f -Minimal Surface and Manifold with Positive m-bakry Émery Ricci Curvature

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1 J Geom Anal 2015) 25: DOI /s f -inimal Surface and anifold with Positive m-bakry Émery Ricci Curvature Haizhong Li Yong Wei Received: 1 ovember 2012 / Published online: 16 August 2013 athematica Josephina, Inc Abstract In this paper, we first prove a compactness theorem for the space of closed embedded f -minimal surfaces of fixed topology in a closed three-manifold with positive Bakry Émery Ricci curvature. Then we give a Lichnerowicz type lower bound of the first eigenvalue of the f -Laplacian on a compact manifold with positive m-bakry Émery Ricci curvature, and prove that the lower bound is achieved only if the manifold is isometric to the n-sphere, or the n-dimensional hemisphere. Finally, for a compact manifold with positive m-bakry Émery Ricci curvature and f -mean convex boundary, we prove an upper bound for the distance function to the boundary, and the upper bound is achieved if and only if the manifold is isometric to a Euclidean ball. Keywords f -ean curvature f -inimal m-bakry Émery Ricci curvature Eigenvalue estimate athematics Subject Classification 2010) 53C42 53C21 Communicated by Ben Andrews. The research of the authors was supported by SFC o H. Li Department of mathematical sciences, and athematical Sciences Center, Tsinghua University, , Beijing, P.R. China hli@math.tsinghua.edu.cn Y. Wei B) Department of mathematical sciences, Tsinghua University, , Beijing, P.R. China wei-y09@mails.tsinghua.edu.cn

2 422 H. Li, Y. Wei 1 Introduction Let n,g) be a smooth Riemannian manifold and f be a smooth function on. We denote by,, and 2 the gradient, Laplacian, and Hessian operator on with respect to g, respectively. In this paper, by the Bakry Émery Ricci curvature we mean Ric f = Ric + 2 f, 1) which is also called -Bakry Émery Ricci curvature, i.e., the m = case of the m-bakry Émery Ricci curvature [3] defined by Ric m f = Ric f 1 m n f f m>n). 2) When m = n, weletf be constant and define Ric m f = Ric. The equation Ric f = κg for some constant κ is just the gradient Ricci soliton equation, which plays an important role in the study of Ricci flow see [6]). The equation Ric m f = κg corresponds to the quasi-einstein equation cf. [7]), which has been studied by many authors. Denote by dv the Riemannian volume form on with respect to g; then n,g,e f dv) is often called a smooth metric measure space. We refer interested readers to [32] for further motivation and examples of metric measure spaces. Let be a hypersurface in and ν the outer unit normal vector to. Define the second fundamental form of by hx, Y ) = X ν,y for any two tangent vector fields X and Y on, and the mean curvature by H = trh). Thef -mean curvature see [32, p. 398]) at a point x with respect to ν is given by H f x) = Hx) fx),νx). 3) is called an f -minimal hypersurface in if its f -mean curvature H f vanishes everywhere. The most well-known example of a metric measure space is the Gaussian soliton, R n,g 0,e 1 4 x 2 dv), where g 0 is the standard Euclidean metric on R n. The Gaussian soliton satisfies Ric f = 1 2 g 0.Thef -minimal hypersurface in the Gaussian soliton is the self-shrinker n 1 R n which satisfies H = 1 2 x,ν. Self-shrinkers play an important role in the mean curvature flow, as they correspond to the self-similar solution to the mean curvature flow, and also describe all possible blow-ups at a given singularity. In [11] and [12], Colding inicozzi and Ding Xin considered the compactness property for the space of self-shrinkers in R 3. In this paper, we prove the following compactness theorem for f -minimal surface, which is a generalization of the classical compactness of minimal surfaces in a closed three manifold with positive Ricci curvature by Choi and Schoen [8].

