1 Introduction
Let \((M, g)\) be an n-dimensional complete Riemannian manifold and f be a smooth function defined on M. Then the triple \((M, g,e^{-f}\, dv)\) is called a smooth metric measure space, where dv denotes the volume element of the metric g and \(e^{-f}\,dv\) is called the weighted measure. On the smooth metric measure space \((M, g,e^{-f}\, dv)\), the m-Bakry–Émery Ricci curvature (see [1‐3]) is defined by
where \(m\geq n\) is a constant, and \(m=n\) if and only if f is a constant. We define
Then \(\mathrm{Ric}_{f}\) can be seen as the ∞-dimensional Bakry–Émery Ricci curvature. However, there are many differences between the m-Bakry–Émery Ricci curvature and the ∞-Bakry–Émery Ricci curvature. For example, there exist complete noncompact Riemannian manifolds which satisfy \(\mathrm{Ric}_{f}=\lambda g\) for some positive constant λ (which is called the shrinking gradient Ricci soliton), but not for \(\mathrm{Ric}_{f}^{m}=\lambda g\). We recall that the f-Laplacian \(\Delta_{f}\) on \((M, g,e^{-f}\,dv)\) is defined by
Since we have the Bochner formula with respect to f-Laplacian:
which is similar to the Bochner formula associated with the Laplacian, many results with respect to the Laplacian have been generalized to those of the f-Laplacian under the m-dimensional Bakry–Émery Ricci curvature. For example, see [4‐7] and the references therein. But for elliptic gradient estimates for f-Laplacian under the ∞-Bakry–Émery Ricci curvature, in order to using the weighted comparison theorem, the assumption \(|\nabla f|\leq\theta\) is forced commonly.
$$ \mathrm{Ric}_{f}^{m}=\mathrm{Ric}+ \nabla^{2}f-\frac{1}{m-n}\,df\otimes df, $$
(1.1)
$$ \mathrm{Ric}_{f}=\mathrm{Ric}+\nabla^{2}f. $$
(1.2)
$$\Delta_{f}=\Delta-\nabla f\nabla. $$
$$\frac{1}{2}\Delta_{f}|\nabla u|^{2}\geq \frac{1}{m}(\Delta_{f}u)^{2}+\nabla u\nabla( \Delta_{f}u)+\mathrm{Ric}_{f}^{m}(\nabla u,\nabla u), $$
In this paper, under the assumption that the ∞-Bakry–Émery Ricci curvature is bounded from below, we consider the following nonlinear elliptic equation:
where a, b are two real constants. Inspired by the ideas of Brighton in [8], we can obtain global gradient estimates for positive solutions to (1.3) without any restriction on \(|\nabla f|\).
$$ \Delta_{f}u+au\log u+bu=0, $$
(1.3)
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Theorem 1.1
Let
\((M, g,e^{-f}\,dv)\)
be an
n-dimensional complete smooth metric measure space with
\(\mathrm{Ric}_{f}(B_{p}(2R))\geq-(n-1)K\), where
\(K\geq0\)
is a constant. Suppose that
u
is a positive solution to (1.3) with
\(u\leq A\)
on
\(B_{p}(2R)\). Then on
\(B_{p}(R)\)
with
\(R>1\), the following inequality holds:
where
C
is a positive constant which depends on the dimension
n, \(\beta=\max_{\{x|d(x,p)=1\}} \Delta_{f} r(x)\)
and
$$ \begin{aligned} |\nabla u|^{2}\leq C A^{2} \biggl[\max \biggl\{ \frac{4}{5}b+a \biggl(1+ \frac {4}{5}L \biggr),0 \biggr\} +K+\frac{|\beta|+1}{R} \biggr], \end{aligned} $$
(1.4)
$$ L= \textstyle\begin{cases} \sup_{B_{p}(2R)}(\log u), &\textit{if } a\geq0, \\ \inf_{B_{p}(2R)}(\log u), &\textit{if } a< 0. \end{cases} $$
(1.5)
Letting \(R\rightarrow\infty\) in (1.4), we obtain the following global estimates on complete noncompact Riemannian manifolds:
Corollary 1.2
Let
\((M, g,e^{-f}\,dv)\)
be an
n-dimensional complete smooth metric measure space with
\(\mathrm{Ric}_{f} \geq-(n-1)K\), where
\(K\geq0\)
is a constant. If
u
is a positive solution to (1.3) with
\(u\leq A\), then we have
where
$$ \begin{aligned} |\nabla u|^{2}\leq C A^{2} \biggl[\max \biggl\{ \frac{4}{5}b+a \biggl(1+ \frac {4}{5}L \biggr),0 \biggr\} +K \biggr], \end{aligned} $$
(1.6)
$$ L= \textstyle\begin{cases} \sup_{M}(\log u), &\textit{if } a\geq0, \\ \inf_{M}(\log u), &\textit{if } a< 0. \end{cases} $$
(1.7)
Using the ideas of the proof of Theorem 1.1, by choosing \(\tilde{h}=\log u\) a gap develops between the constants, and we also establish the following.
