1. Introduction According to a general Brunn-Minkowski inequality, we obtain a proof of the uniqueness of solutions to the following fully nonlinear elliptic Hessian equation: (1)σkuij+uδij=fup-1 on Sn,where u is the support function of convex bodies, uij are the second-order covariant derivations of u with respect to any orthonormal frame e1,e2,…,en on Sn, δij is the standard Kronecker symbol, Sn is the unit sphere of n-dimension, f is a positive function defined on Sn, k∈{1,2,…,n}, p>1, and σk is the kth elementary symmetric function defined as follows: for λ=(λ1,λ2,…,λn)∈Rn, (2)σkλ=∑1⩽i1<i2<⋯<ik⩽nλi1λi2⋯λik.The definition can be extended to any symmetric matrix W∈Rn×n by σk(W)=σk(λ(W)), where λ(W)=(λ1(W),λ2(W),…,λn(W)) is the eigenvalue vector of W.

Equation (1) arrives from the geometry of convex bodies. A compact convex subset of Euclidean (n+1)-space Rn+1 with nonempty interiors is called a convex body. An important concept related to a convex body Q is its support function.

Definition 1. Let M (the boundary of a convex body Q) be a smooth, closed, uniformly convex hypersurface enclosing the origin in Rn+1. Assume that M is parameterized by its inverse Gauss map X:Sn→M⊂Rn+1; the support function u of M (or Q) is defined by (3)ux=x,Xx, ∀x∈Sn,where ·,· denotes the standard inner product in Rn+1.

u is convex after being extended as a function of homogeneous degree 1 in Rn+1. Conversely, any continuous convex function u of homogeneous degree 1 determines a convex body as follows: (4)Q=y∈Rn+1:y·x⩽ux, ∀x∈Sn.From some basic concepts to support function, Minkowski sum [see Definition 4], and mixed volumes [see Definition 5], Minkowski developed a set of theories related to convex bodies. If k=n and p=1, (1) is the Monge-Ampère equation corresponding to the classical Minkowski problem (5)detuij+uδij=f on Sn,which has been solved by Nirenberg [1], Pogorelov [2, 3], Cheng and Yau [4], and many others. When p=1, (1) is the classical Christoffel-Minkowski problem: (6)σkuij+uδij=f on Sn.A necessary condition [3] for (6) to have a solution is (7)∫Snxifxds=0, ∀i=1,2,…,n+1,where ds is the standard area form on Sn. Guan et al. [5] obtained that (7) is sufficient for (6) to have an admissible solution [see Definition 6].

Firey [6] generalized the Minkowski sum to p-sum [see Definition 4] from p=1 to p⩾1 in 1962. Later, Lutwak [7] extended the classical surface area measure to the p-sum cases. Also in [7], Lutwak first introduced the general Minkowski problem, which is called Lp-Minkowski problem thereafter. In the smooth category, Lp-Minkowski problem is equivalent to considering the following Monge-Ampère equation: (8)detuij+uδij=fup-1 on Sn.The uniqueness of Lp-Minkowski problem for p>1 and p≠n+1 (the uniqueness holds up to a dilation if p=n+1) has been solved in [7]. However, the uniqueness for p<1 is difficult and still open. In [8], Jian et al. obtained that, for any -n-1<p<0, there exists a positive function f∈C∞(Sn) to guarantee that (8) has two different solutions, which means that we need more conditions to consider the uniqueness.

When considering cases 1⩽k<n, attention is paid to the generalized Christoffel-Minkowski problem. In the smooth category, we need to study the k-Hessian equation (1).

For (1), Hu et al. [9] got the existence and uniqueness of solutions to (1) when 1⩽k<n and p>k+1 under appropriate conditions. However, the uniqueness of (1) when p<1 has not been solved well. In this paper, we study the uniqueness of (1) for p>1.

Our main result is the following.

Theorem 2. Suppose u is a positive admissible solution of (9)σkuij+uδij=fup0 on Sn,where 1⩽k<n, k∈Z, p0∈R+∖k, and f is a positive function defined on the unit sphere Sn and then the uniqueness holds. If p0=k, the uniqueness holds up to a dilation, which means that if u solves (9), then au:∀a∈R+ are the whole solutions of (9).

Remark 3. Here, we rewrite (1) by (9), where p0=p-1.

The organization of this paper is as follows. In Section 2, we show some basic concepts and lemmas which have been obtained by Guan et al. in [10]. In Section 3, we prove two useful propositions according to the methods in [11]. In the last section, we prove the main theorem.

2. Preliminaries Definition 4. Given two convex bodies Q1 and Q2 in Rn+1 with respective support functions u1, u2, and λ,μ⩾0 (λ+μ>0), the Minkowski sum λQ1+μQ2⊂Rn+1 is defined by the convex body whose support function is λu1+μu2.

For p⩾1, let Q1 and Q2 be two convex bodies containing the origin in Rn+1 in their interiors, and λ,μ⩾0 (λ+μ>0). The convex body λ∘Q1+pμ∘Q2, whose support function is given by λu1p+μu2p1/p, is called Firey’s p-sum of Q1 and Q2, where “+p” means the p-summation and “∘” means Firey’s multiplication.

Definition 5. Let Q1,Q2,…,Qr be convex bodies in Rn+1 and the volume of their Minkowski sum (10)Q=λ1Q1+λ2Q2+⋯+λrQr, λi⩾0,is an (n+1)th degree homogeneous polynomial of the family λ1,λ2,…,λr. Specially, the volume of Q is (11)VolQ=Volλ1Q1+λ2Q2+⋯+λrQr=∑i1,i2,…,in+1=1rλi1λi2⋯λin+1VQi1,Qi2,…,Qin+1,where the functions V are symmetric. Then V(Q1,Q2,…,Qn+1) is called the Minkowski mixed volume of Q1,Q2,…,Qn+1.

