Lecture 20. Upper and lower heat kernel Gaussian bounds

Posted on October 5, 2013 by Fabrice Baudoin

In this short Lecture, as in the previous one, we consider a complete and n-dimensional Riemannian manifold (\mathbb{M},g) with non negative Ricci curvature. The volume doubling property that was proved is closely related to sharp lower and upper Gaussian bounds that are due to P. Li and S.T. Yau. We first record a basic consequence of the volume doubling property whose proof is let to the reader.

Theorem: Let C > 0 be the constant such that for every x \in \mathbb{M}, R \ge 0, \mu(B(x,2R)) \le C \mu (B(x,R)). Let Q = \log_2 C. For any x\in \mathbb{M} and r > 0 one has \mu(B(x,tr)) \ge C^{-1} t^{Q} \mu(B(x,r)),\ \ \ 0\le t\le 1.

We are now in position to prove the main result of the Lecture.

Theorem: For any 0 < \varepsilon < 1 there exists a constant C = C(n,\varepsilon) > 0, which tends to \infty as \varepsilon \to 0^+, such that for every x,y\in \mathbb{M} and t > 0 one has
\frac{C^{-1}}{\mu(B(x,\sqrt t))} \exp \left(-\frac{ d(x,y)^2}{(4-\varepsilon)t}\right)\le p(x,y,t)\le \frac{C}{\mu(B(x,\sqrt t))} \exp \left(-\frac{d(x,y)^2}{(4+\varepsilon)t}\right).

Proof: We begin by establishing the lower bound. First, from the Harnack inequality we obtain for all y \in \mathbb{M}, t > 0, and every 0 < \varepsilon < 1,
p(x,y,t)\ge p(x,x,\varepsilon t) \varepsilon^\frac{n}{2} \exp\left( -\frac{d(x,y)^2}{(4-\varepsilon)t}\right).
We thus need to estimate p(x,x,\varepsilon t) from below. But this has already been done in the proof of the volume doubling property where we established:
p(x,x,\varepsilon t) \ge \frac{C^*}{\mu(B(x,\sqrt{\varepsilon/2} \sqrt t))},\ \ \ \ \ x\in \mathbb{M},\ t > 0.
On the other hand, since \sqrt{\varepsilon/2} < 1, by the trivial inequality \mu(B(x,\sqrt{\varepsilon/2} \sqrt t)) \le \mu(B(x,\sqrt t)), we conclude
p(x,y,t) \geq \frac{C^*}{ \mu(B(x,\sqrt t))} \varepsilon^\frac{n}{2} \exp\left( -\frac{d(x,y)^2}{(4-\varepsilon)t}\right).
This proves the Gaussian lower bound.

For the Gaussian upper bound, we first observe that the following upper bound was proved in a previous lecture:
p(x,y,t)\le \frac{C}{\mu(B(x,\sqrt t))^{\frac{1}{2}} \mu(B(y,\sqrt t))^{\frac{1}{2}}} \exp \left(-\frac{d(x,y)^2}{(4+\varepsilon')t}\right).
At this point, by the triangle inequality and the volume doubling property we find.
\mu(B(x,\sqrt{ t})) \leq \mu(B(y,d(x,y)+\sqrt{ t}))
\leq C_1 \mu(B(y,\sqrt{ t})) \left(\frac {d(x,y)+\sqrt{ t}}{\sqrt t} \right)^Q.
with Q=\log_2 C, where C is the doubling constant.
This gives
\frac{1}{\mu(B(y,\sqrt{ t}))}\leq \frac{C_1}{\mu(B(x,\sqrt{ t}))} \left(\frac {d(x,y)}{\sqrt{ t}}+1 \right)^Q.
Combining this with the above estimate we obtain
p(x,y,t)\le \frac{C_1^{1/2}C}{\mu(B(x,\sqrt t))} \left(\frac {d(x,y)}{\sqrt{ t}}+1 \right)^{\frac{Q}{2}} \exp \left(-\frac{d(x,y)^2}{(4+\varepsilon')t}\right).
If now 0 < \varepsilon < 1, it is clear that we can choose 0 < \varepsilon' < \varepsilon such that
\frac{C_1^{1/2}C}{\mu(B(x,\sqrt t))} \left(\frac {d(x,y)}{\sqrt{ t}}+1 \right)^{\frac{Q}{2}} \exp \left(-\frac{d(x,y)^2}{(4+\varepsilon')t}\right) \le \frac{C^*}{\mu(B(x,\sqrt t))} \exp \left(-\frac{d(x,y)^2}{(4+\varepsilon)t}\right),
where C^* is a constant which tends to \infty as \varepsilon \to 0^+. The desired conclusion follows by suitably adjusting the values of both \varepsilon' and of the constant in the right-hand side of the estimate \square

