Anisotropic flow, entropy, and $L^p$-Minkowski problem

Abstract

We provide a natural simple argument using anistropic flows to prove the existence of weak solutions to Lutwak’s $L^p$-Minkowski problem on $S^n$ which were obtained by other methods.

1 Introduction

For $\alpha>0$ and nonnegative $f\in L^1(\mathbb {S}^n)$ with positive integral, we are interested in finding a weak solution to the Monge–Ampére equation

(1.1) $$ \begin{align} u^{\frac1\alpha}\det(\bar{\nabla}^2_{ij} u+u\bar{g}_{ij})=f, \end{align} $$

or in other words, a weak solution to Lutwak’s $L^p$ -Minkowski problem on $S^n$ when $-n-1<p<1$ for $p=1-\frac 1\alpha $ where $\bar {\nabla }$ is the Levi-Civita connection of $\mathbb {S}^n$ , $\bar {g}_{ij}$ , with $\bar {g}$ being the induced round metric on the unit sphere. By a weak (Alexandrov) solution, we mean the following: Given a nontrivial finite Borel measure $\mu $ on $\mathbb {S}^n$ (for example, $d\mu =f\,d\theta $ for the Lebesgue measure $\theta $ on $S^n$ and the f in (1.1)), find a convex body $\Omega \subset \Bbb R^{n+1}$ with $o\in \Omega $ such that

(1.2) $$ \begin{align} d\mu=u^{\frac1\alpha}\,dS_\Omega, \end{align} $$

where $u(x)=\max _{z\in \Omega }\langle x,z\rangle $ is the support function and $S_\Omega $ is the surface area measure of $\Omega $ (see [45]). If $\partial \Omega $ is $C^2_+$ , then

$$ \begin{align*}dS_\Omega=\det(\bar{\nabla}^2_{ij} u+u\bar{g}_{ij})d\theta=K^{-1}d\theta, \end{align*} $$

where $K(x)$ is the Gaussian curvature at the point of $\partial \Omega $ where $x\in S^n$ is the exterior unit normal (see [45]). Concerning the regularity of the solution of (1.1), if $f\in C^{0,\beta }(S^n)$ and u are positive, then u is $C^{2,\beta }$ according to Caffarelli’s regularity theory in [15, 16]. On the other hand, even if f is positive and continuous for $\alpha>\frac 1n$ , there might exist weak solution where $u(x)=0$ for some $x\in S^n$ and u is not even $C^1$ according to Example 4.2 in [7]. Moreover, even if $f\in C^{0,\beta }(S^n)$ is positive, it is possible that $u(x)=0$ for some $x\in S^n$ for $\alpha>\frac 1n$ , but Choi, Kim, and Lee [19] still managed to obtain some regularity results in this case.

The case $\alpha =\frac 1{n+2}$ of the Monge–Ampére equation (1.1) is the critical case when the left-hand side of (1.1) is invariant under linear transformations of $\Omega $ , and the case $\alpha =1$ is the so-called logarithmic Minkowski problem posed by Firey [23]. Setting $p=1-\frac 1\alpha <1$ , the Monge–Ampére equation (1.1) is Lutwak’s $L^p$ -Minkowski problem

(1.3) $$ \begin{align} u^{1-p}\det(\bar{\nabla}^2_{ij} u+u\bar{g}_{ij})=f. \end{align} $$

In this notation, (1.2) reads as

(1.4) $$ \begin{align} d\mu=u^{1-p}\,dS_\Omega; \end{align} $$

that equation makes sense for any $p\in \Bbb R$ . Within the rapidly developing $L^p$ -Brunn–Minkowski theory (where $p=1$ is the classical case originating from Minkowski’s oeuvre) initiated by Lutwak [39–41], if $p>1$ and $p\neq n+1$ , then Hug, Lutwak, Yang, and Zhang [30] (improving on Chou and Wang [20]) prove that (1.4) has an Alexandrov solution if and only if the $\mu $ is not concentrated onto any closed hemisphere, and the solution is unique. We note that there are examples in [25] (see also [30]) and show that if $1<p<n+1$ , then it may happen that the density function f is a positive continuous in (1.3) and $o\in \partial K$ holds for the unique Alexandrov solution, and actually Bianchi, Böröczky, and Colesanti [7] exhibit an example that $o\in \partial K$ even if the density function f is a positive continuous in (1.3) assuming $-n-1<p<1$ .

In the case $p\in (0,1)$ (or equivalently, $\alpha>1$ ), if the measure $\mu $ is not concentrated onto any great subsphere of $S^n$ , then Chen, Li, and Zhu [17] prove that there exists an Alexandrov solution $K\in \mathcal {K}_o^n$ of (1.4) using a variational argument (see also [8]). We note that for $p\in (0,1)$ and $n\geq 2$ , no complete characterization of $L^p$ -surface area measures is known (see [12] for the case $n=1$ , and [8, 43] for partial results about the case when $n\geq 2$ and the support of $\mu $ is contained in a great subsphere of $S^n$ ).

Concerning the case $p=0$ (or equivalently, $\alpha =1$ ), the still open logarithmic Minkowski problem (1.3) or (1.4) was posed by Firey [23] in 1974. The paper [11] characterized even measures $\mu $ such that (1.4) has an even solution for $p=0$ by the so-called subspace concentration condition (see (a) and (b) in Theorem 1.1). In general, Chen, Li, and Zhu [18] proved that if a nontrivial finite Borel measure $\mu $ on $S^{n-1}$ satisfies the same subspace concentration condition, then (1.4) has a solution for $p=0$ . On the other hand, Böröczky and Hegedus [10] provide conditions on the restriction of the $\mu $ in (1.4) to a pair of antipodal points.

If $-n-1<p<0$ (or equivalently, $\frac 1{n+2}<\alpha <1$ ) and $f\in L_{\frac {n+1}{n+1+p}}(S^{n})$ in (1.3), then (1.3) has a solution according to [8]. For a rather special discrete measure $\mu $ satisfying that $\mu $ is not concentrated on any closed hemisphere and any n unit vectors in the support of $\mu $ are independent, Zhu [47] solves the $L^p$ -Minkowski problem (1.4) for $p<0$ . The $p=-n-1$ (or equivalently, $\alpha =\frac 1{n+2}$ ) case of the $L^p$ -Minkowski problem is the critical case because its link with the $\mathrm {SL}(n)$ invariant centro-affine curvature whose reciprocal is $u^{n+2}\det (\bar {\nabla }^2_{ij} u+u\bar {g}_{ij})$ (see [29] or [38]). For positive results concerning the critical case $p=-n-1$ , see, for example, [28, 34], and for obstructions for a solution, see, for example, [20, 22].

In the super-critical case $p<-n-1$ (or equivalently, $\alpha <\frac 1{n+2}$ ), there is a recent important work by Li, Guang, and Wang [27] proving that for any positive $C^2$ function f, there exists a $C^4$ solution of (1.3). See also [22] for non-existence examples.

The main contribution of this paper is to provide a very natural argument based on anisotropic flows developed by Andrews [4] to handle the case $-n-1<p<1$ , or equivalently, the case $\frac 1{n+2}<\alpha <\infty $ .

Entropy functional. For any convex body $\Omega $ , a fixed positive function f on $\mathbb {S}^n$ and $\alpha \in (0, \infty )$ , define

(1.5) $$ \begin{align} \mathcal{E}_{\alpha, f} (\Omega) := \sup_{z\in\Omega}{\mathcal E}_{\alpha, f}(\Omega,z), \end{align} $$

where

(1.6) $$ \begin{align} {\mathcal E}_{\alpha, f}(\Omega,z) := \begin{cases} \frac{\alpha}{\alpha-1}\log\left(\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n} u_{z}(x)^{1-\frac1\alpha}\,f(x) d\theta(x)\right),&\alpha\neq 1,\\ \hspace{3pt}\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n} \log(u_{z}(x))\, f(x) d\theta(x),&\alpha=1. \end{cases} \end{align} $$

Here, $u_{z}(x):=\sup _{y\in \Omega }\left \langle y-z,x\right \rangle $ is the support function of $\Omega $ in direction x with respect to $z_0$ and $\frac {\ \ }{\ \ }{\hskip -0.4cm}\int _{\mathbb {S}^n} h(x)\, d\theta (x)=\frac {1}{\omega _n} \int _{\mathbb {S}^n} h(x)$ with $\omega _n$ being the surface area of $\mathbb {S}^n$ and $\theta $ is the Lebesgue measure on $S^n$ . When $\alpha =1$ and $f(x)\equiv 1$ , then the above quantity agrees with the entropy in [26], first introduced by Firey [23] for the centrally symmetric $\Omega $ . General integral quantities were studied by Andrews in [2, 4]. Here, we shall assume that $\frac {\ \ }{\ \ }{\hskip -0.4cm}\int _{\mathbb {S}^n} f(x)\, d\theta (x)=1$ , namely, $\frac 1{\omega _n}\,f(x)d\theta (x)$ is a probability measure. For the special case $f\equiv 1$ , $\mathcal {E}_{\alpha , f}(\Omega ) $ becomes the entropy $\mathcal {E}_\alpha (\Omega )$ in [6].