Definition 6. For k∈1,2,…,n, let Γk be the convex cone in Rn which is determined by (12)Γk=λ∈Rn:σ1λ>0,σ2λ>0,…,σkλ>0.A function u∈C2(Sn) is called k-convex if (13)Wx=uijx+uxδij∈Γk, ∀x∈Sn,and u is called an admissible solution to (1) if u is k-convex and satisfies (1).

Definition 7. Let A1,A2,…,Am be symmetric real k×k matrices, λ1,λ2,…,λm∈R; the determinant of λ1A1+⋯+λmAm is a homogeneous polynomial of degree k in λ1,λ2,…,λm. Namely, (14)detλ1A1+⋯+λmAm=∑i1,…,ik=1mλi1⋯λikDkAi1,…,Aik.In fact, the coefficient λi1⋯λik depends only on Ai1,…,Aik; then they are uniquely determined. Dk(A1,…,Ak) is called the mixed discriminant of A1,…,Ak.

For later applications, we collect some results here which have been proved in [10].

Lemma 8. Let u1,u2,…,un+1 be the support function of convex bodies Q1,Q2,…,Qn+1, respectively. Denoting Minkowski mixed volume V(Q1,Q2,…,Qn+1) by V(u1,u2,…,un+1) and (15)Wm=umij+umδij, m=1,2,…,n+1,then(16)Vu1,u2,…,un+1=∫Snu1DnW2,W3,…,Wn+1ds,where Dn(W2,W3,…,Wn+1) is the mixed discriminant [see Definition 7] of W2,W3,…,Wn+1.

Remark 9. For all 1⩽k⩽n, setting uk+2=⋯=un+1=1, then (17)Vu1,…,uk+1,1,…,1≔Vk+1u1,u2,…,uk+1=∫Snu1DkW2,W3,…,Wk+1ds,where Dk(W2,W3,…,Wk+1) is the mixed discriminant of W2,W3,…,Wk+1. Furthermore, if u1=u2=⋯=un+1=u, denote V(u1,u2,…,un+1)≔V(u) and Vk+1(u1,u2,…,uk+1)≔Vk+1(u); then (18)Vu=∫Snudetuij+uδijds,Vk+1u=∫Snuσkuij+uδijds.

Lemma 10. V is a symmetric multilinear form on (C2(Sn))n+1.

Lemma 11. For any function u∈C2(Sn), W=uij+uδij, 1⩽k<n, we have the Minkowski type integral formula, (19)∫SnuσkWds=∫Snσk+1Wds,where ds is the standard area element on Sn.

The following is a form of Alexandrov-Fenchel inequality for positive k-convex functions which comes from [10].

Lemma 12 (Alexandrov-Fenchel inequality). If u1,u2,…,uk are k-convex, u1 is positive, and there exists l∈2,3,…,k such that ul⩾0 on Sn, then, for any v∈C2(Sn), (20)Vk+12v,u1,u2,…,uk⩾Vk+1u1,u1,u2,…,ukVk+1v,v,u2,…,uk,with equality if and only if v=au1+∑i=1n+1aixi for some constants a,a1,…,an+1.

3. Two Important Propositions Now we prove two important propositions. The methods we use are from [11].

Proposition 13. Suppose u0,u1>0 are k-convex; then (21)Vk+11/k+11-tu0+tu1⩾1-tVk+11/k+1u0+tVk+11/k+1u1, ∀t∈0,1,with equality if and only if u0=au1+∑i=1n+1aixi for some constants a,a1,…,an+1.

Proof. We only need to prove that (22)Ft=Vk+11/k+11-tu0+tu1is concave on 0,1. Setting ut=(1-t)u0+tu1, t∈0,1, we have (23)Ft=Vk+11/k+1ut,ut,…,ut︷k+1.By the symmetric multilinear property of V, it is obvious that (24)F′t=Vk+11/k+1-1ut,…,ut︷k+1Vk+1-u0+u1,ut,…,ut︷k,(25)F′′t=kVk+11/k+1-2ut,…,ut︷k+1·Vk+1ut,…,ut︷k+1·Vk+1-u0+u1,-u0+u1,ut,…,ut︷k-1-Vk+12-u0+u1,ut,…,ut︷k⩽0,where the last inequality uses (20); thus F is a concave function on 0,1. The equality condition is checked easily.

Proposition 14 (general Brunn-Minkowski inequality). Supposing u0,u1>0 are k-convex, then(26)∫Snu1σku0ij+u0δijds⩾Vk+11/k+1u1Vk+11-1/k+1u0,with equality if and only if u0=au1+∑i=1n+1aixi for some constants a,a1,…,an+1.

Proof. Setting(27)Ft=Vk+11/k+11-tu0+tu1-1-tVk+11/k+1u0-tVk+11/k+1u1,then F(0)=F(1)=0. By (21), F(t)⩾0; thus F′(0)⩾0; namely, (28)Vk+11/k+1-1u0Vk+1-u0+u1,u0,…,u0︷k+Vk+11/k+1u0-Vk+11/k+1u1⩾0.Then(29)Vk+11/k+1-1u0∫Sn-u0+u1σku0ij+u0δijds+Vk+11/k+1u0⩾Vk+11/k+1u1.By (19), (30)Vk+11/k+1-1u0∫Snu1σku0ij+u0δijds⩾Vk+11/k+1u1,and then(31)∫Snu1σku0ij+u0δijds⩾Vk+11/k+1u1Vk+11-1/k+1u0.