To conclude the lecture, we finally mention without proof, what the previous arguments give in the case where \mathbf{Ric} \ge -K with K \ge 0. We encourage the reader to do the proof by herself/himself as an exercise.

Theorem: Let us assume \mathbf{Ric} \ge -K with K \ge 0. For any 0 < \varepsilon < 1 there exist constants C_1,C_2= C(n,K,\varepsilon) > 0, such that for every x,y\in \mathbb{M} and t > 0 one has
p(x,y,t)\le \frac{C_1}{\mu(B(x,\sqrt t))} \exp \left(-\frac{d(x,y)^2}{(4+\varepsilon)t} +KC_2 (t +d(x,y)^2)\right).
p(x,y,t)\ge \frac{C^{-1}_1}{\mu(B(x,\sqrt t))} \exp \left(-\frac{d(x,y)^2}{(4-\varepsilon)t} -KC_2 (t +d(x,y)^2)\right).

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Lecture 19. Volume doubling property

Posted on October 3, 2013 by Fabrice Baudoin

In this Lecture we consider a complete and n-dimensional Riemannian manifold (\mathbb{M},g) with non negative Ricci curvature. Our goal is to prove the following fundamental result, which is known as the volume doubling property.

Theorem: There exists a constant C=C(n) > 0 such that for every x\in \mathbb{M} and every r > 0 one has \mu(B(x,2r))\le C \mu (B(x,r)).

Actually by suitably adapting the arguments given in this Lecture, the previous result can be extended to the case of negative Ricci curvature as follows:

Theorem: Assume \mathbf{Ric} \ge -K with K \ge 0. There exist positive constants C_1=C_1(n,K), C_2=C_2(n,K) such that for every x\in \mathbb{M} and every r > 0 one has
\mu(B(x,2r))\le C_1e^{KC_2 r^2} \mu (B(x,r)).

For simplicity, we show the arguments in the case K=0 and let the reader work out the arguments in the case K \neq 0.

This result can be obtained from geometric methods as a consequence of the Bishop-Gromov comparison theorem. The proof we give instead only relies on the previous methods and has the advantage to generalize to a much larger class of operators than Laplace-Beltrami on Riemannian manifolds.

The key heat kernel estimate that leads to the doubling property is the following uniform and scale invariant lower bound on the heat kernel measure of balls.

Theorem: There exist an absolute constant K > 0, and A > 0, depending only on n, such that
P_{Ar^2}(\mathbf 1_{B(x,r)})(x) \ge K, \ \ \ \ \ x\in \mathbb{M}, r > 0.

Proof: We first recall the following result that was proved in a previous Lecture: Let a \in C^1([0,T],[0,\infty)) and \gamma \in C((0,T),\mathbb{R}). Given f \ge 0, which is bounded and such that \sqrt{f} is Lipschitz, we have
a(T) P_T \left( f \Gamma (\ln f) \right) -a(0)(P_{T} f) \Gamma (\ln P_{T}f)
\ge \int_0^T \left(a'+2\rho a -\frac{4a\gamma}{n} \right)\Phi (s) ds +\left(\frac{4}{n}\int_0^T a\gamma ds\right)LP_{T} f -\left(\frac{2 }{n}\int_0^T a\gamma^2ds\right)P_T f.