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Theorem 1.3
Let
\((M, g,e^{-f}\,dv)\)
be an
n-dimensional complete smooth metric measure space with
\(\mathrm{Ric}_{f}(B_{p}(2R))\geq-(n-1)K\), where
\(K\geq0\)
is a constant. Suppose that
u
is a positive solution to (1.3) on
\(B_{p}(2R)\)
such that:
Then on
\(B_{p}(R)\)
with
\(R>1\), the following inequality holds:
where
\(\beta=\max_{\{x|d(x,p)=1\}}\Delta_{f}r(x)\).
(1)
either
\(\nabla f \nabla(\log u)-a\log u-b\leq\delta|\nabla (\log u)|^{2}\)
for some
\(0\leq\delta<\frac{1}{2}\);
(2)
or
\(\nabla f \nabla(\log u)-a\log u-b\geq2 |\nabla(\log u)|^{2}\).
$$ \big|\nabla(\log u)\big|^{2}\leq\frac{C_{1}(n,\delta,\beta)}{R}+C_{2}(n, \delta)\max\bigl\{ a+(n-1)K,0\bigr\} , $$
(1.8)
Letting \(R\rightarrow\infty\) in (1.8), we obtain the following global estimates on complete noncompact Riemannian manifolds:
Corollary 1.4
Let
\((M, g,e^{-f}\,dv)\)
be an
n-dimensional complete smooth metric measure space with
\(\mathrm{Ric}_{f} \geq-(n-1)K\), where
\(K\geq0\)
is a constant. Let
u
be a positive solution to (1.3). Then under the assumption of either (1) or (2) as in Theorem
1.3, we have
$$ \begin{aligned} \big|\nabla(\log u)\big|^{2}\leq C(n, \delta)\max\bigl\{ a+(n-1)K,0\bigr\} . \end{aligned} $$
(1.9)
Clearly, if either \(u\leq e^{-(\frac{5}{4}+\frac{b}{a})}\) and \(a>0\), or \(u\geq e^{-(\frac{5}{4}+\frac{b}{a})}\) and \(a<0\), then we have \(\frac {4}{5}b+ a (1+\frac{4}{5}L )\leq0\). This gives the following result.
Corollary 1.5
Let
\((M, g,e^{-f}\,dv)\)
be an
n-dimensional complete smooth metric measure space with
\(\mathrm{Ric}_{f} \geq0\).