3 f -inimal Surface and anifold with Positive Ric m f 423 Theorem 1 Let 3,g,e f dv) be a closed metric measure space with positive Bakry Émery Ricci curvature. Then the space of closed embedded f -minimal surfaces of fixed topological type in is compact in the C k topology for any k 2. Here closed means compact and without boundary. ote that when f is a constant function, we get the classical Choi Schoen s theorem see [8, Theorem 1]). We remark that our approach in Sect. 3 to prove Theorem 1 also works for the positive m-bakry Émery Ricci curvature case, with some slight adjustments of the Bochner formula and Reilly formula; see 8) and 9). But since Ric f Ric m f, we get no extension results of Theorem 1. Recently, Ailana Fraser and artin Li [17] proved a compactness theorem for the space of embedded minimal surfaces with free boundary in a three-manifold with non-negative Ricci curvature and convex boundary. So it is also interesting to get an analogue result of Theorem 1 for f -minimal surfaces with free boundary case. In our proof of Theorem 1, one of the key ingredients is the observation that an f -minimal hypersurface is a minimal hypersurface in with the conformal changed metric g = e n 1 2 f g. This can be easily seen from the first variation formula of the volume. We will use this observation in the next section to get the singular compactness result. However, Theorem 1 cannot directly follow from Choi Schoen s compactness theorem for minimal surfaces in a three-manifold with positive Ricci curvature. In fact, the Ricci curvature of the conformal changed metric g may not have a sign: Recall that for n 3, the scalar curvature R of the conformal changed metric g = e n 1 2 f g is given by cf. [31]) R = e n 1 2 f n 2 ) n 1 f f + R, where R is the scalar curvature of, g). Although the positive Bakry Émery Ricci curvature assumption implies that R + f > 0, we cannot conclude that the scalar curvature and then the Ricci curvature of the conformal metric have a sign. For example, the Gaussian soliton R n,g 0,e 1 4 x 2 dv) has positive Bakry Émery Ricci curvature, while the scalar curvature R of the conformal changed metric g = e 2n 1) 1 x 2 g 0 on R n is R = e 1 2n 1) x 2 n n 2 4n 1) x 2 which is positive when x is small and becomes negative when x is large. Therefore, the Ricci curvature of g does not have a sign. Our proof follows from the standard argument in Choi Schoen s paper: We first need a first eigenvalue estimate of the f -Laplacian f = f for f -minimal surfaces in a manifold with positive Bakry Émery Ricci curvature. This is also one of the key ingredients in the proof of Theorem 1, and was proved by Li a and Sheng-Hua Du [24] recently. In Sect. 3, we show that a Du s result holds under a weaker condition, for example, the orientability assumption is not necessary. Then by considering an f -minimal surface as a minimal surface in, g), and combining ),

