holds if K ad L are dilates. Whe ϕ is strictly covex, equality holds if ad oly if K ad L are dilates. If ϕt = t p, p, the the correspodig results of Theorems -3 i the L p Bru-Mikowski theory are obtaied. This paper is orgaized as follows. Sectio 2 cotais the basic defiitio ad otatios, ad shows that the Orlicz combiatio of covex bodies is also a covex body. Sectio 3 lists the elemetary properties of Orlicz combiatio. I Sectio 4, we prove a geeral case of Theorem usig Steier symmetrizatio, which is oe of the methods to prove the origial Bru-Mikowski iequality. see e.g. [, Chapter 5, Sectio 5] or [57, p ]. However, for the Orlicz case, our proof is quite differet. Sectio 5 gives the proof of Theorem 2 ad Theorem 3.

6 6 D.M. XI, H.L. JIN, AND G.S. LENG Whe we were about to submit our paper, we were iformed that Garder, Hug, ad Weil [5] had also obtaied a Orlicz Bru-Mikowski iequality ad posted their results o the arxiv.org a couple of days before. Please ote that we use a completely differet approach techique of Steier symmetrizatio, although our results coicide with theirs. 2. Preiaries For quick later referece we collect some otatios ad basic facts about covex bodies. Good geeral refereces for the theory of covex bodies are the books of Garder [4], Gruber [6], Leichtweiss [25], ad Scheider [52]. Let S deote the uit sphere, B the uit -ball, ω the volume of B, ad o the origi i the Euclidea -dimesioal space R. Deote by K the class of covex bodies compact, covex sets with o-empty iteriors i R, ad let K o be the class of members of K cotaiig the origi i their iteriors. By ita ad A we deote, respectively, the iterior ad boudary of A R. The sets relita ad relbda are the relative iterior ad relative boudary, that is, the iterior ad boudary of A relative to its affie hull. We say a sequece {ϕ i } C is such that ϕ i ϕ C, provided max t I ϕ it ϕt 0, for every compact iterval I [0,. The support fuctio h K : R R of a compact covex set K R is defied, for x R, by h K x = max{x y : y K}, 2. ad it uiquely determies the compact covex set. Obviously, for a pair of compact covex sets K, L R, we have h K h L if ad oly if K L.

7 THE ORLICZ BRUNN-MINKOWSKI INEQUALITY 7 A fuctio is a support fuctio of a compact covex set if ad oly if it is positively homogeeous of degree oe ad subadditive. Let K K ad x K. Deote by νx a outer ormal vector of K at x. Obviously, h K νx = x νx. The, the hyperplae {y R y νx = h K νx} is a support hyperplae of K at x. We shall use δ to deote the Hausdorff metric o K. If K, L K, the Hausdorff distace δk, L is defied by or equivaletly, K if δk, L = mi{α : K L + αb ad L K + αb }, δk, L = max u S h Ku h L u. A class of covex bodies {K i } is said to coverge to a covex body δk i, K 0, as i. Let K K. The surface area measure S K of K is a measure o S defied by S K ω = x K,νx ω dh x, ω S, where H deotes the -dimesioal Hausdorff measure. The surface area measure has the followig property: K i K S Ki S K weakly. 2.2 Let ϕ C, K, L K o, α > 0 ad β 0. The defiitio of Orlicz Mikowski additio ad Orlicz combiatio are give i Sectio. I the followig, we check that the Orlicz Mikowski combiatio M ϕ α, β; K, L is ideed a covex body cotaiig the origi i its iterior. Set M = M ϕ α, β; K, L; i fact, we eed to show that the fuctio h M is homogeeous of degree oe ad subadditive, ad that h M is positive.

