Alexandrov’s Theorem for Anisotropic Capillary Hypersurfaces in the Half-Space

Abstract

In this paper, we show that any embedded capillary hypersurface in the half-space with anisotropic constant mean curvature is a truncated Wulff shape. This extends Wente’s result (Pac J Math 88:387–397, 1980. https://doi.org/10.2140/pjm.1980.88.387) to the anisotropic case and He–Li–Ma–Ge’s result (Indiana Univ Math J 58(2):853–868, 2009. https://doi.org/10.1512/iumj.2009.58.3515) to the capillary boundary case. The main ingredients in the proof are a new Heintze-Karcher inequality and a new Minkowski formula, which have their own interest.

1 Introduction

Capillary phenomena appear in the study of the equilibrium shape of liquid drops and crystals in a given solid container. The mathematical model has been established through the work of Young, Laplace, Gauss and others, as a variational problem on minimizing a free energy functional under a volume constraint. A modern formulation of Gauss’ model includes a possibly anisotropic surface tension density, which we are interested in. For more detailed description of the isotropic and anisotropic capillary phenomena, we refer to [12] and [6].

For our purposes, we consider the anisotropic capillary problem in the half-space

$$\begin{aligned} \mathbb {R}^{n+1}_+=\{x\in \mathbb {R}^{n+1}: \langle x, E_{n+1}\rangle > 0\}. \end{aligned}$$

Here $E_{n+1}$ denotes the $(n+1)$-coordinate unit vector. Let $\Sigma$ be a compact orientable embedded hypersurface in $\overline{\mathbb {R}^{n+1}_+}$ with boundary $\partial \Sigma$ lying on $\partial {\mathbb {R}}_+^{n+1}$, which, together with $\partial {\mathbb {R}}_+^{n+1}$, encloses a bounded domain $\Omega$. Let $\nu$ be the unit normal of $\Sigma$ pointing outward $\Omega$. We consider the free energy functional

$$\begin{aligned} \mathcal {E}({\Sigma })=\int _{\Sigma } F(\nu ) {\hbox {d}}A+\omega _0\vert \partial \Omega \cap \partial \mathbb {R}^{n+1}_+\vert , \end{aligned}$$

where the term $\int _{\Sigma } F(\nu ) {\hbox {d}}A$ is the anisotropic surface tension and the term $\omega _0\vert \partial \Omega \cap \partial \mathbb {R}^{n+1}_+\vert$ is the wetting energy accounting for the adhesion between the fluid and the walls of the container. Here $F: \mathbb {S}^n\rightarrow \mathbb {R}_+$ is a $C^2$ positive function on $\mathbb {S}^n$, such that $(\nabla ^2 F+F \sigma )>0$, where $\sigma$ is the canonical metric on $\mathbb {S}^n$ and $\nabla ^2$ is the Hessian on $\mathbb {S}^n$, and $\omega _0\in \mathbb {R}$ is a given constant. The Cahn-Hoffman map associated with F is given by

$$\begin{aligned} \Phi :\mathbb {S}^n\rightarrow \mathbb {R}^{n+1},\quad \Phi (x) =\nabla F(x)+F(x)x, \end{aligned}$$

where $\nabla$ denote the gradient on $\mathbb {S}^n$. One easily sees that $\Phi (x)=D{\tilde{F}}(x)$, where ${\tilde{F}}$ is the positive one-homogeneous extension of F to $\mathbb {R}^{n+1}$ and D denotes the Euclidean derivative. The image $\Phi (\mathbb {S}^n)$ of $\Phi$ is a strictly convex, closed hypersurface in $\mathbb {R}^{n+1}$, which is the unit Wulff shape with respect to F, which we denote by $\mathcal {W}_F$.

In the isotropic case $F\equiv 1$, the global minimizer of $\mathcal {E}$ under a volume constraint is characterized as a spherical cap by De Giorgi, which is the solution to the relative isoperimetric problem; see for example [27, Chapter 19]. In the anisotropic case, the global minimizer of $\mathcal {E}$ under volume constraint has been characterized by Winterbottom [11] as a truncated Wulff shape, which is also called a Winterbottom shape, or Winterbottom construction in applied mathematics, especially in material science; see for example [4] and references therein. The Winterbottom construction can be viewed as the capillary counterpart of Wulff construction, which characterizes the global minimizer for purely anisotropic surface tension, see [13, 32, 35]. For anisotropic free energy functionals involving a gravitational potential energy term, the existence, the regularity and boundary regularity of global minimizers have been studied by De Giorgi [7], Almgren, [1] and Taylor [33]; see also the recent work by De Philippis and Maggi [6, 8]. For the symmetry and uniqueness of global minimizers we refer to the work of Baer [5] for a class of F with certain symmetry, following the work of Gonzalez [15] in the isotropic case, via a symmetrization technique.

In this paper, we shall study the rigidity for the stationary surfaces for the free energy functional $\mathcal {E}$ under a volume constraint. Given a variation $\{{\Sigma }_t\}$ of ${\Sigma }$, whose boundary $\partial {\Sigma }_t$ moves freely on $\partial \mathbb {R}^{n+1}_+$ and according to a variational vector field Y such that $Y\mid _{\partial {\Sigma }}\in T(\partial \mathbb {R}^{n+1}_+)$, the first variation formula of $\mathcal {E}$ is given by

$$\begin{aligned} \frac{\hbox {d}}{{\hbox {d}}t}\mid _{t=0}\mathcal {E}({\Sigma }_t)=\int _{\Sigma }H^F\left<Y,\nu \right>{\hbox {d}}A+\int _{\partial {\Sigma }}\left<Y,R(p(\Phi (\nu )))\right> {\hbox {d}}s+\omega _0\int _{\partial \Sigma }\left<Y,\mu \right> {\hbox {d}}s, \end{aligned}$$

where $H^F$ is the anisotropic mean curvature of ${\Sigma }$, p is the projection onto the $\{\nu , E_{n+1}\}$-plane, R is the $\pi /2$-rotation in the $\{\nu , E_{n+1}\}$-plane, $\mu$ is the conormal of $\partial \Sigma \subset \Sigma$, see [22, 23, 30]. For its proof we refer to the one of [23, Proposition 2]. It follows that the stationary points of $\mathcal {E}$ among $C^2$ hypersurfaces under a volume constraint are anisotropic $\omega _0$-capillary hypersurfaces with constant anisotropic mean curvature. In this paper we say a hypersurface in $\overline{\mathbb {R}^{n+1}_+}$ with boundary $\partial {\Sigma }\subset \partial {{\mathbb {R}}}^{n+1}_+$ anisotropic $\omega _0$ -capillary if

$$\begin{aligned} \langle \Phi (\nu ),-E_{n+1}\rangle =\omega _0,\quad \text { on }\partial {\Sigma }. \end{aligned}$$

(1.1)

We emphasize that it is not necessarily a constant anisotropic mean curvature hypersurface. Moreover we are interested in hypersurfaces which intersect with $\partial \mathbb {R}^{n+1}_+$ transversely.

The rigidity of embedded closed constant mean curvature hypersurfaces was obtained by Alexandrov [2] in the celebrated Alexandrov’s theorem, that any embedded closed hypersurface of constant mean curvature in $\mathbb {R}^{n+1}$ must be a sphere. In the proof Alexandrov introduced the famous moving plane method. Wente [34] showed that any embedded compact hypersurface of constant mean curvature with capillary boundary in $\overline{\mathbb {R}^{n+1}_+}$ is a spherical cap. Taking into account of the anisotropy, He–Li–Ma–Ge [17] proved that any embedded closed hypersurface in $\mathbb {R}^{n+1}$ with constant anisotropic mean curvature must be a Wulff shape. See also a related result by Morgan [28] in $\mathbb {R}^2$ for a more general anisotropic function F. We also mention that the Alexandrov-type theorem for general finite perimeter sets has been proved in the isotropic setting by Delgadino-Maggi [9], and in the anisotropic setting by De Rosa-Kolasiński-Santilli [10]. For the closely related work on the stability problem of constant anisotropic mean curvature hypersurface without boundary or with capillary boundary, we refer to [16, 22,23,24,25,26, 30] and references therein.