For positive $f\in C^{\infty }(\mathbb S^n)$ , consider the anisotropic flow for convex hypersurfaces $\tilde X(\cdot , \tau ): M_{\tau }\to \mathbb {R}^{n+1}$ :

(1.7) $$ \begin{align} \frac{\partial}{\partial \tau}\tilde X(x, \tau)= -f^{\alpha} (\nu) \tilde K^{\alpha}(x, \tau)\, \nu(x, \tau), \end{align} $$

where $\nu (x, \tau )$ is the unit exterior normal at $\tilde X(x, \tau )$ of $\tilde M_\tau =\tilde X(M, \tau )$ , and $\tilde K(x,\tau )$ is the Gauss curvature of $\tilde M_\tau $ at $\tilde X(x,\tau )$ . Andrews [4] proved that flow (1.7) contracts to a point under finite time if the initial hypersurface $M_0$ is strictly convex. Under a proper normalization, the normalized anisotropy flow of (1.7) is

(1.8) $$ \begin{align} \frac{\partial}{\partial t}X(x, t)= -\frac{f^\alpha(\nu) K^{\alpha}(x, t)}{\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n}f^\alpha K^{\alpha-1}}\, \nu(x, t) +X(x,t). \end{align} $$

The basic observation is that a critical point for entropy $\mathcal {E}_{\alpha , f} (\Omega )$ defined in (1.5) under volume normalization is a solution to equation (1.1). The entropy is monotone along flow (1.8). One may view (1.1) is an “optimal solution” to this variational problem as the flow (1.8) provides a natural path to reach it. This approach was devised in [5] with the aim to obtain convergence of the normalized flow (1.8). The main arguments in [5] follows those in [6, 26] where convergence of isotropic flows by power of Gauss curvature (i.e., $f=1$ ) was established. Unfortunately, the entropy point estimate in [6, 26] fails for general anisotropic flows except $\frac {1}{n+2}<\alpha \le \frac {1}n$ [4]. The convergence was obtained in [5] assuming $M_0$ and f are invariant under a subgroup G of $O(n+1)$ which has no fixed point. We note that an inverse Gauss curvature flow argument was considered by Bryan, Ivaki, and Scheuer [14] to produce a origin-symmetric solution to (1.1).

Since we are only interested in finding a weak solution to (1.2), one only needs certain “weak” convergence of the flow (1.8). The key steps are to control diameter with entropy under appropriate conditions on measure $\mu =f d\theta $ on $\mathbb S^n$ and use monotonicity of entropy to produce a solution to (1.2). The following is our main result.

Theorem 1.1 For $\alpha>\frac 1{n+2}$ and finite nontrivial Borel measure $\mu $ on $\mathbb {S}^n$ , $n\geq 1$ , there exists a weak solution of (1.2) provided the following holds:

1. (i) If $\alpha>1$ and $\mu $ is not concentrated onto any great subsphere $x^\bot \cap \mathbb {S}^n$ , $x\in \mathbb {S}^n$ .

2. (ii) If $\alpha =1$ and $\mu $ satisfies that for any linear $\ell $ -subspace $L\subset \Bbb R^{n+1}$ with $1\leq \ell \leq n$ , we have

   1. (a) $\displaystyle \mu (L\cap \mathbb {S}^n)\leq \frac {\ell }{n+1}\cdot \mu (\mathbb {S}^n)$ ;

   2. (b) equality in (a) for a linear $\ell $ -subspace $L\subset \Bbb R^{n+1}$ with $1\leq d\leq n$ implies the existence of a complementary linear $(n+1-\ell )$ -subspace $\widetilde {L}\subset \Bbb R^{n+1}$ such that $\mathrm {supp}\,\mu \subset L\cup \widetilde {L}$ .

3. (iii) If $\frac 1{n+2}<\alpha <1$ and $d\mu =f\,d\theta $ for nonnegative $f\in L^{\frac {n+1}{n+2-\frac 1\alpha }}( \mathbb {S}^n)$ with $\int _{\mathbb {S}^n}f>0$ .

Let us briefly discuss what is known about uniqueness of the solution of the $L^p$ -Minkowski problem (1.4). If $p>1$ and $p\neq n$ , then Hug, Lutwak, Yang, and Zhang [30] proved that the Alexandrov solution of the $L^p$ -Minkowski problem (1.4) is unique. However, if $p<1$ , then the solution of the $L^p$ -Minkowski problem (1.3) may not be unique even if f is positive and continuous. Examples are provided by Chen, Li, and Zhu [17, 18] if $p\in [0,1)$ , and Milman [42] shows that for any $C\in \mathcal {K}_{(0)}$ , one finds $q\in (-n,1)$ such that if $p<q$ , then there exist multiple solutions to the $L^p$ -Minkowski problem (1.4) with $\mu =S_{C,p}$ ; or in other words, there exists $K\in \mathcal {K}_{(0)}$ with $K\neq C$ and $S_{K,p}=S_{C,p}$ . In addition, Jian, Lu, and Wang [33] and Li, Liu, and Lu [37] prove that for any $p<0$ , there exists positive even $C^\infty $ function f with rotational symmetry such that the $L^p$ -Minkowski problem (1.3) has multiple positive even $C^\infty $ solutions. We note that in the case of the centro-affine Minkowski problem $p=-n$ , Li [36] even verified the possibility of existence of infinitely many solutions without affine equivalence, and Stancu [46] related unique solution in the cases $p=0$ and $p=-n$ .

The case when f is a constant function in the $L^p$ -Minkowski problem (1.3) has received a special attention since [23]. When $p=-(n+1)$ , (1.3) is self-similar solution of affine curvature flow. It is proved by Andrews that all solutions are centered ellipsoids. If $n=2$ and $p=2$ , the uniqueness was proved by Andrews [3]. For general n and $p>-(n+1)$ , through the work of Lutwak [40], Guan-Ni [26], and Andrews, Guan, and Ni [6], Brendle, Choi, and Daskalopoulos [13] finally classified that the only solutions are centered balls. See also [21, 32, 44] for other approaches. Stability versions of these results have been obtained by Ivaki [31], but still no stability version is known in the case $p\in [0,1)$ if we allow any solutions of (1.3) not only even ones.

Concerning recent versions of the $L^p$ -Minkowski problem, see [9].

The paper is structured as follows: The required diameter bounds are discussed in Section 2. Section 3 verifies the main properties of the Entropy, Section 4 proves our main result (Theorem 4.1) about flows, and finally Theorem 1.1 is proved in Section 5 via weak approximation.

2 Entropy and diameter estimates

For $\delta \in [0,1)$ and linear i-subspace L of $\Bbb R^{n+1}$ with $1\leq \mathrm {dim}\,L\leq n$ , we consider the collar

$$ \begin{align*}\Psi(L\cap \mathbb{S}^n,\delta)=\{x\in \mathbb{S}^n:\langle x,y\rangle\leq \delta\mbox{ for }y\in L^\bot\cap \mathbb{S}^n\}. \end{align*} $$

Let $B(1)\subset \Bbb R^{n+1}$ be the unit ball centered at the origin.

Theorem 2.1 Let $\alpha>\frac 1{n+2}$ , let $\frac {\ \ }{\ \ }{\hskip -0.4cm}\int _{\mathbb {S}^n}f=1$ for a bounded measurable function f on $\mathbb {S}^n$ with $\inf f>0$ , and let $\Omega \subset \Bbb R^{n+1}$ be a convex body such that $|\Omega |=|B(1)|$ and $\mathrm {diam}\, \Omega = D$ . For any $\delta ,\tau \in (0,1)$ , we have

1. (i) if $\alpha>1$ , and $\frac {\ \ }{\ \ }{\hskip -0.4cm}\int _{\Psi (z^\bot \cap \mathbb {S}^n,\delta )}f\leq 1-\tau $ for any $z\in S^n$ , then

   $$ \begin{align*}\exp\left(\frac{\alpha-1}{\alpha}\,\mathcal{E}_{\alpha, f} (\Omega)\right) \geq \gamma_1 \tau\delta^{1-\frac1\alpha}D^{1-\frac1\alpha}, \end{align*} $$
   where $\gamma _1>0$ depends on n and $\alpha $ ;

2. (ii) if $\alpha =1$ , and

   $$ \begin{align*}\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\Psi(L\cap \mathbb{S}^n,\delta)}f< \frac{(1-\tau)i}{n+1}, \end{align*} $$
   for any linear i-subspace L of $\Bbb R^{n+1}$ , $i=1,\ldots ,n$ , then
   $$ \begin{align*}\mathcal{E}_{1, f} (\Omega)\geq\tau\log D +\log\delta-4\log(n+1); \end{align*} $$

3. (iii) if $\frac 1{n+2}<\alpha <1$ , $p=1-\frac 1\alpha $ (where $-n-1<p<0$ ), $\tau \leq \frac 12\frac {\ \ }{\ \ }{\hskip -0.4cm}\int _{\mathbb {S}^n}f\cdot u^{1-\frac 1\alpha }$ and

   (2.1) $$ \begin{align} \frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\Psi(z^\bot\cap \mathbb{S}^n,\delta)}f^{\frac{n+1}{n+1+p}}\leq \tau^{\frac{n+1}{n+1+p}}, \end{align} $$
   for any $z\in S^{n-1}$ , then
   $$ \begin{align*}\mbox{either}\ D\leq 16n^2/\delta^2, \ \ \mbox{or } D\leq \left(\frac12\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n}f\cdot u^{1-\frac1\alpha}\right)^{\frac2{p}}. \end{align*} $$
   Moreover, if $\tau \leq \frac 12\exp \left (\frac {\alpha -1}{\alpha }\,\mathcal {E}_{\alpha , f} (\Omega )\right )$ , then
   $$ \begin{align*}\mbox{either}\ D\leq 16n^2/\delta^2, \ \ \mbox{or } D\leq \left(\frac12\exp\left(\frac{\alpha-1}{\alpha}\,\mathcal{E}_{\alpha, f} (\Omega)\right)\right)^{\frac2{p}}. \end{align*} $$

Remark 2.2 We note that for any $\alpha \ge 1$ , bounded f with $\inf f>0$ and $\frac {\ \ }{\ \ }{\hskip -0.4cm}\int _{\mathbb {S}^n}f=1$ , and $\tau \in (0,1)$ , there exists $\delta \in (0,1)$ such that conditions in (i) and (ii) hold. In the case of $1>\alpha >\frac 1{n+2}$ , (iii) holds if in addition that $\tau \leq \frac 12\exp \left (\frac {1-\alpha }{\alpha }\,\mathcal {E}_{\alpha , f} (\Omega )\right )$ for the convex body $\Omega \subset \Bbb R^{n+1}$ .

