(1)THE Lp MINKOWSKI PROBLEM FOR POLYTOPES GUANGXIAN ZHU Abstract

THE L_p MINKOWSKI PROBLEM FOR POLYTOPES

GUANGXIAN ZHU

Abstract. Necessary and sufficient conditions are given for the existence of solutions to the discreteL_p Minkowski problem for the critical case where 0< p <1.

1. Introduction

The setting for this paper is n-dimensional Euclidean space Rⁿ. A convex body in Rⁿ is a compact convex set that has non-empty interior. If K is a convex body in Rⁿ, then the surface area measure,S_K, of K is a Borel measure on the unit sphere,Sⁿ⁻¹, defined for a Borel ω ⊂Sⁿ⁻¹, by

S_K(ω) = Z

x∈ν_K⁻¹(ω)

dHⁿ⁻¹(x),

whereν_K :∂⁰K →Sⁿ⁻¹ is the Gauss map of K, defined on∂⁰K, the set of points of ∂K that have a unique outer unit normal, and Hⁿ⁻¹ is (n−1)-dimensional Hausdorff measure.

The Minkowski problem is one of the cornerstones of the classical Brunn-Minkowski theory:

What are necessary and sufficient conditions on a finite Borel measure µon Sⁿ⁻¹ so that it is the surface area measure of a convex body in Rⁿ?

More than a century ago, Minkowski himself solved his problem for the case where the given measure is discrete [47]. The complete solution to this problem for arbitrary measures was given by Aleksandrov, and Fenchel and Jessen (see, e.g., [52]): Ifµis not concentrated on a great subsphere of Sⁿ⁻¹, thenµ is the surface area measure of a convex body if and only if

Z

Sⁿ⁻¹

udµ= 0.

In [38], Lutwak showed that there is an L_p analogue of the surface area measure and posed the associatedL_p Minkowski problem which has the classical Minkowski problem as an important case. If p ∈R and K is a convex body inRⁿ that contains the origin in its interior, then the L_p surface area measure,S_p(K,·), of K is a Borel measure on Sⁿ⁻¹ defined for a Borelω ⊂Sⁿ⁻¹, by

S_p(K, ω) = Z

x∈ν_K⁻¹(ω)

(x·ν_K(x))^1−pdHⁿ⁻¹(x).

Obviously, S₁(K,·) is the classical surface area measure ofK. In recent years, the L_p surface area measure appeared in, e.g., [1, 4, 7, 19, 20, 22, 23, 27, 35–37, 40–42, 45, 46, 48–51, 55].

Today, the L_p Minkowski problem is one of the central problems in convex geometric analysis.

It can be stated in the following way:

Lp Minkowski problem: For fixed p, what are necessary and sufficient conditions on a finite Borel measure µon Sⁿ⁻¹ so that µis the L_p surface area measure of a convex body in Rⁿ?

2010Mathematics Subject Classification: 52A40.

Key Words: Polytope,Lp surface area measure,Lp Minkowski problem, Monge-Amp`ere equation.

1

Whenµhas a densityf, with respect to spherical Lebesgue measure, theLp Minkowski problem involves establishing existence for the Monge-Amp`ere type equation:

(1.1) h^1−pdet(h_ij +hδ_ij) = f,

where h_ij is the covariant derivative of hwith respect to an orthonormal frame on Sⁿ⁻¹ and δ_ij is the Kronecker delta.

Obviously, theL₁Minkowski problem is the classical Minkowski problem. Establishing existence and uniqueness for the solution of the classical Minkowski problem was done by Aleksandrov, and Fenchel and Jessen (see, e.g., [52]). When p 6= 1, the Lp Minkowski problem has been studied by, e.g., Lutwak [38], Lutwak and Oliker [39], Guan and Lin [18], Chou and Wang [10], Hug, et al. [30], B¨or¨oczky, et al. [5]. Additional references regarding the L_p Minkowski problem and Minkowski-type problems can be found in [5, 8, 10, 17–21, 28–30, 32–34, 38, 39, 44, 53, 54].

The solutions to the Minkowski problem and the L_p Minkowski problem connect with some important flows (see, e.g., [2, 3, 9, 12, 31]), and have important applications to Sobolev-type in- equalities, see, e.g., Zhang [58], Lutwak, et al. [43], Ciachi, et al. [11], Haberl and Schuster [24–26], and Wang [57].

Most previous work on the L_p Minkowski problem was limited to the case where p > 1. The reason that uniqueness of solutions to the L_p Minkowski problem for p > 1 can be shown is the availability of mixed volume inequalities established by Lutwak [38]. One reason that the L_p Minkowski problem becomes challenging when p < 1 is because little is known about the mixed volume inequalities when p < 1 (see, e.g., [6]). In Rⁿ, necessary and sufficient conditions for the existence of the solution of the even Lp Minkowski problem for the case of 0 < p < 1 was given by Haberl, et al. [21]. Necessary and sufficient conditions for the existence of solutions to the even L₀ Minkowski problem (also called the logarithmic Minkowski problem) was recently established by B¨or¨oczky, et al. [5]. Without the assumption that the measure is even, existence of solution of the PDE (1.1) for the case where p > −n were given by Chou and Wang [10]. In [59, 60], the author established necessary and sufficient conditions for the existence of the solution of the L_p Minkowski problem for the case where p = 0 and p = −n, and µ is a discrete measure whose support-vectors (i.e., vectors in the support of the measure) are in general position.