8 8 D.M. XI, H.L. JIN, AND G.S. LENG First, for γ > 0 we have hk γu hl γu h M γu = if{τ > 0 : αϕ + βϕ } τ τ = γ if{ τ γ > 0 : αϕ hk u hl u + βϕ } τ/γ τ/γ = γh M u. Next, we show that h M is subadditive. Set h M u = τ u ad h M v = τ v ; the we have αϕ h Ku τ u βϕ h Lv τ v =. Furthermore, = τ u hk u αϕ + τ v τ u + τ v τ u hl u + βϕ h Lu τ u = ad αϕ h Kv τ v + hk v αϕ τ u + τ v τ v + τ u βϕ + τ v hl v βϕ τ u + τ v τ u τ u + τ v τ v hk u + h K v hl u + h L v αϕ + βϕ τ u + τ v τ u + τ v hk u + v hl u + v αϕ + βϕ, τ u + τ v τ u + τ v which implies that h M u + v h M u + h M v. Fially, sice αϕ h K u h K u/ϕ α + βϕ h L u h K u/ϕ α, from.8, we have h M u h K u/ϕ α > 0. Thus, M ϕα, β; K, L cotais o i its iterior. 3. properties of Orlicz combiatio Suppose ϕ C, a, b, α > 0, ad β 0. Sice the fuctio z αϕ a + βϕ b is strictly decreasig, we defie a positive fuctio z z C ϕ α, β; a, b by z = C ϕ α, β; a, b, if ad oly if a αϕ z b + βϕ =. 3. z The fuctios C ϕ α, β; a, b have some properties listed i the followig lemma.

9 THE ORLICZ BRUNN-MINKOWSKI INEQUALITY 9 Lemma 3.. Suppose ϕ C ad a, b > 0. Let α > 0, β 0. i If d > 0, the C ϕ α, β; ad, bd = dc ϕ α, β; a, b. ii Suppose ϕ, ϕ 2 C. If ϕ 2 ϕ, the C ϕ2 α, β; a, b C ϕ α, β; a, b. iii Suppose a i, b i > 0 are such that a i a ad b i b. The, C ϕ α, β; a i, b i C ϕ α, β; a, b. iv Suppose {ϕ i } C are such that ϕ i ϕ. The, C ϕi α, β; a, b C ϕ α, β; a, b. v Suppose α i > 0, β i 0 are such that α i α ad β i β. The, C ϕ α i, β i ; a, b C ϕ α, β; a, b. Proof. i Suppose d > 0. By 3., we have ad bd = αϕ + βϕ C ϕ α, β; ad, bd C ϕ α, β; ad, bd a b = αϕ + βϕ, C ϕ α, β; ad, bd/d C ϕ α, β; ad, bd/d ad a = αϕ C ϕ α, β; a, b + βϕ Thus, we have C ϕ α, β; ad, bd = dc ϕ α, β; a, b. b C ϕ α, β; a, b. ii Set C ϕi α, β; a, b = z i, i =, 2. Sice ϕ 2 ϕ, we have a b a b = αϕ 2 + βϕ 2 αϕ + βϕ, z 2 z 2 z 2 z 2 which implies z 2 z. iii Set z i = C ϕ α, β; a i, b i, i =, 2,..., ad z 0 = C ϕ α, β; a, b. We will prove iii by showig that every subsequece of {z i } has a subsequece covergig to C ϕ α, β; a, b. From ai bi ai + b i = αϕ + βϕ < α + βϕ, z i z i z i we have z i < a i + b i /ϕ α+β, ad sice a i a, b i b, there is a costat R > 0 such that z i R, i =, 2,... Let {z i } deote a subsequece of {z i }. The {z i } has a coverget subsequece, also

10 0 D.M. XI, H.L. JIN, AND G.S. LENG deoted by {z i }, ad we suppose that z i z 0. Sice ϕ is cotiuous, we have z 0 > 0 ad αϕ a z 0 which implies z 0 = z 0. + βϕ b z 0 [ ai bi ] = αϕ + βϕ =, i z i z i iv Set τ i = C ϕi α, β; a, b, i =, 2,..., ad τ 0 = C ϕ α, β; a, b. We claim that i ϕ i x = ϕ x, 3.2 for all x > 0. Let η > 0 be arbitrary. Sice ϕ C, we coclude that ϕ is cocave o [0,. Hece ϕ is cotiuous o 0,. The, there exists a δ 0, x, such that ϕ x δ > ϕ x η, 3.3 ϕ x + δ < ϕ x + η. 3.4 Sice ϕ i ϕ uiformly o [ϕ x δ, ϕ x + δ], there exists a N > 0, such that ϕ i ϕ x δ < ϕϕ x δ + δ = x, 3.5 ϕ i ϕ x + δ > ϕϕ x + δ δ = x, 3.6 for all i > N. The, by 3.3, 3.4, 3.5, ad 3.6, we have ϕ x η < ϕ i x < ϕ x + η, for all i > N. Sice η > 0 is arbitrary, we complete the proof of our claim. From a b = αϕ i + βϕ i τ i τ i < α + βϕ i a + b τ i, we have τ i < a + b/ϕ i. By 3.2, there is a costat r > 0, α+β such that ϕ i > r, i =, 2,... Thus, {τ α+β i} is bouded. The,

