VARIATIONS OF A CLASS OF MONGE-AMPÈRE TYPE FUNCTIONALS AND THEIR APPLICATIONS. 1. Introduction

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1 VARIATIONS OF A CLASS OF MONGE-AMPÈRE TYPE FUNCTIONALS AND THEIR APPLICATIONS HAODI CHEN, SHIBING CHEN, AND QI-RUI LI Abstract. In this paper, we study a class of Monge-Ampère type functionals arising from the L p dual Minkowski problem in convex geometry. As an application, we obtain some existence and non-uniqueness results for the problem.. Introduction The characterisation problem of geometric measures in convex geometry has a long history and strong influence on fully nonlinear PDEs. A best known example is the classical Minkowski problem. For a full discussion on this problem and its resolution, one may consult Cheng-Yau [8] and Pogorelov [34]. Other important geometric measures in Brunn-Minkowski theory include curvature measures and area measures, and the associated problems of prescribing curvature and area measures were also intensively studied. See Schneider s book [35] for a comprehensive introduction. Most recently Lutwak-Yang-Zhang [33] introduced the L p dual curvature measures and proposed the associated Minkowski type problem. Let K 0 be the set of all convex bodies (i.e., compact convex sets that have non-empty interior) in R n+ containing the origin in their interiors. Associated to each Ω K 0 are the support function u = u Ω : R and the radial function r = r Ω : R, which are respectively defined by u(x) = max{x z : z Ω}, and r(ξ) = max{λ : λξ Ω}. Let r(ξ) = r Ω (ξ) := r Ω (ξ)ξ. Then Ω = { r(ξ) : ξ }. Denote by ν = ν Ω : Ω the spherical image, namely ν(z) = {x : z x = u Ω (x)}. With these notions in hand, the radial Gauss mapping A = A Ω and the reverse radial Gauss mapping A = A Ω any ω, (.) A (ω) = {ν( r(ξ)) : ξ ω}, A (ω) = {ξ : ν( r(ξ)) ω}. are defined as follows: for 200 Mathematics Subject Classification. Primary 35J20, 35J96; Secondary 52A40. Key words and phrases. Minkowski type problems, Monge-Ampère type functionals, Second variation, Stability, non-uniqueness.

2 In [33] the L p dual curvature measures C p,q (Ω, ), where p, q R, are a two-parameter family of Borel measures on, defined by r q (ξ) (.2) Cp,q (Ω, ω) = u p (A (ξ)) dσ (ξ). A (ω) The associated Minkowski type problem was posed by Lutwak-Yang-Zhang [33]: Given a finite Borel measure µ on, find necessary and sufficient conditions on µ so that it is the L p dual curvature measure of a convex body. If µ is absolutely continuous w.r.t. σ and f = dµ dσ S is the Radon-Nikodym derivative, then, in terms of the support function n u, the problem reduces to the following Monge-Ampère equations (.3) det( 2 u + ui) = ( u 2 + u 2) n+ q 2 u p f(x) on, where is the covariant derivative w.r.t. an orthonormal frame on. The L p dual Minkowski problem includes the classical Minkowski problem as a special case, and unifies the L p -Minkowski problem and dual Minkowski problem introduced in [25, 3]. There is a large number of papers devoted to these problems, see e.g. [4, 7, 6, 7, 9, 2, 30, 32] for the L p -Minkowski problem, and [5, 5, 24, 25, 29, 38] for the dual Minkowski problem. The L p -Minkowski problem amounts to solve (.3) with q = n +. It is of particular interest, as the problem describes the self-similar solutions to the flows by powers of the Gauss curvature [3, 22]: (.4) t X(x, t) = K α (x, t)ν(x, t), where X(, t) is a time-dependent embedding of a family of convex hypersurfaces M t, K(, t) and ν(, t) denote the Gauss curvature and unit outer normal of M t respectively. In fact the self-similar solutions to (.4) satisfy (.3) with f and p = /α. For α =, flow (.4) was first studied by Firey [20] to model the shape change of tumbling stones. It was conjectured that, when α > /(n + 2), flow (.4) deforms each convex hypersurface in R n+ into a round point. Andrews proved the conjecture for the case n = in [2], and for the case n = 2 and α = in []. Very recently Brendle-Choi- Daskalopoulos [8] resolved this conjecture for all dimensions n 2. This shows that u is the only solution to (.3) when q = n +, p ( n, ) and f. However Lutwak-Yang-Zhang s L p dual curvature measure [33] is more general than (.2), as their definition allows a dependence of a fixed star body Q (i.e. a compact star-shaped set about the origin). If Q is taken as the unit ball B R n+, then their conception is formulated by (.2). 2