Our main result in this paper is the following Alexandrov-type theorem for embedded anisotropic capillary hypersurfaces of constant anisotropic mean curvature in $\overline{\mathbb {R}^{n+1}_+}$.

Theorem 1.1

Let $\omega _0\in \left( -F(E_{n+1}),F(-E_{n+1})\right)$. Let ${\Sigma }\subset \overline{\mathbb {R}^{n+1}_+}$ be a $C^2$ embedded compact anisotropic $\omega _0$-capillary hypersurface with constant anisotropic mean curvature. Then ${\Sigma }$ is an $\omega _0$-capillary Wulff shape.

An $\omega _0$-capillary Wulff shape is part of a Wulff shape in $\overline{\mathbb {R}^{n+1}_+}$ such that the anisotropic capillary boundary condition (1.1) holds. We remark that the assumption $\omega _0\in \left( -F(E_{n+1}),F(-E_{n+1})\right)$ is a necessary condition so that Wulff shapes intersect with $\partial \mathbb {R}_+^{n+1}$ transversely, see Remark 2.1.

As mentioned above, Theorem 1.1 for the isotropic case was proved by Wente in [34], where he used Alexandrov’s moving plane method. However, the moving plane method fails in general for the anisotropic case, at least if F has less symmetry. A new proof of Wente’s result has been done by the authors [21] through the establishment of a Heintze-Kacher-type inequality in the capillary problem, which is inspired by the original idea of Heintze-Karcher [18] (see also Montiel-Ros [29]). This method is flexible to the anisotropic case and this is the way we achieve Theorem 1.1.

Following this way we first need to establish a Heintze-Harcher type inequality for anisotropic capillary hypersurfaces. In order to state the inequality, we need a constant vector $E^F_{n+1}\in \mathbb {R}^{n+1}$ defined as

$$\begin{aligned} E^F_{n+1}=\left\{ \begin{array}{lrl} \frac{\Phi (E_{n+1})}{F(E_{n+1})}, \quad &{}\text {if } \omega _0<0,\\ -\frac{\Phi (-E_{n+1})}{F(-E_{n+1})}, \quad &{}\text {if } \omega _0>0. \end{array} \right. \end{aligned}$$

(1.2)

Note that $E^F_{n+1}$ is the unique vector in the direction $\Phi (E_{n+1})$, whose scalar product with $E_{n+1}$ is 1. When $\omega _0=0$, one can define it by any unit vector. This constant vector plays a crucial role in the paper. A hypersurface is said to be strictly anisotropic-mean convex if $H^F>0$. Now we state our anisotropic Heintze-Karcher inequality.

Theorem 1.2

Let $\omega _0\in \left( -F(E_{n+1}),F(-E_{n+1})\right)$ and ${\Sigma }\subset \overline{\mathbb {R}^{n+1}_+}$ be a $C^2$ compact embedded strictly anisotropic-mean convex hypersurface with boundary $\partial {\Sigma }\subset \partial \mathbb {R}_+^{n+1}$ such that

$$\begin{aligned} \langle \Phi (\nu (x)), -E_{n+1}\rangle =\omega (x)\le \omega _0,\quad \hbox { for any }x\in \partial \Sigma . \end{aligned}$$

(1.3)

Then it holds that

$$\begin{aligned} \int _{\Sigma }\frac{F(\nu )+\omega _0\left<\nu ,E^F_{n+1}\right>}{H^F}{\hbox {d}}A\ge \frac{n+1}{n}\vert \Omega \vert . \end{aligned}$$

(1.4)

Equality in (1.4) holds if and only if ${\Sigma }$ is an $\omega _0$-capillary Wulff shape.

We will follow the argument in [21] to prove Theorem 1.2. The main idea is to define suitable parallel hypersurfaces $\zeta _F(\cdot , t)$, in order to sweepout the enclosed domain $\Omega$ and use the area formula to compute the volume. A crucial ingredient is an anisotropic angle comparison principle in Proposition 3.1 which enables us to prove the surjectivity of $\zeta _F$.

Then we need the following anisotropic Minkowski-type formula:

Theorem 1.3

Let $\omega _0\in \left( -F(E_{n+1}),F(-E_{n+1})\right)$ and ${\Sigma }\subset \overline{\mathbb {R}^{n+1}_+}$ be a $C^2$ compact anisotropic $\omega _0$-capillary hypersurface. Let $H_r^F$ be the (normalized) anisotropic r-th mean curvature for some $r\in \{1,\ldots ,n\}$ and $H_0^F\equiv 1$ by convention. Then it holds

$$\begin{aligned} \int _{\Sigma }H^F_{r-1}\left( F(\nu )+\omega _0\left<\nu , E^{F}_{n+1}\right>\right) -H^F_r\left<x,\nu \right>{\hbox {d}}A=0. \end{aligned}$$

(1.5)

In particular,

$$\begin{aligned} \int _{\Sigma }\left( F(\nu )+\omega _0\left<\nu , E^{F}_{n+1}\right>\right) -H^F_1\left<x,\nu \right>{\hbox {d}}A=0. \end{aligned}$$

(1.6)

Remark 1.1

We remark the importance of using the constant vector $E^F_{n+1}$. In fact, (1.4) and (1.5) hold true, if we replace $E^F_{n+1}$ by $E_{n+1}$. However, if one used $E_{n+1}$ instead of $E_{n+1}^F,$ we could only prove our main Theorem 1.1, for a smaller range $\omega _0\in (-1/F^o(E_{n+1}), 1/F^o(-E_{n+1}))$. For one of reasons see Proposition 3.2. The main reason lies in the proof of the Heintze-Karcher inequality. For details, see Remarks 3.1 and 3.2. This is one of the crucial differences between the isotropic case and the anisotropic case.

For the isotropic case $F\equiv 1$, (1.4) and (1.5) were proved by the authors [21]. We refer to [21] and [36] for a historical description of the Heintze-Karcher inequality and the Minkowski formula respectively, and references therein.

The anisotropic Heintze-Karcher inequality and the anisotropic Minkowski formula for closed hypersurfaces have been proved by He–Li–Ma–Ge [17, Theorem 4.4] and He–Li [19]. In this case, our argument provides a slight improvement for the anisotropic Heintze-Karcher inequality, at least when F is even, see Remark 3.3.

Corollary 1.1

Let ${\Sigma }\subset \mathbb {R}^{n+1}$ be a $C^2$ closed embedded strictly anisotropic-mean convex hypersurface. Then it holds that

$$\begin{aligned} \int _{\Sigma }\frac{F(\nu )}{H^F}{\hbox {d}}A\ge \frac{n+1}{n}\vert \Omega \vert + \max \left\{ 0, \max _{e\in \mathbb {S}^n} \int _{\Sigma }\frac{ \langle \nu ,\Phi (e) \rangle }{H^F}\, {\hbox {d}}A \right\} . \end{aligned}$$

(1.7)

Equality holds if and only if ${\Sigma }$ is a Wulff shape.

Finally we follow an argument of Ros [31] by combining Theorems 1.2 and 1.3 to establish the Alexandrov-type theorem for capillary hypersurfaces with constant anisotropic mean curvature, Theorem 1.1, and also the Alexandrov-type theorem for capillary hypersurfaces with constant higher order anisotropic mean curvature whose definition will be given in Sect. 2.

