2 Preliminaries

We consider the warped product $(P,g_p)$ of the form given in the introduction and satisfying the two conditions (H1) and (H2).

Let $\Sigma $ be a closed, connected and oriented hypersurface of $(P^{n+1}, g_P)$ . We consider the function $f=h'$ and the vector field $X=h{\partial }/{\partial t}$ . We will also assume that $\Sigma $ is star-shaped, that is, the function $\langle X,\nu \rangle $ is nonnegative everywhere on $\Sigma $ , where $\nu $ is the outward normal unit vector field. In [Reference Brendle2], Brendle used the fact that X is a conformal vector field, that is, $\mathcal {L}_Xg=2fg$ , to obtain the following Hsiung–Minkowski formula for hypersurfaces of these warped products:

(2.1) $$ \begin{align} \int_{\Sigma}H\langle X,\nu\rangle \,dv_g=\int_{\Sigma}f\,dv_g, \end{align} $$

where g is the induced metric on $\Sigma $ and H the mean curvature of $\Sigma $ .

More generally, there are similar formulas involving higher-order mean curvatures. Before giving these formulas, we recall some results about higher-order mean curvatures. They are extrinsic quantities defined from the second fundamental form and generalising the notion of mean curvature. Up to a normalisation the mean curvature H is the trace of the second fundamental form B,

$$ \begin{align*} H=\frac{1}{n}\mathrm{tr\,}(B). \end{align*} $$

In other words, the mean curvature is

$$ \begin{align*} H=\frac{1}{n}S_1(\kappa_1,\ldots,\kappa_n), \end{align*} $$

where $S_1$ is the first elementary symmetric polynomial and $\kappa _1,\ldots ,\kappa _n$ are the principal curvatures. Higher-order mean curvatures are defined in a similar way for $r\in \{1,\ldots ,n\}$ by

$$ \begin{align*} H_r={\binom{n}{r} }^{-1}S_r(B)={\binom{n}{r} }^{-1}S_r(\kappa_1,\ldots,\kappa_n), \end{align*} $$

where $S_r$ is the rth elementary symmetric polynomial, that is, for any n-tuple $(x_1,\ldots ,x_n)$ ,

$$ \begin{align*} S_r(x_1,\ldots,x_n)=\sum_{1\leqslant i_1<\cdots<i_r \leqslant n}x_{i_1}\cdots x_{i_r}. \end{align*} $$

Note that, of course, $H_1$ is nothing more than the mean curvature H.

We now recall some well-known classical inequalities between the higher-order mean curvatures. First, if $H_r>0,$ then $H_j>0$ for any $j\in \{1,\ldots ,s-1\}$ and

(2.2) $$ \begin{align} H_{r}^{{1}/{r}}\leqslant H_{r-1}^{{1}/{(r-1)}}\leqslant \cdots\leqslant H_2^{1/2}\leqslant H. \end{align} $$

These classical inequalities can be found, for instance, in Li et al. [Reference Li, Wei and Xiong7] where the proof of Barbosa and Colares [Reference Barbosa and Colares1] for space forms is extended to warped products.

Moreover, we also have the classical Newton–Maclaurin inequalities

$$ \begin{align*} H_{i-1}H_j\leqslant H_{j-1}H_i \quad\text{for}\ 1\leqslant i<j\leqslant r. \end{align*} $$

Therefore, since $H_i>0$ and $H_j>0$ ,

(2.3) $$ \begin{align} \dfrac{H_{i-1}}{H_i}\leqslant \dfrac{H_{j-1}}{H_j}. \end{align} $$

Moreover, equality in (2.2) as well as in (2.3) occurs if and only if $\Sigma $ is totally umbilical.

In particular, for $s\in \{1,\ldots ,r\}$ ,

$$ \begin{align*} \dfrac{H_{s-1}}{H_s}\leqslant \dfrac{H_{r-1}}{H_r}, \end{align*} $$

which, since $H_s>0$ and $H_r>0$ , gives

$$ \begin{align*} \dfrac{H_{s-1}}{H_{r-1}}\leqslant \dfrac{H_{s}}{H_r}. \end{align*} $$

Note at this point that the assumption of Theorem 1.1 says that a constant c can be included between those two quantities ${H_{s-1}}/{H_{r-1}}$ and ${H_{s}}/{H_r}$ .

We now recall the following general Hsiung–Minkowski formula in warped products. Namely, if conditions (H1) and (H2) are satisfied and $\Sigma $ is star-shaped, then

(2.4) $$ \begin{align} \int_{\Sigma}fH_{s-1}\,dv_g\leqslant \int_{\Sigma}H_s\langle X,\nu\rangle \,dv_g. \end{align} $$

This inequality was first proved by Brendle and Eichmair [Reference Brendle and Eichmair3] if $\Sigma $ is convex. It turned out that the convexity assumption is not necessary and can be replaced by s-convexity [Reference Li, Wei and Xiong7], that is, $H_s$ is positive everywhere, which is a weaker assumption than convexity (which means that all the principal curvatures are positive).