Proof Given $\alpha>\frac 1{n+2}$ , bounded f with $\inf f>0$ and $\frac {\ \ }{\ \ }{\hskip -0.4cm}\int _{\mathbb {S}^n}f=1$ , and $\tau \in (0,1)$ , the existence of suitable $\delta \in (0,1)$ follows from the fact that the Lebesgue measure is a Borel measure.

Now, we assume that the conditions in (i)–(iii) hold. We may assume that the centroid of $\Omega $ is the origin; thus, Kannan, Lovász, and Simonovics [35] yield the existence of an o-symmetric ellipsoid such that

(2.2) $$ \begin{align} E\subset\Omega\subset (n+1)E,\mbox{ and hence }-\Omega\subset (n+1)\Omega. \end{align} $$

Let u be the support function of $\Omega $ , and let $R=\max \{\|y\|:\,y\in \Omega \}\geq D/2$ and $z_0\in \mathbb {S}^n$ such that $Rz_0\in \partial \Omega $ . We observe that the definition of the entropy yields

$$ \begin{align*} \frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n}fu^{1-\frac1\alpha}&\leq \exp\left(\frac{1-\alpha}{\alpha}\,\mathcal{E}_{\alpha, f} (\Omega)\right) \mbox{ if}\ \alpha>1;\\ \frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n}f\log u&\leq \mathcal{E}_{0, f} (\Omega);\\ \frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n}fu^{1-\frac1\alpha}&\geq \exp\left(\frac{1-\alpha}{\alpha}\,\mathcal{E}_{\alpha, f} (\Omega)\right) \mbox{ if}\ \frac1{n+2}<\alpha<1. \end{align*} $$

Case 1: $\alpha>1$ .

According to the condition in (i), we may choose $\zeta \in \{+1,-1\}$ such that

$$ \begin{align*}\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\Phi}f\geq \frac{\tau}2 \mbox{ for }\Phi=\{x\in \mathbb{S}^n:\langle x,\zeta z_0\rangle>\delta\}, \end{align*} $$

and hence $\frac {R\zeta z_0}{n+1}\in \Omega $ by (2.2). Since $u_\sigma (x)\geq \langle \frac {R\zeta z_0}{n+1},x\rangle \geq \frac {R\delta }{n+1}$ for $x\in \Phi $ , we have

$$ \begin{align*}\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n}fu^{1-\frac1\alpha}\geq\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\Phi}f\left(\frac{R\delta}{n+1}\right)^{1-\frac1\alpha}\geq \frac{\tau}2\cdot\left(\frac{D\delta}{2(n+1)}\right)^{1-\frac1\alpha}. \end{align*} $$

Case 2: $\alpha =1$ .

To simplify notation, we consider the Borel probability measure $\mu (A)=\frac {\ \ }{\ \ }{\hskip -0.4cm}\int _Af$ on $S^n$ . Let $e_1,\ldots ,e_{n+1}\in \mathbb {S}^n$ be the principal directions associated with the ellipsoid E in (2.2), and let $r_1,\ldots ,r_{n+1}>0$ be the half axes of E with $r_ie_i\in \partial E$ where we may assume that $r_1\leq \cdots \leq r_{n+1}$ . In particular, (2.2) yields that

(2.3) $$ \begin{align} (n+1)^{n+1}\prod_{i=1}^{n+1} r_i=\frac{|(n+1)E|}{|B(1)|}\geq \frac{|\Omega|}{|B(1)|}=1. \end{align} $$

We observe that for any $v\in \mathbb {S}^n$ , there exists $e_i$ such that $|\langle v,e_i\rangle |\geq \frac {1}{\sqrt {n+1}}> \frac {\delta }{n+1}$ . For $i=1,\ldots ,n+1$ , we define

$$ \begin{align*}B_i=\left\{v\in \mathbb{S}^n:|\langle v,e_i\rangle|\geq \frac{\delta}{n+1}\mbox{ and } |\langle v,e_j\rangle|<\frac{\delta}{n+1}\mbox{ for }j>i\right\}. \end{align*} $$

In particular, $B_i\subset \Psi (L_i\cap \mathbb {S}^n,\delta )$ for $i=1,\ldots ,n$ and $L_i=\mathrm {lin}\{e_1,\ldots ,e_i\}$ .

It follows that $\mathbb {S}^n$ is partitioned into the Borel sets $B_1,\ldots ,B_{n+1}$ , and as $B_i\subset \Psi (L_i\cap \mathbb {S}^n,\delta )$ for $i=1,\ldots ,n$ , we have

(2.4) $$ \begin{align} \mu(B_1)+\cdots+\mu(B_i)&\leq \frac{i(1-\tau)}{n+1} \mbox{ for}\ i=1,\ldots,n, \end{align} $$

(2.5) $$ \begin{align} \mu(B_1)+\cdots+\mu(B_{n+1})&= 1. \end{align} $$

For $\zeta =\frac {1-\tau }{n+1}$ , we have $0< \zeta <\frac 1{n+1}$ , and define

(2.6) $$ \begin{align} \beta_i&= \mu(B_i)-\zeta\mbox{ for}\ i=1,\ldots,n, \end{align} $$

(2.7) $$ \begin{align} \beta_{n+1}&= \mu(B_{n+1})-\zeta-\tau, \end{align} $$

where (2.4) and (2.5) yield

(2.8) $$ \begin{align} \beta_1+\cdots+\beta_i&\leq 0\mbox{ for}\ i=1,\ldots,m-1, \end{align} $$

(2.9) $$ \begin{align} \beta_1+\cdots+\beta_{n+1}&= 0. \end{align} $$

As $r_ie_i\in \Omega $ , it follows from the definition of $B_i$ that $u(x)\geq \langle x,r_ie_i\rangle \geq r_i\cdot \frac {\delta }{n+1}$ for $x\in B_i$ , $i=1,\ldots ,n+1$ . We deduce from applying (2.3), (2.5)–(2.9), $r_1\leq \cdots \leq r_{n+1}$ , and $\zeta <\frac 1{n+1}$ that

$$ \begin{align*} \int_{\mathbb{S}^n}\log u\,d\mu&= \sum_{i=1}^{n+1}\int_{B_i}\log u\,d\mu\\ &\geq \sum_{i=1}^{n+1}\mu(B_i)\log r_i+\sum_{i=1}^{n+1}\mu(B_i)\log \frac{\delta}{n+1} = \sum_{i=1}^{n+1}\mu(B_i)\log r_i+\log \frac{\delta}{n+1}\\ &= \sum_{i=1}^{n+1}\beta_i\log r_i+\sum_{i=1}^{n+1}\zeta \log r_i +\tau\log r_{n+1}+\log \frac{\delta}{n+1}\\ &\geq \sum_{i=1}^{n+1}\beta_i\log r_i+\zeta\log \frac{1}{(n+1)^{n+1}} +\tau\log r_{n+1}+\log \frac{\delta}{n+1}\\ &= (\beta_1+\cdots+\beta_{n+1})\log r_{n+1}+ \sum_{i=1}^{n}(\beta_1+\cdots+\beta_i)(\log r_i-\log r_{i+1})\\ & -(n+1)\zeta\log (n+1) +\tau\log r_{n+1}+\log \frac{\delta}{n+1}\\ &\geq -\log (n+1)+\tau\log r_{n+1}+\log \frac{\delta}{n+1}. \end{align*} $$

Now, $D\leq (n+1)\mathrm {diam}\,E=2(n+1)r_{n+1}\leq (n+1)^2r_{n+1}$ and $\tau <1$ , and hence

$$ \begin{align*} -\log (n+1)+\tau\log r_{n+1}+\log \frac{\delta}{n+1}&\geq -\log (n+1)+\tau\log \frac{D}{(n+1)^2} +\log \frac{\delta}{n+1}\\ &= \log\left(\delta D^\tau\right)-(2+2\tau)\log (n+1)\\ &\geq \tau\log D +\log\delta-4\log(n+1). \end{align*} $$

In particular, we conclude that

$$ \begin{align*}\mathcal{E}_{1, f} (\Omega)\geq \frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n}f\log u =\int_{\mathbb{S}^n}\log u\,d\mu \geq\tau\log D +\log\delta-4\log(n+1). \end{align*} $$

Case 3: $\frac 1{n+2}<\alpha <1$ .

In this case, $-(n+1)<1-\frac 1\alpha <0$ . We may assume that

$$ \begin{align*}D\geq 16n^2/\delta^2, \end{align*} $$

and we consider

$$ \begin{align*} \Phi_0&= \left\{x\in \mathbb{S}^n:\,u(x)> \sqrt{2R}\right\},\\ \Phi_1&= \left\{x\in \mathbb{S}^n:\,u(x)\leq \sqrt{2R}\right\}. \end{align*} $$

Concerning $\Phi _0$ , we have

(2.10) $$ \begin{align} \frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\Phi_0}f\cdot u^{1-\frac1\alpha}\leq (2R)^{\frac12(1-\frac1\alpha)}\int_{\Phi_0}f\leq D^{\frac12(1-\frac1\alpha)}=D^{\frac{p}2}. \end{align} $$

On the other hand, we have $\pm \frac {R}{(n+1)}\,z_0\in \Omega $ by (2.2), thus any $x\in \Phi _1$ satisfies

$$ \begin{align*}\sqrt{2R}\geq u(x)\geq \left|\left\langle x,\frac{R}{n+1}\,z_0\right\rangle\right|, \end{align*} $$

and hence $|\langle x,z_0\rangle |\leq (n+1)\sqrt {\frac {2}{R}}\leq \frac {4n}{\sqrt {D}}\leq \delta $ ; or in other words,

$$ \begin{align*}\Phi_1\subset \Psi(z_{0}^\bot\cap \mathbb{S}^n,\delta). \end{align*} $$