Theorem 1.4

Let $\omega _0\in \left( -F(E_{n+1}),F(-E_{n+1})\right)$. Let ${\Sigma }\subset \overline{\mathbb {R}^{n+1}_+}$ be a $C^2$ embedded compact anisotropic $\omega _0$-capillary hypersurface with constant r-th anisotropic mean curvature for some $r\in \{2,\ldots ,n\}$. Then ${\Sigma }$ is an $\omega _0$-capillary Wulff shape.

The rest of the paper is organized as follows. In Sect. 2, we provide more details about the anisotropic mean curvature and the higher order anisotropic mean curvature, together with the Wulff shape and the $\omega _0$-capillary Wulff shape. In Sect. 3, we prove the Minkowski-type formula in Theorem 1.3 and the Heintze-Karcher-type inequality in Theorem 1.2. In Sect. 4, we prove the Alexandrov-type theorem, Theorems 1.1 and 1.4.

2 Preliminaries

Let $F: \mathbb {S}^n\rightarrow \mathbb {R}_+$ be a $C^2$ positive function on $\mathbb {S}^n$ such that $(\nabla ^2 F+F \sigma )>0.$ We denote

$$\begin{aligned} A_F=\nabla ^2 F+F\sigma . \end{aligned}$$

Let $F^o:\mathbb {R}^{n+1}\rightarrow \mathbb {R}$ be defined by

$$\begin{aligned} F^o(x)=\sup \left\{ \frac{\left<x,z\right>}{F(z)}\mid z\in \mathbb {S}^n\right\} , \end{aligned}$$

where $\left<\cdot ,\cdot \right>$ denotes the standard Euclidean inner product. We collect some well-known facts on F and $F^o$, see e.g. [17].

Proposition 2.1

For any $z\in \mathbb {S}^n$ and $t>0$, the following statements hold:

• (i) $F^o(tz)=tF^o(z)$.

• (ii) $\left<\Phi (z),z\right>=F(z)$.

• (iii) $F^o(\Phi (z))=1$.

• (iv) The following Cauchy–Schwarz inequality holds:

  $$\begin{aligned} \langle x, z\rangle \le F^o(x)F(z). \end{aligned}$$

  (2.1)

• (v) The unit Wulff shape ${\mathcal {W}}_F$ can be interpreted by $F^o$ as

  $$\begin{aligned} {\mathcal {W}}_F=\{x\in \mathbb {R}^{n+1}:F^o(x)=1\}. \end{aligned}$$

A Wulff shape of radius r centered at $x_0\in \mathbb {R}^{n+1}$ is given by

$$\begin{aligned} {\mathcal {W}}_{r_0}(x_0)=\{x\in \mathbb {R}^{n+1}:F^o(x-x_0)=r_0\}. \end{aligned}$$

Let ${\Sigma }\subset \overline{\mathbb {R}^{n+1}_+}$ be a $C^2$ hypersurface with $\partial {\Sigma }\subset \partial \mathbb {R}^{n+1}_+$, which encloses a bounded domain $\Omega$. Let $\nu$ be the unit normal of ${\Sigma }$ pointing outward $\Omega$. The anisotropic normal of ${\Sigma }$ is given

$$\begin{aligned} \nu _F=\Phi (\nu )=\nabla F(\nu )+F(\nu )\nu , \end{aligned}$$

and the anisotropic principal curvatures $\{\kappa _i^F\}_{i=1}^n$ of ${\Sigma }$ are given by the eigenvalues of the anisotropic Weingarten map

$$\begin{aligned} d\nu _F=A_F(\nu )\circ d\nu : T_p{\Sigma }\rightarrow T_p{\Sigma }. \end{aligned}$$

The eigenvalues are real since $(A_F)$ is positive definite and symmetric. For $r\in \{1,\cdots , n\}$, the (normalized) r-th anisotropic mean curvature is defined by

$$\begin{aligned} H_r^F=\frac{1}{\left( {\begin{array}{c}n\\ r\end{array}}\right) }\sigma _r^F, \end{aligned}$$

where $\sigma _r^F$ be the r-th elementary symmetric function on the anisotropic principal curvatures $\{\kappa _i^F\}_{i=1}^n$, namely,

$$\begin{aligned} \sigma _r^F=\sum _{1\le i_1<\cdots <i_r\le n}\kappa ^F_{i_1}\cdots \kappa ^F_{i_r}, \end{aligned}$$

In particular, $H^F=\sigma _1^F$ is the anisotropic mean curvature and $H_1^F$ the normalized anisotropic mean curvature. Alternatively, the r-th anisotropic mean curvature $H^F_r$ of ${\Sigma }$ can be defined through the identity

$$\begin{aligned} \mathcal {P}_n(t)=\prod _{i=1}^n(1+t\kappa ^F_i)=\sum _{i=0}^n\left( {\begin{array}{c}n\\ i\end{array}}\right) H^F_it^i \end{aligned}$$

(2.2)

for all real numbers t.

It is easy to check that the anisotropic principal curvatures of ${\mathcal {W}}_r(x_0)$ are $\frac{1}{r}$, since

$$\begin{aligned} \nu _F(x)=\frac{x-x_0}{r}, \quad \hbox { on } {\mathcal {W}}_r(x_0). \end{aligned}$$

(2.3)

For the convenience of the reader, we provide a proof of (2.3). We first consider the unit Wulff shape, ${\mathcal {W}}_F.$ Since the unit normal vector $\nu$ of ${\mathcal {W}}_F$ at $x \in {\mathcal {W}}_F$ is given by $\Phi ^{-1}(x)$, then the anisotripic normal is just x. For the general case, one use a translation and a scaling.

A truncated Wulff shape is a part of a Wulff shape cut by a hyperplane, say $\{x_{n+1}= 0\}$. Namely, it is an intersection of a Wulff shape and ${{\mathbb {R}}}^{n+1}_+$. As mentioned above, it was used by Winterbottem [11]. At a first glimpse, it is not very easy to image why the hyperplane intersects a Wulff shape at a “constant angle" as in the isotropic case, namely (1.1) holds. It follows from

$$\begin{aligned} \langle \nu _F, -E_{n+1} \rangle = \langle \frac{x-x_0}{r}, -E_{n+1} \rangle = \langle \frac{x_0}{r}, -E_{n+1} \rangle , \end{aligned}$$

which is a constant.

Remark 2.1

The boundary condition $\langle \Phi (\nu ), -E_{n+1}\rangle =\omega _0$ implies $\omega _0\in (-F(E_{n+1}),F(-E_{n+1}))$. Indeed, by the Cauchy-Schwarz inequality (2.1),

$$\begin{aligned} -F(E_{n+1})&=-F(E_{n+1})F^o(\Phi (\nu ))\le \langle \Phi (\nu ),\nonumber \\&-E_{n+1}\rangle \le F^o(\Phi (\nu ))F(-E_{n+1})=F(-E_{n+1}). \end{aligned}$$

(2.4)

Since ${\Sigma }$ is embedded, ${\Sigma }$ intersects $\partial \mathbb {R}_+^{n+1}$ transversely. It follows that equality in (2.4) cannot hold. Therefore, $\omega _0\in (-F(E_{n+1}),F(-E_{n+1}))$ is a necessary condition for anisotropic $\omega _0$-capillary hypersurfaces.