Remark 1.1
In particular, when \(a=0\), Eq. (1.3) becomes
and (1.6) becomes
In this case, on a complete smooth metric measure space \((M, g,e^{-f}\, dv)\) with \(\mathrm{Ric}_{f} \geq0\), there exists no bounded positive solution to (1.10) with \(b<0\). On the other hand, if \(a=b=0\), our Theorem 1.1 becomes Theorem 1 of Brighton in [8].
$$ \Delta_{f}u+bu=0 $$
(1.10)
$$ \begin{aligned} |\nabla u|^{2}\leq C A^{2} \biggl[\max \biggl\{ \frac{4}{5}b,0 \biggr\} +K \biggr]. \end{aligned} $$
(1.11)
Remark 1.2
Remark 1.3
Some related results for gradient estimates of positive solutions to
can be found in [9‐11]. Moreover, Qian in [10] used a different method to derive similar estimates to (1.12) with constant f. On the other hand, if we assume \(\mathrm{Ric}_{f} \geq -(n-1)K\) and \(|\nabla f|\leq\theta\), then from (1.1), we obtain
Hence, Theorem 1.5 in [11] follows from Theorem 1.1 of [11] immediately. However, our estimates in this paper are not dependent on \(|\nabla f|\).
$$ \Delta_{f}u+au\log u=0 $$
(1.12)
$$\begin{aligned}\mathrm{Ric}_{f}^{m}&= \mathrm{Ric}_{f}-\frac{1}{m-n}\, df\otimes df \\ &\geq-(n-1) \biggl(K+\frac{\theta^{2}}{(m-n)(n-1)} \biggr):=-(n-1)\tilde{K}. \end{aligned} $$
2 Proof of results
We firstly give the following lemma which plays an important role in the proof of main results.
Lemma 2.1
Let
u
be a positive solution to (1.3) with
\(u\leq A\)
and
\(\mathrm{Ric}_{f}\geq-(n-1)K\)
for some positive constant
K. Denote
\(\tilde{u}=u/A\)
and
\(h=\tilde {u}^{\epsilon}\)
for
\(\epsilon\in(0,1)\). If there exists one positive constant
δ
satisfying
then we have
where
$$ \frac{1}{n}+\frac{2(\epsilon-1)}{n\epsilon\delta}\geq0, $$
(2.1)
$$ \begin{aligned}[b] \frac{1}{2}\Delta_{f}| \nabla h|^{2}\geq{}& \biggl(\frac{(\epsilon-1)^{2}}{n\epsilon^{2}}-\frac{\epsilon -1}{\epsilon}+ \frac{2\delta(\epsilon-1)}{n\epsilon} \biggr)\frac{|\nabla h|^{4}}{h^{2}}+\frac{\epsilon-1}{\epsilon }\frac{\nabla h}{h} \nabla\bigl(|\nabla h|^{2}\bigr) \\ &-\bigl[a+\tilde{b}\epsilon+(n-1)K+a\tilde{L}\bigr]|\nabla h|^{2}, \end{aligned} $$
(2.2)
$$ \tilde{L}= \textstyle\begin{cases} \sup_{M}(\log h), &\textit{if } a\geq0, \\ \inf_{M}(\log h), &\textit{if } a< 0. \end{cases} $$
(2.3)
Proof
Under the scaling \(u\rightarrow\tilde{u}=u/A\), it follows from (1.3) that ũ satisfies
where the constant b̃ is given by \(\tilde{b}=b+a\log A\). Let \(h=\tilde{u}^{\epsilon}\), where \(\epsilon\in(0,1)\) is a constant to be determined. Then we have
Since \(0<\tilde{u}\leq1\), we have \(\log h\leq0\) and
which implies
Thus, under the assumption \(\mathrm{Ric}_{f}\geq-(n-1)K\), one has
For any fixed point p, if there exists a positive constant δ such that \(\nabla f\nabla h-ah\log h-\tilde{b}\epsilon h\leq \delta\frac{|\nabla h|^{2}}{h}\), then from (2.8), we can deduce
On the contrary, if \(\nabla f\nabla h-ah\log h-\tilde{b}\epsilon h\geq \delta\frac{|\nabla h|^{2}}{h}\) at the point p, then from (2.8), we can deduce
as long as (2.1) holds.