11 THE ORLICZ BRUNN-MINKOWSKI INEQUALITY each subsequece of {τ i } has a coverget subsequece, also deoted by {τ i }, ad we suppose it coverges to τ 0. Sice = αϕ i a τ i + βϕ i b τ i αϕ i a τ i, the, a τ i ϕ i /α. Thus, by 3.2, we have τ 0 a > 0. ϕ /α By the cotiuity of ϕ i, ad ϕ i ϕ, we have αϕ a τ 0 + βϕ b τ 0 which implies τ 0 = τ 0. [ a b ] = αϕ i + βϕ i =, i τ i τ i v Set µ i = C ϕ α i, β i ; a, b, i =, 2,..., ad µ 0 = C ϕ α, β; a, b. From a b a + b = α i ϕ + β i ϕ > α i + β i ϕ, µ i µ i µ i we obtai µ i < a + b/ϕ α i +β i. Noticig that ϕ is cotiuous, we have that {µ i } is bouded. Hece, each subsequece of {µ i } has a coverget subsequece, deoted also by {µ i }, covergig to some µ 0. By the cotiuity of ϕ, we have µ 0 > 0 ad a b [ a b ] αϕ + βϕ = α µ 0 µ i ϕ + β i ϕ =, 0 i µ i µ i which implies µ 0 = µ 0. Notice that h Mϕ α,β;k,lu = C ϕ α, β; h K u, h L u, ad that the covergece of covex bodies is equivalet to the poitwise covergece of the correspodig support fuctios o S see e.g. [52, p ]. Therefore, we obtai the followig properties of Orlicz combiatio. Lemma 3.2. Suppose ϕ C ad K, L K o. Let α > 0, β 0. i If d > 0, the M ϕ α, β; dk, dl = dm ϕ α, β; K, L. ii Suppose ϕ, ϕ 2 C. If ϕ 2 ϕ, the M ϕ2 α, β; K, L M ϕ α, β; K, L. iii Suppose K i, L i K o are such that K i K ad L i L. The, M ϕ α, β; K i, L i M ϕ α, β; K, L.

12 2 D.M. XI, H.L. JIN, AND G.S. LENG iv Suppose {ϕ i } C are such that ϕ i ϕ. The, M ϕi α, β; K, L M ϕ α, β; K, L. v Suppose α i > 0, β i 0 are such that α i α ad β i β. The, M ϕ α i, β i ; K, L M ϕ α, β; K, L. Properties ii ad iv are ot used i this paper, but Properties i, iii ad v are basic for our proofs. 4. The Orlicz Bru-Mikowski iequality Let K R be a covex body. For u S, deote by K u the image of the orthogoal projectio of K oto u. We write l u K; y : K u R ad l u K; y : K u R for the overgraph ad udergraph fuctios of K i the directio u; i.e. K = {y + tu : l u K; y t l u K; y for y K u }. 4. Thus the Steier symmetral S u K of K K i the directio u ca be defied as the body whose orthogoal projectio oto u is idetical to that of K ad whose overgraph ad udergraph fuctios are give by l u S u K; y = l u S u K; y = 2 [l uk; y + l u K; y ]. 4.2 I this paper, we use the followig otatios: whe u S is fixed, the poit x = x, s always meas x + su, where x u ad s R. We will usually write h K x, s rather tha h K x, s. Suppose K K ad x, x 2 u. By 4., for a, s K, we have the, h K x, = a, s x, = a x + s a x + l u K; a, max {x, a, s} max {x a,s K a a + l u K; a }. K u O the other had, oticig that a, l u K; a K for arbitrary a K u, we have h K x, = max {x, a, s} max {x a,s K a a + l u K; a }. K u