From our work in this paper, one can in fact introduce a notion of “anisotropic contact angle" as follows, which is a natural generalization of the contact angle in the isotropic case. We define $\theta :\partial {\Sigma }\rightarrow (0, \pi )$ by

$$\begin{aligned} - \cos \theta = \left\{ \begin{array}{cl} F(E_{n+1})^{-1} \langle \nu _F, -E_{n+1} \rangle , &{} \quad \hbox { if }\langle \nu _F, -E_{n+1} \rangle <0,\\ 0, &{}\quad \hbox { if } \langle \nu _F, -E_{n+1} \rangle =0, \\ F(-E_{n+1})^{-1} \langle \nu _F, -E_{n+1} \rangle , &{} \quad \hbox { if }\langle \nu _F,- E_{n+1} \rangle >0.\\ \end{array} \right. \end{aligned}$$

If $\theta =\pi /2$, or equivalently $\langle \nu _F, -E_{n+1} \rangle =0$, we call that the anisotropic hypersurface intersects $\partial {{\mathbb {R}}}^{n+1}_+$ perpendicularly, or it is a free boundary anisotropic hypersurface.

3 Minkowski-type formula and Heintze-Karcher-type inequality

3.1 Minkowski-type formula

To prove the Minkowski-type formula, we need the following structural lemma for compact hypersurfaces in $\mathbb {R}^{n+1}$ with boundary, which is well-known and widely use; see, for example [3, 20].

Lemma 3.1

Let ${\Sigma }\subset \mathbb {R}^{n+1}$ be a compact hypersurface with boundary. Then it holds that

$$\begin{aligned} n\int _{{\Sigma }}\nu {\hbox {d}}A=\int _{\partial {\Sigma }}\left\{ \left<x,\mu \right>\nu -\left<x,\nu \right>\mu \right\} {\hbox {d}}s. \end{aligned}$$

(3.1)

In the paper we denote $\mu$ the unit outward co-normal of $\partial {\Sigma }$ in ${\Sigma }$. Recall $E^{F}_{n+1}$ defined in (1.2). It is easy to check that

$$\begin{aligned} \langle E^{F}_{n+1},E_{n+1} \rangle =1. \end{aligned}$$

(3.2)

Proof of Theorem 1.3

We first prove (1.6). We begin by introducing the following $C^1$ vector field along ${\Sigma }$:

$$\begin{aligned} X_F(x)=F(\nu (x))x-\left<x,\nu (x)\right>\nu _F(x). \end{aligned}$$

Observe that $X_F$ is indeed a tangential vector field along ${\Sigma }$, since

$$\begin{aligned} \left<X_F,\nu \right>=F(\nu )\left<x,\nu \right>-\left<x,\nu \right>\left<\nu _F,\nu \right>=0. \end{aligned}$$

Notice also that along ${\Sigma }$ we have

$$\begin{aligned} \textrm{div}_{\Sigma }(X_F)&=nF(\nu )+\langle d\nu (\nabla F),x\rangle -\left<x,d\nu (\nu _F^T)\right>\nonumber \\&-\left<x,\nu \right>H^F=nF(\nu )- H^F\left<x,\nu \right>, \end{aligned}$$

(3.3)

where $\textrm{div}_{\Sigma }$ is the divergence on ${\Sigma }$. In the second equality, we have used the self-adjointness of $d\nu$. Here $\nu ^T_F$ and $x^T$ denote the tangential projection on ${\Sigma }$ of $\nu _F$ and x respectively. In particular, $\nu ^T_F=\nu _F-\langle \nu _F, \nu \rangle \nu =\nabla F(\nu )$. On one hand, integrating (3.3) along ${\Sigma }$ and using the divergence theorem, we find

$$\begin{aligned} \int _{\Sigma }nF(\nu )-H^F(x)\left<x,\nu \right>{\hbox {d}}A=\int _{\partial {\Sigma }}F(\nu ) \left<x,\mu \right>-\left<x,\nu \right>\left<\nu _F,\mu \right>{\hbox {d}}s. \end{aligned}$$

(3.4)

On the other hand, by (3.1) we have

$$\begin{aligned} -n\omega _0\int _{\Sigma }\left<\nu , E^F_{n+1}\right>{\hbox {d}}A=\int _{\partial {\Sigma }}(-\left<x,\mu \right>\left<\nu ,E^F_{n+1}\right> \omega _0+\left<x,\nu \right>\left<\mu ,E^F_{n+1}\right>\omega _0) {\hbox {d}}s. \end{aligned}$$

(3.5)

It is easy to see that at any $x\in \partial {\Sigma }\subset \partial \mathbb {R}^{n+1}_+$

$$\begin{aligned} E_{n+1}=\left<\nu ,E_{n+1}\right>\nu +\left<\mu ,E_{n+1}\right>\mu , \end{aligned}$$

and hence we have

$$\begin{aligned} -\omega _0&=\left<E_{n+1},\nu _F\right>=\left<\nu ,E_{n+1} \right>F(\nu )+\left<\mu ,E_{n+1}\right>\left<\nu _F,\mu \right>, \end{aligned}$$

(3.6)

$$\begin{aligned} 0&=\left<E_{n+1},x\right>=\left<\nu ,E_{n+1}\right>\left<x,\nu \right> +\left<\mu ,E_{n+1}\right>\left<x,\mu \right>. \end{aligned}$$

(3.7)

Moreover by (3.2) we have

$$\begin{aligned} 1=\langle E^F_{n+1},E_{n+1}\rangle =\langle \nu ,E_{n+1}\rangle \langle E^F_{n+1},\nu \rangle +\langle \mu ,E_{n+1}\rangle \langle E^F_{n+1},\mu \rangle . \end{aligned}$$

(3.8)

This yields that

$$\begin{aligned}&-\left<x,\mu \right>\left<\nu ,E^{F}_{n+1}\right>\omega _0 +\left<x,\nu \right>\left<\mu ,E^{F}_{n+1}\right>\omega _0\\&\quad =\left<x,\mu \right>\left<\nu ,E^{F}_{n+1}\right>\left<\nu ,E_{n+1}\right>F(\nu ) -\left<x,\nu \right>\left<\mu ,E^{F}_{n+1}\right>\left<\nu ,E_{n+1}\right>F(\nu )\\&\quad +\left<x,\mu \right>\left<\nu ,E^{F}_{n+1}\right>\left<\mu ,E_{n+1}\right> \left<\nu _F,\mu \right>-\left<x,\nu \right>\left<\mu ,E^{F}_{n+1} \right>\left<\mu ,E_{n+1}\right>\left<\nu _F,\mu \right>\\&\quad =\left<x,\mu \right>\left<\nu ,E^{F}_{n+1}\right>\left<\nu ,E_{n+1}\right>F(\nu ) +\left<x,\mu \right>\left<\mu ,E_{n+1}\right>\left<\mu ,E^{F}_{n+1}\right>F(\nu )\\&\quad -\left<x,\nu \right>\left<\nu ,E_{n+1}\right>\left<\nu ,E^{F}_{n+1}\right>\left<\nu _F,\mu \right> +\left<x,\nu \right>\left<\mu ,E^{F}_{n+1}\right>\left<\mu ,E_{n+1}\right>\left<\nu _F,\mu \right>\\&\quad =F(\nu )\left<x,\mu \right>-\left<x,\nu \right>\left<\nu _F,\mu \right>, \end{aligned}$$

where we have used (3.6) in the first equality, (3.7) in the second equality and (3.8) in the last one. In particular, this identity, together with (3.4) and (3.5), implies

$$\begin{aligned} \int _{\Sigma }nF(\nu )-H^F(x)\left<x,\nu \right>{\hbox {d}}A=-n\int _{\Sigma }\left<\nu ,\omega _0 E^F_{n+1}\right>{\hbox {d}}A, \end{aligned}$$

which is (1.6).