3 for non-constant f, the L p -Minkowski problem admits multiple solutions when p 0 [23, 27, 28, 37]. For general p, q R, the existence of solutions to (.3) was partially addressed in [6, 2, 4, 26], and the uniqueness of solutions was proved for p > q [26, 33]. The main goal of this paper is to show a non-uniqueness result for the L p dual Minkowski problem. We say u C 2 ( ) is uniformly convex if u is the support function of a convex body whose boundary has uniformly positive principal curvatures. following. Theorem.. Let f. Our main result is the Then equation (.3) admits an even, smooth, uniformly convex, positive solution u, provided that p, q R satisfy one of the following (A) q 2n 2 > p 0, (A2) q > 0 and q < p < min{0, q 2n 2}, where q is given in (.9) below. (A3) p + 2n + 2 < q 0. Clearly u is a solution to (.3) for f. Hence our Theorem. shows that if one of (A)-(A3) holds, then besides the unit ball B there is another origin-symmetric convex body Ω whose L p dual curvature measure coincides with the standard spherical measure σ. Since (.5) is not affine-invariant unless q = p = n +, ellipsoids are not solutions to the problem in general. In [24] the authors showed that, if f, n =, p = 0, and q is an even integer no less than 6, then (.3) has a non-constant solution. Our Theorem. (A) extends their result. Theorem. follows from Theorems.2 &.3 below. Both theorems are proved by studying a Monge-Amperè type functional (.5). For any finite Borel measure σ on and integrable function g, we use the following convention: gdσ := gdσ. S σ( n ) Let µ and µ be two finite Borel measures on. Consider the functional: (.5) J p,q,µ,µ (Ω) = Φ p,µ (Ω) + Ψ q,µ (Ω), for Ω K 0, where (.6) Φ p,µ (Ω) = p log u p Ω (x)dµ(x), if p 0, log u Ω (x)dµ(x), if p = 0, 3

4 and q log r q Ω (ξ)dµ (ξ), if q 0, S (.7) Ψ q,µ (Ω) = n log r Ω (ξ)dµ (ξ), if q = 0. Observe that J p,q,µ,µ is homogeneous degree zero, namely (.8) J p,q,µ,µ (tω) = J p,q,µ,µ (Ω), t > 0. For convenience, we shall omit sometimes the subscript µ (or µ ) in (.5)-(.7) if µ (or µ ) is exactly the standard spherical measure. We will see that, up to a rescaling, (.3) is the Euler equation of functional (.5) for dµ = fdσ and dµ = dσ. Let K e 0 K 0 be the set of all origin-symmetric convex bodies. By a Blaschke-Santaló type inequality [3], we are able to use a variational argument to prove Theorem.2 below, which shows the existence of origin-symmetric solutions to the L p dual Minkowski problem. To our best knowledge, even for the symmetric measures, Theorem.2 under condition (B2) gives the first existence result for the problem when p < 0, q > 0 and q n +, hence is of particular interest. We point out that, under condition (B) or (B3), the existence of origin-symmetric solutions was obtained in [2, 26] and in [29] for p = 0. As this existence result is needed in our main result Theorem., we still include a proof in this paper for reader s convenience. Theorem.2. Let dµ = dσ, dµ = fdσ, f be an even function on, and /C f C for some constant C > 0. Assume that p, q R satisfy one of the following (B) p 0 and q 0; (B2) q > 0 and q < p < 0, where q > 0 is defined as q if q > n +, q n n + if q = n +, (.9) q = nq if < q < n +, q + if 0 < q. (B3) p 0 and q 0. 4

5 Then there is a convex body Ω 0 K e 0 such that (.0) J p,q,µ (Ω 0 ) = max{j p,q,µ (Ω) : Ω K e 0}. Moreover Ω 0 is strictly convex and is C,γ for some γ (0, ), and satisfies (.) Cp,q (Ω 0, ω) = λ Ω0 fdσ, for any Borel set ω, where (.2) λ Ω0 = ω r q dσ u p dµ. If f is additionally smooth, then the support function u Ω0 convex, positive solution to (.3) with f replaced by λ Ω0 f. is an even, smooth, uniformly For p q, Ω 0 := λ p q Ω 0 Ω 0 solves the L p dual Minkowski problem, namely Ω 0 satisfies (.3) Cp,q ( Ω 0, ω) = fdσ, for any Borel set ω, and u Ω0 ω is an even, smooth, uniformly convex, positive solution to (.3), provided f is additionally smooth. Remark.. In [26], the existence of solutions to (.3) when f is not necessarily even was obtained for p > q. When p q, the existence result, without evenness assumption on f, becomes much more difficult. It was available for p > n and q = n + [9], and for p > and q > 0 [6]. In a subsequent paper [4], we will prove that, for p > 0 and all q R, the problem admits a weak solution if the prescribed measure µ is not concentrated on any closed hemisphere, while the evenness of µ is not required. Remark.2. As in [4], we are able to prove by approximation the L p dual Minkowski problem admits an origin-symmetric solution when p, q satisfy condition (B2) in Theorem.2, and f is an even and nonnegative function on, fdσ > 0, and L q q +p - integrable (when q + ) or L s -integrable for some s > (when q = + ). See Theorem 3.2 in Section 3 below. We then show that, if µ = µ = σ and q > p + 2n + 2, then the unit ball B is not a maximiser of (.0). This together with Theorem.2 proves Theorem.. Theorem.3. Let µ = µ = σ. If q > p + 2n + 2, then there is an even function η C ( ), and a small ε > 0, such that the convex body Ω t K e 0, whose support 5