One reason that the Minkowski and the L_p Minkowski problem for polytopes are important is because the Minkowski problem and the L_p Minkowski problem (for p >1) for measures can be solved by an approximation argument by first solving the polytopal case (see, e.g., Hug, et al. [30]

and Schneider [52], pp. 392-393).

A polytope in Rⁿ is the convex hull of a finite set of points in Rⁿ provided that it has positive n-dimensional volume. The convex hull of a subset of these points is called afacet of the polytope if it lies entirely on the boundary of the polytope and has positive (n−1)-dimensional volume. If a polytopeP contains the origin in its interior withN facets whose outer unit normals areu₁, ..., u_N, and such that if the facet with outer unit normal u_k has area a_k and distance from the origin h_k for all k ∈ {1, ..., N}. Then,

S_p(P,·) =

N

X

k=1

h^1−p_k a_kδ_uk(·).

where δ_uk denotes the delta measure that is concentrated at the point u_k. It is the aim of this paper to establish:

Theorem. If p∈(0,1), and µ is a discrete measure on the unit sphere, then µ is the L_p surface area measure of a polytope if and only if the support ofµis not concentrated on a closed hemisphere.

This paper is organized as follows. In Section 2, we recall some basic facts about convex bodies.

In Section 3, we study an extremal problem related to the L_p Minkowski problem. In Section 4, we prove the main theorem of this paper.

For the case where p > 1 with p 6= n, necessary and sufficient conditions for the existence of solutions to the discrete L_p Minkowski problem were established by Hug, et al. [30]. In Section 5, we give a new proof of this condition. The proof presented in this paper includes a new approach to the classical Minkowski problem.

2. Preliminaries

In this section, we collect some basic definitions and facts about convex bodies. For general references regarding convex bodies see, e.g., [13, 15, 16, 52, 56].

The sets of this paper are subsets of the n-dimensional Euclidean space Rⁿ. For x, y ∈Rⁿ, we write x·y for the standard inner product of x and y, |x| for the Euclidean norm of x and Bⁿ for the unit ball ofRⁿ.

ForK₁, K₂ ⊂Rⁿ and c₁, c₂ ≥0, the Minkowski combination, c₁K₁+c₂K₂, is defined by c₁K₁+c₂K₂ ={c₁x₁+c₂x₂ :x₁ ∈K₁, x₂ ∈K₂}.

The support function h_K :Rⁿ →R of a compact convex setK is defined, for x∈Rⁿ, by h(K, x) = max{x·y:y∈K}.

Obviously, for c≥0 and x∈Rⁿ,

h(cK, x) =h(K, cx) =ch(K, x).

The Hausdorff distance between two compact setsK, L in Rⁿ is defined by δ(K, L) = inf{t≥0 :K ⊂L+tBⁿ, L⊂K +tBⁿ}.

It is easily shown that the Hausdorff distance between two convex bodies, K and L, is δ(K, L) = max

u∈Sⁿ⁻¹|h(K, u)−h(L, u)|.

For a convex body K in Rⁿ, and u ∈ Sⁿ⁻¹, the support hyperplane H(K, u) in direction u is defined by

H(K, u) ={x∈Rⁿ:x·u=h(K, u)}, the half-space H⁻(K, u) in direction u is defined by

H⁻(K, u) ={x∈Rⁿ:x·u≤h(K, u)}, and thesupport set F(K, u) in direction u is defined by

F(K, u) =K∩H(K, u).

For a compact K ∈Rⁿ, the diameter of K is defined by

d(K) = max{|x−y|:x, y ∈K}.

LetP be the set of polytopes inRⁿ. If the unit vectorsu₁, ..., u_N (N ≥n+1) are not concentrated on a closed hemisphere, letP(u₁, ..., u_N) be the subset ofP such that a polytopeP ∈ P(u₁, ..., u_N) if

P =

N

\

k=1

H⁻(P, u_k).

Obviously, if P ∈ P(u₁, ..., u_N), then P has at most N facets, and the outer unit normals of P are a subset of{u1, ..., uN}. LetPN(u1, ..., uN) be the subset of P(u1, ..., uN) such that a polytope P ∈ PN(u1, ..., uN) if,P ∈ P(u1, ..., uN) and P has exactly N facets.

3. An extreme problem related to the Lp Minkowski problem

Supposep∈(0,1), α₁, ..., α_N ∈R⁺, the unit vectorsu₁, ..., u_N (N ≥n+ 1) are not concentrated on a closed hemisphere, and P ∈ P(u₁, ..., u_N). Define the function, Φ_P :P →R, by

ΦP(ξ) =

N

X

k=1

αk(h(P, uk)−ξ·uk)^p. In this section, we study the extremal problem

(3.0) inf{sup

ξ∈Q

Φ_Q(ξ) :Q∈ P(u₁, ..., u_N) and V(Q) = 1}.

We will prove that Φ_P(ξ) is strictly concave on P and that there exists a unique ξ_p(P)∈Int (P) such that

Φ_P(ξ_p(P)) = sup

ξ∈P

Φ_P(ξ).

Moreover, we will prove that there exists a polytope with u₁, ..., u_N as its outer unit normals, and the polytope solving problem (3.0).

We first prove the concavity of Φ_P(ξ).