4 424 H. Li, Y. Wei with Yang Yau s inequality [33], we get an a priori upper bound on the weighted area of the f -minimal surface in terms of the topology. This together with the Gauss equation and Gauss Bonnet theorem gives an upper bound for the total curvature of an f -minimal surface. Finally, by the singular compactness proposition and a contradiction argument, we get the smooth compactness Theorem 1. As a corollary of Theorem 1, we get the following curvature estimates. The proof is by using a contradiction argument as in Choi Schoen s paper [8]. Corollary 2 Let 3,g,e f dv) be a closed metric measure space with positive Bakry Émery Ricci curvature. There exists a constant C depending only on and an integer χ such that if is a closed embedded f -minimal surface of Euler characteristic χ in, then max h C, where h is the norm of the second fundamental form of. ext, in Sect. 4, we will use the Reilly formula to give a Lichnerowicz type lower bound for the first eigenvalue of the f -Laplacian on a compact manifold with positive m-bakry Émery Ricci curvature. The classical Lichnerowicz theorem [23] says that for an n-dimensional closed Riemannian manifold with Ricci curvature Ric n 1)K > 0, the first eigenvalue of the Laplacian on satisfies λ 1 ) nk. Obata [26] then proved that the equality holds only when is isometric to the n- sphere of radius 1/ K. This was generalized by Reilly [28] to a compact manifold with mean-convex boundary for the Dirichlet problem, and by Escobar [14] to a compact manifold with convex boundary for the eumann problem. The same lower bound for λ 1 ) holds and the equality holds when is isometric to the n- dimensional hemisphere of radius 1/ K. The following theorem shows that a similar result also holds for a manifold with positive m-bakry Émery Ricci curvature and with some suitable boundary condition. We remark that the result λ 1 mk in Theorem 3 was essentially proved in [24], just with some slightly different expressions. Our contribution is the rigidity result when the equality λ 1 = mk holds. Theorem 3 Let n,g) be an n-dimensional Riemannian manifold possibly with boundary ) and f be a smooth function on. Assume that the m-bakry Émery Ricci curvature satisfies Ric m f m 1)K > 0. Furthermore, if the boundary is nonempty, for the Dirichlet problem we assume the f -mean curvature on is nonnegative; for the eumann problem we assume the boundary is weakly convex, i.e., the second fundamental form h 0 on. Then the first eigenvalue λ 1 of the f -Laplacian on satisfies λ 1 mk. 4) oreover, equality is attained only when m = n, f is constant, and Ric m f = Ric. In this case, if has no boundary, then is the n-sphere of radius 1/ K; if has nonempty boundary, then is the n-dimensional hemisphere of radius 1/ K.

5 f -inimal Surface and anifold with Positive Ric m f 425 One can compare Theorem 3 with Bakry Qian s eigenvalue comparison results in [4], which states that the first eigenvalue λ 1 with eumann boundary condition when the boundary is nonempty) of the f -Laplacian is bounded from below by the first eigenvalue of a one-dimensional model. See also [2] and [18] for more recent results about the first eigenvalue of the f -Laplacian on manifolds with positive Bakry Émery curvature. We remark that when Ric m f m 1)K > 0, n,g) is automatically compact and the diameter of satisfies diam) π/ K;see[27, Theorem 5]. So we do not need to assume is compact in Theorem 3. In[30], Ruan proved that when diam) is equal to π/ K, then, g) is isometric to the n-sphere of radius 1/ K. In Sect. 5, we will prove a similar result for manifolds with nonnegative m-bakry Émery Ricci curvature and f -mean convex boundary i.e., the f -mean curvature on is positive). Theorem 4 Let n,g) be an n-dimensional complete Riemannian manifold with nonempty boundary and f be a smooth function on. Assume that the m-bakry Émery Ricci curvature is nonnegative on, and the f -mean curvature of the boundary satisfies H f m 1)K > 0 for some constant K>0. Let d denote the distance function on. Then sup dx, ) 1 x K. 5) oreover, if we assume that is compact, then is also compact and equality holds in 5) only when is isometric to an n-dimensional Euclidean ball of radius 1/K. Theorem 4 is an analogue result of Theorem 1.1 in [22], where the manifold with nonnegative Ricci curvature and with mean convex boundary was considered. Our proof of Theorem 4 follows the arguments in [22], with some adjustments. As in [22], we also conjecture that the uniform boundary convexity could make be compact and hence would also be compact. 2 Reilly Formula on a etric easure Space In this section, we first exhibit the Reilly formulas on a metric measure space, which are the important tools to prove our main theorems. Let n,g,e f dv) be a compact metric measure space with boundary.the f -Laplacian f = f on is self-adjoint with respect to the weighted measure e f dv. A simple calculation gives the following Bochner formula see [24, 25, 32]) for any function u C 3 ): 1 2 f u 2 = 2 u 2 + Ric f u, u) + g u, f u). 6) Using the Bochner formula 6) and integration by parts, Li a and Sheng-Hua Du [24] obtained the following Reilly formula:

6 426 H. Li, Y. Wei 0 = Ricf u, u) f u u 2 ) e f dv + f u + H f u ) u ν ν u, u ) + h u, u) e f dμ. 7) ν Here, Ric f is the Bakry Émery Ricci tensor of ; dv and dμ are volume forms on and, respectively. f,, and 2 are the f -Laplacian, gradient, and Hessian on, respectively; f = f and are the f -Laplacian and gradient operators on ; ν is the unit outward normal of ; H f and h are the f -mean curvature and second fundamental form of in with respect to ν, respectively. The Bochner formula 6) looks similar to the classic Bochner formula. However, we have a difficulty that tr 2 u) f u. One way to deal with this is to consider the Bochner formula for m-bakry Émery Ricci curvature. When m>n,letz = m n, and by a basic algebraic inequality a + b) 2 a2 z z 1 b2 for z>1, we have see [21]) 2 u 2 1 n u) 2 = 1 n f u + f u) 2 1 n n m f u) 2 n ) m n f u) 2 = 1 m f u) 2 1 m n f u) 2. Substituting this into 6), 7) and using the definition 2) ofm-bakry Émery Ricci curvature, we get and f u 2 1 m f u) 2 + Ric m f u, u) + g u, f u) 8) Ric m f u, u) m 1 f u + H f u ν ) m f u 2 ) u ν e f dv u, u ) + h u, u) e f dμ. 9) ν ote that the Bochner formula 8) looks very similar to the Bochner formula for the Ricci tensor of an m-dimensional manifold. This seems to be Bakry Émery s motivation [3] for the definition of the m-bakry Émery Ricci tensor and for their more general curvature dimension inequalities for diffusion operators. 3 The Space of f -inimal Surfaces In this section, we assume that n,g,e f dv) is a closed metric measure space with positive Bakry Émery Ricci curvature Ric f. We will prove Theorem 1. First, we need the following lemma of Frankel type, which was stated in G. Wei and W. Wylie s

7 f -inimal Surface and anifold with Positive Ric m f 427 paper see Theorem 7.4 in [32]). Here we give an alternative proof using the Reilly formula 7). Lemma 5 Let n,g,e f dv) be a closed metric measure space with positive Ric f. Then any two closed embedded f -minimal hypersurfaces Σ 1 and Σ 2 in must intersect, i.e., Σ 1 Σ 2 Ø. So that any closed embedded f -minimal hypersurface in is connected. Proof The proof is motivated by Fraser Li s paper [17]. Suppose Σ 1 and Σ 2 are disjoint. Let Ω be the domain bounded by Σ 1 and Σ 2 ; then Ω is a compact manifold with boundary Σ 1 Σ 2. Consider the following boundary value problem on Ω f u = 0, on Ω u = 0, on Σ 1 10) u = 1, on Σ 2 Let û = u ϕ, where ϕ C Ω) satisfying ϕ = 0onΣ 1 and ϕ = 1onΣ 2. Then the above problem is equivalent to the following { f û = f ϕ, on Ω û = 0, on Σ 1 Σ 2 11) Since f ϕ C Ω), the classical results for elliptic equations with homogeneous boundary value imply that 11) has a solution û C Ω), and therefore u =û + ϕ C Ω) is a solution to 10). Applying u and Ω to the Reilly formula 7), we obtain 0 Ric f u, u)e f dv. 12) Ω The boundary terms for Σ 1 Σ 2 vanish since Σ 1 and Σ 2 are f -minimal and u is constant on Σ 1 and Σ 2. Since Ric f is positive, 12) implies u is constant on Ω, which is a contradiction since u = 0onΣ 1 and u = 1onΣ 2. In [19], Lawson proved that for a closed embedded minimal hypersurface in a closed manifold with positive Ricci curvature, if both and are orientable, then \ consists of two components Ω 1 and Ω 2. The following lemma is a generalization of this result to the f -minimal case. Lemma 6 Let n,g,e f dv) be a closed metric measure space with positive Ric f, and let be a closed embedded f -minimal hypersurface. If both and are orientable, then \ consists of two components Ω 1 and Ω 2. Proof First we observe that for a compact connected metric measure space Ω, g, e f dv) with boundary Ω,ifRic f of Ω is positive and the f -mean curvature of the boundary Ω is nonnegative, then Ω is connected. This can be proved by an argument similar to that for Lemma 5: Suppose Ω is not connected. Let Σ be one of its components. Choose an f -harmonic function u i.e., f u = 0onΩ) which is