6 function is u(x, t) = + tη(x), satisfies (.4) J p,q (B ) < J p,q (Ω t ), for t (0, ε). This paper is organised as follows. In Section 2, we calculate the first and second variations of functional (.5). We show that B is an unstable critical point of the functional J p,q provided q > p + 2n + 2, which consequently proves Theorem.3. In Section 3, we prove Theorem.2 via variational argument, and then complete the proof of Theorem.. The Poincáre inequality on is related to the stability of B under the functional (.5). It can be obtained by studying the eigenvalues of the spherical Laplace operator [36]. In Section 4, we provide an alternative proof for the Poincáre inequality with sharp constant via the uniqueness of the self-similar solution to the flow (.4) when α > n+2 [, 2, 8], which makes our paper self-contained. 2. Second variation for Monge-Ampère type functional (.5) Let u and r be respectively the support function and radial function of Ω K 0. Given any η C 0 ( ), there is an ε > 0, depending on min u and max η, such that u(x) + tη(x) > 0 for all x and t < ε. Consider a family of convex bodies (2.) Ω t = {z : z x u(x) + tη(x), x }, for t < ε. Let u(x, t) and r(x, t) be the support function and radial function of Ω t. Lemma 2.. Suppose that Ω is C and strictly convex at z 0 Ω. Then the limits below exist u(x 0, t) u(x 0, 0) u(x 0 ) := lim, t 0 t r(ξ 0, t) r(ξ 0, 0) ṙ(ξ 0 ) := lim, t 0 t where x 0 is the unit outer normal of Ω at z 0 and ξ 0 = z 0 / z 0 = AΩ (x 0). Furthermore (2.2) u(x 0 ) = η(x 0 ), and (2.3) ṙ r (ξ 0) = u u (x 0). Proof. By (2.) and the definition of support function, we have (2.4) u(x, t) u(x) + tη(x), for all x, t < ε. 6

7 Therefore u(x 0, t) u(x 0, 0) (2.5) lim sup t 0 + t η(x 0 ). On the other hand, let u z0 (x) = u(x) z 0 x and u z0 (x, t) = u(x, t) z 0 x. Since Ω is C at z 0, one infers that there exists a x t so that (2.6) u z0 (x 0, t) = (u z0 + tη)(x t ) with x t x 0 as t 0. For this, as u z0 (x 0, 0) = u z0 (x 0 ) = 0, if x tk x then u z0 (x ) = 0 and so x is a unit outer normal at z 0. Hence x must coincide with x 0. Therefore, by u z0 (x t ) 0, (2.7) lim inf t 0 + u(x 0, t) u(x 0, 0) t For t 0, (2.4) and (2.6) give respectively lim inf t 0 u(x 0, t) u(x 0, 0) t Hence (2.2) follows. u z0 (x 0, t) u z0 (x 0 ) = lim inf t 0 + t u z0 (x t ) + tη(x t ) = lim inf t 0 + t η(x 0 ). u(x 0, t) u(x 0, 0) η(x 0 ), and lim sup t 0 t η(x 0 ). Next we prove (2.3). For this, let h(x) = (η/u)(x) C 0 ( ). By (2.2), we have (2.8) u(x 0, t) = u(x 0 ) + tη(x 0 ) + o(t). It follows that (2.9) 0 = log r(ξ 0 ) + log u(x 0 ) log(ξ 0 x 0 ) = log r(ξ 0 ) + log u(x 0, t) log(ξ 0 x 0 ) + ( log u(x 0 ) log u(x 0, t) ) log r(ξ 0 ) + log r(ξ 0, t) th(x 0 ) + o(t). On the other hand, since Ω is strictly convex at z 0, there is a ξ t such that log r(ξ t, t) = log(ξ t x 0 ) log u(x 0, t) with ξ t ξ 0 as t 0. This together with (2.8) shows that (2.0) 0 log r(ξ t ) + log u(x 0 ) log(ξ t x 0 ) = log r(ξ t ) + log u(x 0 ) ( log u(x 0, t) log r(ξ t, t) ) = log r(ξ 0 ) + log r(ξ 0, t) th(x 0 ) + o(t). We complete the proof by (2.9) and (2.0). 7

8 In the rest of this section, we always assume that u is uniformly convex. Then the radial Gauss mapping A and the reverse radial Gauss mapping A, defined by (.), are one-to-one mappings. Given ω, we consider the cone-like region inside Ω C := {z R n+ : z = λν (x), λ [0, ], x ω}, where ν denotes the spherical image of Ω. It is well-known that the volume element of C can be expressed by dvol(c) = u(x) n + K(ν (x)) dσ (x) = n + rn+ (ξ)dσ (ξ), where K(z) is the Gauss curvature of Ω at z. It follows that the determinants of the Jacobian of the mappings A and A are given by (2.) JacA (x) = u(x) r n+ (A (x))k(ν (x)), JacA (ξ) = rn+ (ξ)k( r(ξ)). u(a (ξ)) Let η C ( ). By the uniform convexity of u, there is a small ε > 0 such that, for all t < ε, (i) Ω t defined by (2.) lies in K 0 ; (ii) u(x, t) := u(x) + tη is the support function of Ω t ; and (iii) u(x, t) is uniformly convex. Let us compute the first and second variations of functional (.5). Proposition 2.. Let Ω K 0 be a convex body whose support function u is uniformly convex. Given η C, let Ω t be the convex bodies defined by (2.). Let dµ = fdσ dµ = f dσ. Denote by α = n + q, β = p. Then d (2.2) J p,q,µ,µ (Ω t ) = J dt t=0 r S q dµ p,q,µ,µ (x)η(x)dσ (x), n where J p,q,µ,µ (x) = f A Ω (r A Ω )α K λuβ f(x), with λ = and K is the Gauss curvature of Ω calculated at ν Ω (x). 8 r q dµ u p dµ, and