13 Thus, we get that THE ORLICZ BRUNN-MINKOWSKI INEQUALITY 3 I a similar way, we get that h K x, = max a K u {x a + l u K; a }. 4.3 h K x 2, = max a K u {x 2 a + l u K; a }. 4.4 The followig lemma will be used i the proof of our theorem. Lemma 4.. [47, Lemma.2] Suppose K Ko ad u S. For y relitk u, the overgraph ad udergraph fuctios of K i directio u are give by ad l u K; y = mi x u {h K x, x y } l u K; y = mi x u {h K x, x y }. We refer to [47] for a proof. See [4] for a applicatio i the proof of the L p Busema-Petty cetroid iequality. I additio to Lemma 4., ote the followig elemetary fact: give a covex body K ad a directio u S, for each y relitk u, every outer ormal vector at the upper boudary poit y, l u K; y ca be writte as x,, while every outer ormal vector at the lower boudary poit y, l u K; y ca be writte as x 2,, where x, x 2 u. The followig lemma will be used i the proofs of our theorems. Lemma 4.2. Suppose K K. Let u S ad x, x 2 u. The, ad h K x, + h K x 2, 2h SuK x + x 2,, h K x, + h K x 2, 2h Su K x + x 2, Proof. For arbitrary a 0 K u, oticig that a 0, l u S u K; a 0 K, we have h K x, = max {x, a, s} x a a 0 + l u K; a ,s K

14 4 D.M. XI, H.L. JIN, AND G.S. LENG I a similar way, we have h K x 2, = The, max {x 2, a, s} x a 2 a 0 + l u K; a ,s K h K x, + h K x 2, x + x 2 a 0 + [l u K; a 0 + l u K; a 0], 4.9 for all a 0 K u. By 4.2, 4.3, ad 4.4, we have h Su K x + x 2, = max { x + x 2 2 a K u 2 ad h Su K x + x 2, = max 2 { x + x 2 a K u 2 a + l uk; a + l u K; a }, a + l uk; a + l u K; a } Sice 4.9 holds for all a 0 K u, equatios 4.0 ad 4. imply that 4.5 ad 4.6 hold. Lemma 4.3. Let ϕ C, α > 0, β 0, ad u S. If K, L K o, the M ϕ α, β; S u K, S u L S u M ϕ α, β; K, L. Proof. Set M = M ϕ α, β; K, L ad M S = M ϕ α, β; S u K, S u L. By Lemma 4., for arbitrary y relitm u, there are poits x, x 2 u such that ad l u M; y = h M x, x y l u M; y = h M x 2, x 2 y. Suppose z = h M x, ad z 2 = h M x 2,. The, hk x αϕ, hl x + βϕ, =, 4.2 z z ad hk x αϕ 2, hl x + βϕ 2, =. 4.3 z 2 z 2

15 THE ORLICZ BRUNN-MINKOWSKI INEQUALITY 5 By addig 4.2 multiplied with z ad 4.3 multiplied with z 2, usig the covexity of ϕ, ad by Lemma 4.2, we get hk x z + z 2 = z αϕ, hk x + z 2 αϕ 2, z z 2 hl x + z βϕ, hl x + z 2 βϕ 2, z z 2 [ hk x z + z 2 αϕ, + h K x 2, z + z 2 hl x + βϕ, + h L x 2, ] z + z 2 [ 2hSu K x +x 2 z + z 2 αϕ Therefore, we obtai hsu K x +x 2, 2 αϕ z + z 2 /2 which implies that, 2 z + z 2 hsu L x +x 2 + βϕ 2, z + z 2 / hSu L x +x 2, ] 2 + βϕ. z + z 2 4.5, 4.6 z + z 2 h MS x + x 2, Now 4.7 ad Lemma 4. show that l u S u M; y = 2 l um; y + 2 l um; y = 2 z x y + 2 z 2 x 2 y h MS x + x 2, x + x mi {h MS x, x y } x u = l u M S ; y. y I the same way, we obtai l u S u M; y l u M S ; y. Sice y relitm u is arbitrary, this completes the proof of the lemma.