$$ \Delta_{f} \tilde{u}+a\tilde{u}\log \tilde{u}+\tilde{b} \tilde{u}=0, $$
(2.4)
$$ \log h=\epsilon\log\tilde{u}. $$
(2.5)
$$ \begin{aligned}[b] \Delta_{f} h&= \Delta_{f}\bigl(\tilde{u}^{\epsilon}\bigr)=\epsilon(\epsilon-1) \tilde {u}^{\epsilon-2}|\nabla \tilde{u}|^{2}+\epsilon \tilde{u}^{\epsilon-1}\Delta_{f} \tilde{u} \\ &=\epsilon(\epsilon-1)\tilde{u}^{\epsilon-2}|\nabla\tilde {u}|^{2}-a \epsilon\tilde{u}^{\epsilon}\log\tilde{u}-\tilde{b}\epsilon \tilde{u}^{\epsilon}\\ &=\frac{\epsilon-1}{\epsilon} \frac{|\nabla h|^{2}}{h}-ah\log h-\tilde {b}\epsilon h, \end{aligned} $$
(2.6)
$$ \begin{aligned}[b] \nabla h\nabla\Delta_{f} h&= \nabla h\nabla \biggl(\frac{\epsilon-1}{\epsilon} \frac{|\nabla h|^{2}}{h}-ah\log h-\tilde{b} \epsilon h \biggr) \\ &=\frac{\epsilon-1}{\epsilon}\nabla h\nabla\frac{|\nabla h|^{2}}{h}-a\nabla h\nabla(h\log h)- \tilde{b}\epsilon|\nabla h|^{2} \\ &=\frac{\epsilon-1}{\epsilon}\frac{\nabla h}{h}\nabla\bigl(|\nabla h|^{2} \bigr)-\frac{\epsilon-1}{\epsilon}\frac{|\nabla h|^{4}}{h^{2}} \\ &\quad{}-ah\log h\frac{|\nabla h|^{2}}{h}-(a+\tilde{b}\epsilon)|\nabla h|^{2}. \end{aligned} $$
(2.7)
$$ \begin{aligned}[b] \frac{1}{2}\Delta_{f}| \nabla h|^{2}&=|\nabla^{2}h|^{2}+\nabla h\nabla \Delta_{f} h+\mathrm{Ric}_{f} (\nabla h,\nabla h) \\ &\geq\frac{1}{n}(\Delta h)^{2}+\nabla h\nabla \Delta_{f} h-(n-1)K|\nabla h|^{2} \\ &=\frac{1}{n} \biggl(\frac{\epsilon-1}{\epsilon} \frac{|\nabla h|^{2}}{h}+\nabla f\nabla h-ah\log h-\tilde{b}\epsilon h \biggr)^{2}+\frac{\epsilon-1}{\epsilon } \frac{\nabla h}{h}\nabla \bigl(|\nabla h|^{2}\bigr) \\ &\quad{}-\frac{\epsilon-1}{\epsilon}\frac{|\nabla h|^{4}}{h^{2}} -(ah\log h)\frac{|\nabla h|^{2}}{h}-\bigl[a+ \tilde{b}\epsilon+(n-1)K\bigr]|\nabla h|^{2} \\ &= \biggl(\frac{(\epsilon-1)^{2}}{n\epsilon^{2}}-\frac{\epsilon-1}{\epsilon } \biggr)\frac{|\nabla h|^{4}}{h^{2}} + \frac{2(\epsilon-1)}{n\epsilon}\frac{|\nabla h|^{2}}{h} (\nabla f\nabla h-ah\log h-\tilde{b}\epsilon h ) \\ &\quad{}+\frac{1}{n} (\nabla f\nabla h-ah\log h-\tilde{b}\epsilon h )^{2}+\frac{\epsilon-1}{\epsilon }\frac{\nabla h}{h}\nabla\bigl(|\nabla h|^{2}\bigr) \\ &\quad{}-\bigl[a+\tilde{b}\epsilon+(n-1)K+a\log h\bigr]|\nabla h|^{2}. \end{aligned} $$
(2.8)