9 If Ω is a convex body satisfying J p,q,µ,σ 0, then (2.3) = d 2 t=0 J dt 2 p,q,µ,σ(ω t ) { u S p dµ n S n β u β η 2 dµ + ( h ij η ij + Hη) u β ηdµ α p q u S p dµ n ( ) 2 } u β ηdµ, u β uη + u η ηdµ (r AΩ )2 where {h ij } is the inverse matrix of {u ij + uδ ij }, and H = h ii is the mean curvature of Ω. Proof. Note that, for all t < ε, Ω t are C 2 and strictly convex, with uniformly convex support function u(x, t) = u(x) + tη. Denote by r = r(ξ, t) the radial function of Ω t. Hence, by (2.) and Lemma 2., we compute the first variation of (.5) as follows d dt J p,q,µ,µ (Ω t) = ffl u β ηdµ + ffl r q ṙ u S p dµ n r S q dµ n S r dµ n = u β Sn f ηdµ + AΩ t u S p dµ n S r n S q dµ (r A n Ω t ) α K ηdσ { Sn f = AΩ t r S q dµ (r A n Ω t ) α K ηdσ r q dµ } (2.4) u β fηdσ u S p, dµ n where the geometric quantities above are of Ω t. Taking t = 0, we get (2.2). Note that r AΩ t = u 2 Ω t + u Ωt 2. Letting µ = σ in (2.4) and then differentiating (2.4) w.r.t. t again, we further calculate, by the assumption J p,q,µ,σ 0, d 2 t=0 (2.5) J dt 2 p,q,µ,σ(ω t ) { = S ij η r S q n (η ij + ηδ ij ) dσ n S (r A dσ n Ω )α uη + u η α S (r A n Ω )α+2 K ηdσ λβ u β fη 2 dσ q ( η )( ) u S p dµ n S (r A n Ω )α K dσ u β r q dσ ( S ) S ηdµ + p n 2 } ( ) 2 u β ηdµ u S p dµ n n { = u β( h ij η u S p ij + Hη ) β uη + u η ηdµ α u ηdµ dµ n S r n 2 β u β η 2 dµ + p q ( ) 2 } u β ηdµ. S u n S p dµ n This finishes the proof. 9

10 By virtue of Proposition 2., we are able to prove Theorem.3. Proof of Theorem.3. Let η C ( ) be an even function. As the unit ball B is uniformly convex, there is a small ε = ε η > 0, depending on η, such that, for all t < ε, Ω η t := {z R n+ : x z + tη(x), x } has support function u(x, t) = + tη(x), which is positive and uniformly convex. Clearly Ω η t K e 0. By Proposition 2., (2.6) and d 2 dt 2 t=0 J p,q (Ω η t ) = d J p,q (Ω η t ) = 0, dt t=0 ( ) ( η η + (n α β)η 2 dσ + (p q) ( ) 2dσS (2.7) = (q p) η η n where η := ffl ηdσ is the mean value of η. η 2 dσ, ηdσ By (i) in Theorem 4. below, there is an η 0 C ( ), with η 0 = 0, η 0 0, such that ) 2 where (2n δ p,q) η 2 0dσ η 0 2 dσ. δ p,q := q p 2n 2 > 0. Then for Ω t := Ω η 0 t, whose support function is + tη 0, one has by (2.7) d 2 t=0 (2.8) J dt 2 p,q (Ω t ) 2 δ p,q η0dσ 2 > 0. For ε 0 = ε η0 > 0 very small, one knows that Ω t is smooth and uniformly convex for all t < ε 0. Hence by (2.7) and (2.8) J p,q (Ω t ) = J p,q (B ) + t d J p,q (Ω t ) + d2 t=0 t2 J dt t=0 2 dt 2 p,q (Ω t ) + o(t 2 ) > J p,q (B ), for t (0, ε 0), provided ε 0, depending on η 0, is sufficiently small. Remark 2.. When p = 0 and q = n +, the second variation of the functional was obtained in [22]. 0

11 3. Proof of Theorems. &.2 In this section, we first prove Theorem.2. This together with Theorem.3 shows Theorem.. Given Ω K e 0, the polar set of Ω is defined as follows Ω = {y R n+ : y x x Ω}. The following generalised Blaschke-Santaló inequality was proved in [3]. Theorem 3. (Blaschke-Santaló type inequality [3]). Given q > 0, let q > 0 be the number given by (.9). For γ (0, q ], γ +, there is a constant C n,q,γ > 0 such that, (3.) ( r q Ω dσ ) ( ) q r γ Ω dσ γ Sn C n,q,γ, Ω K e 0. Theorem 3. enables us to solve the optimisation problem (.0). Proposition 3.. Under the assumptions of Theorem.2, there is a convex body Ω 0 K e 0 solving the maximisation problem (.0). Proof. We prove Proposition 3. when either (B) or (B2) holds. For the case (B3), we use a dual argument. Assume that either (B) or (B2) is satisfied. We have p 0 and q > 0. By the homogeneity (.8), it suffices to show there is a Ω 0 K0, e with Ψ q,σ(ω 0 ) = 0, such that (3.2) Φp,µ (Ω 0 ) = max { Φp,µ (Ω) : Ψ Ω K q,σsn(ω) = 0}, 0 e where u p Ωdµ, if p 0, p S Φ p,µ (Ω) := n log u Ω dµ, if p = 0. For (3.2), let {Ω j } K e 0, with Ψ q,σ(ω j ) = 0, be a maximising sequence. We denote u j = u Ωj and r j = r Ωj for convenience. We claim (3.3) max u j C, for some C > 0, independent of j. We next prove (3.3) (under the assumption (B) or (B2)) case by case: (i) p < 0; (ii) p = 0; (iii) p > 0.