It follows from $|\Omega |=|B(1)|$ and the Blaschke–Santaló inequality (cf. [45]) that

$$ \begin{align*}\int_{\mathbb{S}^n} u^{-(n+1)}\leq (n+1)|B(1)|=\omega_{n},\mbox{ and hence } \frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n} u^{-(n+1)}\leq 1. \end{align*} $$

For $p=1-\frac 1\alpha \in (-n-1,0)$ , Hölder’s inequality and $\int _{\Phi _1}f^{\frac {n+1}{n+1+p}}< \tau ^{\frac {n+1}{n+1+p}}$ yield

$$ \begin{align*}\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\Phi_1}f\cdot u^{1-\frac1\alpha}\leq \left(\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\Phi_1}f^{\frac{n+1}{n+1+p}}\right)^{\frac{n+1+p}{n+1}} \left(\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\Phi_1} u_\sigma^{-(n+1)}\right)^{\frac{|p|}{n+1}}\leq \left(\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\Phi_1}f^{\frac{n+1}{n+1+p}}\right)^{\frac{n+1+p}{n+1}}\leq \tau. \end{align*} $$

Finally, adding the last estimate to (2.10) yields

$$ \begin{align*}\exp\left(\frac{\alpha-1}{\alpha}\,\mathcal{E}_{\alpha, f} (\Omega)\right)\leq \frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n}f\cdot u^{1-\frac1\alpha}\leq D^{\frac{p}2}+\tau, \end{align*} $$

and hence the conditions either $\tau \leq \frac 12\frac {\ \ }{\ \ }{\hskip -0.4cm}\int _{\mathbb {S}^n}f\cdot u^{1-\frac 1\alpha }$ or $\tau \leq \frac 12\exp \left (\frac {1-\alpha }{\alpha }\,\mathcal {E}_{\alpha , f} (\Omega )\right )$ on $\tau $ implies (iii).

3 Anisotropic flows and monotonicity of entropies

The following theorem was proved by Andrews in [4] (see also for a discussion of contracting of non-homogeneous fully nonlinear anisotropic curvature flows in [24]).

Theorem 3.1 [4]

For any $\alpha>0$ and positive $f\in C^{\infty }(\mathbb S^n)$ and any initial smooth, strictly convex hypersurface $\tilde M_0\subset \mathbb R^{n+1}$ , the hypersurfaces $\tilde M_{\tau }$ given by the solution of (1.7) exist for a finite time T and converge in Hausdorff distance to a point $p \in \mathbb R^{n+1}$ as $\tau $ approaches T.

Assuming

$$\begin{align*}\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb S^n}f=1, \quad |\Omega_0|=|B(1)|,\end{align*}$$

solution (1.7) yields a smooth convex solution to the normalized flow (1.8) with volume preserved.

Set

(3.1) $$ \begin{align} h_z(x,t)\doteqdot f(x) u_z^{-\frac{1}{\alpha}}(x,t)K(x,t), \quad d\sigma_t(x) =\frac{u_z(x,t)}{K(x,t)}d\theta(x).\end{align} $$

Note that $\frac {\ \ }{\ \ }{\hskip -0.4cm}\int _{\mathbb {S}^n} d\sigma _t(x) =\frac {\ \ }{\ \ }{\hskip -0.4cm}\int _{\mathbb {S}^n} d\theta (x) =1$ .

Since the un-normalized flow (1.7) shrinks to a point in finite time, we may assume that it is the origin. Then the support function $u(x,t)$ is positive for the normalized flow (1.8).

Lemma 3.2

1. (a) The entropy $ \mathcal {E}_{\alpha , f}(\Omega _{t})$ defined in (1.5) is monotonically decreasing,

   (3.2) $$ \begin{align} \mathcal{E}_{\alpha, f}(\Omega_{t_2})\le \mathcal{E}_{\alpha, f}(\Omega_{t_1}), \quad \forall t_1\le t_2 \in [0, \infty).\end{align} $$

2. (b) There is $D>0$ depending only on $\inf f, \sup f, \alpha , \Omega _0$ such that

   (3.3) $$ \begin{align} \mathrm{diam}\,\Omega_t=D(t)\le D, \ \forall t\ge 0.\end{align} $$

3. (c) $\forall t_0\in [0, \infty )$ ,

   (3.4) $$ \begin{align} \mathcal{E}_{\alpha, f}(\Omega_{t_0}, 0)\ge \mathcal{E}_{\alpha, f, \infty} +\int_{t_0}^{\infty}\left(\frac{\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n} h^{\alpha+1}(x,t)\, d\sigma_t} {\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n} h(x,t) \, d\sigma_t \cdot \frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n} h^{\alpha}(x,t)\, d\sigma_t}-1\right)\, dt,\end{align} $$

   where

   $$\begin{align*}h(x,t)=h_0(x, t), \ \mathcal{E}_{\alpha, f, \infty}\doteqdot\lim_{t\to \infty} \mathcal{E}_{\alpha, f}(\Omega_{t}).\end{align*}$$

Proof

1. (a) We follow argument in [26]. For each $T_0>$ fixed, pick $T> T_0$ . Let $a^{T}=(a^{T}_1,\ldots , a^{T}_{n+1})$ be an interior point of $\Omega _T$ . Set $u^T=u- e^{t-T}\sum _{i=1}^{n+1} a^T_ix_i$ ; it satisfies equation

   (3.5) $$ \begin{align} \frac{\partial}{\partial t}u^T(x, t)= -\frac{f^\alpha(x) K^{\alpha}(x, t)}{\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n}f^\alpha K^{\alpha-1}} +u^T(x,t).\end{align} $$

   Note that since $a^T$ is an interior point of $\Omega _T$ and $u(x,T)$ is the support function of $\Omega _T$ with respect to $a^T$ , $u^T(x, T)> 0, \forall x\in \mathbb S^n$ . We claim

   $$\begin{align*}u^T(x, t)>0, \ \forall t\in [0, T).\end{align*}$$
   Suppose $u^T(x_0, t')\le 0$ for some $0<t'<T, x_0\in \mathbb S^n$ , and equation (3.5) implies $u^T(x_0, t)<0$ for all $t>t'$ , which contradicts to $u^T(x, T)> 0$ .

   Set $a^T(t)=e^{t-T}a^T$ . By the claim, $a^T(t)$ is in the interior of $\Omega _t, \ \forall t\le T$ . Denote

   $$\begin{align*}d\sigma_{T,t}=u^T(x,t)K^{-1}(x,t)d\theta,\end{align*}$$
   we rewrite equation (3.3) as
   (3.6) $$ \begin{align} \frac{\partial}{\partial t}u_{a^T(t)}(x,t)= -\frac{f^\alpha(x) K^{\alpha}(x, t)}{\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n}h_{a^T(t)}^{\alpha}(x,t)\, d\sigma_{T,t}} +u_{a^T(t)}(x,t).\end{align} $$
   We have
   $$\begin{align*}\frac{\partial}{\partial t} \mathcal{E}_{\alpha, f}(\Omega_{t}, a^T(t))=\frac{-\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n} h_{a^T(t)}^{\alpha+1}(x,t)\, d\sigma_{T,t}} {\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n} h_{a^T(t)}(x,t) \, d\sigma_{T,t} \cdot \frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n} h_{a^T(t)}^{\alpha}(x,t)\, d\sigma_{T,t}}+1.\end{align*}$$

   Thus, $\forall t<T$ ,

   (3.7) $$ \begin{align} &\mathcal{E}_{\alpha, f}(\Omega_{t}, a^T(t))-\mathcal{E}_{\alpha, f}(\Omega_T, a^T)\\ \nonumber &= \int_{t}^{T}\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb S^n} \left(\frac{\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n} h_{a^T(t)}^{\alpha+1}(x,t)\, d\sigma_{T,t}} {\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n} h_{a^T(t)}(x,t) \, d\sigma_{T,t} \cdot \frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n} h_{a^T(t)}^{\alpha}(x,t)\, d\sigma_{T,t}}-1\right)\, dt\ge 0. \end{align} $$
   Therefore,
   $$\begin{align*}\mathcal{E}_{\alpha, f}(\Omega_{t})\ge\mathcal{E}_{\alpha, f}(\Omega_T, a^T), \ \forall t<T.\end{align*}$$
   Since $a^T$ is arbitrary, (3.2) is proved.

2. (b) The boundedness of $D(t)$ follows from Theorem 2.1 combined with the estimate $\mathcal {E}_{\alpha , 1}(\Omega _{t})\leq \mathcal {E}_{\alpha , 1}(B(1))$ from (a) (see also [6, 26]). The only nontrivial case is when $\frac 1{n+2}<\alpha <1$ because we have to choose a $\tau $ independent of t. However, we may choose any $\tau \in (0,1)$ with $\tau \leq \frac 12\exp \left (\frac {1-\alpha }{\alpha }\,\mathcal {E}_{\alpha , f} (B(1))\right )$ according to $\mathcal {E}_{\alpha , 1}(\Omega _{t})\leq \mathcal {E}_{\alpha , 1}(B(1))$ .