$$\begin{aligned} \frac{1}{2}\Delta_{f}| \nabla h|^{2}\geq{}& \biggl(\frac{(\epsilon-1)^{2}}{n\epsilon ^{2}}-\frac{\epsilon-1}{\epsilon} \biggr) \frac{|\nabla h|^{4}}{h^{2}} +\frac{2(\epsilon-1)}{n\epsilon}\frac{|\nabla h|^{2}}{h} \biggl(\delta \frac {|\nabla h|^{2}}{h} \biggr) \\ &+\frac{1}{n} (\nabla f\nabla h-ah\log h-\tilde{b}\epsilon h )^{2}+\frac{\epsilon-1}{\epsilon }\frac{\nabla h}{h}\nabla\bigl(|\nabla h|^{2}\bigr) \\ &-\bigl[a+\tilde{b}\epsilon+(n-1)K+a\log h\bigr]|\nabla h|^{2} \\ \geq{}& \biggl(\frac{(\epsilon-1)^{2}}{n\epsilon^{2}}-\frac{\epsilon-1}{\epsilon }+\frac{2\delta(\epsilon-1)}{n\epsilon} \biggr) \frac{|\nabla h|^{4}}{h^{2}}+\frac{\epsilon-1}{\epsilon }\frac{\nabla h}{h}\nabla\bigl(|\nabla h|^{2}\bigr) \\ &-\bigl[a+\tilde{b}\epsilon+(n-1)K+a\tilde{L}\bigr]|\nabla h|^{2}. \end{aligned}$$
(2.9)
$$ \begin{aligned}[b] \frac{1}{2}\Delta_{f}| \nabla h|^{2}\geq{}& \biggl(\frac{(\epsilon-1)^{2}}{n\epsilon ^{2}}-\frac{\epsilon-1}{\epsilon} \biggr) \frac{|\nabla h|^{4}}{h^{2}} +\frac{2(\epsilon-1)}{n\epsilon\delta} (\nabla f\nabla h-ah\log h-\tilde{b} \epsilon h )^{2} \\ &+\frac{1}{n} (\nabla f\nabla h-ah\log h-\tilde{b}\epsilon h )^{2}+\frac{\epsilon-1}{\epsilon }\frac{\nabla h}{h}\nabla\bigl(|\nabla h|^{2}\bigr) \\ &-\bigl[a+\tilde{b}\epsilon+(n-1)K+a\log h\bigr]|\nabla h|^{2} \\ ={}& \biggl(\frac{(\epsilon-1)^{2}}{n\epsilon^{2}}-\frac{\epsilon-1}{\epsilon } \biggr)\frac{|\nabla h|^{4}}{h^{2}} + \biggl(\frac{1}{n}+\frac{2(\epsilon-1)}{n\epsilon\delta} \biggr) (\nabla f\nabla h-ah\log h- \tilde{b}\epsilon h )^{2} \\ &+\frac{\epsilon-1}{\epsilon }\frac{\nabla h}{h}\nabla\bigl(|\nabla h|^{2} \bigr)-\bigl[a+\tilde{b}\epsilon +(n-1)K+a\log h\bigr]|\nabla h|^{2} \\ \geq{}& \biggl[\frac{(\epsilon-1)^{2}}{n\epsilon^{2}}-\frac{\epsilon-1}{\epsilon} + \biggl(\frac{1}{n}+ \frac{2(\epsilon-1)}{n\epsilon\delta} \biggr)\delta^{2} \biggr]\frac{|\nabla h|^{4}}{h^{2}}+ \frac{\epsilon-1}{\epsilon }\frac{\nabla h}{h}\nabla\bigl(|\nabla h|^{2}\bigr) \\ &-\bigl[a+\tilde{b}\epsilon+(n-1)K+a\log h\bigr]|\nabla h|^{2} \\ \geq{}& \biggl(\frac{(\epsilon-1)^{2}}{n\epsilon^{2}}-\frac{\epsilon-1}{\epsilon }+\frac{2\delta(\epsilon-1)}{n\epsilon} \biggr) \frac{|\nabla h|^{4}}{h^{2}}+\frac{\epsilon-1}{\epsilon }\frac{\nabla h}{h}\nabla\bigl(|\nabla h|^{2}\bigr) \\ &-\bigl[a+\tilde{b}\epsilon+(n-1)K+a\tilde{L}\bigr]|\nabla h|^{2} \end{aligned} $$
(2.10)
2.1 Proof of Theorem 1.1