8 428 H. Li, Y. Wei equal to 0 on Σ and is equal to one on Ω \ Σ. The existence of u is by the classical results for elliptic equations as in the proof of Lemma 5. Then the Reilly formula 7) implies that u is a constant, which is a contradiction. To prove Lemma 6, we follow the argument in [19]. Let D = \. For any p we have a neighborhood U and local coordinates x 1,...,x n ) on U such that U corresponds to the hyperplane x 1 = 0. Then we get local coordinates for the boundary points of D = D D by first considering x 1 0 and then x 1 0. ote that D has positive Ric f and the f -mean curvature of the boundary is nonnegative since is f -minimal. If D were connected, then the boundary of D would be connected by the previous paragraph. However, since is orientable and connected by Lemma 5, we have that D has two components. If follows that D has two components D + and D and that D is the disjoint union of D + and D. This completes the proof. We remark that although an f -minimal hypersurface can be characterized as a minimal hypersurface in, g), Lemma 6 cannot follow directly from Lawson s result [19, Theorem 2], since we may not have a sign about the Ricci curvature of the conformal changed metric g. In the following, we will give a lower bound of the first eigenvalue of the f -Laplacian on an f -minimal hypersurface in a closed metric measure space with positive Ric f.let be an f -minimal hypersurface in n,g,e f dv). Denote by dμ the volume form on with respect to the metric induced from, g). The f -Laplacian f = f is a self-adjoint operator on with respect to e f dμ. The first eigenvalue λ 1 of f is the lowest nonzero real number which satisfies f u = λ 1 u with Dirichlet or eumann boundary condition if the boundary of is not empty. By the variational characterization, when is closed or for the eumann problem when has boundary) we also have λ 1 = inf ue f dμ=0 u 2 e f dμ u2 e f dμ. 13) For the Dirichlet problem when has boundary, the infimum in 13) istaken among all smooth functions which vanish on the boundary. Using the Reilly formula, Li a and Sheng-Hua Du [24, Theorem 3] proved that for a closed embedded f -minimal hypersurface in a closed orientable metric measure space n,g,e f dv) with Ric f κ>0, if divides into two components, then the first eigenvalue of the f -Laplacian on satisfies λ 1 κ/2, which generalized a result of Choi and Wang [9]. Here, using the universal covering space argument and Lemma 5, we show that a Du s theorem holds under a weaker assumption. Theorem 7 Let be a closed embedded f -minimal hypersurface in a closed metric measure space n,g,e f dv) with Ric f κ>0. Then the first eigenvalue λ 1 of the f -Laplacian on satisfies λ 1 κ/2.

9 f -inimal Surface and anifold with Positive Ric m f 429 Proof Let Ñ be the universal cover of. Then Ñ satisfies the same curvature assumption as. Since a compact manifold with positive Ric f has finite fundamental group π 1 ) see, e.g., [16, 20, 32]), Ñ is compact and π : Ñ is a finite covering. Let be the lifting of. Since is embedded and Ñ is simply connected, both Ñ and are orientable and then divides Ñ into two components by Lemma 6. By Theorem 3 in [24], λ 1 ) κ/2. But the pullback of the first eigenfunction of into is again an eigenfunction of. Therefore, λ 1 ) λ 1 ) κ/2. By combining Theorem 7 with the classical Yang Yau result see [31, 33]), we get the following volume estimates. Corollary 8 Let be a closed embedded f -minimal surface of genus g in a closed metric measure space 3,g,e f dv) with Ric f κ>0. Then d μ 16π κ 1 + g)e min f, 14) where d μ = e f dμ is the volume form on with respect to the induced metric from, g). Proof Let g = e f g, and denote by, the gradient and Laplacian on with respect to the induced metric from 3, g). Then the first eigenvalue λ 1 of the Laplacian satisfies λ 1 = inf ud μ=0 e min f u 2 d μ u2 d μ inf u 2 e f dμ u2 e f dμ ue f dμ=0 = λ 1 e min f κ 2 emin f. Here we used the relation u 2 = e f u 2 in the first inequality; the second equality is due to the variational characterization 13) for the first eigenvalue of the f -Laplacian; the second inequality is due to Theorem 7 and that f is non-negative. Then from the classical Yang Yau inequality λ 1 d μ 8π1 + g), we get the inequality 14). From the Gauss equation and the minimality of 2 in 3, g),wehave 1 2 h 2 = K K, 15)

