\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2006(2006), No. 15, pp. 1--9.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2006 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2006/15\hfil A Liouville theorem for $F$-harmonic maps] {A Liouville theorem for $F$-harmonic maps with finite $F$-energy} \author[M. Kassi\hfil EJDE-2006/15\hfilneg] {M'hamed Kassi} \address{M'hamed Kassi \hfill\break Equipe d'Analyse Complexe \\ Laboratoire d'Analyse Fonctionnelle, Harmonique et Complexe\\ D\'epartement de Math\'ematiques\\ Facult\'e des Sciences \\ Universit\'e Ibn Tofail \\ K\'enitra, Maroc} \email{mhamedkassi@yahoo.fr} \date{} \thanks{Submitted March 24, 2005. Published January 31, 2006.} \subjclass[2000]{58E20, 53C21, 58J05} \keywords{$F$-harmonic maps; Liouville propriety; Stokes formula; \hfill\break\indent comparison theorem} \begin{abstract} Let $(M,g)$ be a $m$-dimensional complete Riemannian manifold with a pole, and $(N,h)$ a Riemannian manifold. Let $F : \mathbb{R}^{+}\to \mathbb{R}^{+} $ be a strictly increasing $C^{2}$ function such that $F(0)=0$ and $d_{F}:=\sup(tF'(t)(F(t))^{-1}) < \infty$. We show that if $d_{F} < m/2$, then every $F$-harmonic map $ u : M\to N$ with finite $F$-energy (i.e a local extremal of $E_{F}(u):= \int_{M} F(\vert du\vert^{2}/2)dV_{g}$ and $E_{F}(u)$ is finite) is a constant map provided that the radial curvature of $M$ satisfies a pinching condition depending to $d_{F}$. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \allowdisplaybreaks \section{Introduction and statement of result} Let $(M,g)$ and $(N,h)$ be two Riemannian manifolds and $F$ be a given $C^{2}$ function $F : \mathbb{R}^{+} \to \mathbb{R}^{+} $. Then, a map $u : M \to N$ of class $C^{2}$ is said to be $F$-harmonic if for every compact $K$ of $M$, the map $u$ is extremal of $F$-energy: $$ E_{F}(u):=\int_{K}F({\vert du\vert^{2}\over 2})dV_{g}. $$ In a normal coordinate system, the tension field associated with $E_{F}(u)$ by the Euler-Lagrange equations is $$ \tau_{F}(u):= \sum_{i=1}^{m}(\nabla_{e_{i}}(F'({\vert du\vert^{2}\over 2})du))e_{i}= F'({\vert du\vert^{2}\over 2})\tau(u) + du.\Bigl\{\mathop{\rm grad}\bigl(F'({\vert du\vert^{2}\over 2})\bigr)\Bigr\} $$ where $\tau(u)$ is the usual tension field of $u$ defined by $$ \tau(u)_{k} = \Delta_{M}u^{k} + \sum_{\beta,\gamma;i,j}^{n;m}{}^{N}\Gamma_{\alpha\gamma}^{k}(u)g^{ij}{\partial u^{\beta}\over\partial x_{i}}{\partial u^{\gamma}\over\partial x_{j}},\quad k=1,\dots,n \,. $$ Then, the map $u$ is $F$-harmonic if $\tau_{F}(u) = 0$. For further properties of $F$-harmonic maps, we refer the