We choose a(t)=\tau+T-t,and \gamma(t)=-\frac{n}{4(\tau+T-t)} where \tau > 0 will later be optimized. Noting that we presently have
a' = 1,\ \ \ a\gamma = - \frac{n}{4},\ \ \ \ a\gamma^2 = \frac{n^2}{16(\tau + T - t)^2},
we obtain the inequality
\tau P_T(f \Gamma(\ln f)) -(T+\tau) P_T f \Gamma(\ln P_T f) \ge -T L P_T f -\frac{n}{8} \ln \left( 1+\frac{T}{\tau}\right) P_T f
In what follows we consider a bounded function f on \mathbb{M} such that \Gamma(f) \le 1 almost everywhere on \mathbb{M}. For any \lambda \in \mathbb R we consider the function \psi defined by
\psi(\lambda,t) = \frac{1}{\lambda} \log P_t(e^{\lambda f}), \ \ \ \text{or alternatively}\ \ \ P_t(e^{\lambda f}) = e^{\lambda \psi}.
Notice that Jensen’s inequality gives \lambda \psi \ge \lambda P_t f, and so we have P_t f \le \psi. We now apply the previous inequality to the function e^{\lambda f}, obtaining
\lambda^2 \tau P_T\left(e^{\lambda f} \Gamma(f)\right) - \lambda^2 (T+\tau) e^{\lambda \psi} \Gamma(\psi) \ge - T L P_T(e^{\lambda f}) - \frac{n}{8} e^{\lambda \psi} \ln\left(1+\frac{T}{\tau}\right).
Keeping in mind that \Gamma(f) \le 1, we see that P_T(e^{\lambda f} \Gamma(f)) \le e^{\lambda \psi}. Using this observation in combination with the fact that
L \left(P_t (e^{\lambda f})\right) = \frac{\partial}{\partial t} \left(P_t (e^{\lambda f})\right) = \frac{\partial e^{\lambda \psi}}{\partial t} = \lambda e^{\lambda \psi} \frac{\partial \psi}{\partial t} ,
and switching notation from T to t, we infer
\lambda^2 \tau \ge \lambda^2 (t+\tau) e^{\lambda \psi} \Gamma(\psi)- \lambda t \frac{\partial \psi}{\partial t} - \frac{n}{8} \ln\left(1+\frac{t}{\tau}\right).
The latter inequality finally gives
\frac{\partial \psi}{\partial t} \ge - \frac{\lambda}{t}\left(\tau + \frac{n}{8\lambda^2} \ln\left(1+\frac{t}{\tau}\right)\right)\ge - \frac{\lambda}{t}\left(\tau + \frac{nt}{8\lambda^2\tau} \right).
We now optimize the right-hand side of the inequality with respect to \tau. We notice explicitly that the maximum value of the right-hand side is attained at \tau_0 = \sqrt{\frac{nt}{8 \lambda^2} }. We find therefore
\frac{\partial \psi}{\partial t} \ge -\sqrt{\frac{n}{2t} }
We now integrate the inequality between s and t, obtaining
\psi(\lambda,s) \le \psi(\lambda,t) +\sqrt{ \frac{n}{2}} \int_{s}^t \frac{d\tau}{\sqrt{\tau}}.
We infer then
P_s(\lambda f) \le \lambda \psi(\lambda,t) + \lambda \sqrt{ 2nt}.
Letting s\to 0^+ we conclude
\lambda f \le \lambda \psi(\lambda,t) + \lambda \sqrt{ 2nt}.

At this point we let B = B(x,r) = \{x\in \mathbb{M}\mid d(y,x) < r\}, and consider the function f(y) = - d(y,x). Since we clearly have e^{\lambda f} \le e^{-\lambda r} \mathbf 1_{B^c} + \mathbf 1_B, it follows that for every t > 0 one has e^{\lambda \psi(\lambda,t)(x)} = P_t(e^{\lambda f})(x) \le e^{-\lambda r} + P_t(\mathbf 1_B)(x). This gives the lower bound P_t(\mathbf 1_B)(x) \ge e^{\lambda \psi(\lambda,t)(x)} - e^{-\lambda r}. To estimate the first term in the right-hand side of the latter inequality, we use the previous estimate which gives P_{t}(\mathbf 1_B)(x) \ge e^{-\lambda \sqrt{2nt}} - e^{-\lambda r}. To make use of this estimate, we now choose \lambda = \frac{1}{r}, t = Ar^2, obtaining P_{Ar^2}(\mathbf 1_B)(x) \ge e^{-A\sqrt{2n}} - e^{-1}. The conclusion follows then easily \square

We now turn to the proof of the volume doubling property. We first recall the following basic result which is a straightforward consequence of the Li Yau inequality.