12 Case I: p < 0. We follow an argument in [9]. Let δ > 0 be a fixed small constant. Set S j = {u j δ}, S j 2 = {δ < u j < /δ} and S j 3 = {u j /δ}. Since Ω j is origin-symmetric, we conclude that (3.4) S j 0, S j 2 0, as L j := max u j. Let Ω be the polar set of Ω, and r j = r Ω j. It is well known that r j = /u j, see e.g. [35]. Denote γ = p > 0. By condition (B2), γ < q. We have Φ p,µ (Ω j ) = u γ γ S j j fdσ Sj 2 Sj 3 C r γ S j j dσ + C δ S2 j + Cδ γ ( ) γ q C dσ S j γ q + C δ S2 j + Cδ γ, r q j for any γ < q < q. Since ffl r q j dσ =, if L j, then by (3.) and (3.4) lim sup j Φ p,µ (Ω j ) Cδ γ. As Φ p,µ (Ω j ) Φ p,µ (B ) > 0, we arrive a contradiction by letting δ 0. Case II: p = 0. Let l j = min r j and L j = max r j. By a rotation of coordinates we may assume that L j = r j (e ). Since Ω j is origin-symmetric, the points ±L j e Ω j. Hence (3.5) u j (x) = max{z x : x Ω j } L j x e, x. Therefore Φ p,µ (Ω j ) (log L j )/C log x e f(x)dσ (x) (log L j )/C + C, which implies that Φ p,µ (Ω j ), if L j. This cannot occur as {Ω j } is a maximising sequence. Case III: p > 0. Again let L j = max r j. As in Case II, we have (3.5). Hence Φ p,µ (Ω j ) u p p {x : x e 2 } j dµ Lp j /C if L j. Since {Ω j } is a maximising sequence, one infers L j C. Combining Case I-III, we have proved (3.3) under the assumption (B) or (B2). 2

13 ( Let w + j = max x uj (x) + u j ( x) ) ( and w j = min x uj (x) + u j ( x) ) be the maximum and the minimum of the width of Ω j. We next show that (3.6) w j /C, for some C > 0, independent of j. This estimate together with (3.3) means that Ω j is of uniformly good shape. For 0 < q < n +, we have ( ) n+ = r q j dσ q which shows (3.6) by using (3.3). For q n +, we have = r q j dσ = (w+ j )q ( rj w + j Again, (3.6) follows from (3.3). r n+ j dσ CVolume(Ω j ) (w + j )n w j, ) qdσs n C(w + j )q n Volume(Ω j ) C(w + j )q w j. As above, l j = min r j. Assume without loss of generality that l j = r j (e ). By the symmetry of Ω j, For q = 0, we thus have 0 = l j r j (ξ) ξ e, ξ. log r j dσ log l j log ξ e dσ log l j + C. This shows that l j δ for some δ > 0 uniformly, and so (3.6) follows. In virtue of (3.3) and (3.6), we conclude by the Blaschke selection theorem that Ω j, after passing to a subsequence, converges to a Ω 0 K0 e in Hausdorff distance, thus completing the proof under the assumption (B) or (B2). For case (B3), let {Ω j } K e 0 be a maximising sequence of functional J p,q,µ. p = q and q = p, and Ω j be the polar set of Ω j. One easily sees that (3.7) J p,q,µ,σ(ω) = J p,q,σ,µ(ω ) Ω K e 0. It then follows that {Ω j} is a maximising sequence of J p,q,σ,µ. Observe that if p, q satisfy (B3), then p, q satisfy (B). Hence, by our previous argument for (B), it is not hard to conclude that, after a proper rescaling, t j Ω j converges to a Ω 0 K e 0 such that J p,q,σ,µ(ω 0) = max{j p,q,σ,µ(ω) : Ω K e 0} 3 Let

14 By (3.7), we conclude that Ω 0 = Ω 0 satisfies (.0). Note that if p, q satisfy (B2), then p, q also satisfy (B2). Hence we would not get more by applying the above dual argument to (B2). We then show that, after a dilation, the maximiser Ω 0 obtained in Proposition 3. is a solution to the L p dual Minkowski problem, under an additional assumption: Ω 0 is C and strictly convex. Proposition 3.2. If Ω 0 is C and strictly convex, then Ω 0 satisfies (.). Proof. Let u and r be respectively the support function and radial function of Ω 0. For any even function η C 0 ( ), let Ω t K e 0 be the convex bodies given by (2.), with u replaced by u 0. Denote by u(x, t) and r(x, t) the support function and radial function of Ω t. By Lemma 2., we have as in proof of Proposition 2. d ( Sn J p,q,µ (Ω t ) = λ dt t=0 r S q Ω0 u p r q ) ηdµ + η A Ω0 dσ. dσ n S u A n Ω0 By [33, Lemma 5.], we further calculate d ( (3.8) J p,q,µ (Ω t ) = λ dt t=0 r S q Ω0 u p ηdµ + u p ηd dσ C ) p,q (Ω 0, ). n Since Ω 0 is the maximiser and η is arbitrary, we deduce that gd C p,q (Ω 0, ) = λ Ω0 gdµ, even function g C 0 ( ), thus completing the proof by the evenness of f. Proposition 3.3. Let Ω 0 be the maximiser obtained in Proposition 3.. Then Ω 0 is strictly convex and is C,γ for some γ (0, ) Proof. Let u be the support function of Ω 0 and ū = ū Ω0 extension, namely ū : R n+ R, defined by be its homogeneous degree one ū(y ) = sup Z Ω 0 Y Z. The face of Ω 0 with outer normal Y R n+ is then given by F Ω0 (Y ) = {Z Ω 0 : ū(y ) = Y Z}, 4