reader to \cite{a1,a2}. For the particular case of $F(t)=t$, the Liouville problem for harmonic maps with finite energy have been studied in \cite{e1,k1,k2,s1,t1}. While for $F(t) = \frac{2}{p}t^{p/2}$, with $ p\geq 2$, this is the problem of $p$-harmonic maps with finite $p$-energy (corollary \ref{cor1}. If $F(t)= \sqrt{1+2t} -1$ corresponding to the minimal graph (corollary \ref{cor2}). In this paper, we study the same problem for $F$-harmonic maps with finite $F$-energy without condition on the curvature for the target manifold. We assume that $F$ is strictly increasing, $F(0)=0$, and $d_{F}=\sup{tF'(t)\over F(t)} < \infty$, ``the degree of $F$''. For $x$ in $M$, we set $r(x) = d_{g}(x,x_0)$. \begin{theorem} \label{thm1} Let $(M,g)$ be a $m$-dimensional complete Riemannian manifold, $m>2$, with a pole $x_0$, and let $(N,h)$ be a Riemannian manifold. If $d_F< m/2$, then every $F$-harmonic map of $M$ into $N$ with finite $F$-energy is constant provided that the radial curvature $K_r$ of $M$ satisfies one of the following two conditions: \begin{itemize} \item[(i)] $-\alpha^{2}\leq K_{r} \leq -\beta^{2}$ with $\alpha > 0, \beta > 0$ and $1+(m-1)\beta - 2d_{F}\alpha > 0$ \item[(ii)] $-{\alpha\over 1+r^{2}} \leq K_{r}\leq {\beta\over 1+r^{2}} $ with $\alpha\geq 0$ and $\beta\in[0,{1\over 4}]$ such that $2+(m-1)(1+\sqrt{1-4\beta})-2d_{F}(1+\sqrt{1+4\alpha}) > 0$. \end{itemize} \end{theorem} Furthermore, we have the following corollaries. \begin{corollary} \label{cor1} Let $(M,g)$ and $(N,h)$ be as in the theorem. Then, every $C^{2}$ $p$-harmonic map of $M$ into $N$ with finite $p$-energy, for $p2$, every $C^{2}$ map $u$ of $M$ into $N$, with finite energy, solution of $$ {\tau(u)\over\sqrt{1+\vert du\vert^{2}}} + du.\Bigl\{\mathop{\rm grad}\bigl({1\over\sqrt{1+\vert du\vert^{2}}}\bigr)\Bigr\} = 0 $$ is constant. \end{corollary} For $m=2$, the statement of the theorem is false in general. In fact, for the case (i), there exist holomorphic maps of the hyperbolic disc with finite energy \cite{t1}. While for the case (ii) there exist holomorphic maps of $\mathbb{C}$ into $\mathbb{P}^1$ with finite energy \cite{s1}. \section{Proof of Theorem \ref{thm1}} Let $X$ and $Y$ be two vector fields on $M$. It is well-known \cite{b1,k1}, that the stress-energy for harmonic maps is $$ S_{u} := \frac{|du|^{2}}{2}\langle X,Y\rangle _{g} - \langle du(X),du(Y)\rangle_{h} $$ and satisfies $$ (\mathop{\rm div}S_{u})(X) = -\langle \tau(u),du(X)\rangle _{h}. $$ Following \cite{a2}, we define the stress-energy of $F$-harmonic maps by $$ S_{F,u}(X,Y):= F({\vert du\vert^{2}\over 2})\langle