16 6 D.M. XI, H.L. JIN, AND G.S. LENG x x 2 x ξ o 2 ξ x 4 Figure. Method to fid the chords. x 3 We say a chord [x, y] of a covex body K is a iterior chord of K if x, y itk, where x, y deotes the relative iterior of [x, y]. We say a chord is a boudary chord of a covex body if it is cotaied i the boudary of this covex body. It ca be cocluded from [52, Theorem..8] that a chord of a covex body K is a iterior chord if ad oly if there is a iterior poit of K that lies i this chord. Therefore, a chord of a covex body is either a iterior chord or a boudary chord. I order to get the equality coditio of.7, we eed the followig elemetary observatio. Lemma 4.4. Suppose K K. If x, x 2 K are two distict boudary poits of K, the x ad x 2 ca be coected by k iterior chords with k 3. Proof. Sice K is a covex body, we ca suppose x o is a iterior poit of K. Suppose 2, sice whe = it is obvious. Next, we describe how to fid the k iterior chords see Figure. i If [x, x 2 ] is a iterior chord of K, the [x, x 2 ] is the chord which we are searchig for. ii If we suppose [x, x 2 ] is ot a iterior chord of K, the [x, x 2 ] K. There exists a uique poit x 3 K, such that x o x, x 3. If

17 THE ORLICZ BRUNN-MINKOWSKI INEQUALITY 7 [x 3, x 2 ] is a iterior chord of K, the [x, x 3 ], [x 3, x 2 ] are the chords which we are searchig for. iii If we suppose [x, x 2 ] ad [x 3, x 2 ] are ot iterior chords of K, the [x 3, x 2 ] K. There exists a uique poit x 4 K, such that x o x 2, x 4. If [x 4, x ] is a iterior chord of K, the [x, x 4 ], [x 4, x 2 ] are the chords which we are searchig for. iv We suppose [x, x 2 ], [x 3, x 2 ] ad [x 4, x ] are ot iterior chords of K. By our costructio, the poits x o, x, x 2, x 3, x 4 lie i a 2-dimesioal plae. Let ξ be the midpoit of the chord [x, x 4 ]. The [ξ, x 2 ] is a iterior chord of K because ξ, x 2 x, x 3. There exists a uique poit ξ 2 x 2, x 3 such that x o ξ, ξ 2. So, ξ 2 K. It is clear that the chords [ξ, ξ 2 ] ad [x, ξ 2 ] are iterior chords of K. The, [x, ξ 2 ], [ξ 2, ξ ], [ξ, x 2 ] are the chords which we are searchig for. Suppose ϕ C is strictly covex; the followig lemma gives the ecessary equality coditio i the iequality of Lemma 4.3. Lemma 4.5. Suppose ϕ C is strictly covex. Let K, L Ko, ad α, β > 0. If M ϕ α, β; S u K, S u L = S u M ϕ α, β; K, L 4.8 for all u S, the K ad L are dilates. Proof. Set M = M ϕ α, β; K, L. Suppose [ξ, ξ 2 ] is a arbitrary iterior chord of the covex body M. Let u = ξ ξ 2 / ξ ξ 2 S, where deotes the Euclidea orm. The, [ξ, ξ 2 ] is parallel to u, ξ is the upper boudary poit, ad ξ 2 is the lower boudary poit. Thus, there exists y relitm u, such that ξ = y, l u MM; y, ad ξ 2 = y, l u M; y. Sice y relitm u, each outer ormal vector of M at ξ ca be writte as x,, ad each outer ormal vector at ξ 2 ca be writte as x 2,, where x, x 2 u. The, we have h M x, = x, y, l u M; y,