Next we prove (1.5) for general r by using (1.6) as in [21]. Consider a family of hypersurfaces ${\Sigma }_t$ with boundary for small $t>0$, defined by

$$\begin{aligned} \varphi _t(x)=x+t(\nu _F(x)+\omega _0E^{F}_{n+1})\quad x\in {\Sigma }. \end{aligned}$$

We claim that ${\Sigma }_t$ is also an anisotropic $\omega _0$-capillary hypersurface in $\overline{\mathbb {R}^{n+1}_+}$. On one hand, the $\omega _0$-capillarity condition and (3.2) yield that for any $x\in \partial {\Sigma }$,

$$\begin{aligned} \left<\nu _F(x)+\omega _0E^{F}_{n+1},-E_{n+1}\right>=\omega _0-\omega _0=0. \end{aligned}$$

Hence, $\varphi _t(x)\in \partial \mathbb {R}_+^{n+1}$ for $x\in \partial {\Sigma }$ which means $\partial {\Sigma }_t\subset \partial \mathbb {R}_+^{n+1}$. On the other hand, denoting by $e^F_i$ an anisotropic principal vector at $x\in {\Sigma }$ corresponding to $\kappa _i^F$ for $i=1,\cdots , n$, we have

$$\begin{aligned} (\varphi _t)_*(e^F_i)=(1+t\kappa ^F_i)e^F_i,\quad i=1,\ldots ,n. \end{aligned}$$

(3.9)

We see from (3.9) that $\nu ^{{\Sigma }_t}(\varphi _t(x))=\nu (x)$, and in turn $\nu _F^{{\Sigma }_t}(\varphi _t(x))=\nu _F(x)$. Here $\nu ^{{\Sigma }_t}$ and $\nu _F^{{\Sigma }_t}$ denote the outward normal and anisotropic normal to ${\Sigma }_t$ respectively. In view of this, we have

$$\begin{aligned} \left<\nu _F^{{\Sigma }_t}(\varphi _t(x)),-E_{n+1}\right>=\left<\nu _F(x),-E_{n+1}\right>=\omega _0. \end{aligned}$$

Therefore, ${\Sigma }_t$ is also an anisotropic $\omega _0$-capillary hypersurface in $\overline{\mathbb {R}^{n+1}_+}$ and hence (1.6) holds for ${\Sigma }_t$ for any small t. Exploiting (1.6) for every such ${\Sigma }_t$, we find that

$$\begin{aligned} \int _{{\Sigma }_t=\varphi _t({\Sigma })} \left( F(\nu _t)+\omega _0\left<\nu _t,E^F_{n+1}\right>\right) -H_1^F(t)\mid _y\left<y,\nu _t\right> {\hbox {d}}A_t(y)=0. \end{aligned}$$

(3.10)

By (3.9), the tangential Jacobian of $\varphi _t$ along ${\Sigma }$ at x is just

$$\begin{aligned} \textrm{J}^{\Sigma }\varphi _t(x)=\prod _{i=1}^n(1+t\kappa _i^F(x))=\mathcal {P}_n(t), \end{aligned}$$

(3.11)

where $\mathcal {P}_n(t)$ is the polynomial defined in (2.2). Moreover, using (3.9) again, we see that the corresponding anisotropic principal curvatures are given by

$$\begin{aligned} \kappa ^F_i(\varphi _t(x))=\frac{\kappa ^F_i(x)}{1+t\kappa ^F_i(x)}. \end{aligned}$$

(3.12)

Hence fix $x\in {\Sigma }$, the anisotropic mean curvature of ${\Sigma }_t$ at $\varphi _t(x)$, say $H^F(t)$, is given by

$$\begin{aligned} H^F(t) =\frac{\mathcal {P}'_n(t)}{\mathcal {P}_n(t)} =\frac{\sum _{i=0}^{n}i\left( {\begin{array}{c}n\\ i\end{array}}\right) H^F_it^{i-1}}{\mathcal {P}_n(t)}, \end{aligned}$$

(3.13)

where $H^F_i=H^F_i(x)$ is the i-th mean curvature of ${\Sigma }$ at x.

Using the area formula, (3.11) and (3.13), we find from (3.10) that

$$\begin{aligned}&\int _{\Sigma }n\left( F(\nu )+\omega _0\left<\nu ,E_{n+1}^F\right>\right) \mathcal {P}_n(t)-t\left( \left<x,\nu _F(x)\right>\right. \nonumber \\&\quad \left. +\omega _0\left<x,E_{n+1}^F\right>\right) \mathcal {P}'_n(t)-\mathcal {P}'_n(t)\left<x,\nu \right>{\hbox {d}}A_x=0. \end{aligned}$$

As the left hand side in this equality is a polynomial in the time variable t, this shows that all its coefficients vanish, and hence a direct computation yields (1.5). $\square$

We remark that the definition of the family of capillary hypersurfaces ${\Sigma }_t$ was inspired by [21]. These are the parallel hypersurfaces in the case of capillary boundary.

Remark 3.1

If we replace $E^F_{n+1}$ by $E_{n+1}$ in the proof, every step above is valid and we achieve that

$$\begin{aligned} \int _{\Sigma }H^F_{r-1}\left( F(\nu )+\omega _0\left<\nu , E_{n+1}\right>\right) -H^F_r\left<x,\nu \right>{\hbox {d}}A=0. \end{aligned}$$

Alternatively, we can prove directly that

$$\begin{aligned} \int _{\Sigma }H^F_{r-1}\left<\nu , E^F_{n+1}\right>{\hbox {d}}A=\int _{\Sigma }H^F_{r-1}\left<\nu , E_{n+1}\right>{\hbox {d}}A. \end{aligned}$$

Since we do not need it in this paper, we omit the proof.

3.2 Heintze-Karcher-type inequality

To prove the Heintze-Karcher-type inequality, we need the following key proposition, which amounts to be an anisotropic angle comparison principle. It is clear that in the isotropic case it is trivial. However in the anisotropic case it is non-trivial.

Proposition 3.1

Let $x,z\in \mathbb {S}^n$ be two distinct points and $y\in \mathbb {S}^n$ lie in a length-minimizing geodesic joining x and z in $\mathbb {S}^n$, then we have

$$\begin{aligned} \left<\Phi (x),z\right>\le \left<\Phi (y),z\right>. \end{aligned}$$

Equality holds if and only if $x=y$.

Proof

We denote $d_0=d_{\mathbb {S}^n}(x,z)$ and $d_1=d_{\mathbb {S}^n}(x,y)$, where $d_{\mathbb {S}^n}$ denotes the intrinsic distance on $\mathbb {S}^n$. If $y\ne x$, clearly $0<d_1\le d_0$. Let $\gamma :[0,d_0]\rightarrow \mathbb {S}^n$ be the arc-length parameterized geodesic with $\gamma (0)=x$, $\gamma (d_0)=z$. Considering the function

$$\begin{aligned} f=\langle \Phi (\gamma (t)),z\rangle , \quad t\in [0,d_0], \end{aligned}$$

we have that

$$\begin{aligned}&\langle \Phi (y),z\rangle -\langle \Phi (x),z\rangle =f(d_1)-f(0)\nonumber \\&\quad = \int _0^{d_1} \left<\frac{\hbox {d}}{{\hbox {d}}t}\Phi (\gamma (t)),z\right>{\hbox {d}}t=\int _0^{d_1}\langle D_{\dot{\gamma }(t)}\Phi (\gamma (t)), z\rangle {\hbox {d}}t, \end{aligned}$$

(3.14)

where D is the Euclidean covariant derivative. Since $\gamma$ is length-minimizing, it is easy to see that

$$\begin{aligned} \langle \dot{\gamma }(t), z\rangle \ge 0, \quad \forall t\in (0, d_0). \end{aligned}$$

Thus z can be expressed as $z=\sin s\dot{\gamma }(t)+\cos s\gamma (t)$ with some $s\in (0,\pi )$. It follows that

$$\begin{aligned} \langle D_{\dot{\gamma }(t)}\Phi (\gamma (t)), z\rangle =\sin s (\nabla ^2 F+F I)(\dot{\gamma }(t),\dot{\gamma }(t)), \end{aligned}$$