X,Y\rangle _{g} - F'({\vert du\vert^{2}\over 2})\langle du(X),du(Y)\rangle _{h}\,. $$ When $F(t):= t$ we have $S_{F,u} := S_{u}$. Also $(\mathop{\rm div}S_{F,u})(X) = -\langle\tau_{F}(u),du(X)\rangle _{h}$ thanks to the following lemma. \begin{lemma} \label{lem1} For every vector field $X$ on $M$, we have \begin{gather} (\mathop{\rm div}S_{F,u})(X) = - \langle \tau_{F}(u),du(X) \rangle_{h}, \\ \begin{aligned} &\mathop{\rm div}(F({\vert du\vert^{2}\over 2})X)\\ & =\mathop{\rm div}(F'({\vert du\vert^{2}\over 2})\langle du(X),du(e_{i}) \rangle_{h}e_{i}) - \langle \tau_{F}(u),du(X) \rangle_{h} + [S_{F,u},X], \end{aligned} \end{gather} where $$ [S_{F,u},X](x) = \sum_{i,j=1}^{m}\Bigl(F({\vert du\vert^{2}\over 2})\delta_{ij} - F'({\vert du\vert^{2}\over 2})\langle du(e_{i}),du(e_{j}) \rangle_{h}\Bigr)\langle \nabla_{e_{i}}X,e_{j} \rangle_{g}\,. $$ In particular, if $u$ is $F$-harmonic and $D\subset\subset M$ is a $C^{1}$ boundary domain, then we have $$ \int_{\partial D}S_{F,u}(X,\nu)d\sigma_{g} = \int_{D}[S_{F,u},X]dV_{g} $$ where $\nu$ is the normal to $\partial D$. \end{lemma} \begin{proof} Let $x \in M$. Chose a normal coordinate system such that at $x$. $ g_{ij}(x) = \delta_{ij}$ $ dg(x) = 0$, where $(e_1, \dots, e_m)$ being a normal basis, we have $\nabla_{e_{j}}e_{k}=0$ for all $j,k$ and \begin{align*} &(\mathop{\rm div}S_{F,u})(X)\\ &= \sum_{i=1}^{m}\Big\{\nabla_{e_{i}}S_{F,u}(e_{i},X) - S_{F,u}(e_{i},\nabla_{e_{i}}X) - S_{F,u}(\nabla_{e_{i}}e_{i},X)\Big\}\\ &= \sum_{i=1}^{m}\Big\{ \nabla_{e_{i}}\bigl(F({\vert du\vert^{2}\over 2})\langle e_{i},X\rangle - \langle F'({\vert du\vert^{2}\over 2})du(e_{i}),du(X)\rangle \bigr) - F({\vert du\vert^{2}\over 2}) \langle e_{i},\nabla_{e_{i}}X\rangle\\ &\quad + F'({\vert du\vert^{2}\over 2})\langle du(e_{i}),du(\nabla_{e_{i}}X)\rangle - S_{F,u}(\nabla_{e_{i}}e_{i},X)\Big\}\\ &= \sum_{i=1}^{m}\Big\{\nabla_{e_{i}}\bigl(F({\vert du\vert^{2}\over 2})\langle e_{i},X\rangle \bigr)\\ &\quad - \nabla_{e_{i}}\bigl(\langle F'({\vert du\vert^{2}\over 2})du(e_{i}),du(X)\rangle \bigr)-F({\vert du\vert^{2}\over 2})\langle e_{i},\nabla_{e_{i}}X\rangle \\ &\quad + F'({\vert du\vert^{2}\over 2})\langle du(e_{i}),du(\nabla_{e_{i}}X)\rangle - S_{F,u}(\nabla_{e_{i}}e_{i},X)\Big\}\\ &= \sum_{i=1}^{m}\Big\{\big(\sum_{j=1}^{m} F'({\vert du\vert^{2}\over 2})\langle \nabla_{e_{i}}(du(e_{j})),du(e_{j})\rangle \big)\langle e_{i},X\rangle\\ &\quad + F({\vert du\vert^{2}\over 2})\nabla_{e_{i}}\langle e_{i},X\rangle - \langle \nabla_{e_{i}}(F'({\vert du\vert^{2}\over 2})du(e_{i})),du(X)\rangle \\ &\quad - F'({\vert du\vert^{2}\over 2})\langle