3. (c) $\forall \epsilon>0, \ \forall t_0$ fixed, pick $T>T_0>t_0$ . As $ \mathcal {E}_{\alpha , f}(\Omega _{T})$ is bounded by (a), $\exists a^T$ inside $\Omega _T$ such that $ \mathcal {E}_{\alpha , f}(\Omega _{T})\le \mathcal {E}_{\alpha , f}(\Omega _{T}, a^T)+\epsilon $ . By (3.7),

   $$ \begin{align*} &\mathcal{E}_{\alpha, f}(\Omega_{t_0}, a^T(t_0))-\mathcal{E}_{\alpha, f}(\Omega_{T})\\ &\ge \int_{t_0}^{T_0}\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb S^n} \left(\frac{\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n} h_{a^T(t)}^{\alpha+1}(x,t)\, d\sigma_{T,t}} {\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n} h_{a^T(t)}(x,t) \, d\sigma_{T,t} \cdot \frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n} h_{a^T(t)}^{\alpha}(x,t)\, d\sigma_{T,t}}-1\right)\, dt-\epsilon. \end{align*} $$

   As $|a^T|\le D, \ \forall T$ , let $T\to \infty $ ,

   $$\begin{align*}a^T(t)\to 0, \ u^T(x,t)\to u(x,t), \ \ \mbox{ uniformly for}\ 0\le t\le T_0, x\in \mathbb S^n. \end{align*}$$

   We obtain $\forall t_0<T_0$ ,

   $$ \begin{align*} \mathcal{E}_{\alpha, f}(\Omega_{t_0}, 0)-\mathcal{E}_{\alpha, f, \infty}\ge \int_{t_0}^{T_0}\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb S^n} \left(\frac{\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n} h^{\alpha+1}(x,t)\, d\sigma_t} {\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n} h(x,t) \, d\sigma_t \cdot \frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n} h^{\alpha}(x,t)\, d\sigma_t}-1\right)\, dt-\epsilon.\end{align*} $$

   Then let $T_0\to \infty $ , as $\epsilon>0$ is arbitrary, we obtain (3.4).

4 Weak convergence

The goal of this section is to prove the following statement.

Theorem 4.1 For a $C^\infty $ function $f:\mathbb {S}^n\to (0,\infty )$ and $\alpha>\frac 1{n+2}$ with $ \frac {\ \ }{\ \ }{\hskip -0.4cm}\int _{\mathbb {S}^n}f=1$ , there exist $\lambda>0$ and a convex body $\Omega \subset \Bbb R^{n+1}$ with $o\in \Omega $ whose support function u is a (possibly weak) solution of the Monge–Ampère equation

(4.1) $$ \begin{align} u^{\frac1\alpha}\det(\bar{\nabla}^2_{ij} u+u\bar{g}_{ij})= f \end{align} $$

and $\Omega $ satisfies that

(4.2) $$ \begin{align} \mathcal{E}_{\alpha, f} (\lambda\Omega)\leq \mathcal{E}_{\alpha, f} (B(1)), \quad |\lambda\Omega|=|B(1)|, \end{align} $$

where $C^{-1}<\lambda <C$ for a $C>1$ depending only on the $\alpha , \tau , \delta $ in Theorem 2.1 such that f satisfies the conditions in Theorem 2.1.

From now on, we will assume that the f in Theorem 4.1 satisfies the corresponding condition in Theorem 2.1 and $\Omega _0=B(1)$ in (1.8). We note that for any $z\in B(1)$ , $v_z\leq 2$ for the support function $v_z$ of $B(1)$ at z, and hence if $\alpha>\frac 1{n+2}$ , then

(4.3) $$ \begin{align} \mathcal{E}_{\alpha, f_k} (B(1))\leq\left\{ \begin{array}{rl} \frac{\alpha}{\alpha-1}\cdot\log 2^{1-\frac1\alpha},&\mbox{ if }\alpha\neq 1,\\ \log 2,&\mbox{ if }\alpha=1. \end{array} \right. \end{align} $$

The following is a consequence of Theorem 2.1 and Lemma 3.2.

Lemma 4.2 There exist $C_{\alpha , \tau , \delta }>0, D_{\alpha , \tau , \delta }>0$ , and $c_{\alpha , \tau , \delta }\in \mathbb R$ depending only on constants $\alpha , \tau , \delta $ in Theorem 2.1 such that, along (1.8), we have

(4.4) $$ \begin{align} D(t)\le D_{\alpha, \tau, \delta},\ \mathcal{E}_{\alpha, f}(\Omega_{t}, 0)\ge c_{\alpha, \tau, \delta}, \ \frac{1}{C_{\alpha, \tau, \delta}}\le \frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb S^n} h(x,t)d\sigma_t\le C_{\alpha, \tau, \delta}.\end{align} $$

Proof For each $\alpha>\frac 1{n+2}$ fixed with condition on f as in Theorem 2.1, $\mathcal {E}_{\alpha , f}(\Omega _{t})$ is bounded from below in terms of the diameter $D(t)$ . Since $|\Omega _t|=|B(1)|$ , we have $D(t)\ge 2$ by the Isodiametric Inequality (cf. [45]). By Theorem 2.1, $\mathcal {E}_{\alpha , f}(\Omega _{t})$ is bounded from below by a constant $c_{\alpha , \tau , \delta }>0$ , and hence $\mathcal {E}_{\alpha , f, \infty } \ge c_{\alpha , \tau , \delta }$ . It follows from Lemma 3.2 that $\mathcal {E}_{\alpha , f}(\Omega _{t})\le \mathcal {E}_{\alpha , f}(B(1))$ , and this estimate combined with (4.3) and Theorem 2.1 yields $D(t)\le D_{\alpha , \tau , \delta }$ where $D_{\alpha , \tau , \delta }$ depends only on constants in condition on f in Theorem 2.1. Finally, the inequalities follow from Lemma 3.2.

Set

(4.5) $$ \begin{align} \eta(t)=\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n} h(x,t) \, d\sigma_{t}. \end{align} $$

We note that $\frac {\ \ }{\ \ }{\hskip -0.4cm}\int _{\mathbb {S}^n} h(x,t) \, d\sigma _{t} $ is monotone and bounded from below and above by Lemma 4.2, and hence we have

(4.6) $$ \begin{align}C_{\alpha, \tau, \delta}\ge \lim_{t\to \infty} \frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n} h(x,t)=\eta\ge \frac1{C_{\alpha, \tau, \delta}}.\end{align} $$

By Lemma 3.2 and Corollary 4.2,

(4.7) $$ \begin{align} \int_0^{\infty} \left(\frac{\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n} h^{\alpha+1}(x,t)\, d\sigma_t} {\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n} h(x,t) \, d\sigma_t \cdot \frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n} h^{\alpha}(x,t)\, d\sigma_t}-1\right)\, dt<\infty.\end{align} $$

Since the integrand is nonnegative, $\exists t_k\to \infty $ such that

(4.8) $$ \begin{align} \lim_{k\to \infty} \left( \frac{\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n}h^{\alpha+1}(x,t_k)\, d\sigma_{t_k}} {\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n} h(x,t_k) \, d\sigma_{t_k} \cdot \frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n} h^{\alpha}(x,t_k)\, d\sigma_{t_k}}-1\right)=0.\end{align} $$

This implies

(4.9) $$ \begin{align} \lim_{k\to \infty}\frac{\left(\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n}h^{\alpha+1}(x,t_k)\, d\sigma_{t_k}\right)^{\frac{1}{1+\alpha}}}{\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n} h(x,t_k) \, d\sigma_{t_k} }= \lim_{k\to \infty}\frac{\left(\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n}h^{\alpha+1}(x,t_k)\, d\sigma_{t_k}\right)^{\frac{\alpha}{1+\alpha}}}{\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n} h^{\alpha}(x,t_k)\, d\sigma_{t_k}}= 1.\end{align} $$

After considering a subsequence, we may assume that

(4.10) $$ \begin{align} \Omega_{t_k}\to \Omega, \quad u(x,t_k)\to u(x),\end{align} $$

where u is the support function of $\Omega $ . In view of (4.9) and (4.6),

(4.11) $$ \begin{align} \lim_{k\to \infty} \frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n}h^{\alpha+1}(x,t_k)\, d\sigma_{t_k}=\eta^{1+\alpha},\ \lim_{k\to \infty}\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n} h^{\alpha}(x,t_k)\, d\sigma_{t_k}= \eta^{\alpha}.\end{align} $$

The following lemma is crucial for the weak convergence, which is a refined form of the classical Hölder inequality. ¹

Lemma 4.3 Let $p,\ q\in \mathbb R^+$ with $\frac 1p+\frac 1q=1$ , and set $\beta =\min \{\frac 1p, \frac 1q\}$ . Let $(M,\mu )$ be a measurable space; $\forall F\in L^p, \ G\in L^q$ ,

(4.12) $$ \begin{align} \int_M |FG| d\mu \le \|F\|_{L^p}\|G\|_{L^q}\left(1-\beta\int_M \left(\frac{|F|^{\frac{p}2}}{(\int_M |F|^p d\mu)^{\frac12}}-\frac{|G|^{\frac{q}2}}{(\int_M |G|^qd\mu )^{\frac12}}\right)^2\right).\end{align} $$

Proof We first prove the following Claim. $\forall s, t\in \mathbb R$ ,

(4.13) $$ \begin{align} e^{\frac{s}{p}+\frac{t}{q}}\le \frac{e^s}{p}+\frac{e^t}{q}-\beta(e^{\frac{s}2}-e^{\frac{t}2})^2.\end{align} $$

We may assume $t\ge s$ , set $\tau =t-s$ , and (4.13) is equivalent to

(4.14) $$ \begin{align} e^{\frac{\tau}{q}}\le \frac{1}{p}+\frac{e^{\tau}}{q}-\beta(1-e^{\frac{\tau}2})^2, \ \forall \tau\ge 0.\end{align} $$

Set

$$\begin{align*}\xi(\tau)=\frac{1}{p}+\frac{e^{\tau}}{q}-\beta(1-e^{\frac{\tau}2})^2-e^{\frac{\tau}{q}}.\end{align*}$$

We have $\xi (0)=0$ ,

$$\begin{align*}\xi'(\tau)=\frac{e^{\frac{\tau}q}}{q}\rho, \ \mbox{where} \ \rho(\tau)=e^{\frac{\tau}{p}}(1-\beta q)+q\beta e^{\frac{\tau}2-\frac{\tau}q}-1.\end{align*}$$

If $\beta =\frac 1q$ , then $\frac 1q\le \frac 12$ ; since $\tau \ge 0$ ,

$$\begin{align*}\rho(\tau)=e^{\frac{\tau}{p}}(1-\beta q)+q\beta e^{\frac{\tau}2-\frac{\tau}q}-1=e^{\frac{\tau}2-\frac{\tau}q}-1\ge 0.\end{align*}$$

If $\beta =\frac 1p$ , then $\frac 1q\ge \frac 12$ ; we have

$$ \begin{align*} \rho'(\tau)&=e^{\frac{\tau}{p}}\left(\frac{1-\beta q}p+\beta q(\frac12-\frac1q)e^{\frac{\tau}2-\frac{\tau}q}\right)\\ & \ge e^{\frac{\tau}{p}}\left(\frac{1-\beta q}p+\beta q(\frac12-\frac1q)\right)\\ &\ge e^{\frac{\tau}{p}}\beta q(\frac12-\frac1p)\ge 0.\end{align*} $$

We conclude that

$$\begin{align*}\rho(\tau)\ge 0, \ \forall \tau\ge 0.\end{align*}$$

In turn,

$$\begin{align*}\xi'(\tau)\ge 0, \ \forall \tau\ge o.\end{align*}$$

This yields (4.14) and (4.13). The Claim is verified.