Lemma 3.1. If 0 < p < 1, α₁, ..., α_N ∈ R⁺, the unit vectors u₁, ..., u_N (N > n+ 1) are not concentrated on a closed hemisphere and P ∈ P(u₁, ..., u_N), then Φ_P(ξ) is strictly concave on P. Proof. For 0< p <1, t^p is strictly concave on [0,+∞). Thus, for 0< λ <1 and ξ₁, ξ₂ ∈P,

λΦ_P(ξ₁) + (1−λ)Φ_P(ξ₂) = λ

N

X

k=1

α_k(h(P, u_k)−ξ₁·u_k)^p+ (1−λ)

N

X

k=1

α_k(h(P, u_k)−ξ₂·u_k)^p

=

N

X

k=1

α_k[λ(h(P, u_k)−ξ₁·u_k)^p + (1−λ)(h(P, u_k)−ξ₂·u_k)^p]

≤

N

X

k=1

α_k[h(P, u_k)−(λξ₁+ (1−λ)ξ₂)·u_k]^p

= Φ_P(λξ₁+ (1−λ)ξ₂),

with equality if and only ifξ₁·u_k =ξ₂·u_k for allk = 1, ..., N. Sinceu₁, ..., u_N are not concentrated on a closed hemisphere, Rⁿ= Span {u₁, ..., u_N}. Thus, ξ₁ =ξ₂. Therefore, Φ_P is strictly concave

onP.

The following lemma is needed.

Lemma 3.2. If 0 < p < 1, α₁, ..., α_N ∈ R⁺, the unit vectors u₁, ..., u_N (N > n+ 1) are not concentrated on a closed hemisphere and P ∈ P(u1, ..., uN), then there exists a unique ξp(P) ∈ Int (P) such that

Φ_P(ξ_p(P)) = max

ξ∈P Φ_P(ξ), where Φ_P(ξ) = PN

k=1α_k(h(P, u_k)−ξ·u_k)^p.

Proof. From Lemma 3.1, for 0 < p < 1, Φ_P(ξ) is strictly concave on P. From this and the fact that P is a compact convex set, we have, there exists a uniqueξ_p(P)∈P such that

ΦP(ξp(P)) = max

ξ∈P ΦP(ξ).

We next prove that ξ_p(P)∈Int (P). Otherwise, suppose ξ_p(P)∈∂P with h(P, uk)−ξp(P)·uk = 0

for k ∈ {i1, ..., im} and

h(P, u_k)−ξ_p(P)·u_k >0

for k ∈ {1, ..., N}\{i₁, ..., i_m}, where 1 ≤ i₁ < ... < i_m ≤ N and 1 ≤ m ≤ N − 1. Choose x₀ ∈Int (P), let

u₀ = x₀−ξ_p(P)

|x0−ξp(P)|, and let

[h(P, u_k)−(ξ_p(P) +δu₀)·u_k]−[h(P, u_k)−ξ_p(P)·u_k] = (−u₀·u_k)δ

=c_kδ, (3.1)

where ck =−u0·uk. Since h(P, uk)−ξp(P)·uk= 0 for k ∈ {i1, ..., im} and x0 is an interior point of P, ck =−u0·uk >0 fork ∈ {i1, ..., im}. Let

(3.2) c₀ = min

h(P, u_k)−ξ_p(P)·u_k :k ∈ {1, ..., N}\{i₁, ..., i_m} >0, and choose δ >0 small enough so that ξ_p(P) +δu₀ ∈Int (P) and

(3.3) min

h(P, u_k)− ξ_p(P) +δu₀

·u_k :k ∈ {1, ..., N}\{i₁, ..., i_m} > c₀ 2. Obviously, for 0< p < 1 andx₀, x₀+ ∆x∈(^c₂⁰,+∞),

|(x₀+ ∆x)^p−x^p₀|< p(c₀

2)^p−1|∆x|.

From this, the fact thath(P, u_k) = ξ_p(P)·u_kfork ∈ {i₁, ..., i_m}, the factc_k >0 fork ∈ {i₁, ..., i_m}, and Equations (3.1), (3.2) and (3.3), we have

Φ_P(ξ_p(P) +δu₀)−Φ_P(ξ_p(P)) =

N

X

k=1

α_k

h(P, u_k)−(ξ_p(P) +δu₀)·u_kp

− h(P, u_k)−ξ_p(P)·u_kp

≥ X

k∈{i₁,...,im}

α_k(c_kδ)^p− X

k∈{1,...,N}\{i₁,...,im}

α_k

h(P, u_k)−ξ_p(P)·u_k +c_kδp

− h(P, u_k)−ξ_p(P)·u_kp

≥

X

k∈{i₁,...,im}

α_kc^p_k

δ^p − X

k∈{1,...,N}\{i₁,...,im}

α_kp(c₀

2)^p−1|c_kδ|

=

X

k∈{i1,...,im}

α_kc^p_k− X

k∈{1,...,N}\{i1,...,im}

α_kp(c₀

2)^p−1|c_k|δ^1−p

δ^p. Thus, there exists a small enough δ₀ >0 such that ξ_p(P) +δ₀u₀ ∈Int (P) and

Φ_P(ξ_p(P) +δ₀u₀)>Φ_P(ξ_p(P)).

This contradicts the definition of ξ_p(P). Therefore, ξ_p(P)∈Int (P).

By definition, for λ >0, p∈(0,1) andP ∈ P(u₁, ..., u_N),

(3.4) ξ_p(λP) =λξ_p(P).