15 which lies in Ω 0 provided Y 0, and (3.9) ū(y ) = F Ω0 (Y ), where ū(y ) := {X R n+ : ū(z) ū(y )+ X, Z Y, Z R n+ } is the subgradient of ū at Y. See Schneider s book [35] for all this. For e, let L e be the hyperplane in R n+ which is tangential to at e. Denote by π = π e : R n the radial projection from L e to, π(y) = y + e. + y 2 Let v = v e : R n R be the restriction of ū on L e, that is (3.0) v(y) = ū(y + e) = + y 2 u(π(y)). It is not hard to check by (3.9) and (3.0) that (3.) v(y) = { X (X e)e : X ū(y + e) }. Let H n denotes the n-dimensional Hausdorff measure. Recall that the surface area measure S(Ω 0, ) is defined as (3.2) S(Ω 0, ω) = H n (ν Ω 0 (ω)), for Borel set ω. It follows from (3.9)-(3.) that for any D R n (3.3) M v (D) = x, e ds(ω 0, x), π(d) where M v (D) := H n ( v(d)) is the Monge-Ampère measure associated to v. We claim that, S(Ω 0, ) is absolutely continuous w.r.t. σ, and there is a C > 0, such that (3.4) /C ϱ Ω0 := ds(ω 0, ) dσ C. Note that ϱ Ω0 is the reciprocal Gauss curvature if Ω 0 is C 2 smooth. Once (3.4) is proved, we deduce by (3.3) and det Dπ(y) = ( + y 2 ) n+ 2 that ϱ Ω0 π (3.5) dm v = ( + y 2 ) n+2 2 dy. For (3.5), one may consult [8, 35] for a full discussion. By (3.4) and (3.5), the density of the Monge-Ampère measure of v in a compact set is bounded between two constants. For a given y 0 R n, let l y0 of (3.), the contact set C y0 be the support function of v(y) at y 0. In view := {y R n : v(y) = l y0 (y)} cannot contain a straight line in R n. Hence we conclude by [9, ] that v is strictly convex and C,γ 5 loc for some

16 γ (0, ). See also [9]. This implies that Ω 0 is strictly convex. Let ϕ : D R be the convex function such that {(x, ϕ(x)) : x D } Ω 0, where D is a closed convex domain, containing the origin, lying in Ω 0 {X R n+ : X e = 0}. One can check that ϕ is exactly the Legendre transform of v. Therefore, by(3.4) and (3.5), dm ϕ /dx is bounded between two positive constants. By [9, ], ϕ is C,γ for some γ (0, ). It remains to show (3.4). For η C 0 ( ), consider Ω t = {z R n+ : x z e tη(x) u(x), x }, which is a perturbation of Ω 0. Denote by u t = u(x, t) and r t = r(ξ, t) the support and radial functions of Ω t. It follows from [33, Theorem 6.4] that d (3.6) Ψ q (r t ) = ηd dt t=0 r S q dσ C q (Ω 0, ). n Since u t e tη u, one has (3.7) lim t 0 + u t (x) u(x) t By (3.6) and (3.7), we obtain where λ Ω0 ηu(x). J p,q (Ω t ) J p,q (Ω 0 ) 0 lim t 0 + t { λ r S q Ω0 u p ηdµ + ηd dσ C } q (Ω 0, ) n { } = λ r S q Ω0 u p fηdσ + (r AΩ dσ 0 ) q n uηds(ω 0, ), n is given by (.2), and the last equality is due to [25, Lemma 3.7]. Since η is arbitrary, u and r are bounded between two positive constants, we get ϱ Ω0 C. Let Ω 0 be the polar set of Ω 0. Then r = /u, see e.g. [35]. For η C 0 ( ), consider Ω t = conv{e tη(x) r (x)x : x }, and Ω t = (Ω t ). Denote by u t = u (ξ, t) and r t = r (x, t) the support and radial function of Ω t, by u t = u(x, t) and r t = r(ξ, t) the support and radial function of Ω t. Since r t e tη r, one gets u t e tη u. Therefore u t (x) u(x) (3.8) lim t 0 + t 6 ηu(x).

17 By [33, Theorem 6.], d (3.9) Ψ q (r t ) = ηd dt t=0 r S q dσ C q (Ω 0, ). n It follows by (3.8) and (3.9) J p,q (Ω t ) J p,q (Ω 0 ) 0 lim t 0 + t } {λ r S q Ω0 u p fηdσ (r AΩ dσ 0 ) q n uηds(ω 0, ), n which shows that ϱ Ω0 /C. This completes the proof. Remark 3.. To see the maximiser of the optimisation problem (.0) is a solution to the L p dual Minkowski problem (.), one can also follow the argument in [25, Lemma 5.]. We are at the position to finish the proof of Theorem.2. Proof of Theorem.2. By Proposition , it remains to show u = u Ω0 is smooth and uniformly convex. Note that for p q, it is not hard to see Ω 0 := λ p q Ω 0 satisfies (.3), and u Ω0 solves (.3). By the homogeneity (.8), Ω 0 is also a maximiser for (.0). By [25, Lemma 3.7] and [33, Proposition 5.4], it follows from (3.8) that, η C 0 ( ), d ( ) J p,q,µ (Ω t ) = λ dt t=0 r S q Ω0 u p ηfdσ + (r AΩ dσ 0 ) q n ηds(ω 0, ). n Since Ω 0 is the maximiser of (.0), we obtain (3.20) ds(ω 0, ) dσ = λ Ω0 (r A Ω 0 ) n+ q u p f. Given any e, let v and ϕ = v (the Legendre transform of v) be as in Proposition 3.3. Then (3.2) det D 2 v = λ Ω0 ( + y 2 ) n++p 2 v p ( Dv 2 + (Dv y v) 2 ) n+ q 2 f π, 7 Ω 0