Back to the proof of the lemma. We may assume

$$\begin{align*}F\ge 0, \ g\ge 0, \ \int F^p>0, \ \int G^q>0.\end{align*}$$

Set

$$\begin{align*}e^s=\frac{F^p}{\int F^p}, \quad e^t=\frac{G^q}{\int G^q}.\end{align*}$$

Put them into (4.13) and integrate, as $\frac 1p+\frac 1q=1$ ,

$$\begin{align*}\frac{\int FG}{(\int F^p)^{\frac1{p}}(\int G^q)^{\frac1{q}}}\le \left(1-\beta\int (\frac{F^{\frac{p}2}}{(\int F^p)^{\frac12}}-\frac{G^{\frac{q}2}}{(\int G^q)^{\frac12}})^2\right).\\[-34pt] \end{align*}$$

We prove weak convergence.

Proposition 4.4 $\forall \alpha>\frac {1}{n+2}$ , suppose that (4.10) and (4.11) hold. Denote

$$\begin{align*}u_{k}=u(x, t_k), \ \sigma_{n,k}=\sigma_n(u_{ij}(x,t_k)+u(x,t_k)\delta_{ij}).\end{align*}$$

Then

(4.15) $$ \begin{align} \lim_{k\to \infty} \frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb S^n} |u_k^{\frac{1}\alpha}\sigma_{n,k}-\frac{f}{\eta}| d\theta =0,\end{align} $$

where $\eta $ is defined in (4.5) which is bounded from below and above in (4.6). As a consequence, there is a convex body $\Omega \subset \mathbb R^{n+1}$ with $o\in \Omega $ ,

$$\begin{align*}|\Omega|=|B(1)|, \quad \mathcal{E}_{\alpha, f}(\Omega_{t})\le \mathcal{E}_{\alpha, f}(B(1)),\end{align*}$$

and its support function u satisfies

(4.16) $$ \begin{align} u^{\frac1{\alpha}} S_{\Omega}=\frac1{\eta} fd \theta. \end{align} $$

Proof We only need to verify (4.15). By (4.11), it is equivalent to prove

(4.17) $$ \begin{align} \lim_{k\to \infty} \frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb S^n} |u_k^{\frac{1}\alpha}\sigma_{n,k}-f\eta^{-1}(t_k)| d\theta =0.\end{align} $$

Since $D(t_k)$ is bounded,

$$\begin{align*}\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb S^n} u_k^{\frac{1}{\alpha^2}}\sigma_{n,k} d\theta\le (D(t_k))^{\frac1{\alpha^2}}\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb S^n} u_k^{\frac{1}{\alpha^2}}\sigma_{n,k} d\theta\le (D(t_k))^{\frac1{\alpha^2}}|\partial \Omega_{t_k}|\le C.\end{align*}$$

(4.18) $$ \begin{align} \frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb S^n} |u_k^{\frac{1}\alpha}\sigma_{n,k}-f\eta^{-1}(t_k)| d\theta &=\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb S^n} |\frac{f}{\eta(t_k)u_k^{\frac{1}\alpha}\sigma_{n,k}}-1 |u_k^{\frac{1}\alpha}\sigma_{n,k} d\theta\nonumber\\ &\le \left(\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb S^n} |\frac{f}{\eta(t_k)u_k^{\frac{1}\alpha}\sigma_{n,k}}-1 |^{1+\alpha} d\sigma_{t_k}\right)^{\frac{1}{1+\alpha}}\left(\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb S^n} u_k^{(\frac1{\alpha}-1)\frac{1+\alpha}{\alpha}}d\sigma_{t_k}\right)^{\frac{\alpha}{1+\alpha}}\nonumber \\ &= \left(\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb S^n} |\frac{f}{\eta(t_k)u_k^{\frac{1}\alpha}\sigma_{n,k}}-1 |^{1+\alpha} d\sigma_{t_k}\right)^{\frac{1}{1+\alpha}}\left(\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb S^n} u_k^{\frac{1}{\alpha^2}}\sigma_{n,k} d\theta\right)^{\frac{\alpha}{1+\alpha}}\nonumber \\ &\le C \left(\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb S^n} |f\eta^{-1}(t_k)u_k^{-\frac{1}\alpha}\sigma^{-1}_{n,k}-1 |^{1+\alpha} d\sigma_{t_k}\right)^{\frac{1}{1+\alpha}}. \end{align} $$

By (4.8), (4.11), and Lemma 4.3, with $p=\alpha +1$ , $F^{\frac {1}{1+\alpha }}=h(x,t_k)$ , $G=1$ ,

(4.19) $$ \begin{align} \lim_{k\to \infty} \frac{\ \ }{\ \ }{\hskip -0.4cm}\int \left((\frac{h(x,t_k)}{\eta(t_k)})^{\frac{1+\alpha}2}-1\right)^2d\sigma_{t_k}=0.\end{align} $$

For $t_k$ fixed, let

$$\begin{align*}\gamma_k(x)=f\eta^{-1}(t_k)u_k^{-\frac{1}\alpha}\sigma^{-1}_{n,k}=h(x,t_k)\eta^{-1}(t_k)\end{align*}$$

and set

$$\begin{align*}\Sigma_{k}=\left\{x\in \mathbb S^n \ | \ |\gamma_k(x)-1|\le \frac12\right\}.\end{align*}$$

It is straightforward to check that $\exists A_{\alpha }\ge 1$ depending only on $\alpha $ such that

$$ \begin{align*} A_{\alpha}|\gamma^{\frac{1+\alpha}2}_k(x)-1|&\ge |\gamma_k(x)-1|, \ \forall x\in \Sigma_k, \\ A_{\alpha} |\gamma^{\frac{1+\alpha}2}_k(x)-1|^{2}&\ge |\gamma_k(x)-1|^{1+\alpha}, \ \forall x\in \Sigma_{k}^c.\end{align*} $$

Since $ |\gamma ^{\frac {1+\alpha }2}_k(x)-1|\le 2^{1+\alpha }, \ \forall x\in \Sigma _k$ , let $\delta =\min \{1+\alpha , 2\}$ ,

$$ \begin{align*} \ \frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb S^n} |\gamma_k(x)-1 |^{1+\alpha} d\sigma_{t_k}=&\frac{1}{\omega_n}\left(\int_{\Sigma_k} |\gamma_k(x)-1 |^{1+\alpha} d\sigma_{t_k}+\int_{\Sigma_k^c} |\gamma_k(x)-1 |^{1+\alpha} d\sigma_{t_k}\right)\\ \le & \frac{A^{1+\alpha}_{\alpha}}{\omega_n} \left(\int_{\Sigma_k} |\gamma_k^{\frac{1+\alpha}2}(x)-1 |^{1+\alpha} d\sigma_{t_k}+\int_{\Sigma_k^c} |\gamma_k^{\frac{1+\alpha}2}(x)-1 |^{2} d\sigma_{t_k}\right)\\ \le & \frac{(2A_{\alpha})^{1+\alpha}}{\omega_n} \left(\int_{\Sigma_k} |\gamma_k^{\frac{1+\alpha}2}(x)-1 |^{\delta} d\sigma_{t_k}+\int_{\Sigma_k^c} |\gamma_k^{\frac{1+\alpha}2}(x)-1 |^{2} d\sigma_{t_k}\right)\\ \le & (2A_{\alpha})^{1+\alpha} \left(\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb S^n} |\gamma_k^{\frac{1+\alpha}2}(x)-1 |^{\delta} d\sigma_{t_k}+\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb S^n} |\gamma_k^{\frac{1+\alpha}2}(x)-1 |^{2} d\sigma_{t_k}\right)\\ \le & (2A_{\alpha})^{1+\alpha} \left((\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb S^n} |\gamma_k^{\frac{1+\alpha}2}(x)-1 |^{2} d\sigma_{t_k})^{\frac{\delta}2}+\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb S^n} |\gamma_k^{\frac{1+\alpha}2}(x)-1 |^{2} d\sigma_{t_k}\right).\end{align*} $$

By (4.19),

$$\begin{align*}\lim_{k\to \infty}\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb S^n} |\gamma_k^{\frac{1+\alpha}2}(x)-1 |^{2} d\sigma_{t_k}=0.\end{align*}$$

Hence,

(4.20) $$ \begin{align} \lim_{k\to \infty} \frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb S^n} |\gamma_k(x)-1 |^{1+\alpha} d\sigma_{t_k}=0.\end{align} $$

Now, (4.17) follows from (4.18)–(4.20).

Proof Proof of Theorem 4.1.

It follows from Proposition 4.4 after a proper rescaling as $\eta $ satisfies (4.6) and (4.16).

5 The general Monge–Ampère equations – proof of Theorem 1.1

In order to prove Theorem 1.1, we need weak approximation in the following sense.