Since $(\nabla ^2 F+F I)>0$, we get $\langle D_{\dot{\gamma }(t)}\Phi (\gamma (t)), z\rangle >0$ for any $t\in (0,d_1)$. This fact, together with (3.14), leads to the assertion. $\square$

We note that when $y=z$, Proposition 3.1 is nothing but the Cauchy-Schwarz inequality (2.1), since one readily observes from Proposition 2.1(ii)(iii) that

$$\begin{aligned} \left<\Phi (x),z\right>\le F^o(\Phi (x))F(z)=\left<\Phi (z),z\right>. \end{aligned}$$

The Cauchy-Schwarz inequality (2.1) also impies the following property:

Proposition 3.2

For $\omega _0\in \left( -F(E_{n+1}),F(-E_{n+1})\right)$, if holds that

$$\begin{aligned} F(z)+\omega _0\left<z,E^F_{n+1}\right>>0,\quad \text {for any }z\in \mathbb {S}^n. \end{aligned}$$

Proof

It is clear that we need not to consider the case $\omega _0=0$. If $\omega _0<0$, we only need to consider the points z satisfying $\langle z,E^F_{n+1}\rangle >0$. At any such z, since $\omega _0>-F(E_{n+1})$, we have

$$\begin{aligned} F(z)+\omega _0\left<z, E^F_{n+1}\right>>&F(z)-F(E_{n+1})\left<z, \frac{\Phi (E_{n+1})}{F(E_{n+1})}\right>\\&\ge F(z)-F(z)F^o(\Phi (E_{n+1}))=0, \end{aligned}$$

where we have used the Cauchy-Schwarz inequality for the second inequality.

For the case $\omega _0>0$, we just need to consider the points z such that $\langle z,E^F_{n+1}\rangle <0$. Since $\omega _0<F(-E_{n+1})$, using the Cauchy-Schwarz inequality again, we find

$$\begin{aligned} F(z)+\omega _0\left<z, E^F_{n+1}\right>>&F(z)+F(-E_{n+1})\left<z,-\frac{\Phi (-E_{n+1})}{F(-E_{n+1})}\right>\\&\ge F(z)-F(z)F^o(\Phi (-E_{n+1}))=0. \end{aligned}$$

The proposition is thus proven. $\square$

Now we can start to prove Theorem 1.2.

Proof of Theorem 1.2

Let ${\Sigma }\subset \overline{\mathbb {R}^{n+1}_+}$ be an anisotropic capillary hypersurface satisfying (1.3). For any $x\in {\Sigma }$, let $\kappa ^F_i(x)$ be the anisotropic principal curvature and $e^F_i(x)$ be the corresponding anisotropic principal vector of ${\Sigma }$ at x such that $\vert e^F_1\wedge e^F_2\wedge \cdots \wedge e^F_n\vert =1$. Since ${\Sigma }$ is strictly anisotropic mean convex,

$$\begin{aligned} \max _i{\kappa ^F_i(x)}\ge \frac{1}{n} H^F(x)>0, \hbox { for }x\in {\Sigma }. \end{aligned}$$

We define

$$\begin{aligned} Z=\left\{ (x,t)\in {\Sigma }\times \mathbb {R}:0<t\le \frac{1}{\max {\kappa ^F_i(x)}}\right\} , \end{aligned}$$

and

$$\begin{aligned}&\zeta _F: Z\rightarrow \mathbb {R}^{n+1},\end{aligned}$$

(3.15)

$$\begin{aligned}&\zeta _F(x,t)=x-t\left( \nu _F(x)+\omega _0 E^{F}_{n+1}\right) . \end{aligned}$$

(3.16)

Claim: $\Omega \subset \zeta _F(Z)$.

Recall ${\mathcal {W}}_r(x_0)$ is the Wulff shape centered at $x_0$ with radius r. For any $y\in \Omega$, we consider a family of Wulff shapes $\{ {\mathcal {W}}_r(y+r\omega _0 E^F_{n+1})\}_{r\ge 0}$. Since $y\in \Omega$ is an interior point, we definitely have ${\mathcal {W}}_r(y+r\omega _0 E^F_{n+1})\subset \Omega$ for r small enough. On the other hand, by the assumption $-F(E_{n+1})<\omega _0< F(-E_{n+1})$, the definition (1.2) of $E_{n+1}^F$ and Proposition 2.1(i)(iii), it is easy to see that

$$\begin{aligned} F^o(-\omega _0 E^F_{n+1})< 1. \end{aligned}$$

It follows that

$$\begin{aligned} F^o\big ( y-(y+r\omega _0 E^F_{n+1})\big )=r F^o(-\omega _0 E^F_{n+1})< r, \end{aligned}$$

which implies that for any small $r>0$, y is always in the domain bounded by the Wulff shape ${\mathcal {W}}_r(y+r\omega _0 E^{F}_{n+1})$. Hence ${\mathcal {W}}_r(y+r\omega _0E^F_{n+1})$ must touch ${\Sigma }$ as we increase the radius r. Consequently, for any $y\in \Omega$, there exists $x\in {\Sigma }$ and $r_y>0$, such that ${\mathcal {W}}_{r_y}(y+r_y\omega _0 E^F_{n+1})$ touches ${\Sigma }$ for the first time, at some point $x\in {\Sigma }$. In terms of the touching point, only the following two cases are possible:

Case 1. $x\in \mathring{\Sigma }$.

In this case, since $x\in \mathring{{\Sigma }}$, the Wulff shape ${\mathcal {W}}_{r_y}(y+r_y\omega _0 E^F_{n+1})$ is tangent to ${\Sigma }$ at x from the interior. Hence

$$\begin{aligned} \nu (x)=\nu ^{{\mathcal {W}}}(x), \end{aligned}$$

(3.17)

where $\nu ^{{\mathcal {W}}}$ denotes the outward unit normal of ${\mathcal {W}}_{r_y}(y+r_y\omega _0 E^F_{n+1})$. Moreover, since the touching of ${\mathcal {W}}_{r_y}(y+r_y\omega _0 E^F_{n+1})$ with ${\Sigma }$ is from interior, we see that

$$\begin{aligned} d\nu \le d\nu ^{\mathcal {W}}, \end{aligned}$$

(3.18)

in the sense that the coefficient matrix of the difference of two classical Weingarten operators $d\nu -d\nu ^{\mathcal {W}}$ is semi-negative definite. It follows from (3.17) and (3.18) that

$$\begin{aligned} A_F(\nu )\circ d\nu \le A_F(\nu ^{\mathcal {W}}) \circ d\nu ^{\mathcal {W}}. \end{aligned}$$

(3.19)

Since the anisotropic principal curvatures of ${\mathcal {W}}_{r_y}(y-r_y\omega _0 E^F_{n+1})$ are equal to $\frac{1}{r_y}$, we see from (3.19) that

$$\begin{aligned} \max _{1\le i\le n}{\kappa ^F_{i}(x)}\le \frac{1}{r_y}. \end{aligned}$$

Invoking the definition of Z and $\zeta _F$, we find that $y\in \zeta _F(Z)$ in this case.

Case 2. $x\in \partial {\Sigma }$.