10 430 H. Li, Y. Wei where h 2 is the squared norm of the second fundamental form of in 3, g). K and K are sectional curvature of and Gauss curvature of, with respect to g and the induced metric from g, respectively. Integrating 15) over with respect to d μ and applying the Gauss Bonnet theorem, we get h 2 d μ = 2 K d μ 2 K d μ C d μ 4πχ) C 16π 1 + g) + 8πg 1). 16) κ We will use the next proposition, which shows us how to use the uniform bounds 14) and 16) to obtain a singular compactness result see [10, Proposition 7.14], and [8, 11]). Proposition 9 Let 3 be a closed Riemannian three-manifold and i asequence of closed embedded minimal surfaces of genus g with Area i ) C 1, and h i 2 dμi C 2, i where Area i ) is the volume of i with respect to the induced metric from, and h i is the second fundamental form of i, C 1,C 2 are two constants independent of i. Then there exists a finite set of points S and a subsequence i that converges uniformly in C k any k 2) topology on compact subsets of \ S to a smooth embedded minimal surface possibly with multiplicity). ow we are in position to complete the proof of Theorem 1. Let i be any sequence of closed embedded f -minimal surfaces of fixed genus, which is also a sequence of closed embedded minimal surfaces of fixed genus in n, g).from14) and 16), we have uniform area and total curvature bounds of i n, g) in terms of the genus. Then Proposition 9 implies that i have a subsequence i which converges away from finitely many points to a smooth embedded minimal surface in n, g). ote that is f -minimal in n,g). It remains to show that the convergence holds across these points, i.e., the convergence is smooth everywhere. By Allard s regularity theorem [1], this follows from showing that the convergence is of multiplicity one. ote that since Ric f > 0on, has finite fundamental group π 1 ), after passing a finite cover, we may assume is simply connected. From the proof of Corollary 8, the first eigenvalue λ 1 of of the subsequence i has a positive lower bound κ/2. If the convergence is not multiplicity one, for large i,we can construct a test function to show that λ 1 i ) tends to zero, which violates the lower bound of λ 1. The detailed argument is just the same as Choi Schoen s paper [8] see also [10]), which we omit here. 4 First Eigenvalue of the f -Laplacian on a anifold with Positive m-bakry Émery Ricci Curvature In this section, we will use the Reilly formula 9) to prove Theorem 3. Let f u = λ 1 u, i.e., u is the first eigenfunction of the f -Laplacian. When m>n, from 9) and