18 8 D.M. XI, H.L. JIN, AND G.S. LENG hece, Similarly, we have l u M; y = h M x, x y. l u M; y = h M x 2, x 2 y. By the same argumet as that for Lemma 4.3, we ca establish iequalities 4.4, 4.5, ad 4.7. If 4.8 holds for all u S, the 4.4, 4.5, ad 4.7 are all equalities. Sice ϕ is strictly covex, 4.4 is a equality if ad oly if h K x, z = h Kx 2, z 2, ad h Lx, z = h Lx 2, z 2, ad the there is a positive costat c 0 such that c 0 = h Kx, h L x, = h Kx 2, h L x 2,. 4.9 For every directio v S, there is a poit ξ 3 M, such that v is a outer ormal vector at ξ 3. If ξ 3 ξ, by Lemma 4.4, there are k 3 iterior chords of M, such that they coect ξ to ξ 3. Clearly, for each iterior chord of M, there is a similar equality as 4.9. The, we obtai that c 0 = h Kx, h L x, = h Kv h L v. If ξ 3 = ξ, the v is a ormal vector of M at ξ. Sice 4.9 holds for each ormal vector of M at ξ, we have c 0 = h Kx 2, h L x 2, = h Kv h L v. Therefore K ad L are dilates because v is arbitrary. From Lemma 4.3 ad Lemma 4.5 we get the followig theorem, which is ideed a origial versio of Orlicz Bru-Mikowski iequality. Theorem 4.6. Suppose ϕ C, K, L K o, ad α, β > 0. Let V K = a ω, ad V L = b ω, the V M ϕ α, β; K, L C ϕ α, β; a, b ω. 4.20

19 THE ORLICZ BRUNN-MINKOWSKI INEQUALITY 9 Equality holds if K ad L are dilates. Whe ϕ is strictly covex, equality holds if ad oly if K ad L are dilates. Proof. There is a sequece of directios {u i }, such that the sequece {K i } coverges to ab ad {L i } coverges to bb, where the sequeces {K i } ad {L i } are defied by K i = S ui S u K, ad L i = S ui S u L. Sice the Steier symmetrizatio preserves the volume, by Lemma 4.3 we have V M ϕ α, β; K, L V M ϕ α, β; ab, bb. From the defiitio of Orlicz combiatio of covex bodies, we get that M ϕ α, β; ab, bb is a -ball with radius C ϕ α, β; a, b. This implies If K ad L are dilates, there exists a covex body A Ko whose volume is ω, such that A, K, ad L are dilates. That is, K = aa, ad L = ba. By.9, we have aha u bha u αϕ + βϕ =, for all u S, τ u τ u where τ u = h Mϕ α,β;k,lu. This implies that h Mϕ α,β;k,lu = C ϕ α, β; a, bh A u, for all u S. Therefore, V M ϕ α, β; K, L = C ϕ α, β; a, b ω. Suppose ϕ is strictly covex. If equality holds i 4.20, the M ϕ α, β; S u K, S u L = S u M ϕ α, β; K, L, for all u S. By Lemma 4.5, we coclude that K ad L are dilates. The followig theorem is the geeral versio of Theorem. Theorem 4.7. Suppose ϕ C ad K, L Ko. If α, β > 0, the V K V L αϕ + βϕ. 4.2 V M ϕ α, β; K, L V M ϕ α, β; K, L

20 20 D.M. XI, H.L. JIN, AND G.S. LENG Equality holds if K ad L are dilates. Whe ϕ is strictly covex, equality holds if ad oly if K ad L are dilates. Proof. Let V K = a ω ad V L = b ω with a, b > 0. By 4.20, we have V M ϕ α, β; K, L Cϕ α, β; a, bω. Sice V K = aω, we get V K V M ϕ α, β; K, L a C ϕ α, β; a, b. Therefore, V K ϕ V M ϕ α, β; K, L ϕ a C ϕ α, β; a, b. I the same way, we also get Hece, V L ϕ V M ϕ α, β; K, L ϕ b C ϕ α, β; a, b a b = αϕ + βϕ C ϕ α, β; a, b C ϕ α, β; a, b V K V L αϕ + βϕ V M ϕ α, β; K, L V M ϕ α, β; K, L.. The equality coditio ca be obtaied as i Theorem 4.6. Takig α = β = i Theorem 4.7, we obtai Theorem. 5. The Orlicz mixed volumes I this sectio, we study the Orlicz mixed volume, which is defied by 3. Sice ϕ C, the left derivative ϕ l ad right derivative ϕ r exist, ϕ l is left-cotiuous ad ϕ r is right-cotiuous o [0, +. Furthermore, ϕ l ad ϕ r are positive o 0, +.