du(e_{i}),\nabla_{e_{i}}(du(X))\rangle \\ &\quad - F({\vert du\vert^{2}\over 2})\langle e_{i},\nabla_{e_{i}}X \rangle +F'({\vert du\vert^{2}\over 2})\langle du(e_{i}),du(\nabla_{e_{i}}X)\rangle \\ &\quad - S_{F,u}(\nabla_{e_{i}}e_{i},X)\Big\}\,. \end{align*} Thus \begin{align*} (\mathop{\rm div}S_{F,u})(X) &= \sum_{i,j=1}^{m}\Big\{F'({\vert du\vert^{2}\over 2})\langle \nabla_{e_{i}}(du(e_{j})),du(e_{j})\rangle X_{i}\Big\}\\ &\quad + \sum_{i=1}^{m}\Big\{ F({\vert du\vert^{2}\over 2})\langle \nabla_{e_{i}}e_{i},X\rangle + F({\vert du\vert^{2}\over 2})\langle e_{i},\nabla_{e_{i}}X\rangle \\ &\quad - \langle \nabla_{e_{i}}(F'({\vert du\vert^{2}\over 2})du(e_{i})),du(X)\rangle\\ &\quad - F'({\vert du\vert^{2}\over 2})\langle du(e_{i}),\nabla_{e_{i}}(du(X))\rangle -F({\vert du\vert^{2}\over 2})\langle e_{i},\nabla_{e_{i}}X \rangle \\ &\quad + F'({\vert du\vert^{2}\over 2}) \langle du(e_{i}),du(\nabla_{e_{i}}X) \rangle - S_{F,u}(\nabla_{e_{i}}e_{i},X)\Big\}\\ &= \sum_{i,j=1}^{m}\Big\{F'({\vert du\vert^{2}\over 2}) \langle X_{i}\nabla_{e_{i}}(du(e_{j})),du(e_{j})\rangle \Big\}\\ &\quad - \sum_{i=1}^{m}\Big\{ F'({\vert du\vert^{2}\over 2})\langle du(e_{i}),\nabla_{e_{i}}(du(X))\rangle \\ &\quad + F'({\vert du\vert^{2}\over 2}) \langle du(e_{i}),du(\nabla_{e_{i}}X) \rangle +F({\vert du\vert^{2}\over 2}) \langle \nabla_{e_{i}}e_{i},X \rangle\\ &\quad + F({\vert du\vert^{2}\over 2})\langle e_{i},\nabla_{e_{i}}X\rangle -F({\vert du\vert^{2}\over 2})\langle e_{i},\nabla_{e_{i}}X\rangle\\ &\quad - \langle \nabla_{e_{i}}(F'({\vert du\vert^{2}\over 2})du(e_{i})),du(X) \rangle - S_{F,u}(\nabla_{e_{i}}e_{i},X)\Big\}. \end{align*} Since $ \nabla_{e_{i}}e_{i} = 0$, with $(\nabla_{e_{i}}du)(X)=\nabla_{e_{i}}(du(X))- du(\nabla_{e_{i}}X)$ and by symmetry $(\nabla_{e_{i}}du)(X)= (\nabla_{X}du)(e_{i})$, we have \begin{align*} \mathop{\rm div}(S_{F,u})(X) &= \sum_{j=1}^{m}\Big\{ F'({\vert du\vert^{2}\over 2})\langle \nabla_{X}(du(e_{j}),du(e_{j})\rangle \Big\}\\ &\quad - \sum_{i=1}^{m}\Big\{ F'({\vert du\vert^{2}\over 2}) \langle du(e_{i}),\nabla_{e_{i}}(du(X)) -du(\nabla_{e_{i}}X)\rangle \\ &\quad - \langle \nabla_{e_{i}}(F'({\vert du\vert^{2}\over 2})du(e_{i})),du(X) \rangle \Big\}. \end{align*} Finally, $$ \mathop{\rm div}(S_{F,u})(X) = -\langle\tau_{F}(u),du(X)\rangle . $$ Also \begin{align*} \mathop{\rm div}(F({\vert du\vert^{2}\over 2})X) &= \sum_{i=1}^{m}\langle \nabla_{e_{i}}(F({\vert du\vert^{2}\over 2})X),e_{i}\rangle \\ &= \sum_{i=1}^{m}\Big\{ \langle \nabla_{e_{i}}(F({\vert du\vert^{2}\over 2}))X,e_{i} \rangle + F({\vert du\vert^{2}\over 2}) \langle \nabla_{e_{i}}X,e_{i}\rangle \Big\} \\ &= \nabla_{X}F({\vert du\vert^{2}\over 2})+ \sum_{i=1}^{m}F({\vert du\vert^{2}\over 2}) \langle \nabla_{e_{i}}X,e_{i}\rangle . \end{align*} Then, by straightforward computation, we obtain \begin{align*} \nabla_{X}F({\vert du\vert^{2}\over 2}) &= \sum_{i=1}^{m}{1\over 2}F'({\vert du\vert^{2}\over 2})\nabla_{X}\langle du(e_{i}),du(e_{i})\rangle \\ &= \sum_{i=1}^{m}F'({\vert du\vert^{2}\over 2})\langle \nabla_{X}(du(e_{i})),du(e_{i})\rangle \\ &= \sum_{i=1}^{m} F'({\vert du\vert^{2}\over 2}) \langle (\nabla_{X}du)(e_{i})+ du(\nabla_{X}e_{i}),du(e_{i})\rangle \\ &= \sum_{i=1}^{m}F'({\vert du \vert^{2}\over 2})\langle (\nabla_{X}du)(e_i),du(e_i)\rangle \\ &= \sum_{i=1}^{m}F'({\vert du\vert^{2}\over 2})\langle (\nabla_{e_{i}}du)(X),du(e_{i})\rangle \quad \text{(by symmetry)} \\ &= \sum_{i=1}^{m}\Big\{\langle \nabla_{e_{i}}(du(X)),F'({\vert du\vert^{2}\over 2})du(e_{i})\rangle \\ &\quad - F'({\vert du\vert^{2}\over 2}) \langle du(\nabla_{e_{i}}X),du(e_{i})\rangle \Big\} \end{align*} Thus \begin{align*} \nabla_{X}F({\vert du\vert^{2}\over 2}) &= \sum_{i=1}^{m}\Big\{\nabla_{e_{i}}\langle du(X),F{'}({\vert du\vert^{2}\over 2})du(e_{i})\rangle\\ &\quad - \langle du(X),\nabla_{e_{i}}(F'({\vert du\vert^{2}\over 2})du(e_{i}))\rangle \\ &\quad - F'({\vert du\vert^{2}\over 2})\rangle du(\nabla_{e_{i}}X),du(e_{i})\rangle \Big\}\\ &= \sum_{i=1}^{m}\Big\{ \mathop{\rm div}\bigl(F'({\vert du\vert^{2}\over 2}) \rangle du(X),du(e_{i})\rangle e_{i}\bigr)\\ &\quad + \langle du(X),-\nabla_{e_{i}}(F'({\vert du\vert^{2}\over 2})du(e_{i}))\rangle \\ &\quad - F'({\vert du\vert^{2}\over 2}) \langle du(\nabla_{e_{i}}X),du(e_{i})\rangle \Big\} \\ &= \sum_{i=1}^{m}\Big\{ \mathop{\rm div}\bigl(F'({\vert du\vert^{2}\over 2}) \langle du(X),du(e_{i})\rangle e_{i}\bigr)\Big\}\\ &\quad - \langle du(X),\tau_{F}(u)\rangle - \sum_{i=1}^{m} F'({\vert du\vert^{2}\over 2})\langle du(\nabla_{e_{i}}X),du(e_{i})\rangle \end{align*} Thus \begin{align*} \mathop{\rm div}(F({\vert du\vert^{2}\over 2})X) &= \sum_{i=1}^{m}\Big\{ \mathop{\rm div}\bigl(F'({\vert du\vert^{2}\over 2})\langle du(X),du(e_{i})\rangle e_{i}\bigr)\Big\}\\ &\quad - \langle du(X),\tau_{F}(u)\rangle + [S_{F,u},X] \end{align*} with $$ [S_{F,u},X] = \sum_{i,j=1}^{m}\Bigl(F({\vert du\vert^{2}\over 2})\delta_{ij} - F'({\vert du\vert^{2}\over 2}) \langle du(e_{i}),du(e_{j})\rangle_{h}\Bigr)\langle \nabla_{e_{i}}X,e_{j}\rangle_{g} $$ because $\nabla_{e_{i}}X = \langle \nabla_{e_{i}}X,e_{j}\rangle e_{j}$. If $D\subset\subset M$ is a $C^{1}$ boundary domain, we get by the use of Stokes formula \begin{align*} &\int_{D}(\mathop{\rm