18 and (3.22) det D 2 ϕ = λ Ω 0 (+ Dϕ 2 ) n++p 2 (Dϕ x ϕ) p ( x 2 +ϕ 2 ) n+ q Dϕ, 2 /f( ), + Dϕ 2 in the Aleksandrov sense. By Proposition 3.3, v and ϕ are strictly convex and are C,γ for some γ (0, ). If f is Hölder, then the right hand sides of (3.2) and (3.22) are both Hölder continuous. By [0], v and ϕ are both C 2,γ for some γ (0, ). Smoothness of v and vφ then follows from the standard theory of uniformly elliptic equations, provided f C ( ). Hence Ω 0 is smooth. By (3.20), u C ( ) solves (.3) with f replaced by λ Ω0 f. The following result improves Theorem.2 under condition (B2). Theorem 3.2. Let p, q satisfy condition (B2) in Theorem.2, and dµ = fdσ, where f is an even, non-negative function and fdσ > 0. Assume that f L q q +p ( ) if q +, or f L s ( ) for some s > if q = +. Then there is a convex body Ω K0 e such that C p,q (Ω, ω) = µ(ω) for all Borel set ω. Proof. We use an approximation argument similar to [4]. For positive integers j, let dµ j = f j dσ be a sequence of measures, where f j is a truncation of f, j if f(x) j, f j (x) = f(x) if /j < f(x) < j, /j if f(x) /j. Recall that J p,q,µj satisfies (.8). Hence by Theorem.2, there is a Ω j K e 0 such that, (3.23) Cp,q ( Ω j, ω) = µ j (ω), for any Borel set ω, and if r j = and ũ r Ωj j = then u Ωj ( ) ( ) (3.24) ũ p j dµ p j r q j dσ q = exp J p,q,µj ( Ω j ) exp J p,q,µj (B ) /C f,n,p, Sn for a positive constant C f,n,p > 0, independent of j. ( ffl ) Let Ω j = λ j Ωj, where λ j = r q j dσ q so that (3.25) r q Ω j dσ =. 8

19 Let u j = u Ωj, r j = r Ωj and L j := max u j. As in the proof of Proposition 3., for a small constant δ > 0, let S j = {u j δ}, S j 2 = {δ < u j < /δ} and S j 3 = {u j /δ}. It is not hard to see that (3.26) S j 0 and S j 2 0, if L j. First let us consider the case q. Denote γ = p > 0 and r j = r Ω j, the radial function of Ω j (the polar set of Ω j ). We have u p j dµ j = u γ S j j f j dσ Sj 2 Sj 3 ( ) γ ( q (3.27) C dσ f (3.28) r q j Cδ γ, if L j, S j q q γ j dσ ) q γ q + C δ S j 2 f j dσ + Cδ γ where f L q q +p ( ), (3.), (3.25) and (3.26) are used for the last line. As the LHS of (3.24) is rescaling invariant, its value is unchanged if ũ j, r j are replaced by u j, r j. We conclude that L j are uniformly bounded, by (3.25), (3.28) and sending δ 0. As in the proof of Proposition 3., we also deduce from (3.25) that l j := min u j stay uniformly away from zero. In view of (3.23), for q +. (3.29) ( ) γ ( r q j dσ = ũ p j dµ j = r γ j f jdσ r q q j dσ The first equality in (3.29) together with (3.24) shows that r q j dσ /C f,n,p,q > 0. q q f γ j dσ ) q γ q. While the inequality in (3.29), (3.) and f L q q +p ( ) give r q j dσ C f,n,p,q. Hence /C f,n,p,q λ j C f,n,p,q, for a constant C f,n,p,q > 0 only depending on f, n, p, q. The above estimates for L j, l j, λ j imply max u Ωj and min u Ωj are uniformly bounded from above and below. By the Blaschke selection theorem, Ω j converges, after passing to a subsequence, to a Ω K e 0 in Hausdorff distance. By the weak convergence of L p dual curvature measures [33, Proposition 5.2], it follows from (3.23) that C p,q (Ω, ω) = µ(ω), thus completing the proof for q +. 9

20 When q = +, (3.27) and (3.29) still hold if q is replaced by α = sγ. Hence we s can finish the proof by the same discussion as above. Next we prove Theorem.. Proof of Theorem.. By Theorem.2 and by the homogeneity (.8), there is a Ω 0 K e 0 such that u Ω0 is a solution to the equation (.3) with f and J p,q (Ω 0 ) = max{j p,q (Ω) : Ω K e 0}. We deduce from Theorem.3 that J p,q (B ) < J p,q (Ω 0 ). Therefore u Ω0 u B. While u B and u Ω0 both solve (.3) when f. 4. Sharp Poincaré inequality on This section is devoted to the Poincaré inequality on. This inequality is wellknown and has many applications. It can be proved by studying the eigenvalues of the spherical Laplace operator [36]. We prove it by the stability of the unit ball B under the functional (.5) and the uniqueness of the self-similar solution to the powered Gauss curvature flow (.4). Theorem 4.. We have (i) inf (ii) { ffl η 2 dσ ffl η 2 dσ inf { ffl η 2 dσ ffl η 2 dσ : η C ( ) is even, : η C ( ), } ηdσ = 0, η 0 = 2n + 2; } ηdσ = 0, η 0 = n. Remark 4.. By approximation, Theorem 4. holds for η W,2 ( ). Proof of (i) in Theorem 4.. Let Ω t = {z R n+ : z x + tη(x)} K e 0. Consider J p (Ω t ) := J p,n+ (Ω t ). 20