21 THE ORLICZ BRUNN-MINKOWSKI INEQUALITY 2 Lemma 5.. Let ϕ C satisfy ϕ =, let a, b > 0, ad β 0. The, C ϕ, β; a, b is differetiable at β = 0, ad C ϕ, 0; a, b = β 0 + C ϕ, β; a, b C ϕ, 0; a, b β = a b ϕ l ϕ. a Proof. Set z β = C ϕ, β; a, b, y β = ϕ a z β, for all β 0. Obviously, z 0 = a ad y 0 =. It follows by Lemma 3. v that z β a + ad y β as β 0 +. Sice ϕ l ad ϕ r are positive o 0, +, we have ϕ lt = ad ϕ ϕ l ϕ t rt = b Sice y β = βϕ z β, we have a z β β 0 + β Hece, we get = β 0 + y β β ϕ rϕ t, t 0, +. a z β b a z = ϕ β b = ϕ β 0 + y β a y β y β a ϕ l. C ϕ, 0; a, b = β 0 + z β z 0 β a = z z β β β 0 + β 0 + = a ϕ l ϕ b a. β The followig lemma shows that h Mϕ,ϵ;K,Lu is uiformly differetiable at ϵ = 0. This fact plays a key role i the proof of Theorem 2. Lemma 5.2. Let ϕ C satisfy ϕ =, ad let K, L K o. The the covergece i is uiform o S. h Mϕ,ϵ;K,Lu h K u = h Ku ϕ l ϕ hl u h K u 5.

22 22 D.M. XI, H.L. JIN, AND G.S. LENG Proof. Set K ϵ = M ϕ, ϵ; K, L for all ϵ 0. From Lemma 3.2 v, K ϵ K. Sice h Kϵ u = C ϕ, ϵ, h K u, h L u for each u S, by Lemma 5., we have h Kϵ u h K u = h Ku ϕ l ϕ hl u. h K u Let g : [0, + [0, + be a cocave fuctio, ad let x > y > 0. The, g lxx y < gx gy < g lyx y. 5.2 Let y ϵ u = ϕ hk u h K. The, y ϵ u ϵ u as ϵ 0 +. From ϕ C we coclude that ϕ is cocave o [0, +. By substitutig g = ϕ ito 5.2, ad the facts that y 0 u = ad ϕ y 0 u =, we have ϕ l y ϵu ϕ y ϵ u Notice that ad Therefore, we have h Kϵ u h K u ϵ ϕ l ϕ y ϵ u y ϵu. = h Kϵ u ϕ y ϵ u, ϵ hl u y ϵ u = ϵϕ. h Kϵ u h Kϵ u ϕ l ϕ hl u h K ϵ u h K u h Kϵ u ϵ h Kϵ u ϕ l h Ku/h Kϵ u ϕ hl u. h Kϵ u 5.3 Sice h Kϵ u h K u as ϵ 0 + uiformly o S, we have h L /h Kϵ coverges to h L /h K uiformly, ad h K /h Kϵ coverges to uiformly. Thus, h L /h Kϵ ad h K /h Kϵ are uiformly bouded ad they lie i a compact iterval I, ad ϕt is uiformly cotiuous o I. So the left side of 5.3 coverges to h Ku ϕ hl u ϕ l h K uiformly. u Notice that ϕ l t is left-cotiuous at t =, h K/h Kϵ coverges to uiformly, ad h K /h Kϵ. For arbitrary η > 0, there exists a δ > 0, such that ϕ l t ϕ l < η for all δ < t. For this δ, there