Obviously, if Pi ∈ P(u1, ..., uN) andPi converges to a polytopeP, then P ∈ P(u1, ..., uN).

Lemma 3.3. If p ∈(0,1), α1, ..., αN are positive, the unit vectors u1, ..., uN (N ≥n+ 1) are not concentrated on a closed hemisphere, P_i ∈ P(u₁, ..., u_N), and P_i converges to a polytope P. Then, lim_i→∞ξ_p(P_i) =ξ_p(P) and

i→∞lim Φ_Pi(ξ_p(P_i)) = Φ_P(ξ_p(P)), where Φ_P(ξ) = PN

k=1α_k(h(P, u_k)−ξ·u_k)^p.

Proof. Since P_i → P and ξ_p(P_i) ∈ Int (P_i), ξ_p(P_i) is bounded. Suppose ξ_p(P_i) does not converge to ξ_p(P), then there exists a subsequence P_ij of P_i such that P_ij converges to P,ξ_p(P_ij)→ξ₀ but ξ₀ 6=ξ_p(P). Obviously,ξ₀ ∈P and

j→∞lim Φ_Pij(ξ_p(P_ij)) = Φ_P(ξ₀)

<Φ_P(ξ_p(P))

= lim

j→∞Φ_Pij(ξ_p(P)).

This contradicts the fact that

Φ_Pij(ξ_p(P_ij))≥Φ_Pij(ξ_p(P)).

Therefore, lim_i→∞ξ_p(P_i) = ξ_p(P) and thus,

i→∞lim Φ_Pi(ξ_p(P_i)) = Φ_P(ξ_p(P)).

The following lemma is useful in the proving of the compactness of problem (3.0).

Lemma 3.4. Suppose p > 0, α₁, ..., α_N are positive, and the unit vectors u₁, ..., u_N (N ≥ n+ 1) are not concentrated on a hemisphere. If P_k ∈ P(u₁, ..., u_N), o ∈ P_k, and R(P_k) is not bounded, then

N

X

i=1

αih(Pk, ui)^p is not bounded.

Proof. Without loss of generality, we can assume

k→∞lim R(P_k) =∞.

Let

f(u) =

N

X

i=1

α_i|u·u_i|^p, where u∈Sⁿ⁻¹.

Sinceu₁, ..., u_N are not contained in a closed hemisphere,Rⁿ = Span{u₁, ..., u_N}. Thus,f(u)>

0 for all u∈Sⁿ⁻¹. On the other hand, f(u) is continuous on Sⁿ⁻¹. Thus, there exists a constant a₀ >0 such that

N

X

i=1

α_i|u·u_i|^p ≥a₀ for all u∈Sⁿ⁻¹.

Choose uk ∈Sⁿ⁻¹ such that R(Pk)uk∈Pk. Since o∈Pk,

N

X

i=1

α_ih(P_k, u_i)^p ≥

N

X

i=1

α_i|R(P_k)u_k·u_i|^p

=R(Pk)^p

N

X

i=1

αi|uk·ui|^p

≥a₀R(P_k)^p →+∞.

The following lemma will be needed.

Lemma 3.5. If P is a polytope in Rⁿ and v₀ ∈Sⁿ⁻¹ with V_n−1(F(P, v₀)) = 0, then there exists a δ₀ >0 such that for0≤δ < δ₀

V(P ∩ {x:x·v₀ ≥h(P, v₀)−δ}) = c_nδⁿ+...+c₂δ², where c_n, ..., c₂ are constants that depend on P and v₀.

Proof. It is known (e.g., [14], Proposition 3.1) that

g(δ) =Vn−1(P ∩ {x:x·v₀ =h(P, v₀)−δ})

is a piecewise polynomial function of degree at most n−1. By conditions, g(0) = 0. Thus, there exists a δ₀ >0 andc⁰_n−1, ..., c⁰₁ (depend on P and v₀) such that when 0≤δ < δ₀

g(δ) =c⁰_n−1δⁿ⁻¹+...+c⁰₁δ.

Therefore, when 0≤δ < δ₀,

V(P ∩ {x:x·v₀ ≥h(P, v₀)−δ}) = Z δ

0

g(t)dt

=c_nδⁿ+...+c₂δ²,

where c_n =c⁰_n−1/n, ..., c₂ =c⁰₁/2 are constants that depend on P and v₀. We next solve problem (3.0).

Lemma 3.6. If 0 < p < 1, α1, ..., αN are positive, and the unit vectors u1, ..., uN (N ≥ n + 1) are not concentrated on a hemisphere, then there exists a P ∈ PN(u1, ..., uN) such that ξp(P) =o, V(P) = 1, and

Φ_P(o) = inf{max

ξ∈Q Φ_Q(ξ) :Q∈ P(u₁, ..., u_N) and V(Q) = 1}, where Φ_Q(ξ) = PN

k=1α_k(h(Q, u_k)−ξ·u_k)^p.

Proof. Obviously, forP, Q∈ P(u₁, ..., u_N), if there exists a x∈Rⁿ such that P =Q+x, then Φ_P(ξ_p(P)) = Φ_Q(ξ_p(Q)).

Thus, we can choose a sequenceP_i ∈ P(u₁, ..., u_N) withξ_p(P_i) = oand V(P_i) = 1 such that Φ_Pi(o) converges to

inf{max

ξ∈Q Φ_Q(ξ) :Q∈ P(u₁, ..., u_N) and V(Q) = 1}.