div}S_{F,u})(X)+\int_{D}[S_{F,u},X]\\ &= \int_{D}\mathop{\rm div}(F({\vert du\vert^{2}\over 2})X) - \int_{D} \sum_{i=1}^{m}\mathop{\rm div}\bigl(F'({\vert du\vert^{2}\over 2})e_{i}\bigr) \\ &= \int_{\partial D}F({\vert du\vert^{2}\over 2}) \langle X,\nu\rangle - \int_{\partial D}F'({\vert du\vert^{2}\over 2}) \langle du(X),du(\nu)\rangle \,. \end{align*} Thus, if $u$ is $F$-harmonic: $$ \int_{\partial D}\Bigl(F({\vert du\vert^{2}\over 2}) \langle X,\nu \rangle - F'({\vert du\vert^{2}\over 2}) \langle du(X),du(\nu) \rangle \bigr) = \int_{D}[S_{F,u},X]. $$ This completes the proof. \end{proof} \begin{lemma} \label{lem2} Let $u : M\to N$ be a $F$-harmonic with finite $F$-energy and $X$ a vector field on $M$ such that $\vert X\vert \leq \phi(r)$ for $\phi : \mathbb{R}^{+} \to \mathbb{R}^{+}$ satisfying $$ \int_{1}^{+\infty}{dt\over\phi(t)}=+\infty. $$ Then there exists an increasing strictly sequence $(R_{n})$ such that $$ \lim_{n\to\infty}\int_{B(x_{0},R_{n})}[S_{F,u},X]dV_{g} = 0. $$ \end{lemma} \begin{proof} Since $tF'(t)\leq d_{F}F(t)$ we have \begin{align*} &\Big\vert\int_{B(x_{0},R)}[S_{F,u},X]\Big\vert \\ & \leq \Big\vert \int_{\partial B(x_{0},R)}F({\vert du\vert^{2}\over 2})\langle X,\nu\rangle\Big\vert + \Big\vert \int_{\partial B(x_{0},R)}F'({\vert du\vert^{2}\over 2})\rangle du(X),du(\nu)\rangle\Big\vert \\ & \leq \int_{\partial B(x_{0},R)}F({\vert du\vert^{2}\over 2})\vert\langle X,\nu\rangle\vert + \int_{\partial B(x_{0},R)}F'({\vert du\vert^{2}\over 2}) \vert\rangle du(X),du(\nu)\rangle\vert \\ & \leq (1+2d_{F})\int_{\partial B(x_{0},R)}F({\vert du\vert^{2}\over 2})\vert X\vert\,. \end{align*} By the Co-area formula and $\vert X\vert \leq \phi(r(x))$, \begin{align*} \int_{0}^{\infty}{1\over\phi(t)}\Bigl(\int_{\partial B(x_{0},t)} F({\vert du\vert^{2}\over 2})\vert X\vert\Bigr)dt &= \int_{M}{\vert X\vert\vert\nabla r\vert\over\phi(r)}F({\vert du\vert^{2}\over 2}) \\ & \leq \int_{M}F({\vert du\vert^{2}\over 2}) < \infty \end{align*} Since $\int_{1}^{\infty}{dt\over\phi(t)}=\infty$, there exists a increasing strictly sequence $(R_{n})$ such that $\lim_{n\to\infty}\int_{\partial B(x_{0},R_{n})}F({\vert du\vert^{2}\over 2})\vert X\vert =0$. Hence $$ \lim_{n\to\infty}\int_{B(x_{0},R_{n})}[S_{F,u},X]dV_{g} = 0. $$ This completes the proof of Lemma \ref{lem2}. \end{proof} For the theorem, it suffices to choose $X$ satisfying Lemma \ref{lem2} and the condition $[S_{F,u},X]\geq cF(\vert du\vert^{2}/2)$ where $c>0$ is a constant. For that we take $X=r\nabla r$ and using the comparison theorem of the Hessian \cite{g1}. \begin{theorem}[Comparison