¹Faculté de Sciences et Techniques, Université de Zinder, Zinder, Niger

²Département de Mathématiques et Informatique, Université Abdou Moumouni, Niamey, Niger

Copyright © 2017 by authors and Scientific Research Publishing Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

Received: August 24, 2016; Accepted: October 14, 2016; Published: October 17, 2016

The original online version of this article (Issa Allassane Kaboye, Bazanfaré Mahaman (2016) Manifolds with Bakry-Emery Ricci Curvature bounded below 6, 754-764. http://dx.doi.org/10.4236/apm.2016.611061 unfortunately contains a mistake. The author wishes to correct the errors.

Lemma 3.5. Let (M, g, e^−fdvolg) be a complete smooth metric measure space with $Ri{c}_f\ge 0$ ; fix $p\in M$ ; if there exists c so that $\left|f\left(x\right)\right|\le c$ then for $R\ge r>0$

$\frac{Vo{l}_f\left(B\left(p,R\right)\right)}{Vo{l}_f\left(B\left(p,r\right)\right)}\le e^{4c}{\left(\frac{R}{r}\right)}^n$

Proof

Let x be a point in M and let $\gamma :\left[0,r\right]\to M$ be a minimal geodesic joining p to x and ${\left(e_i\left(t\right)\right)}_{1\le i\le n-1}$ be a parallel orthonormal vector fields along $\gamma$ . Set $Y_i\left(t\right)=\frac{t}{r}{e}_i\left(t\right)$ .

By the second variation formula we have:

$\begin{array}{c}m\left(r\right)=\Delta r\le {\displaystyle \sum _{i=1}^{n-1}I\left(Y_i,Y_i\right)}\\ ={\displaystyle {\int }_0^r\left({\displaystyle \sum _{i=1}^{n-1}{\Vert {Y^{\prime }}_i\left(t\right)\Vert }^2-\langle R\left(Y_i\left(t\right),{\gamma }^{\prime }\left(t\right)\right){\gamma }^{\prime }\left(t\right),Y_i\left(t\right)\rangle }\right)\text{d}t}\\ \le \frac{1}{r^2}{\displaystyle {\int }_0^r\left(n-1-t^{2}Ric(\gamma \text{'}(t))\right)}dt\end{array}$

$\frac{n-1}{r}+{\displaystyle {\int }_0^r\frac{t^2}{r^2}Hess\left(f\right)\left({\gamma }^{\prime },{\gamma }^{\prime }\right)\text{d}t}$

$\begin{array}{l}\frac{n-1}{r}+{\displaystyle {\int }_0^r\frac{t^2}{r^2}{\left(f\circ \gamma \right)}^{\prime \text{}\prime }\text{d}t}\\ =\frac{n-1}{r}+\frac{1}{r^2}{\displaystyle {\int }_0^r\frac{\text{d}}{\text{d}t}}\left(t^2{\left(f\circ \gamma \right)}^{\prime }\left(t\right)\right)\text{d}t-\frac{2}{r^2}{\displaystyle {\int }_0^{r}t{\left(f\circ \gamma \right)}^{\prime }\left(t\right)\text{d}t}\\ =\frac{n-1}{r}+{\partial }_{r}f-\frac{2}{r}f\left(x\right)+\frac{2}{r^2}{\displaystyle {\int }_0^r\left(f\circ \gamma \right)\left(t\right)\text{d}t}\end{array}$

Hence

$\begin{array}{c}{m}_f\left(r\right)=\Delta r-{\partial }_{r}f=\frac{\partial }{\partial r}\left(\ln \left(A_f\left(r,\theta \right)\right)\right)\\ \le \frac{n-1}{r}+{\partial }_{r}f-\frac{2}{r}f\left(x\right)+\frac{2}{r^2}{\displaystyle {\int }_0^r\left(f\circ \gamma \right)\left(t\right)\text{d}t}\end{array}$

For all positive reals r and s, integrating this relation we have:

$\begin{array}{c}{\displaystyle {\int }_r^s{m}_f\left(t\right)\text{d}t}=\ln \left(\frac{A_f\left(s,\theta \right)}{A_f\left(r,\theta \right)}\right)\le {\left(\frac{s}{r}\right)}^{n-1}+2{\displaystyle {\int }_r^s\left(\frac{1}{t^2}{\displaystyle {\int }_0^{t}f\left(u\right)\text{d}u}-\frac{1}{t}f\left(t\right)\right)\text{d}t}\\ =\ln {\left(\frac{s}{r}\right)}^{n-1}-{\left(\frac{2}{t}{\displaystyle {\int }_0^{t}f\text{d}t}\right)|}_r^s+2{\displaystyle {\int }_r^s\frac{1}{t}}f\text{d}t-2{\displaystyle {\int }_r^s\frac{1}{t}}f\text{d}t\\ =\ln {\left(\frac{s}{r}\right)}^{n-1}-{\left(\frac{2}{t}{\displaystyle {\int }_0^{t}f\text{d}t}\right)|}_r^s\le \ln {\left(\frac{s}{r}\right)}^{n-1}+4c\end{array}$

Therefore we have $r^{n-1}{A}_f\left(s,\theta \right)\le e^{4c}{A}_f\left(r,\theta \right)s^{n-1}$ . Hence

${\displaystyle {\int }_0^R{\displaystyle {\int }_{S^{n-1}}{r}^{n-1}{A}_f\left(s,\theta \right)\text{d}\theta \text{d}r}}\le e^{4c}{\displaystyle {\int }_0^R{\displaystyle {\int }_{S^{n-1}}{A}_f\left(r,\theta \right)\text{d}\theta \text{d}r}}$

which implies

$\frac{R^n}{n}{\displaystyle {\int }_{S^{n-1}}{A}_f\left(s,\theta \right)\text{d}\theta }\le e^{4c}{s}^{n-1}{\displaystyle {\int }_0^R{\displaystyle {\int }_{S^{n-1}}{A}_f\left(r,\theta \right)\text{d}\theta \text{d}r}}=e^{4c}{s}^{n-1}vo{l}_f\left(B\left(p,R\right)\right)$

and integrating from 0 to R' with respect to s we obtain the conclusion.