Proposition: Let p(x,y,t) be the heat kernel on \mathbb{M}. For every x,y, z\in \mathbb{M} and every 0 < s < t < \infty one has
p(x,y,s) \le p(x,z,t) \left(\frac{t}{s}\right)^{\frac{n}{2}} \exp\left( \frac{d(y,z)^2}{4(t-s)} \right).

We are now in position to prove the doubling.

From the semigroup property and the symmetry of the heat kernel we have for any y\in \mathbb{M} and t > 0 p(y,y,2t) = \int_\mathbb{M} p(y,z,t)^2 d\mu(z). Consider now a function h\in C^\infty_0(\mathbb{M}) such that 0\le h\le 1, h\equiv 1 on B(x,\sqrt{t}/2) and h\equiv 0 outside B(x,\sqrt t). We thus have
P_t h(y) = \int_\mathbb{M} p(y,z,t) h(z) d\mu(z) \le \left(\int_{B(x,\sqrt t)} p(y,z,t)^2 d\mu(z)\right)^{\frac{1}{2}} \left(\int_\mathbb{M} h(z)^2 d\mu(z)\right)^{\frac{1}{2}}
\le p(y,y,2t)^{\frac{1}{2}} \mu(B(x,\sqrt t))^{\frac{1}{2}}.
If we take y=x, and t =r^2, we obtain
P_{r^2} \left(\mathbf 1_{B(x,r)}\right)(x)^2 \le P_{r^2} h(x)^2 \leq p(x,x,2r^2)\ \mu(B(x,r)).
At this point we use the crucial previous theorem, which gives for some 0 < A = A(n) < 1
P_{Ar^2}(\mathbf 1_{B(x,r)})(x) \ge K, \ \ \ \ \ x\in \mathbb{M}, r > 0. Combining the latter inequality with the Harnack inequality, we obtain the following on-diagonal lower bound
p(x,x,2r^2) \ge \frac{K^*}{\mu(B(x,r))},\ \ \ \ \ x\in \mathbb{M},\ r> 0.
Applying the Harnack inequality for every y\in B(x,\sqrt t) we find
p(x,x,t) \le C(n) p(x,y,2t).
Integration over B(x,\sqrt t) gives
p(x,x,t)\mu(B(x,\sqrt t)) \le C(n) \int_{B(x,\sqrt t)}p(x,y,2t)d\mu(y) \le C(n), where we have used P_t1\le 1. Letting t = r^2, we obtain from this the on-diagonal upper bound p(x,x,r^2) \le \frac{C(n)}{\mu(B(x,r))}. We finally obtain
\mu(B(x,2r)) \le \frac{C}{p(x,x,4r^2)} \le \frac{C^*}{p(x,x,2r^2)} \le C^{**} \mu(B(x,r)),
where we have used once more the Harnack inequality,
which gives \frac{p(x,x,2r^2)}{p(x,x,4r^2)}\le C.

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Richard Courant: From Göttingen to New-York.

Posted on September 29, 2013 by Fabrice Baudoin

An interesting look into the life and career of Richard Courant:

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Curvature inequalities in sub-Riemannian geometry

Posted on September 26, 2013 by Fabrice Baudoin

An important part of my research from the past few years has been to try to understand the notion of Ricci curvature lower bound in sub-Riemannian geometry. On this problem, my point of view is more the one of an analyst or even a probabilist than a geometer: My main device is a second order differential operator, a sub-Laplacian, and my main tools to understand global geometric properties are heat flow techniques. In sub-Riemannian geometry, this point of view can be fruitful in some aspects that I will describe now.

Although during the last three decades there have been several advances in the study of sub Riemannian spaces and the closely connected theory of sub-elliptic pde’s, most of the developments to date are of a local nature. As a consequence, the theory presently lacks a body of results which, similarly to the Riemannian case, connect properties of solutions of the relevant pde’s to the geometry of the ambient manifold. To illustrate this point, let me discuss some of the most fundamental local estimates concerning the subelliptic Hörmander‘s type operators.