Lemma 5.1 For $\delta ,\varepsilon \in (0,\frac 12)$ and a Borel probability measure $\mu $ on $\mathbb {S}^n$ , $n\geq 1$ , there exists a sequence $d\mu _k=\frac 1{\omega _n}\,f_k\,d\theta $ of Borel probability measures whose weak limit is $\mu $ and $f_k\in C^\infty ( \mathbb {S}^n)$ satisfies $f_k>0$ and the following properties:

1. (i) If $\mu \left (\Psi (z^\bot \cap \mathbb {S}^n,2\delta )\right )\leq 1-\varepsilon $ for any $z\in S^{n-1}$ , then

   (5.1) $$ \begin{align} \frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\Psi(z^\bot\cap \mathbb{S}^n,\delta)}f_k\leq 1-\varepsilon \mbox{ for any}\ z\in S^{n-1}. \end{align} $$

2. (ii) If $\mu (\Psi (L\cap \mathbb {S}^n,2\delta ))<(1-2\delta )\cdot \frac {\ell }{n+1}$ for any linear $\ell $ -subspace L of $\Bbb R^{n+1}$ , $\ell =1,\ldots ,n$ , then

   (5.2) $$ \begin{align} \mu_k\left(\Psi\left(L\cap \mathbb{S}^n,\delta\right)\right)<(1-\delta)\cdot \frac{\ell}{n+1}. \end{align} $$

3. (iii) If $d\mu =\frac 1{\omega _n}\,f\,d\theta $ for $f\in L^{r}(\mathbb {S}^n)$ where $r>1$ , and

   (5.3) $$ \begin{align} \frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\Psi(z^\bot\cap \mathbb{S}^n,2\delta)}f^r\leq \varepsilon \end{align} $$

   for any $z\in S^{n-1}$ , then

   (5.4) $$ \begin{align} \int_{\Psi(z^\bot\cap \mathbb{S}^n,\delta)}f_k^r\leq 2^r\varepsilon \mbox{ for any}\ z\in S^{n-1}. \end{align} $$

Proof For $k\geq 1$ , let $\{B_{k,i}\}_{i=1,\ldots ,m(k)}$ be a partition of $S^n$ into spherically convex Borel measurable sets $B_{k,i}$ with $\mathrm {diam}B_{k,i}\leq \frac 1k$ and $\theta (B_{k,i})>0$ . For each $B_{k,i}$ , we choose a $C^\infty $ function $h_{k,i}:\mathbb {S}^n\to [0,\infty )$ such that for $M_{k,i}=\max h_{k,i}$ and the probability measure $d\tilde {\theta }=\frac 1{\omega _n}\,d\theta $ , we have:

• • $h_{k,i}=0$ if $x\not \in B_{k,i}$ ;

• • $M_{k,i}\leq (1+\frac 1k)\cdot \frac {\mu (B_{k,i})}{\tilde {\theta }(B_{k,i})}$ ;

• • $\theta \left (\left \{x\in B_{k,i}:h_{k,i}(x)<M_{k,i}\right \}\right )<\frac 1k\,\theta (B_{k,i})$ ;

• • $\int _{B_{k,i}}h_{k,i}\,d\tilde {\theta }=\mu (B_{k,i})$ .

We consider the positive $C^\infty $ function $\tilde {f}_k\hspace{-0.5pt}=\hspace{-0.5pt}\frac 1k\hspace{-0.5pt}+\hspace{-0.5pt}\sum _{i=1}^{m(k)}h_{k,i}$ , and hence $f_k\hspace{-0.5pt}=\hspace{-0.5pt}\left (\hspace{2pt} \frac {\ \ }{\ \ }{\hskip -0.4cm}\int _{\mathbb {S}^n}\tilde {f}_k\right )^{-1}\tilde {f}$ satisfies that the probability measure $d\mu _k=f_k\,d\tilde {\theta }$ tends weakly to $\mu $ , and for large $k\geq 1/\delta $ , $\mu _k$ satisfies (i), and if (ii) holds, then $\mu _k$ also satisfies (5.2).

Turning to (iii), we assume that $d\mu =f\,d\tilde {\theta }$ for $f\in L^{r}(\mathbb {S}^n)$ where $r>1$ , and f satisfies (5.3). For any large k and $i=1,\ldots ,m(k)$ , we deduce from the Hölder inequality that

$$ \begin{align*} \frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{B_{k,i}} \tilde{f}_k^r&=\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{B_{k,i}} \left(h_{k,i}+\frac1k\right)^r\leq 2^{r-1}\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{B_{k,i}} h_{k,i}^r+2^{r-1}\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{B_{k,i}} \frac1{k^r}\\ &\leq 2^{r-1}\tilde{\theta}(B_{k,i})M_{k,i}^r+2^{r-1}\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{B_{k,i}} \frac1{k^r}\\ &\leq 2^{r-1}\left(1+\frac1k\right)^r\tilde{\theta}(B_{k,i}) \left(\frac{\int_{B_{k,i}} f}{\tilde{\theta}(B_{k,i})}\right)^r+2^{r-1}\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{B_{k,i}} \frac1{k^r}\\ &\leq 2^{r-1}\left(1+\frac1k\right)^r\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{B_{k,i}} f^r+2^{r-1}\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{B_{k,i}}\frac1{k^r}. \end{align*} $$

Summing this estimate up for large k and all $B_{k,i}$ with $B_{k,i}\cap \Psi (z^\bot \cap \mathbb {S}^n,\delta )\neq \emptyset $ , and using that $\frac {\ \ }{\ \ }{\hskip -0.4cm}\int _{\mathbb {S}^n}\tilde {f}_k\geq 2^{-1/2}$ for large k, we deduce that

$$ \begin{align*}\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\Psi(z^\bot\cap \mathbb{S}^n,\delta)} f_k^r\leq \sqrt{2} \frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\Psi(z^\bot\cap \mathbb{S}^n,\delta)} \tilde{f}_k^r\leq \sqrt{2}\cdot2^{r-1}\left(1+\frac1k\right)^r\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\Psi(z^\bot\cap \mathbb{S}^n,2\delta)} f^r+ \sqrt{2}\cdot\frac{2^{r-1}}{k^r}\leq 2^r\varepsilon. \end{align*} $$

For $\alpha>0$ and $p=1-\frac 1\alpha $ , the $L^p$ -surface area $dS_{\Omega ,p}=u^{1-p}dS_\Omega $ was introduced in the seminal works [39–41] for a convex body $\Omega \subset \Bbb R^{n+1}$ with $o\in \Omega $ and support function u. Since the surface area measure is weakly continuous for $p<1$ , and if $K\subset \Bbb R^{n+1}$ is an at most n-dimensional compact convex set, then $S_{K,p}\equiv 0$ for $p<1$ , we have the following statement.

Lemma 5.2 If convex bodies $\Omega _m\subset \Bbb R^{n+1}$ tend to a compact convex set $K\subset \Bbb R^{n+1}$ where $o\in \Omega _m,K$ , and $\liminf _{m\to \infty }S_{\Omega _m,p}>0$ , then $\mathrm {int}K\neq \emptyset $ and $S_{\Omega _m,p}$ tends weakly to $S_{K,p}$ .

For the reader’s sake, let us recall Theorem 1.1.

Theorem 5.3 For $\alpha>\frac 1{n+2}$ and finite nontrivial Borel measure $\mu $ on $\mathbb {S}^n$ , $n\geq 1$ , there exists a weak solution of (1.2) provided the following holds:

1. (i) If $\alpha>1$ and $\mu $ is not concentrated onto any great subsphere $x^\bot \cap \mathbb {S}^n$ , $x\in \mathbb {S}^n$ .

2. (ii) If $\alpha =1$ and $\mu $ satisfies that for any linear $\ell $ -subspace $L\subset \Bbb R^{n+1}$ with $1\leq \ell \leq n$ , we have:

   1. (a) $\displaystyle \mu (L\cap \mathbb {S}^n)\leq \frac {\ell }{n+1}\cdot \mu (\mathbb {S}^n)$ ;

   2. (b) equality in (a) for a linear $\ell $ -subspace $L\subset \Bbb R^{n+1}$ with $1\leq d\leq n$ implies the existence of a complementary linear $(n+1-\ell )$ -subspace $\widetilde {L}\subset \Bbb R^{n+1}$ such that $\mathrm {supp}\,\mu \subset L\cup \widetilde {L}$ .

3. (iii) If $\frac 1{n+2}<\alpha <1$ , assume $d\mu =fd\theta $ for nonnegative $f\in L^{\frac {n+1}{n+2-\frac 1\alpha }}( \mathbb {S}^n)$ with $\int _{\mathbb {S}^n}f>0$ .

Proof Let $\alpha>\frac 1{n+2}$ . After rescaling, we may assume that the $\mu $ in (1.2) is a probability measure. We consider the sequence $d\mu _k=\frac 1{\omega _n}f_k\,d\theta $ of Lemma 5.1 of Borel probability measures whose weak limit is $\mu $ and $f_k\in C^\infty ( \mathbb {S}^n)$ satisfies $f_k>0$ . For each $f_k$ , let $\Omega _k\subset \Bbb R^{n+1}$ be the convex body with $o\in \Omega _k$ provided by Theorem 4.1 whose support function $u_k$ is the solution of the Monge–Ampère equation

(5.5) $$ \begin{align} u_k^{\frac1\alpha}\,dS_{\Omega_k}=f _k\,d\theta; \end{align} $$

$\exists \lambda _k>0$ under control, with $|\lambda _k\Omega |=|B(1)|$ , $\Omega _k$ satisfies that

(5.6) $$ \begin{align} \mathcal{E}_{\alpha, f_k} (\lambda_k\Omega_k)\leq \mathcal{E}_{\alpha, f_k} (B(1)). \end{align} $$

We also need the observations that

(5.7) $$ \begin{align} |\Omega_k|=\frac1{n+1}\int_{\mathbb{S}^n}u_k\,dS_{\Omega_k}, \end{align} $$

and if $p=1-\frac 1\alpha $ , then

(5.8) $$ \begin{align} S_{\Omega_k,p}(\mathbb{S}^n)=\int_{\mathbb{S}^n}u_k^{1-\frac1\alpha}\,dS_{\Omega_k} =\omega_n\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n}f_k=\omega_n. \end{align} $$

We claim that if there exists $\Delta>0$ depending on n, $\alpha $ , and $\mu $ such that

(5.9) $$ \begin{align} \mathrm{diam}\Omega_k\leq \Delta,\ \mbox{then Theorem 5.3 holds.} \end{align} $$

To prove this claim, we note that (5.9) yields the existence of a subsequence of $\{\Omega _k\}$ tending to a compact convex set $\Omega $ with $o\in \Omega $ , which is a convex body by (5.8) and Lemma 5.2. Moreover, Lemma 5.2 also yields that $\Omega $ is an Alexandrov solution of (1.2), verifying the claim (5.9).