Note that (2.1) and (2.4) are the generalisations in this class of warped products of the so-called Hsiung–Minkowski formulas [Reference Hsiung5] for space forms (see, for instance, [Reference Li, Wei and Xiong7]). A crucial difference is that, in space forms, they are equalities, which is not the case here except for identity (2.1). However, the error term is well known (see [Reference Brendle and Eichmair3, Reference Li, Wei and Xiong7, Reference Wu and Xia9]). In order to give this error term explicitly, we need to introduce additional quantities. For $j\in \{1,\ldots ,n\}$ , we set $\Lambda _j=(\kappa _1,\ldots ,\kappa _{j-1},\kappa _{j+1},\ldots ,\kappa _n)$ , that is, the ordered set of principal curvatures without the jth one $\kappa _j$ , and we define

$$ \begin{align*}H_{r;j}={\binom{n}{r} }^{-1}S_r(\Lambda_j),\end{align*} $$

for any $j,r\in \{1,\ldots ,n\}$ . It is well known (see, for instance, [Reference Chen, Guan, Li and Scheuer4]) that if $H_r>0$ , then $H_{r;j}>0$ for any $j\in \{1,\ldots ,n\}$ . We can now recall the Hsiung–Minkowski formula with explicit error term in (2.4). To be precise (see [Reference Brendle and Eichmair3, Reference Li, Wei and Xiong7]),

(2.5) $$ \begin{align} \int_{\Sigma}fH_{s-1}\,dv_g= \int_{\Sigma}H_s\langle X,\nu\rangle \,dv_g-(s-1)\displaystyle\sum_{j=1}^n\int_{\Sigma}A_jH_{s-2;j}\,dv_g, \end{align} $$

where $A_j=-{(n(n-1))}^{-1}\langle X,e_j\rangle \mathrm {Ric}(e_j,\nu ).$ Moreover,

$$ \begin{align*}\mathrm{Ric}(e_j,\nu)=-\dfrac{n-1}{h(r)^2}\bigg(\dfrac{h"(r)}{h(r)}+\dfrac{k-h'(r)^2}{h(r)^2}\bigg)\langle X,e_j\rangle\langle X,\nu\rangle,\end{align*} $$

so that if M is star-shaped and assumption (H2) holds, then $A_j\geqslant 0$ for any $j\in \{1,\ldots ,n\}$ .

We finish this section by recalling the following inequalities whose proof can be found in [Reference Wu and Xia9]: for any $k,l\in \{2,\ldots ,r\}$ with $l<k$ and any $j\in \{1\ldots ,n\}$ ,

(2.6) $$ \begin{align} (k-1)H_{k-2;j}H_{l-1}>(l-1)H_{k-1}H_{l-2;j}. \end{align} $$

We now have all the ingredients to prove Theorem 1.1.

3 Proof of Theorem 1.1

Proof. From the assumptions, $H_{s-1}\leqslant cH_{r-1}$ so that

(3.1) $$ \begin{align} \int_{\Sigma}fH_{s-1}\,dv_g&\leqslant c \int_{\Sigma}H_{r-1}\,dv_g\nonumber\\ &= c\int_{\Sigma}H_r\langle X,\nu\rangle \,dv_g-c(r-1)\displaystyle\sum_{j=1}^n\int_{\Sigma}A_jH_{r-2;j}\,dv_g, \end{align} $$

where we have used the Hsiung–Minkowski formula (2.5) involving $H_{r-1}$ and $H_r$ . By assumption, we also have $cH_r\leqslant H_s$ and since $\langle X,\nu \rangle $ is nonnegative, (3.1) yields

$$ \begin{align*} \int_{\Sigma}fH_{s-1}\,dv_g \leqslant\displaystyle\int_{\Sigma}H_s\langle X,\nu\rangle \,dv_g-c(r-1)\sum_{j=1}^n\int_{\Sigma}A_jH_{r-2;j}\,dv_g. \end{align*} $$

We now use the Hsiung–Minkowski formula a second time, but involving $H_{s-1}$ and $H_s$ , to obtain

(3.2) $$ \begin{align} \int_{\Sigma}fH_{s-1}\,dv_g&\leqslant \displaystyle\int_{\Sigma}fH_{s-1}\,dv_g + (s-1)\sum_{j=1}^n\int_{\Sigma}A_jH_{s-2;j}\,dv_g \notag\\ & \quad -c(r-1)\sum_{j=1}^n\int_{\Sigma}A_jH_{r-2;j}dv_g. \end{align} $$