11 f -inimal Surface and anifold with Positive Ric m f 431 the boundary condition of,wehave m 1 m λ2 1 u 2 e f dv Ric m f u, u)e f dv m 1)K u 2 e f dv. 17) Dividing by u2 e f dv and using the fact that λ 1 = u 2 e f dv/ u2 e f dv implies λ 1 mk. When m = n, since f is constant and Ric m f = Ric, the inequality 4) is due to the classic results by Lichnerowicz [23], Reilly [28], and Escobar [14]. ext, we consider the rigidity when the equality holds in 4). When m>n,we show that the inequality 4) cannot assume equality. Since if λ 1 = mk, then 17) becomes equality and then the Reilly formula 9) also attains equality. Since the algebraic inequality a + b) 2 a2 z z 1 b2 assumes the equality if and only if z 1)a + zb = 0 for z>1). We have that 0 = f u + m m n f u = u + n m n f u 18) holds everywhere on. ultiplying 18) with u and integrating on with respect to e m n n f dv give that 0 = = = u u + n ) m n f u e m n n f dv u 2 e n m n f dv + u 2 e n m n f dv, u u ν e n m n f dμ where the third equality is due to the boundary condition of u. Therefore, we have that u is a constant function on, which is a contradiction since u is the first eigenfunction of the f -Laplacian and cannot be a constant. So we conclude that the equality holds in 4) only when m = n, f is constant, and Ric m f = Ric. Then by Obata [26], Reilly [28], and Escobar [14], we complete the proof of Theorem 3. 5 anifolds with onnegative Ric m f and f -ean Convex Boundary In this section, we modify the argument in [22] to give the proof of Theorem 4. For any point x, since is complete, there exists a geodesic γ :[0,d] parameterized by arc length with γ0) = x, γd) and d = dx, ). We need to prove d 1/K. Choose an orthonormal basis e 1,...,e n 1 for T γd) and let

12 432 H. Li, Y. Wei e i s) be the parallel transport of e i along γ.letv i s) = ϕs)e i s) with ϕ0) = 0 and ϕd) = 1. From the first variation formula, we have that for each 1 i n 1 0 = δγv i ) = γ d d), e i ϕs) γ s), e i s) ds = γ d), e i, 0 which implies that γ d) is orthogonal to at γd). The second variation formula gives that see a related argument in [15]) n 1 δ 2 γv i,v i ) = i=1 d 0 n 1)ϕ s) 2 ϕs) 2 Ric γ s), γ s) )) ds H γd) ) 0. By the definition of m-bakry Émery Ricci curvature, we have 0 d n 1)ϕ s) 2 ϕs) 2 Ric m f γ s), γ s) )) ds H γd) ) 0 d + 0 ϕs) 2 2 f γ s), γ s) ) 1 f γs) ),γ s) ) 2 ds. m n Since Ric m f is nonnegative, by using the facts that and that d ds f γs) ) = f γs) ),γ s) d 2 ds 2 f γs) ) = 2 f γ s), γ s) ), and by integration by parts, we deduce that d 0 n 1)ϕ s) 2 2ϕs)ϕ s) f γs) ),γ s) 0 1 m n ϕs)2 f γs) ),γ s) ) 2 ds H f γd) ). Using the Cauchy Schwarz inequality we get that 0 d 0 m 1)ϕ s) 2 ds H f γd) ). 19) Choose ϕs) = s d and note that H f m 1)K on ; from 19) we have that d 1/K. Since the point x is arbitrary, we have proved the inequality 5). ow we assume that is compact. Then 5) implies that is also compact. By an argument similar to that in the proof of Lemma 5, we can prove that is connected: Suppose not, and let Σ be one of its components. Choose an f -harmonic function u on, which is equal to zero on Σ and is equal to one on \ Σ. Then