Choose a fixed P₀ ∈ P(u₁, ..., u_N) withV(P₀) = 1, then inf{max

ξ∈Q Φ_Q(ξ) :Q∈ P(u₁, ..., u_N) and V(Q) = 1} ≤Φ_P0(ξ_p(P₀)).

We claim that Pi is bounded. Otherwise, from Lemma 3.4, ΦPi(ξp(Pi)) converges to +∞. This contradicts the previous inequality. Therefore, P_i is bounded.

From Lemma 3.3 and the Blaschke selection theorem, there exists a subsequence of P_i that converges to a polytopeP such that P ∈ P(u₁, ..., u_N),V(P) = 1, ξ(P) = o and

(3.6) Φ_P(o) = inf{max

ξ∈Q Φ_Q(ξ) :Q∈ P(u₁, ..., u_N) and V(Q) = 1}.

We next prove thatF(P, u_i) are facets for alli= 1, ..., N. Otherwise, there exists ai₀ ∈ {1, ..., N} such that

F(P, u_i0) is not a facet of P.

Choose δ >0 small enough so that the polytope

P_δ=P ∩ {x:x·u_i0 ≤h(P, u_i0)−δ} ∈ P(u₁, ..., u_N).

and (by Lemma 3.5)

V(P_δ) = 1−(c_nδⁿ+...+c₂δ²), where cn, ..., c2 are constants that depend onP and direction ui0.

From Lemma 3.3, for any δ_i →0 it always true thatξ_p(P_δi)→o. We have,

δ→0limξ_p(P_δ) = o.

Letδ be small enough so that h(P, uk)> ξp(Pδ)·uk+δ for all k ∈ {1, ..., N}, and let λ=V(Pδ)⁻¹ⁿ = (1−(cnδⁿ+...+c2δ²))⁻ⁿ¹.

From this and Equation (3.4), we have Φ_λPδ(ξ_p(λP_δ)) =

N

X

k=1

α_k h(λP_δ, u_k)−ξ_p(λP_δ)·u_kp

=λ^p

N

X

k=1

α_k h(P_δ, u_k)−ξ_p(P_δ)·u_kp

=λ^p

N

X

k=1

α_k h(P, u_k)−ξ_p(P_δ)·u_kp

−α_i0λ^p h(P, u_i0)−ξ_p(P_δ)·u_i0p

+α_i0λ^p h(P, u_i0)−ξ_p(P_δ)·u_i0 −δp

=

N

X

k=1

αk h(P, uk)−ξp(Pδ)·uk

p

+ (λ^p−1)

N

X

k=1

αk h(P, uk)−ξp(Pδ)·uk

p

+α_i0λ^p

h(P, u_i0)−ξ_p(P_δ)·u_i0 −δp

− h(P, u_i0)−ξ_p(P_δ)·u_i0p

= Φ_P(ξ_p(P_δ)) +B(δ), (3.7)

where

B(δ) = (λ^p −1)

N

X

k=1

α_k h(P, u_k)−ξ_p(P_δ)·u_kp

!

+α_i0λ^p

h(P, u_i0)−ξ_p(P_δ)·u_i0 −δp

− h(P, u_i0)−ξ_p(P_δ)·u_i0p

=h

(1−(c_nδⁿ+...+c₂δ²))^−pn −1i X^N

k=1

α_k(h(P, u_k)−ξ_p(P_δ)·u_k)^p

!

+α_i0λ^p

h(P, u_i0)−ξ_p(P_δ)·u_i0 −δp

− h(P, u_i0)−ξ_p(P_δ)·u_i0p .

From the facts thatd₀ =d(P)> h(P, u_i0)−ξ_p(P_δ)·u_i0 > h(P, u_i0)−ξ_p(P_δ)·u_i0−δ >0, 0< p <1 and the function f(t) =t^p is concave on [0,∞), we have

h(P, u_i0)−ξ_p(P_δ)·u_i0 −δp

− h(P, u_i0)−ξ_p(P_δ)·u_i0p

<(d₀−δ)^p−d^p₀. Then,

B(δ) = (λ^p−1)

N

X

k=1

αk h(P, uk)−ξp(Pδ)·uk

p

!

+α_i0λ^p

h(P, u_i0)−ξ_p(P_δ)·u_i0 −δp

− h(P, u_i0)−ξ_p(P_δ)·u_i0p

<h

(1−(c_nδⁿ+...+c₂δ²))^−np −1i X^N

k=1

α_k(h(P, u_k)−ξ_p(P_δ)·u_k)^p

!

+α_i0λ^p

(d₀−δ)^p−d^p₀ . (3.8)

On the other hand,

(3.9) lim

δ→0 N

X

k=1

α_k h(P, u_k)−ξ_p(P_δ)·u_kp

=

N

X

k=1

α_kh(P, u_k)^p,

(3.10) (d₀−δ)^p−d^p₀ <0,

and

limδ→0

(1−(c_nδⁿ+...+c₂δ²))^−np −1 (d₀−δ)^p−d^p₀

= lim

δ→0

(−_n^p)(1−(c_nδⁿ+...+c₂δ²))^−np−1(−nc_nδⁿ⁻¹−...−2c₂δ)

p(d₀−δ)^p−1(−1) = 0.