In order to obtain the upper bound of \(|\nabla h|\) by using the maximum principle for (2.2), we need to choose ϵ, δ such that the coefficient of \(\frac{|\nabla h|^{4}}{h^{2}}\) in (2.2) is positive. That is, we need
In particular, by choosing \(\epsilon=\frac{4}{5}\) and letting \(\delta\rightarrow\frac{1}{2}\), we find that the inequality (2.1) holds and (2.2) becomes
$$ \frac{(\epsilon-1)^{2}}{n\epsilon^{2}}-\frac{\epsilon-1}{\epsilon}+\frac {2\delta(\epsilon-1)}{n\epsilon}>0. $$
(2.11)
$$ \begin{aligned}[b] \frac{1}{2}\Delta_{f}| \nabla h|^{2}\geq{}&\frac{4n-3}{16n}\frac{|\nabla h|^{4}}{h^{2}}- \frac{1}{4}\frac {\nabla h}{h}\nabla\bigl(|\nabla h|^{2}\bigr) \\ &-\bigl[a+\tilde{b}\epsilon+(n-1)K+a\tilde{L}\bigr]|\nabla h|^{2}. \end{aligned} $$
(2.12)
As in [8], we define a cut-off function \(\psi\in C^{2}([0,+\infty))\) by
satisfying \(\psi(t)\in[0,1]\) and
where C is a positive constant. Let
Using Eq. (2.19) in [8] (see Eq. (4.5) in [5] or [12, Theorem 3.1]), we obtain
and
Denote by \(B_{p}(R)\) the geodesic ball centered at p with radius R. Let \(G=\phi|\nabla h|^{2}\). Assume G achieves its maximum at the point \(x_{0}\in B_{p}(2R)\) and assume \(G(x_{0})>0\) (otherwise the proof is trivial). Then, at the point \(x_{0}\),
and
where in the second inequality, we used (2.12). Multiplying both sides of (2.17) by \(\frac{\phi}{G}\), we obtain
Substituting the Cauchy inequality
into (2.18) gives
where \(C_{1}\), \(C_{2}\) are two positive constants depending on n. Hence, on \(B_{p}(R)\) with \(R>1\), it follows from (2.19) that
In particular, the estimate (2.20) gives
which finishes the proof of Theorem 1.1.
$$ \psi(t)= \textstyle\begin{cases} 1, & t\in[0,R];\\ 0, & t\in[2R,+\infty], \end{cases} $$
(2.13)
$$ -\frac{C}{R}\leq\frac{\psi'(t)}{\sqrt{\psi}}\leq0,\qquad \big| \psi''(t)\big|\leq\frac {C}{R^{2}}, $$
(2.14)
$$\phi=\psi\bigl(d(x,p)\bigr). $$
$$ \Delta_{f}\phi\geq-\frac{C\beta}{R}- \frac{C(n-1)K(2R-1)}{R}-\frac{C}{R^{2}} $$
(2.15)
$$ \frac{|\nabla\phi|^{2}}{\phi}\leq\frac{C}{R^{2}}. $$
(2.16)
$$\Delta_{f} G\leq0,\qquad \nabla\bigl(|\nabla h|^{2}\bigr)=- \frac{|\nabla h|^{2}}{\phi} \nabla\phi $$