We divide the rest of the argument verifying Theorem 5.3 into three cases.

Case 1: $\alpha>1$ .

Since $\mu $ is not concentrated to any great subsphere, there exist $\delta \in (0,\frac 12)$ depending on $\mu $ such that $\mu \left (\Psi (z^\bot \cap \mathbb {S}^n,2\delta )\right )\leq 1-2\delta $ for any $z\in S^{n-1}$ . It follows from Lemma 5.1 that we may assume that

(5.10) $$ \begin{align} \frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\Psi(z^\bot\cap \mathbb{S}^n,\delta)}f_k\leq 1-\delta \mbox{ for any}\ z\in S^{n-1}. \end{align} $$

Now, Theorem 4.1 implies that $\lambda _k\geq c$ for a constant $c>0$ depending on n, $\delta $ , and $\alpha $ , and in turn Theorem 4.1, (4.3), and $\frac 1{\alpha }-1<0$ yield that

$$ \begin{align*}\mathcal{E}_{\alpha, f} (\Omega_k)=\frac{\alpha}{\alpha-1}\cdot\log\lambda_k^{\frac1{\alpha}-1} +\mathcal{E}_{\alpha, f} (\lambda_k\Omega_k)\leq \frac{\alpha}{\alpha-1}\cdot\log\lambda_k^{\frac1{\alpha}-1} +\mathcal{E}_{\alpha, f} (B(1))\leq C \end{align*} $$

for a constant $C>0$ depending on n, $\delta $ , and $\alpha $ . Therefore, Theorem 2.1 and (5.10) imply that the sequence $\{\Omega _k\}$ is bounded, and in turn the claim (5.9) implies Theorem 5.3 if $\alpha>1$ .

Case 2: $\alpha =1$ .

The argument is by induction on $n\geq 0$ where we do not put any restriction on the probability measure $\mu $ in the case $n=0$ . For the case $n=0$ , we observe that any finite measure $\mu $ on $S^0$ can be represented in the form $d\mu =u\,dS_\Omega $ for a suitable segment $\Omega \subset \Bbb R^1$ .

For the case $n\geq 1$ , assuming that we have verified Theorem 5.3(ii) in smaller dimensions, we consider a Borel measure probability $\mu $ on $S^n$ satisfying (a) and (b).

Case 2.1: There exists a linear $\ell $ -subspace $L\subset \Bbb R^{n+1}$ with $1\leq \ell \leq n$ and $\mu (L\cap \mathbb {S}^n)= \frac {\ell }{n+1}\cdot \mu (\mathbb {S}^n)$ .

Let $\widetilde {L}\subset \Bbb R^{n+1}$ be the complementary linear $(n+1-\ell )$ -subspace with $\mathrm {supp}\,\mu \subset L\cup \widetilde {L}$ , and hence $\mu (\widetilde {L}\cap \mathbb {S}^n)= \frac {n+1-\ell }{n+1}\cdot \mu (\mathbb {S}^n)$ . It follows by induction that there exist an $\ell $ -dimensional compact convex set $K'\subset L$ and an $(n+1-\ell )$ -dimensional compact convex set $\widetilde {K}'\subset \widetilde {L}$ such that and . Finally, for $K=\widetilde {L}^\bot \cap (K'+L^\bot )$ and $\widetilde {K}=L^\bot \cap (\widetilde {K}'+\widetilde {L}^\bot )$ , there exist $\alpha ,\tilde {\alpha }>0$ such that

$$ \begin{align*}\mu=(n+1)V_{\alpha K+\tilde{\alpha}\widetilde{K}}. \end{align*} $$

Case 2.2: $\mu (L\cap \mathbb {S}^n)< \frac {\ell }{n+1}\cdot \mu (\mathbb {S}^n)$ for any linear $\ell $ -subspace $L\subset \Bbb R^{n+1}$ with $1\leq \ell \leq n$ .

It follows by a compactness argument that there exists $\delta \in (0,\frac 12)$ depending on $\mu $ such that $\mu (\Psi (L\cap \mathbb {S}^n,2\delta ))<(1-2\delta )\cdot \frac {\ell }{n+1}$ for any linear $\ell $ -subspace L of $\Bbb R^{n+1}$ , $\ell =1,\ldots ,n$ . We consider the sequence of probability measures $d\mu _k=\frac 1{\omega _n}f_k\,d\theta $ of Lemma 5.1 tending weakly to $\mu $ such that $f_k>0$ , $f_k\in C^\infty (\mathbb {S}^n)$ , and

(5.11) $$ \begin{align} \mu_k\left(\Psi\left(L\cap \mathbb{S}^n,\delta\right)\right)<(1-\delta)\cdot \frac{\ell}{n+1} \end{align} $$

for any linear $\ell $ -subspace L of $\Bbb R^{n+1}$ , $\ell =1,\ldots ,n$ .

For each $f_k$ , let $\Omega _k\subset \Bbb R^{n+1}$ with $o\in \Omega _k$ be the convex body provided by Theorem 4.1 whose support function $u_k$ is the solution of the Monge–Ampère equation (4.1) and satisfies (4.2) with $f=f_k$ and $\lambda =\lambda _k$ where $|B(1)|=|\lambda _k\Omega _k|$ for $\lambda _k>0$ , and

$$ \begin{align*} |\Omega_k|&=\frac1{n+1}\,\int_{\mathbb{S}^n}u_k\det(\bar{\nabla}^2_{ij} u_k+u_k\bar{g}_{ij})\,d\theta= \frac{\omega_n}{n+1}\,\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n}u_k\det(\bar{\nabla}^2_{ij} u_k+u_k\bar{g}_{ij})\\ &=|B(1)|\,\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n}f_k=|B(1)|, \end{align*} $$

and hence $\lambda _k=1$ . In particular, (4.3) yields

$$ \begin{align*}\mathcal{E}_{1, f_k} (\lambda_k\Omega_k)\leq \mathcal{E}_{1, f_k} (B(1))\leq \log 2. \end{align*} $$

Since $\mathcal {E}_{1, f_k} (\Omega _k)$ is bounded, (5.11) and Theorem 2.1 imply that the sequence $\Omega _k$ stays bounded, as well. Therefore, the claim (5.9) yields Theorem 5.3 if $\alpha =1$ .

Case 3: $\frac 1{n+2}<\alpha <1$ .

We set $p=1-\frac 1\alpha \in (-n-1,0)$ and $r=\frac {n+1}{n+1+p}>1$ , and

(5.12) $$ \begin{align} \tau= \frac12\cdot 2^{-\frac{|p|(n+1)}{|p|+n}}, \end{align} $$

and choose $\delta \in (0,\frac 12)$ such that

$$ \begin{align*}\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\Psi(z^\bot\cap \mathbb{S}^n,2\delta)}f^r\leq \frac{\tau^r}{2^r} \end{align*} $$

for any $z\in S^{n-1}$ . We deduce from Lemma 5.1 that if $z\in S^{n-1}$ , then

(5.13) $$ \begin{align} \frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\Psi(z^\bot\cap \mathbb{S}^n,\delta)}f_k^r\leq \tau^r. \end{align} $$

We deduce from (5.5), (5.7), and $|\lambda _k\Omega _k|=|B(1)|=\frac {\omega _n}{n+1}$ that

(5.14) $$ \begin{align} \frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n}u_k^{p}f_k=\frac{n+1}{\omega_n} \int_{\mathbb{S}^n}u_k\,dS_{\Omega_k}= \frac{n+1}{\omega_n}\,|\Omega_k|=\lambda_k^{-n-1}. \end{align} $$

In particular, (4.3) and the upper bound on the entropy yield that

(5.15) $$ \begin{align} \nonumber 2^{p}&\leq \exp\left(p\cdot \mathcal{E}_{\alpha, f_k} (B(1))\right) \leq \exp\left(p\cdot \mathcal{E}_{\alpha, f} (\lambda_k\Omega_k)\right) \leq \frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n}(\lambda_ku_k)^{p}f_k\\ &=\lambda_k^{p}\int_{\mathbb{S}^n}u_k\,dS_{\Omega_k}= \lambda_k^{p-n}\cdot\frac{n+1}{\omega_n}\cdot|\lambda_k\Omega_k|=\lambda_k^{p-n}. \end{align} $$

It follows from (5.15) that $\lambda _k\leq 2^{\frac {|p|}{|p|+n}}$ , and in turn (5.14) yields that

$$ \begin{align*}\frac{\ \ }{\ \ }{\hskip -0.4cm}\int_{\mathbb{S}^n}u_k^{p}f_k\geq 2^{-\frac{|p|(n+1)}{|p|+n}}. \end{align*} $$

Therefore, $\tau \leq \frac 12\frac {\ \ }{\ \ }{\hskip -0.4cm}\int _{\mathbb {S}^n}u_k^{p}f_k$ (cf. (5.12)), (5.13), and Theorem 2.1 yield that the sequence $\{\Omega _k\}$ is bounded, and in turn the claim (5.9) implies Theorem 5.3 if $\frac 1{n+2}<\alpha <1$ .