23 exists a θ > 0, such that THE ORLICZ BRUNN-MINKOWSKI INEQUALITY 23 δ < h Ku h Kϵ u, for all u S ad 0 ϵ < θ. The, ϕ hk u l ϕ h Kϵ u l < η, for all u S ad 0 ϵ < θ. Therefore, ϕ l h K/h Kϵ coverges uiformly to ϕ l, ad the right side of 5.3 coverges to h Ku ϕ hl u ϕ l h K u uiformly. Thus the covergece i 5. is uiform. Applyig the method i Lutwak [35] see also [20, Lemma ], we get the proof of Theorem 2 by Lemma 5.2. Proof of Theorem 2: Set K ϵ = M ϕ, ϵ; K, L, ϵ 0. By Property v i Lemma 3.2, we have that K ϵ K as ϵ 0 +, which implies that the surface area measure S Kϵ S K weakly. Sice the measures S Kϵ Lemma 5.2, we have ϵ 0 + ad ϵ 0 + h Kϵu h K u S ϵ h Kϵu h K u S ϵ Hece, we have V K ϵ V K ϵ, K are fiite, covergig weakly to S K, by ds Kϵ u = ds K u = S h K u ϕ l ϕ hl u h K u S h K u ϕ l ϕ hl u h K u V K, K ϵ V K = = hl u ϕ l h K uϕ S h K u ds K u, ds K u. ds K u. 5.4 Set l = ϕ l ϕ S hl u h K u h K uds K u. 5.5

24 24 D.M. XI, H.L. JIN, AND G.S. LENG From 5.4 ad.4, we have ad Thus, we obtai V K ϵ V K ϵ, K l = V K ϵ V Kϵ V K if = V K V K ϵ V K if, V K, K ϵ V K l = V K V Kϵ V K sup ϵ 0 + ϵ = V K V K ϵ V K sup. Therefore, l = V K V K ϵ V K. V K ϵ V K = ϵ 0 + = V K V K ϵ V K i=0 V K ϵ V K ϵ V K ϵ i V K i = l. 5.6 Combiig with 5.6 ad 5.5, we complete the proof of Theorem 2. Based o Theorem 2, we give two proofs of Theorem 3. The first uses the Orlicz Bru-Mikowski iequality, while the secod uses Jese s iequality. However, the first proof oly establishes the iequality, while the equality coditio ca be obtaied i the secod proof.

25 THE ORLICZ BRUNN-MINKOWSKI INEQUALITY 25 First Proof of Theorem 3: By Theorem 2, the followig it exists: V ϕ K, L = ϕ l V M ϕ, ϵ; K, L V K. By the covexity of ϕ ote that ϕ =, we have ϕ l x ϕx. 5.7 By Theorem 2, 5.6, 5.7, ad Theorem 4.7, we have V ϕ K, L = ϕ l V K ϵ V K = ϕ lv K = ϕ lv K V K ϵ V K V K V K ϵ V K V K if ϕ V K ϵ V L V K ϕ ϵ 0 + V K ϵ V L = V Kϕ. V K Thus we have established iequality.. The secod Proof of Theorem 3 uses the Jese s iequality. Secod Proof of Theorem 3: By Theorem 2, we have V ϕ K, L = ϕ l ϕ S hl u h K u h K uds K u. Sice S h K uds K u = V K,

26 26 D.M. XI, H.L. JIN, AND G.S. LENG h K S K V K.4, we have is a probability measure o S. By Jese s iequality ad V ϕ K, L V K hl u hk uds K u = ϕ S h K u V K h L u h K uds K u ϕ S h K u V K V K, L = ϕ V K V L ϕ. V K If K, L are dilates, it is easy to see that equality holds i.. Now suppose ϕ is strictly covex. If equality holds, the, by the equality coditio of Jese s iequality, there exists a s > 0 such that h L u = sh K u for almost every u S with respect to the measure h K S K. The, we have V K V ϕ K, L V K V L = ϕs = ϕ. V K Thus, s = V L / /V K /. Furthermore, the equality coditio of.4 implies that K ad L are homothetic. The, L = sk + t for some t R. Sice K has iterior poits, the support of the measure h K S K V K caot be cotaied i the great sphere of S orthogoal to t. Hece t = 0, which implies that K, L are dilates. Ackowledgemets The authors are grateful to the referees for their valuable suggestios ad commets. Refereces [] J. Bastero ad M. Romace, Positios of covex bodies associated to extremal problems ad isotropic measures, Adv. Math ,

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