On the other hand, from (2.6),

$$ \begin{align*} (s-1)H_{r-1}H_{s-2;j}< (r-1)H_{s-1}H_{r-2;j}. \end{align*} $$

From the assumption $H_{s-1}\leqslant cH_{r-1}$ and the fact that $H_{r-2;j}>0$ , we have

(3.3) $$ \begin{align} (s-1)H_{r-1}H_{s-2;j}< c(r-1)H_{r-1}H_{r-2;j}, \end{align} $$

and so, since $H_{r-1}>0$ ,

$$ \begin{align*} (s-1)H_{s-2;j}< c(r-1)H_{r-2;j}. \end{align*} $$

Moreover, since all the $A_j$ are nonnegative,

$$ \begin{align*} (s-1)\sum_{j=1}^n\int_{\Sigma}A_jH_{s-2;j}\,dv_g-c(r-1)\sum_{j=1}^n\int_{\Sigma}A_jH_{r-2;j}\,dv_g\leqslant 0. \end{align*} $$

On the other hand, (3.2) gives

(3.4) $$ \begin{align} (s-1)\sum_{j=1}^n\int_{\Sigma}A_jH_{s-2;j}\,dv_g-c(r-1)\sum_{j=1}^n\int_{\Sigma}A_jH_{r-2;j}\,dv_g\geqslant 0. \end{align} $$

Hence, (3.4) is an equality, that is,

(3.5) $$ \begin{align} \displaystyle\int_{\Sigma}\sum_{j=1}^nA_j( (s-1)H_{s-2;j}-c(r-1)H_{r-2;j})\,dv_g=0. \end{align} $$

But from (3.3) and $A_j\geqslant 0$ for any $j\in \{1,\ldots ,n\}$ , each term of the sum in the integral of (3.5) is nonnegative, so that for any $j\in \{1,\ldots ,n\}$ ,

$$ \begin{align*} A_j((s-1)H_{s-2;j}-c(r-1)H_{r-2;j})=0. \end{align*} $$

Finally, since $(s-1)H_{s-2;j}-c(r-1)H_{r-2;j}<0$ and $A_j\geqslant 0$ , we deduce that $A_j=0$ for any $j\in \{1,\ldots ,n\}$ .

We now recall that $A_j=-{(n(n-1))}^{-1} \langle X,e_j\rangle \mathrm {Ric}(e_j,\nu )$ with

$$ \begin{align*}\mathrm{Ric}(e_j,\nu)=-\dfrac{n-1}{h(r)^2}\bigg(\dfrac{h"(r)}{h(r)}+\dfrac{k-h'(r)^2}{h(r)^2}\bigg)\langle X,e_j\rangle\langle X,\nu\rangle,\end{align*} $$

that is,

(3.6) $$ \begin{align} A_j=\dfrac{1}{nh(r)^2}\bigg(\dfrac{h"(r)}{h(r)}+\dfrac{k-h'(r)^2}{h(r)^2}\bigg)\langle X,e_j\rangle^2\langle X,\nu\rangle. \end{align} $$

From (3.6), $A_j=0$ for any $j\in \{1,\ldots ,n\}$ implies that, at any point $x\in \Sigma $ , the vector $X(x)$ is either normal or tangent. Since X does not vanish, by continuity, X is tangent everywhere or normal everywhere. But, if X is tangent, then the support function $\langle X,\nu \rangle $ vanishes and so the Hsiung–Minkowski (2.5) formula becomes

$$ \begin{align*}\int_{\Sigma}fH_{r-1}\leqslant 0,\end{align*} $$

which is a contradiction since f and $H_{r-1}$ are positive. Hence, $\nu $ is parallel to ${\partial }/{\partial t}$ and so $\Sigma $ is a slice $\{t_1\}\times N$ , which concludes the proof.

We finish by noting that $t_1$ can be explicitly given by the constant c. Indeed, since $\Sigma $ is a slice, we have equality in the sequence of inequalities in (2.2), that is,

$$ \begin{align*}H_{r}^{{1}/{r}}= H_{r-1}^{{1}/{(r-1)}}= \cdots= H_2^{1/2}= H.\end{align*} $$

Hence, the assumption

$$ \begin{align*}\dfrac{H_{s-1}}{H_s}\geqslant c\geqslant \dfrac{H_{s-2}}{H_{s-1}}\end{align*} $$

gives $c={1}/{H^{r-s}}$ . Moreover, the Brendle identity (2.4) becomes

$$ \begin{align*}f(t_1)\mathrm{Vol}(\Sigma)=Hh(t_1)\mathrm{Vol}(\Sigma) \quad\text{on}\ \Sigma=\{t_1\}\times N,\end{align*} $$

and so $t_1$ is the unique positive real number so that

$$ \begin{align*}\dfrac{h'(t_1)}{h(t_1)}=\bigg(\dfrac{1}{c}\bigg)^{{1}/{(r-s)}}.\end{align*} $$