(3.11)

From Equations (3.8), (3.9), (3.10), (3.11), and the fact that 0 < p < 1, we have B(δ) < 0 for small enough δ > 0. From this and Equation (3.7), there exists a δ₀ > 0 such that P_δ0 ∈ P(u₁, ..., u_N) and

Φ_λ0Pδ

0(ξ_p(λ₀P_δ0))<Φ_P(ξ_p(P_δ0))≤Φ_P(ξ_p(P)) = Φ_P(o),

whereλ₀ =V(P_δ0)⁻ⁿ¹. LetP₀ =λ₀P_δ0−ξ_p(λ₀P_δ0), thenP₀ ∈ Pⁿ(u₁, ..., u_N),V(P₀) = 1,ξ_p(P₀) = o and

(3.12) Φ_P0(o)<Φ_P(o).

This contradicts Equation (3.6). Therefore,P ∈ P_N(u₁, ..., u_N).

4. The Lp Minkowski problem for polytopes (0< p <1)

In this section, we prove the main theorem of this paper. We only need prove the following:

Theorem 4.1. If p∈ (0,1), α₁, ..., α_N ∈R⁺, and the unit vectors u₁, ..., u_N (N ≥ n+ 1) are not concentrated on a closed hemisphere, then there exists a polytope P₀ such that

S_p(P₀,·) =

N

X

k=1

α_kδ_uk(·).

Proof. From Lemma 3.6, there exists a polytopeP ∈ P_N(u₁, ..., u_N) with ξ_p(P) = oand V(P) = 1 such that

Φ_P(o) = inf{max

ξ∈Q Φ_Q(ξ) :Q∈ P(u₁, ..., u_N) and V(Q) = 1}, where Φ_Q(ξ) =PN

k=1α_k(h(Q, u_k)−ξ·u_k)^p.

Forδ₁, ..., δ_N ∈R, choose |t| small enough so that the polytope P_t defined by P_t =

N

\

i=1

{x:x·u_i ≤h(P, u_i) +tδ_i} has exactly N facets. Then,

V(P_t) =V(P) +t

N

X

i=1

δ_ia_i

!

+o(t), where a_i is the area of F(P, u_i). Thus,

limt→0

V(P_t)−V(P)

t =

N

X

i=1

δiai.

Letλ(t) =V(P_t)⁻ⁿ¹, then λ(t)P_t ∈ P_Nⁿ(u₁, ..., u_N), V(λ(t)P_t) = 1 and

(4.1) λ⁰(0) =−1

n

N

X

i=1

δ_iS_i. Letξ(t) = ξ_p(λ(t)P_t), and

Φ(t) = max

ξ∈λ(t)Pt

N

X

k=1

α_k(λ(t)h(P_t, u_k)−ξ·u_k)^p

=

N

X

k=1

αk(λ(t)h(Pt, uk)−ξ(t)·uk)^p. (4.2)

From Equation (4.2) and the fact that ξ(t) is an interior point ofλ(t)P_t, we have (4.3)

N

X

k=1

α_k uk,i

[λ(t)h(P_t, u_k)−ξ(t)·u_k]^1−p = 0, for i= 1, ..., n, where u_k= (u_k,1, ..., u_k,n)^T. As a special case when t= 0,

N

X

k=1

αk

u_k,i

h(P, u_k)^1−p = 0,

for i= 1, ..., n. Therefore, (4.4)

N

X

k=1

α_k u_k

h(P, uk)^1−p = 0.

Let

F_i(t, ξ₁, ..., ξ_n) =

N

X

k=1

α_k u_k,i

[λ(t)h(P_t, u_k)−(ξ₁u_k,1 +...+ξ_nu_k,n)]^1−p for i= 1, ..., n. Then,

∂F_i

∂ξ_j (0,...,0)

=

N

X

k=1

(1−p)α_k

h(P, u_k)^2−pu_k,iu_k,j. Thus,

∂F

∂ξ (0,...,0)

!

n×n

=

N

X

k=1

(1−p)α_k

h(P, u_k)^2−puk·u^T_k, where u_ku^T_k is an n×n matrix.

Sinceu₁, ..., u_N are not contained in a closed hemisphere,Rⁿ= Span {u₁, ..., u_N}. Thus, for any x∈Rⁿ with x6= 0, there exists a u_i0 ∈ {u₁, ..., u_N} such thatu_i0 ·x6= 0. Then,

x^T ·

N

X

k=1

(1−p)α_k

h(P, u_k)^2−puk·u^T_k

!

·x=

N

X

k=1

(1−p)α_k

h(P, u_k)^2−p(x·uk)²

≥ (1−p)αi0

h(P, u_i0)^2−p(x·u_i0)² >0.

Thus, (^∂F_∂ξ

(0,...,0)) is positive defined. From this, Equations (4.3) and the inverse function theorem, we have

ξ⁰(0) = (ξ₁⁰(0), ..., ξ⁰_n(0)) exists.

From the fact that Φ(0) is an extreme value of Φ(t) (in Equation (4.2)), Equation (4.1) and Equation (4.4), we have

0 = Φ⁰(0)/p

=

N

X

k=1

α_kh(P, u_k)^p−1(λ⁰(0)h(P, u_k) +δ_k−ξ⁰(0)·u_k)

=

N

X

k=1

α_kh(P, u_k)^p−1

"

−1 n

N

X

i=1

a_iδ_i

!

h(P, u_k) +δ_k

#

−ξ⁰(0)·

" _N X

k=1

α_k u_k h(P, u_k)^1−p

#

=

N

X

k=1

α_kh(P, u_k)^p−1δ_k−

N

X

i=1

a_iδ_i

!PN

k=1α_kh(P, u_k)^p n

=

N

X

k=1

α_kh(P, u_k)^p−1− PN

j=1αjh(P, uj)^p

n a_k

! δ_k. Since δ₁, ..., δ_N are arbitrary,

PN

j=1αjh(P, uj)^p

n h(P, u_k)^1−pa_k =α_k,

for all k = 1, ..., N. Let

P₀ = PN

j=1α_jh(P, u_j)^p n

!_n−p¹ P, we have

S_p(P₀,·) =

N

X

k=1

α_kδ_uk(·).