We will rule out this case by the capillarity assumption (1.3). Let $\nu _F^{\mathcal {W}}(x)$ be the outward anisotropic normal to ${\mathcal {W}}_{r_y}(y+r_y\omega _0 E^F_{n+1})$. It is easy to see that

$$\begin{aligned} \nu _F^{\mathcal {W}}(x)=\Phi (\nu ^{\mathcal {W}}(x))=\frac{x-(y+r_y\omega _0 E^{F}_{n+1})}{r_y}. \end{aligned}$$

Recall that y lies in the interior of $\Omega$. Thus $\left<y,E_{n+1}\right>>0$. On one hand, in view of (1.3) and (3.2) we have

$$\begin{aligned} \left<\nu _F^{\mathcal {W}}(x),-E_{n+1}\right>&=1/r_y\left<y,E_{n+1}\right>+\omega _0\langle E^{F}_{n+1},E_{n+1}\rangle \nonumber \\&>\omega _0\ge \omega (x)=\left<\nu _F(x),-E_{n+1}\right>. \end{aligned}$$

(3.20)

On the other hand, since the Wulff shape ${\mathcal {W}}_{r_y}(y+r_y\omega _0 E^F_{n+1})$ touches ${\Sigma }$ from the interior, we have

$$\begin{aligned} \left\langle \nu (x),-E_{n+1}\right>\ge \left<\nu ^{\mathcal {W}}(x),-E_{n+1}\right>. \end{aligned}$$

Since $\nu$, $\nu ^{\mathcal {W}}$ and $-E_{n+1}$ lie on the two-plane orthogonal to $T_x(\partial {\Sigma })$, we see that $\nu$ lies actually in the geodesic joining $\nu ^{\mathcal {W}}$ and $-E_{n+1}$ in $\mathbb {S}^n$. It then follows from the angle comparison principle Proposition 3.1 that

$$\begin{aligned} \left\langle \Phi (\nu (x)),-E_{n+1}\right>\ge \left<\Phi (\nu ^{\mathcal {W}}(x)),-E_{n+1}\right>. \end{aligned}$$

This is a contradiction to (3.20). The Claim is thus proved.

By a simple computation, we find that

$$\begin{aligned} \partial _t \zeta _F(x, t)&=-\left( \nu _F(x)+\omega _0 E^F_{n+1}\right) , \\ D_{e^F_i} \zeta _F(x, t)&=\left( 1-t\kappa ^F_i(x)\right) e^F_i(x). \end{aligned}$$

Thanks to Proposition 3.2, a classical computation yields that the tangential Jacobian of $\zeta _F$ along Z at (x, t) is just

$$\begin{aligned} \textrm{J}^Z\zeta _F(x,t)= (F(\nu )+\omega _0\langle \nu , E^F_{n+1}\rangle )\prod _{i=1}^n(1-t\kappa ^F_i). \end{aligned}$$

By virtue of the fact that $\Omega \subset \zeta _F(Z)$, the area formula yields

$$\begin{aligned} \vert \Omega \vert \le \vert \zeta _F(Z)\vert \le&\int _{\zeta _F(Z)}\mathcal {H}^0(\zeta _F^{-1}(y))dy =\int _Z \textrm{J}^Z\zeta _F d\mathcal {H}^{n+1}\\&=\int _\Sigma {\hbox {d}}A\int _0^{\frac{1}{\max \left\{ \kappa ^F_i(x)\right\} }}\left( F(\nu )+\omega _0\left<\nu ,E^F_{n+1}\right>\right) \prod _{i=1}^n(1-t\kappa ^F_i(x)){\hbox {d}}t. \end{aligned}$$

By the AM-GM inequality, and the fact that $\max \left\{ \kappa ^F_i(x)\right\} _{i=1}^n\ge \frac{1}{n} H^F(x)$, we obtain

$$\begin{aligned} \vert \Omega \vert \le&\int _{\Sigma }{\hbox {d}}A\int _0^{\frac{1}{\max \left\{ \kappa ^F_i(x)\right\} }} \left( F(\nu )+\omega _0\langle \nu , E^F_{n+1}\rangle \right) \left( \frac{1}{n}\sum _{i=1}^n \left( 1-t\kappa ^F_i(x)\right) \right) ^n {\hbox {d}}t\nonumber \\ \le&\int _{\Sigma }\left( F(\nu )+\omega _0\langle \nu , E^F_{n+1}\rangle \right) {\hbox {d}}A\int _0^{\frac{n}{H^F(x)}} \left( 1-t\frac{H^F(x)}{n}\right) ^n {\hbox {d}}t\nonumber \\&=\frac{n}{n+1}\int _{\Sigma }\frac{F(\nu )+\omega _0\langle \nu , E^F_{n+1}\rangle }{H^F} {\hbox {d}}A, \end{aligned}$$

which gives (1.4).

If equality in (1.4) holds, then from the above argument, we see $\kappa _1^F(x)=\ldots =\kappa _n^F(x)$ for all $x\in {\Sigma }$. It follows from [17, Lemma 2.3] that ${\Sigma }$ must be a part of a Wulff shape ${\mathcal {W}}_{r_0}(x_0)$ for some $r_0$ and some point $x_0$. Hence $H^F$ is a constant $\frac{n}{r_0}$ and since $\nu _F(x)=\frac{x-x_0}{r_0}$, we get, for $x\in \partial {\Sigma }$,

$$\begin{aligned} \omega (x)=\langle \nu _F(x), -E_{n+1}\rangle =\frac{1}{r_0}\langle x_0,E_{n+1}\rangle :={\tilde{\omega }}_0, \end{aligned}$$

which is a constant.

From the equality in Heintze-Karcher inequality and the Minkowski-type formula (1.5), taking into account that $H_F$ is a constant, we deduce that

$$\begin{aligned} \omega _0\int _{\Sigma }\left<\nu ,E_{n+1}^F\right\rangle {\hbox {d}}A={\tilde{\omega }}_0\int _{\Sigma }\left<\nu ,E_{n+1}^F\right\rangle {\hbox {d}}A. \end{aligned}$$

By the divergence theorem and (3.2),

$$\begin{aligned} \int _{\Sigma }\left<\nu ,E_{n+1}^F\right\rangle {\hbox {d}}A=\vert \partial \Omega \cap \partial \mathbb {R}_+^{n+1}\vert \ne 0. \end{aligned}$$

It follows that ${\tilde{\omega }}_0=\omega _0$ which means ${\Sigma }$ is a $\omega _0$-capillary Wulff shape.

Conversely, for any $\omega _0$-capillary Wulff shape, we can see easily from the Minkowski-type formula (1.5) and the fact of constant anisotropic mean curvature that equality holds in (1.4). This completes the proof. $\square$

Remark 3.2

We may use in the proof another foliation of Wulff shapes $\{{\mathcal {W}}_r(y+r\omega _0E_{n+1})\}_{r\ge 0}$. To ensure that ${\mathcal {W}}_r(y+r\omega _0E_{n+1})$ intersects with ${\Sigma }$ for large r, we need to assume

$$\begin{aligned} \omega _0\in \left( -\frac{1}{F^o(E_{n+1})}, \frac{1}{F^o(-E_{n+1})}\right) . \end{aligned}$$

(3.21)

We can follow the proof to achieve that

$$\begin{aligned} \int _{\Sigma }\frac{F(\nu )+\omega _0\left<\nu ,E_{n+1}\right>}{H^F}{\hbox {d}}A\ge \frac{n+1}{n}\vert \Omega \vert . \end{aligned}$$

(3.22)

under the assumption (3.21). On the other hand, by virtue of the Cauchy-Schwarz inequality, we see that (3.21) is in general more restrictive than the natural assumption $\omega _0\in (-{F(E_{n+1}), {F(-E_{n+1})}}).$ This is the reason why we introduce $E^F_{n+1}$.

Proof of Corollary 1.1

It is clear that any closed hypersurface can be seen as a capillary surface in a half space with a empty boundary. For any $e\in \mathbb {S}^n$ we can see e as $E_{n+1}$ and apply Theorem 1.2.