13 f -inimal Surface and anifold with Positive Ric m f 433 since Ric m f 0on and H f m 1)K > 0on, the Reilly formula 9) implies that u ν = 0on, where ν is the outer unit normal to. By integration by parts, we have that u 2 e f dv = u f ue f dv + u u ν e f dμ = 0. Therefore, u is a constant function on, which is a contradiction since u = 0onΣ but u = 1on \ Σ. Suppose the equality holds in 5); we will show that is isometric to an n- dimensional Euclidean ball. By rescaling the metric of, we may assume K = 1. Since is compact, there exists some point x 0 in the interior of such that dx 0, ) = 1. It is clear that the geodesic ball B 1 x 0 ) of radius 1 centered at x 0 is contained in. We claim that is just the geodesic ball B 1 x 0 ). In fact, let ρ = dx 0, ) be the distance function from x 0. Since the m-bakry Émery Ricci curvature of is nonnegative, the f -Laplacian of ρ satisfies see equation 4) in [27]) f ρ) m 1, 20) ρ in the sense of distribution. Let Σ ={q : ρq) = 1}, which is clearly a closed set in by the continuity of ρ. Since is connected, to show Σ =, it suffices to show that Σ is also open in, that is, for any q Σ, there is an open neighborhood U of q in such that ρ 1onU. Ifq is not a conjugate point to x 0 in, then the geodesic sphere B 1 x 0 ) is a smooth hypersurface near q in. Since ρ and ρ are the mean curvature and outer unit normal of the geodesic sphere, we have f ρ) = H f. ote that ρ = 1 on the geodesic sphere and by the f -Laplacian comparison inequality 20), the f -mean curvature of the geodesic sphere is at most m 1. However, by the assumption of Theorem 4, thef -mean curvature of is at least m 1. Then from the maximum principle see [13]), we have that and B 1 x 0 ) coincide in a neighborhood of q. This implies that Σ is open near any q which is not a conjugate point. A similar process in [22] see also Calabi [5]) makes us work through the argument to conclude that ρ is constant near q in, when q is a conjugate point of x 0. This proves that is just the geodesic sphere B 1 x 0 ) and is the geodesic ball B 1 x 0 ). ext we show that is isometric to the Euclidean ball of radius one. Since any q can be joined by a minimizing geodesic γ parameterized by arc-length from x 0 to q, and γ is orthogonal to = B 1 x 0 ) at q, which implies that γ is uniquely determined by q and q is not in the cut locus of x 0. So that ρ = dx 0, ) is smooth up to the boundary. Then the f -Laplacian comparison inequality 20) holds in the classical sense. Let = e f dv and = e f dμ be volumes of and with respect to the weighted measure. From the facts that ρ =1on, ρ = 1, and ρ ν = 1on, integration by parts implies that = ρ ρ ν e f dμ ρ 2 e f dv = ρ f ρ)e f dv m 1).

14 434 H. Li, Y. Wei This implies m. On the other hand, by an argument similar to that in the proof of Theorem 1 in [29], we can prove that m :Letu be a smooth solution of the Dirichlet problem { f u = 1, in, u = 0, on. Integration by parts gives that = f ue f u dv = ν e f dμ. 21) Since Ric m f 0in and H f m 1) on note that we have assumed K = 1), substituting u into the Reilly formula 9) gives that ) u 2 m e f dμ. 22) ν From 21), Hölder s inequality, and 22), it follows that ) 2 u 2 = ν e f dμ ) u 2 e f dμ ν /m and we have m. Therefore, we get the equality =m. Then the equality holds in 22) and therefore the Reilly formula 9) assumes equality too. By an argument similar to that in the last part in Sect. 4, we get m = n, f is constant, and Ric m f = Ric. Then Theorem 1in[29] implies that is isometric to a Euclidean ball. This completes the proof of Theorem 4. Acknowledgement The authors would like to thank the referee for helpful suggestions. References 1. Allard, W.K.: On the first variation of a varifold. Ann. ath. 95, ) 2. Andrews, B., i, L.: Eigenvalue comparison on Bakry Émery manifolds. Commun. Partial Differ. Equ. 37, ) 3. Bakry, D., Émery,.: Diffusions hypercontractives. In: Séminaire de Probabilités, XIX, 1983/84. Lecture otes in ath., vol. 1123, pp Springer, Berlin 1985) 4. Bakry, D., Qian, Z.: Some new results on eigenvectors via dimension, diameter, and Ricci curvature. Adv. ath. 155, ) 5. Calabi, E.: An extension of E. Hopf s maximum principle with an application to Riemannian geometry. Duke ath. J. 25, ) 6. Cao, H.-D.: Recent progress on Ricci solitons. In: Recent Advances in Geometric Analysis. Adv. Lect. ath., vol. 11, pp International Press, Somerville 2010)

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