5. The L_p Minkowski problem for polytopes (p≥1 with p6=n)

In [30], Hug, et al. established a necessary and sufficient condition of the existence of the solution of the discrete L_p Minkowski problem for the case where p > 1 with p 6= n. In this section, we prove it by a different method. Moreover, this proof also includes a new approach to the classical Minkowski problem.

Let p ≥ 1, α₁, ..., α_N ∈ R⁺, the unit vectors u₁, ..., u_N (N ≥ n+ 1) are not concentrated on a closed hemisphere (in addition PN

i=1α_iu_i = 0 for the case where p = 1), P ∈ P(u₁, ..., u_N), and o∈P. Define

Ψ(P) =

N

X

i=1

α_ih(P, u_i)^p. Consider the extreme problem

(5.0) inf{Ψ(Q) :Q∈ P(u₁, ..., u_N), V(Q) = 1, o∈Q}.

In this section, we prove that there exists a polytope P with u₁, ..., u_N as its unit facet normal vectors and o ∈ Int (P), which is the solution of problem (5.0). Moreover, we prove that a dilatation of P is the solution of the corresponding discreteLp Minkowski problem.

The following lemma will be needed.

Lemma 5.1. Suppose the unit vectors u₁, ..., u_N (N ≥ n+ 1) are not concentrated on a closed hemisphere,P ∈ P(u1, ..., uN)ando ∈P. If there exists ani0 (1≤i0 ≤N) such thath(P, ui0) = 0 and |F(P, ui0)|>0, then there exists a δ0 >0 so that when 0< δ < δ0 the polytope

P_δ =

\

i6=i0

H⁻(P, u_i)

\{x:x·u_i0 ≤δ} ∈ P(u₁, ..., u_N) and

V(Pδ) =V(P) + (cnδⁿ+...+c2δ²+c1δ), where cn, ..., c2 are constants that depend on P and ui0, and c1 6= 0.

Proof. By condition P₁ =

\

i6=i0

H⁻(P, u_i)

\{x:x·u_i0 ≤1} ∈ P(u₁, ..., u_N).

Thus, (e.g., [14], Proposition 3.1) for δ∈R,

g(δ) =Vn−1(P₁∩ {x:x·v₀ =δ})

is a piecewise polynomial function of degree at most n−1. By conditions, g(0) 6= 0. Thus, there exists a δ0 >0 andc⁰_n−1, ..., c⁰₁, c⁰₀ (depend on P and ui0) such that c⁰₀ 6= 0 and when 0≤δ < δ0

g(δ) = c⁰_n−1δⁿ⁻¹+...+c⁰₁δ+c⁰₀.

Therefore, when 0≤δ < δ0,

V(P_δ) =V(P) + Z δ

0

g(t)dt

=V(P) + (c_nδⁿ+...+c₂δ²+c₁δ),

where c_n =c⁰_n−1/n, ..., c₂ =c⁰₁/2, c₁ =c⁰₀ are constants that depend on P and u_i0, and c₁ 6= 0.

The following two lemmas solve problem (5.0).

Lemma 5.2. Let p ≥ 1, α₁, ..., α_N ∈ R⁺, and the unit vectors u₁, ..., u_N (N ≥ n + 1) are not concentrated on a closed hemisphere (in addition, PN

i=1αiui = 0 if p = 1). Then, there exists a P ∈ P(u₁, ..., u_N) with o ∈Int (P) such that

Ψ(P) = inf{Ψ(Q) :Q∈ P(u₁, ..., u_N), V(Q) = 1, o∈Q}. Proof. Choose aP₀ ∈ P(u₁, ..., u_N) with V(P₀) = 1 and o ∈P, then

(5.1) inf{Ψ(Q) :Q∈ P(u₁, ..., u_N), V(Q) = 1, o ∈Q} ≤Ψ(P₀).

Choose a sequence P_k∈ P(u₁, ..., u_N) with V(P_k) = 1 and o∈P_k such that Ψ(P_k) converges to inf{Ψ(Q) :Q∈ P(u₁, ..., u_N), V(Q) = 1, o∈Q}.

We claim thatP_kis bounded. Otherwise from Lemma 3.4, Ψ(P_k) is not bounded from above. This contradicts Equation (5.1). Therefore, P_k is bounded. From the Blaschke section theorem, there exists a subsequence ofP_k that converges to a polytope P such that P ∈ P(u₁, ..., u_N),V(P) = 1, o∈P and

(5.2) Ψ(P) = inf{Ψ(Q) :Q∈ P(u₁, ..., u_N), V(Q) = 1, o∈Q}.

From conditions, if p = 1, P, Q contain the origin, and P = Q+x for some x ∈ Rⁿ, then Ψ(P) = Ψ(Q). Thus, we can assume o is an interior point of P (in (5.2)).