theorem] \label{thm2} Let $(M,g)$ be a complete Riemannian manifold with a pole $x_0$ and $k_1$, $k_2$ be two continuous functions on $\mathbb{R}^{+}$ such that $k_{2}(r)\leq K_{r} \leq k_{1}(r)$, where $K_r$ is the radial curvature of $M$, i.e., the sectional curvature of the tangent planes containing the radial vector $\nabla r$. Also, let $J_{i}$ ($i=1,2$) be the solution of classical Jacobi equation $$ J''_{i} + k_{i}J_{i} = 0; \quad J_{i}(0) = 0\quad\hbox{and}\quad J'_{i}(0) = 1 . $$ Then, if $J_{1} > 0$ on $\mathbb{R}^{+}$, we have on $M\setminus\{x_{0}\}$ $$ {J'_{1}(r)\over J_{1}(r)}(g - dr\otimes dr) \leq \mathop{\rm Hess}(r) \leq {J'_{2}(r)\over J_{2}(r)}(g - dr\otimes dr)\,. $$ \end{theorem} Case (i) of Theorem \ref{thm2}: With $k_{1}(r)=-\beta^{2}$ and $k_{2}(r)=-\alpha^{2}$, we have $$ \beta\coth(\beta r)(g - dr\otimes dr)\leq\mathop{\rm Hess}(r)\leq\alpha\coth(\alpha r)(g - dr\otimes dr)\,. $$ Case (ii) of Theorem \ref{thm2}: With $k_{1}(r)={\beta\over r^{2}}$ and $k_{2}(r)=-{\alpha\over r^{2}}$, and the fact that on $M\setminus\{x_{0}\}$, $$ -\frac{\alpha}{r^{2}} \leq -\frac{\alpha}{1+ r^{2}} \leq K_{r} \leq \frac{\beta}{1+ r^{2}} \leq \frac{\beta}{r^{2}} $$ we have $$ \Bigl({1+\sqrt{1-4\beta}\over 2r}\Bigr)(g - dr\otimes dr)\leq\mathop{\rm Hess}(r)\leq\Bigl({1+\sqrt{1+4\alpha}\over 2r}\Bigr)(g - dr\otimes dr)\,. $$ \begin{lemma} \label{lem3} Under hypothesis of Theorem \ref{thm2}, in case (1), we have $$ [S_{F,u},X] \geq (1+(m-1)\beta-2d_{F}\alpha)F({\vert du\vert^{2}\over 2}) $$ and in case (ii), $$ [S_{F,u},X] \geq {1\over 2}(2+(m-1) (1+\sqrt{1-4\beta})-2d_{F} (1+\sqrt{1+4\alpha}) F({\vert du\vert^{2}\over 2}). $$ \end{lemma} \begin{proof} First note that $$ [S_{F,u},X]=\sum_{i,j=1}^{m}\Bigl(F({\vert du\vert^{2}\over 2})\delta_{ij} - F'({\vert du\vert^{2}\over 2})\langle du(e_{i}),du(e_{j})\rangle_{h}\Bigr)<\nabla_{e_{i}}X,e_{j}\rangle_{g} , $$ where $(e_{1},\dots,e_{m-1},{\partial\over\partial r})$ with $e_{m} = {\partial \over \partial r}$, being a normal basis on $B(x_{0},R)$. Then, since $X=r{\partial\over\partial r}$, it follows that $\nabla_{\partial\over\partial r}X={\partial\over\partial r}$ and so we get \begin{gather*} \langle \nabla_{\partial\over\partial r}X,{\partial\over\partial r} \rangle_{g} = 1\,, \\ \langle \nabla_{e_{i}}X,e_{i}\rangle_{g} = r\mathop{\rm Hess}(r)(e_{i},e_{i}), \quad \text{for } i=1, \dots, m-1, \\ \nabla_{e_{i}}X = \sum_{j=1}^{m-1}r\mathop{\rm Hess}(r)(e_{i},e_{j})e_{j}, \quad \text{for } i=1, \dots, m-1. \end{gather*} Therefore, \begin{align*} [S_{F,u},X] &= F({\vert du\vert^{2}\over 2})(1+\sum_{i=1}^{m-1}r \mathop{\rm Hess}(r)(e_{i},e_{i}))\\ &\quad - \sum_{i,j=1}^{m-1}F'({\vert du\vert^{2}\over 2}) \langle du(e_{i}),du(e_{j})\rangle_{h} \langle \nabla_{e_{i}}X,e_{j}\rangle_{g} \\ &\quad - F'({\vert du\vert^{2}\over 2}) \langle du({\partial\over\partial r}),du({\partial\over\partial r})\rangle_{h} \langle \nabla_{\partial\over\partial r}X, {\partial\over\partial r}\rangle_{g}\\ &\quad - \sum_{j=1}^{m-1}F'({\vert du\vert^{2}\over 2}) \langle du({\partial\over\partial r}),du(e_{j})\rangle_{h} \langle \nabla_{\partial\over\partial r}X,e_{j}\rangle_{g} \\ &\quad - \sum_{i=1}^{m-1}F'({\vert du\vert^{2}\over2}) \langle du(e_{i}),du({\partial\over\partial r})\rangle_{h} \langle \nabla_{e_{i}}X,{\partial\over\partial r}\rangle_{g} \\ &= F({\vert du\vert^{2}\over 2})(1+\sum_{i=1}^{m-1}r\mathop{\rm Hess} (r)(e_{i},e_{i}))\\ &\quad - \sum_{i,j=1}^{m-1}F'({\vert du\vert^{2}\over 2}) \langle du(e_{i}),du(e_{j})\rangle r\mathop{\rm Hess}(r)(e_{i},e_{j}) \\ &\quad - F'({\vert du\vert^{2}\over 2}) \langle du({\partial\over\partial r}),du({\partial\over\partial r})\rangle \end{align*} For the case (i), we have \begin{align*} [S_{F,u},X]& \geq F({\vert du\vert^{2}\over 2})+ (m-1)(\beta r)\coth(\beta r)F({\vert du\vert^{2}\over 2})\\ &\quad - F'({\vert du\vert^{2}\over 2})\vert du\vert^{2}(\alpha r) \coth(\alpha r) \\ &\quad + F'({\vert du\vert^{2}\over 2})((\alpha r)\coth(\alpha r)-1) \langle du({\partial\over\partial r}),du({\partial\over\partial r})\rangle\\ & \geq F({\vert du\vert^{2}\over 2}) + F({\vert du\vert^{2}\over 2})((m-1)(\beta r)\coth(\beta r)- 2d_{F}(\alpha r)\coth(\alpha r))\\ & \geq F({\vert du\vert^{2}\over 2}) + F({\vert du\vert^{2}\over 2}) r\coth(\beta r)((m-1)\beta -2d_{F}\alpha \frac{\coth(\alpha r)}{\coth(\beta r)}). \end{align*} Since the function $\coth(x)$ is decreasing and, $x\coth(x)$ is bounded below by a positive constant in $\mathbb{R}^{+}$, we have $$ [S_{F,u},X] \geq (1+(m-1)\beta - 2d_{F}\alpha)F({\vert du\vert^{2}\over 2}) $$ For the case (ii), we have \begin{align*} [S_{F,u},X]& \geq F({\vert du\vert^{2}\over 2})+(m-1)aF({\vert du\vert^{2}\over 2})-bF'({\vert du\vert^{2}\over 2})\vert du\vert^{2} \\ &\quad + (b-1)F'({\vert du\vert^{2}\over 2})\langle du({\partial\over\partial r}),du({\partial\over\partial r})\rangle\\ & \geq (1+(m-1)a-2d_{F}b)F({\vert du\vert^{2}\over 2}), \end{align*} where we have set $$ a = {1+\sqrt{1-4\beta}\over 2}\quad\hbox{and}\quad b = {1+\sqrt{1+4\alpha}\over 2}\geq 1. $$ \end{proof} \subsection*{Acknowledgements} I would like to express my gratitude to Professor S. 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