$$ \begin{aligned} [b]0\geq{}& \Delta_{f}G \\ ={}&\phi\Delta_{f}\bigl(|\nabla h|^{2}\bigr)+|\nabla h|^{2}\Delta_{f}\phi+2\nabla\phi\nabla |\nabla h|^{2} \\ ={}&\phi\Delta_{f}\bigl(|\nabla h|^{2}\bigr)+ \frac{\Delta_{f}\phi}{\phi}G -2\frac{|\nabla\phi|^{2}}{\phi^{2}}G \\ \geq{}&\frac{\Delta_{f}\phi}{\phi}G -2\frac{|\nabla\phi|^{2}}{\phi^{2}}G+2\phi \biggl[\frac{4n-3}{16n} \frac{|\nabla h|^{4}}{h^{2}}-\frac{1}{4}\frac{\nabla h}{h}\nabla\bigl(|\nabla h|^{2}\bigr) \\ &-\bigl[a+\tilde{b}\epsilon+(n-1)K+a\tilde{L}\bigr]|\nabla h|^{2} \biggr] \\ ={}&\frac{\Delta_{f}\phi}{\phi}G -2\frac{|\nabla\phi|^{2}}{\phi^{2}}G+\frac {4n-3}{8n} \frac{G^{2}}{\phi h^{2}}+\frac{G}{2\phi}\nabla\phi\frac{\nabla h}{h} \\ &-2\bigl[a+\tilde{b}\epsilon+(n-1)K+a\tilde{L}\bigr]G, \end{aligned} $$
(2.17)
$$ \begin{aligned}[b] \frac{4n-3}{8n}\frac{G}{h^{2}} \leq{}&{-}\frac{1}{2}\nabla\phi\frac{\nabla h}{h}+2\bigl[a+\tilde{b} \epsilon+(n-1)K+a\tilde{L}\bigr]\phi \\ &-\Delta_{f}\phi+2\frac{|\nabla\phi|^{2}}{\phi}. \end{aligned} $$
(2.18)
$$\begin{aligned}-\frac{1}{2}\nabla\phi\frac{\nabla h}{h}\leq{}& \frac{1}{2}|\nabla\phi|\frac{|\nabla h|}{h} \\ \leq{}&\frac{n}{4n-3}\frac{|\nabla\phi|^{2}}{\phi} +\frac{4n-3}{16n}\phi \frac{|\nabla h|^{2}}{h^{2}} \\ ={}&\frac{n}{4n-3}\frac{|\nabla\phi|^{2}}{\phi} +\frac{4n-3}{16n}\frac{G}{h^{2}} \end{aligned} $$
$$ \begin{aligned}[b] \frac{4n-3}{16n}\frac{G}{h^{2}} \leq{}&2\bigl[a+\tilde{b}\epsilon+(n-1)K+a\tilde {L}\bigr]\phi-\Delta_{f} \phi +\frac{9n-6}{4n-3}\frac{|\nabla\phi|^{2}}{\phi} \\ \leq{}&2\bigl[a+\tilde{b}\epsilon+(n-1)K+a\tilde{L}\bigr]+\frac {C_{1}[(n-1)K(2R-1)+\beta]}{R}+ \frac{C_{2}}{R^{2}}, \end{aligned} $$
(2.19)
$$ \begin{aligned}[b] \frac{4n-3}{16n}G(x)\leq{}& \frac{4n-3}{16n}G(x_{0}) \\ \leq{}&h^{2}(x_{0}) \biggl[2\bigl[a+\tilde{b} \epsilon+(n-1)K+a\tilde{L}\bigr] \\ &+\frac{C_{1}[(n-1)K(2R-1)+\beta]}{R}+\frac{C_{2}}{R^{2}} \biggr]. \end{aligned} $$
(2.20)
$$ \begin{aligned} |\nabla u|^{2}\leq{}&C A^{2} \biggl[\max \biggl\{ \frac{4}{5}b+a \biggl(1+ \frac {4}{5}L \biggr),0 \biggr\} +K+\frac{|\beta|+1}{R} \biggr], \end{aligned} $$
(2.21)
2.2 Proof of Theorem 1.3
We define \(\tilde{h}=\log u\). Then we have
where, in the last equality of (2.22), we used Eq. (1.3). Using the Bochner formula with respect to the f-Laplacian, we have
Moreover, by virtue of (2.22), we have
If the assumption (1) holds, then (2.24) yields
On the other hand, if the assumption (2) holds, then (2.24) shows
Therefore, in these two cases, we have
and (2.23) gives
Following the proof of Theorem 1.1 line by line, we obtain on \(B_{p}(R)\) with \(R>1\),
where δ is taken to zero in the second assumption.