First consider $\omega _0 \in (-F(E_{n+1}), 0)$. Together with the definition of $E_{n+1}^F$ (1.4) gives us

$$\begin{aligned}&\int _{\Sigma }\frac{F(\nu )}{H^F}{\hbox {d}}A\ge \frac{n+1}{n}\vert \Omega \vert - \omega _0 \int _{\Sigma }\frac{\langle \nu , E^F_{n+1} \rangle }{H^F} \, {\hbox {d}}A =\frac{n+1}{n}\vert \Omega \vert \\&\quad - \frac{\omega _0}{F(E_{n+1})} \int _{\Sigma }\frac{\langle \nu , \Phi (E_{n+1}) \rangle }{H^F} \, {\hbox {d}}A. \end{aligned}$$

It follows that

$$\begin{aligned} \int _{\Sigma }\frac{F(\nu )}{H^F}{\hbox {d}}A\ge \frac{n+1}{n}\vert \Omega \vert + \max \left\{ 0, \int _{\Sigma }\frac{\langle \nu , \Phi (E_{n+1}) \rangle }{H^F} \, {\hbox {d}}A \right\} . \end{aligned}$$

Then we consider $\omega _0 \in (0, F(-E_{n+1}))$. Similarly, in this case (1.4) gives us

$$\begin{aligned} \int _{\Sigma }\frac{F(\nu )}{H^F}{\hbox {d}}A&\ge \frac{n+1}{n}\vert \Omega \vert +\max \left\{ 0, \int _{\Sigma }\frac{\langle \nu , \Phi (-E_{n+1}) \rangle }{H^F} \, {\hbox {d}}A.\right\} . \end{aligned}$$

This completes the proof of (1.7). $\square$

Remark 3.3

When F is even, i.e., $F(x)=F(-x)$, we have

$$\begin{aligned} \int _{\Sigma }\frac{F(\nu )}{H^F}{\hbox {d}}A\ge \frac{n+1}{n}\vert \Omega \vert + \max _{e\in \mathbb {S}^n} \int _{\Sigma }\frac{ \langle \nu ,\Phi (e) \rangle }{H^F}\, {\hbox {d}}A \, \left( \ge \frac{n+1}{n}\vert \Omega \vert \right) , \end{aligned}$$

since in this case $\Phi (-E_{n+1})=-\Phi (E_{n+1})$. Hence

$$\begin{aligned} \hbox { either }\int _{\Sigma }\frac{\langle \nu , \Phi (E_{n+1}) \rangle }{H^F} \, {\hbox {d}}A\ge 0\hbox { or }\int _{\Sigma }\frac{\langle \nu , \Phi (-E_{n+1}) \rangle }{H^F} \, {\hbox {d}}A\ge 0. \end{aligned}$$

4 Alexandrov-type Theorem

We first prove a result on the existence of an elliptic point for an anisotropic capillary hypersurface.

Proposition 4.1

Let $\omega _0\in \left( -F(E_{n+1}),F(-E_{n+1})\right)$ and let ${\Sigma }\subset \overline{\mathbb {R}^{n+1}_+}$ be a $C^2$ compact embedded anisotropic $\omega _0$-capillary hypersurface, then ${\Sigma }$ has at least one elliptic point, i.e. a point where all the anisotropic principal curvatures are positive.

Proof

We fix a point $y\in \textrm{int}(\partial \Omega \cap \partial \mathbb {R}_+^{n+1})$. Consider the family of Wulff shapes ${\mathcal {W}}_r(y+r\omega _0 E^F_{n+1})$. Observe that for any $x\in \partial {\Sigma }$ and any $r>0$, there holds

$$\begin{aligned} \left<\nu _F^{\mathcal {W}}(x),E_{n+1}\right>=\left<\frac{x-y}{r} +\omega _0E^F_{n+1},E_{n+1}\right>=\omega _0=\left<\nu _F(x),E_{n+1}\right>. \end{aligned}$$

(4.1)

Since ${\Sigma }$ is compact, for r large enough, ${\Sigma }$ lies inside the domain bounded by the Wulff shape ${\mathcal {W}}_r(y+r\omega _0 E^F_{n+1})$. Hence we can find the smallest r, say $r_0>0$, such that ${\mathcal {W}}_{r_0}(y+r_0\omega _0 E^F_{n+1})$ touches ${\Sigma }$ at a first time at some $x_0\in {\Sigma }$ from exterior.

If $x_0\in \mathring{{\Sigma }}$, then ${\Sigma }$ and ${\mathcal {W}}_{r_0}(y+r_0\omega _0 E^F_{n+1})$ are tangent at x. If $x_0\in \partial {\Sigma }$, from (4.1), we conclude again that ${\Sigma }$ and ${\mathcal {W}}_{r_0}((y+r_0\omega _0 E^F_{n+1}))$ are tangent at x. In both cases, by a similar argument as in the proof of Theorem 1.2, we have that the anisotropic principal curvatures of ${\Sigma }$ at $x_0$ are larger than or equal to $\frac{1}{r_0}$. $\square$

Proof of Theorem 1.1and Theorem 1.4 We begin by recalling that $\omega _0\in \left( -F(E_{n+1}),\right. \left. F(-E_{n+1})\right)$ ensures the non-negative of $F(\nu )+\omega _0\left<\nu ,E^F_{n+1}\right>$ pointwisely along ${\Sigma }$, thanks to Proposition 3.2.

On one hand, by virtue of Proposition 4.1 and Gärding’s argument [14] (see also [17, Lemma 2.1]), we know that $H^F_j$ are positive, for $j\le r$ and for any $x\in {\Sigma }$. Applying Theorem 1.2 and using the Maclaurin inequality $H_1^F\ge (H^F_r)^{1/r}$ and the constancy of $H^F_r$, we have

$$\begin{aligned}&(n+1)(H^F_r)^{1/r}\vert \Omega \vert \le (H^F_r)^{1/r}\int _{\Sigma }\frac{F(\nu )+\omega _0\left<\nu ,E^F_{n+1}\right>}{H_1^F}{\hbox {d}}A \nonumber \\&\quad \le \int _{\Sigma }\left( F(\nu )+\omega _0\left<\nu ,E^F_{n+1}\right>\right) {\hbox {d}}A. \end{aligned}$$

(4.2)

On the other hand, using the Minkowski-type formula (1.5) and the Maclaurin inequality $H_{r-1}^F\ge (H^F_r)^{\frac{r-1}{r}}$, we have

$$\begin{aligned} 0&=\int _{\Sigma }H^F_{r-1}\left( F(\nu )+\omega _0\left<\nu , E^F_{n+1}\right>\right) -H^F_r\left<x,\nu \right>{\hbox {d}}A\\ \ge&\int _{\Sigma }(H^F_r)^{\frac{r-1}{r}}\left( F(\nu )+\omega _0\left<\nu , E^F_{n+1}\right>\right) -H^F_r\left<x,\nu \right>{\hbox {d}}A\\&=(H^F_r)^{\frac{r-1}{r}}\int _{\Sigma }\left( F(\nu )+\omega _0\left<\nu , E^F_{n+1}\right>\right) -(H^F_r)^{\frac{1}{r}}\left<x,\nu \right>{\hbox {d}}A\\&=(H^F_r)^{\frac{r-1}{r}}\left\{ \int _{\Sigma }\left( F(\nu )+\omega _0\left<\nu , E^F_{n+1}\right>\right) {\hbox {d}}A-(n+1)(H^F_r)^{\frac{1}{r}}\vert \Omega \vert \right\} , \end{aligned}$$

where in the last equality we have used that

$$\begin{aligned} (n+1)\vert \Omega \vert =\int _\Omega \textrm{div}{x} \, dx=\int _{\Sigma }\left<x,\nu \right>{\hbox {d}}A. \end{aligned}$$

Thus equality in (4.2) holds, and hence ${\Sigma }$ is an anisotropic $\omega _0$-capillary Wulff shape. This completes the proof. $\square$