We next prove that when p > 1 the origin is an interior point of P. Otherwise, there exists an i₀ (1≤i₀ ≤N) such that h(P, u_i0) = 0 and |F(P, u_i0)|>0.

By Lemma 5.1, we can choose δ >0 small enough so that the polytope P_δ =

\

i6=i0

H⁻(P, u_i)

\{x:x·u_i0 ≤δ} ∈ P(u₁, ..., u_N)

and

V(P_δ) = 1 + (c_nδⁿ+...+c₂δ²+c₁δ), where cn, ..., c1 are constants that depend onP and ui0, and c1 6= 0.

Letλ =V(P_δ)⁻ⁿ¹. From the hypothesis h(P, u_i0) = 0, Ψ(λP_δ) =

N

X

i=1

α_ih(λP_δ, u_i)^p

=λ^p

" _N X

i=1

α_ih(P_δ, u_i)^p

#

=λ^p

" _N X

i=1

α_ih(P, u_i)^p

#

+λ^pα_i0 h(P, u_i0) +δp

−λ^pα_i0h(P, u_i0)^p

=

N

X

i=1

α_ih(P, u_i)^p+ (λ^p−1)

N

X

i=1

α_ih(P, u_i)^p+λ^pα_i0δ^p

= Ψ(P) +B₁(δ), where

B₁(δ) = (λ^p−1)

N

X

i=1

α_ih(P, u_i)^p+λ^pα_i0δ^p

=h

1 + (c_nδⁿ+...+c₂δ²+c₁δ)−^p

n −1iX^N

i=1

α_ih(P, u_i)^p+λ^pα_i0δ^p. Since c1 6= 0 and

δ→0lim

δ^p

(1 +c_nδⁿ+...+c²δ²+c₁δ)^−pn −1

= lim

δ→0

pδ^p−1

(−_n^p)(1 +cnδⁿ+...+c²δ²+c1δ)^−np−1(ncnδⁿ⁻¹+...+c1) = 0, B₁(δ)<0 for small enough positive δ. Thus

Ψ(λPδ)<Ψ(P)

for small enough positive δ. This contradicts Equation (5.2). Therefore, the origin is an interior

point of P.

Lemma 5.3. The minimizing polytope in Lemma 5.2 has N facets.

Proof. If the statement of the Lemma is not true, then there exists an i₀ ∈ {1, ..., N}such that F(P, u_i0)

is not a facet of P.

By Lemma 3.5, we can choose δ >0 small enough so that the polytope Pδ=P ∩ {x:x·ui0 ≤h(P, ui0)−δ} ∈ P(u1, ..., uN).

and

V(P_δ) = 1−(c_nδⁿ+...+c₂δ²), where c_n, ..., c₂ are constants that depend onP and direction u_i0.

Let

λ=λ(δ) = V(P_δ)⁻ⁿ¹ = 1−(c_nδⁿ+...+c₂δ²)−¹

n,

then

Ψ(λP_δ) =

N

X

i=1

α_ih(λP_δ, u_i)^p

=λ^p

N

X

i=1

αih(Pδ, ui)^p

=

N

X

i=1

α_ih(P, u_i)^p+ (λ^p−1)

N

X

i=1

α_ih(P, u_k)^p+α_i0λ^p[(h(P, u_i0)−δ)^p−h(P, u_i0)^p]

= Ψ(P) +B₂(δ), where,

B₂(δ) = (λ^p−1)

N

X

i=1

α_ih(P, u_i)^p+α_i0λ^p[(h(P, u_i0)−δ)^p −h(P, u_i0)^p]

=h

1−(c_nδⁿ+...+c₂δ²)−^p_n

−1i X^N

i=1

α_ih(P, u_i)^p

!

+α_i0λ^p[(a₀−δ)^p−a^p₀], and a₀ =h(P, u_i0).

Since (a₀−δ)^p−a^p₀ <0 for small positive δ and

δ→0lim

1−(c_nδⁿ+...+c₂δ²)−^p

n −1 (a₀−δ)^p−a^p₀

= lim

δ→0

−^p_n 1−(c_nδⁿ+...+c₂δ²)−^p_n−1

(−nc_nδⁿ⁻¹ −...−2c₂δ)

p(a₀−δ)^p−1(−1) = 0,

there exists a δ₀ >0 such that B₂(δ)<0 for all 0< δ < δ₀. Thus, Ψ(λP_δ)<Ψ(P)

for all 0< δ < δ₀. This contradicts Equation (5.2). Therefore, P ∈ P_N(u₁, ..., u_N).

Suppose P is a polytope with N facets whose outer unit normals are u₁, ..., u_N and such that the facet with outer normal u_k has areaa_k. Obviously, for anyu∈Sⁿ⁻¹

X

uk·u≥0

(uk·u)ak=− X

uk·u≤0

(uk·u)ak,

and both equal to the (n −1)-dimensional volume of the projection of P on u^⊥. Thus, for all u∈Sⁿ⁻¹

N

X

k=1

(u_k·u)a_k = 0.

Therefore, (5.3)

N

X

k=1

a_ku_k = 0.

Equation (5.3) is a necessary condition for the existence of the solution of the discrete classical Minkowski problem. The following theorem shows that it is also the sufficient condition. Moreover, the following theorem is the necessary and sufficient conditions for the existence of the solution of the discrete Lp Minkowski problem for the case wherep > 1 withp6=n.