Let (\mathbb M,g) be a smooth and connected Riemannian manifold and \mu be a smooth measure on \mathbb{M}. Let us assume that there exists on \mathbb{M} a family of vector fields \{ X_1, \cdots X_d \} that satisfy the bracket generating condition. We are interested in the subelliptic operator L=-\sum_{i=1}^d X_i^* X_i which is symmetric with respect to the measure \mu. In order to study L, the Riemannian distance d_R of \mathbb M is most of the times confined to the background. There is another distance on \mathbb M, that was introduced by Carathéodory, which plays a central role. A piecewise C^1 curve \gamma:[0,T]\to \mathbb M is called subunitary at x if for every \xi\in T_x\mathbb M one has g(\gamma'(t),\xi)^2 \le \sum_{i=1}^d g(X_i(\gamma(t)),\xi)^2. We define the subunit length of \gamma as L_s(\gamma) = T. If we indicate with S(x,y) the family of subunit curves such that \gamma(0) = x and \gamma(T) = y, then thanks to the fundamental accessibility theorem of Chow-Rashevsky the connectedness of \mathbb M implies that S(x,y) \not= \varnothing for every x,y\in \mathbb M. This allows to define the sub-Riemannian distance on \mathbb M as follows d(x,y) = \inf \{L_s(\gamma)\mid \gamma\in S(x,y)\}.
Another elementary consequence of the Chow-Rashevsky theorem is that the identity map i:(\mathbb M,d) \hookrightarrow (\mathbb M,d_R) is continuous and thus, the topologies of d_R and d coincide. Several fundamental properties of the metric d have been discussed in the seminal paper by Nagel, Stein and Wainger. In particular, the following result provides a uniform local control of the growth of the metric balls in (\mathbb{M},d). It is known as the local doubling condition.

Theorem 1: For any x\in \mathbb{M} there exist constants C(x), R(x) > 0 such that with Q(x) = \log_2 C(x) one has
\mu(B(x,tr)) \ge C(x)^{-1} t^{Q(x)} \mu(B(x,r)),\ \ \ 0\le t\le 1,\ 0 < r\le R(x).
Moreover, given any compact set K\subset \mathbb{M} one has
\underset{x\in K}{\inf}\ C(x) >0,\ \ \ \underset{x\in K}{\inf}\ R(x) > 0.

This remarkable theorem is a fundamental local property of the metric space (\mathbb{M},d) and may be connected to several local estimates related to the operator L. We mention in particular the Poincaré inequality on balls:

Theorem 2: (D. Jerison) Let x \in \mathbb{M} and R > 0. There exists a constant C=C(x,R) > 0 such that for every 0 < r < R and f \in C^1 (B(x,r)),
\int_{B(x,r)} | f -f_{B(x,r)}|^ 2 d\mu \le C r^2 \int_{B(x,r)} \sum_{i=1}^d (X_if)^2 d\mu,
where f_{B(x,r)} =\frac{1}{\mu(B(x,r))} \int_{B(x,r)} f d\mu.

We stress again that these theorems and the methods to prove them are local in nature. This naturally raises the following question:

Question 1: Can we find conditions on the operator L ensuring that global versions of theorems like Theorems 1 and 2 hold true ?

By global versions of Theorems 1 and 2, we mean that the estimates should hold true on non necessarily compact manifolds and with constants uniformly controlled by the sub-Riemannian geometry of \mathbb{M}. To find a way to tackle this question, let us first discuss what could be a satisfying answer in the simplest non trivial case: The case where L is elliptic, that is \{ X_1,\cdots,X_d \} form a basis of the tangent space at each point. In this case, there is a definite satisfying answer to Question 1: Global versions of the Theorems 1 and 2 may be obtained under the assumption that L satisfies a curvature dimension inequality.

The notion of curvature dimension inequality originates from the analysis of the Laplace-Beltrami operator on a Riemannian manifold and more precisely from the Bochner’s formula which states that if \mathbb{M} is a Riemannian manifold with Laplacian \Delta, for any f\in C^\infty(\mathbb{M}) one has
\Delta(|\nabla f|^2) = 2 ||\nabla^2 f||^2 + 2 (\nabla f, \nabla \Delta f) + 2 \text{Ric}(\nabla f,\nabla f).