21 By Proposition 2., or more precisely by letting q = n + in (2.7), we have (4.) d 2 d 2 t=0 J p (Ω t ) = (n + p) (η η) 2 dσ η 2 dσ. When f and q = n +, (.3) is the equation of the self-similar solutions to the flow (.4) with p = /α. By [, 2, 8], u is the only solution for p ( n, ) 2. By [3, 9], the equation (.3) with p > n and q = n + admits a solution which maximises the functional J p. We therefore conclude that d 2 d 2 t=0 J p (Ω t ) 0, p > n. This together with (4.) implies, by letting p n, (4.2) (η η) 2 dσ 2n + 2 η 2 dσ. We next show (4.2) is sharp. Assume not, then for sufficiently small ε > 0, there is an even η ε 0, η ε = 0, such that η ε 2 dσ = (2n + 2 2ε) η 2 εdσ. Let Ω ηε t = {z R n+ : z x + tη ε }. Then for p = n + ε we have by (4.), d 2 d 2 t=0 J n +ε (Ω ηε t ) = ε η 2 εdσ > 0. This means, by virtue of [3, 9], there is another convex body Ω B maximising J n +ε among K e 0, and u Ω solves (.3) with f, p = n + ε, and q = n +, contradicting with the uniqueness of the solution. Proof of (ii) in Theorem 4.. Consider the functional (4.3) Jp (Ω, z) = p log u p zdσ + n + log r n+ z dσ, where z Int Ω and u z, r z are the support and radial function of Ω w.r.t. the centre z, namely u z (x) = max{(y z) x : y Ω} and r z (ξ) = max{λ : λξ + z Ω}. This functional was used by Andrews-Guan-Ni [3] in the study of the flow (.4) with α = ( p). Note that the second term on the RHS of (4.3) is independent of z, as dσ = Volume(Ω). n+ r n+ z 2 In fact u is also the unique solution for p > and p n +, see e.g. [32]. 2

22 Given Ω, let z e = z e (Ω) be the entropy point of Ω, namely z e minimises z J p (Ω, z), among all z Int Ω. For p <, it was proved in [3] that for each bounded convex Ω with Int Ω, there exists a unique entropy point z e IntΩ, and one readily sees x (4.4) uz p e (x) dσ Sn(x) = 0. Let Ω t = {z R n+ : z x + tη} K 0. Denote by z(t) = z e (Ω t ), the entropy point of Ω t. By Lemma 2., we compute, for t very small, d (4.5) dt J ( η p (Ω t, z(t)) = r + n z dσ S K dσ r n+ z dσ u p n S zdσ n uz p (η ż x)dσ where u z, r z are support and radial function of Ω t w.r.t. z = z(t), and K is the Gauss curvature of Ω t. Differentiating (4.5) again, we obtain ), (4.6) = d 2 dt J 2 p (Ω t, z(t)) { h ij (η ij + ηδ ij) r n+ S ηdσ r + β n z dσ n z dσ S K u p n S zdσ n n + η u p S zdσ n S K dσ u p r S z ηdσ + p n z n+ dσ ( ( ) 2 n u p S zdσ n u p 2 z uz p η(η ż x)dσ ηdσ ) 2 }, where β = p and h ij is the inverse matrix of u ij + uδ ij. By (4.4), one has u p z ż xdσ = 0, t. Differentiate this identity w.r.t. t to get (4.7) u p 2 z ηż xdσ = u p 2 z (ż x) 2 dσ. Hence one infers by plugging (4.7) in (4.6) and by Ω 0 = B, z(0) = 0, d 2 dt 2 t=0 Jp (Ω t, z(t)) = (4.8) = ( η + nη)ηdσ β ( (n + p) ηdσ ) 2 η 2 dσ + (n + p) 22 η 2 dσ + β (ż x)ηdσ (η η) 2 dσ + β (ż x) 2 dσ.

23 It is straightforward to see (4.9) (ż x) 2 dσ = ż 2 ( ż ż x)2 dσ = ż 2 x 2 dσ. Since z(t) is determined by (4.4), hence depends on p. For clarification, let us denote it by z p (t). Then we have by differentiating (4.4) 0 = (p ) Sending t = 0, we have u p 2 z p(t) (η ż p x)xdσ, for all p <. ηxdσ = (ż p (0) x)xdσ = ż p (0) (e p x)xdσ, where e p = ż p (0)/ ż p (0). Multiplying e p at both sides one deduces ( ) (4.0) ż p (0) = x 2 dσ (x e p )ηdσ C η L ( ), for all p <. As B maximises J p among all Ω K 0 [, 2, 3, 8], we have by plugging (4.9) in (4.8) 0 η 2 dσ + (n + p) Sending p, we get by (4.0) (η η) 2 dσ + (p ) ż p (0) 2 x 2 dσ. (4.) (η η) 2 dσ n η 2 dσ. It remains to show (4.) is sharp. If not, then for sufficiently small ε > 0, there is an η ε 0, η ε = 0, such that η ε 2 dσ = (n ε) η 2 εdσ. Let Ω ηε t = {z R n+ : z x + tη ε }. Then for p = ε, by (4.8) and (4.0) d 2 d 2 t=0 J ε (Ω ηε t, z(t)) = ( ε + ε) η 2 εdσ ε ż ε (0) 2 C ε η ε 2 L ( ) Cε η ε 2 L ( ) > 0, x 2 dσ provided ε sufficiently small. This implies that B is not a maximiser of J p, thus arriving a contradiction. 23

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