$$ \begin{aligned}[b] \Delta\tilde{h}-\nabla f\nabla \tilde{h}&=\Delta_{f}\tilde{h} \\ &=\frac{\Delta_{f}u}{u}-\big|\nabla(\log u)\big|^{2} \\ &=-|\nabla\tilde{h}|^{2}-a\tilde{h}-b, \end{aligned} $$
(2.22)
$$ \begin{aligned} [b] \frac{1}{2}\Delta_{f}| \nabla\tilde{h}|^{2}={}&\big|\nabla^{2}\tilde{h}\big|^{2}+ \nabla \tilde{h}\nabla\Delta_{f} \tilde{h}+\mathrm{Ric}_{f} (\nabla\tilde {h},\nabla \tilde{h}) \\ \geq{}&\frac{1}{n}(\Delta\tilde{h})^{2}+\nabla \tilde{h}\nabla \Delta_{f} \tilde{h}-(n-1)K|\nabla\tilde{h}|^{2}. \end{aligned} $$
(2.23)
$$ \begin{aligned}[b] (\Delta\tilde{h})^{2}={}& \bigl({-}|\nabla\tilde{h}|^{2}+\nabla f\nabla\tilde {h}-a\tilde{h}-b \bigr)^{2} \\ ={}&|\nabla\tilde{h}|^{4}-2|\nabla\tilde{h}|^{2}(\nabla f \nabla\tilde {h}-a\tilde{h}-b)+(\nabla f\nabla\tilde{h}-a\tilde{h}-b)^{2}. \end{aligned} $$
(2.24)
$$ \begin{aligned} [b](\Delta\tilde{h})^{2}\geq{}&| \nabla\tilde{h}|^{4}-2\delta|\nabla\tilde {h}|^{4}+(\nabla f \nabla\tilde{h}-a\tilde{h}-b)^{2} \\ \geq{}&(1-2\delta)|\nabla\tilde{h}|^{4}. \end{aligned} $$
(2.25)
$$ \begin{aligned}[b] (\Delta\tilde{h})^{2}\geq{}&| \nabla\tilde{h}|^{4}-(\nabla f\nabla\tilde {h}-a\tilde{h}-b)^{2}+( \nabla f\nabla\tilde{h}-a\tilde{h}-b)^{2} \\ ={}&|\nabla\tilde{h}|^{4} \\ \geq{}&(1-2\delta)|\nabla\tilde{h}|^{4}. \end{aligned} $$
(2.26)
$$ (\Delta\tilde{h})^{2}\geq(1-2\delta)|\nabla \tilde{h}|^{4}, $$
(2.27)
$$ \begin{aligned} \frac{1}{2}\Delta_{f}| \nabla\tilde{h}|^{2}\geq{}&\frac{1-2\delta}{n}|\nabla \tilde{h}|^{4}- \nabla \tilde{h}\nabla\bigl(|\nabla\tilde{h}|^{2}\bigr)-\bigl[a+(n-1)K \bigr]|\nabla\tilde{h}|^{2}. \end{aligned} $$
(2.28)
$$ |\nabla\tilde{h}|^{2}\leq\frac{C_{1}(n,\delta,\beta)}{R}+C_{2}(n, \delta)\max \bigl\{ a+(n-1)K,0\bigr\} , $$
(2.29)
We completed the proof of Theorem 1.3.
Acknowledgements
The authors want to thank the referee for helpful suggestions, which made the paper more readable. The research of the author is supported by NSFC No. 11401179.
Competing interests
The authors declare that they have no competing interests.
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