Using the Cauchy-Schwarz inequality, which gives \| \nabla^2 f \|_2^2\ge \frac{1}{n} (\Delta f)^2, we thus see that the assumption that the Riemannian Ricci tensor on \mathbb{M} be bounded from below by \rho_1 \in \mathbb{R} implies
\frac{1}{2}\Delta(|\nabla f|^2) - (\nabla f, \nabla \Delta f) \ge \frac{1}{n} (\Delta f)^2 + \rho_1 \| \nabla f \|^2,\ \ \ \ f\in C^\infty(\mathbb{M}).

What is remarkable is that this inequality perfectly captures the notion of Ricci lower bounds and dimensional upper bound: On a finite dimensional Riemannian manifold \mathbb{M} the inequality is actually equivalent to Ric \ge \rho_1 and \mathbf{dim}(\mathbb{M}) \le n.

This observation leads Bakry to define the notion of curvature dimension inequality for arbitrary second-order differential operators. Let L be, as before, a second order differential operator. Consider the following bilinear differential forms on functions f, g \in C^\infty(\mathbb{M}),
\Gamma(f,g) =\frac{1}{2}(L(fg)-fL g-gL f),
and
\Gamma_{2}(f,g) = \frac{1}{2}\big[L \Gamma(f,g) - \Gamma(f, L g)-\Gamma (g,L f)\big].
When f=g, we simply write \Gamma(f) = \Gamma(f,f), \Gamma_2(f) = \Gamma_2(f,f). A straightforward computation shows that if, in a local chart,
L=\sum_{i,j=1}^n a_{ij} (x) \frac{\partial^2}{ \partial x_i \partial x_j} +\sum_{i=1}^n b_i (x)\frac{\partial}{\partial x_i},
then, in the same chart
\Gamma (f,g)=\sum_{i,j=1}^n a_{ij} (x) \frac{\partial f}{\partial x_i} \frac{\partial g}{\partial x_j}.
As a consequence, for every smooth function f, \Gamma(f) \ge 0. The bilinear form \Gamma_2 is of second order and has no sign in general. We observe that if L is a Laplace-Beltrami operator, then Bochner’s formula writes
\Gamma_2(f)= ||\nabla^2 f||^2 + \ \text{Ric}(\nabla f,\nabla f).

Definition: The operator L is said to satisfy the curvature dimension inequality CD(\rho_1,n), \rho_1 \in \mathbb{R}, n >0, if for every function f \in C^\infty(\mathbb{M}):
\Gamma_{2}(f) \ge \frac{1}{n} (L f)^2 + \rho_1\Gamma(f).

As we just pointed it out, if L is a Laplace-Beltrami operator, then it satisfies CD(\rho_1,n) if and only if Ric \ge \rho_1 and \mathbf{dim}(\mathbb{M}) \le n. But, what is interesting here is that the curvature dimension inequality may be satisfied without L necessarily being a Laplace-Beltrami operator. In that case, the curvature dimension inequality is equivalent to a lower bound on the Bakry-Emery tensor of L. We have then the following, a priori, satisfying answer to our Question 1:

Fact: If L satisfies CD(\rho_1,n), with \rho_1 \ge 0 and n >0, then we have global versions of Theorems 1 and 2.

More precisely, we have the following result:

Theorem: Let L be a subelliptic second order operator that satisfies CD(\rho_1,n), with \rho_1 \ge 0 and n > 0. Then, there exist constants C_d, C_p > 0, depending only on \rho_1 and n, for which one has for every x\in \mathbb{M} and every r > 0:
\mu(B(x,2r)) \le C_d\ \mu(B(x,r));
and
\int_{B(x,r)} |f - f_B|^2 d\mu \le C_p r^2 \int_{B(x,r)} \Gamma(f) d\mu,
for every f\in C^1( B(x,r)).

Unfortunately, this fact only has a limited interest for us, since it is possible to prove that if a subelliptic operator L satisfies CD(\rho_1,n), with \rho_1 \in \mathbb{R} and n > 0, then L is actually elliptic ! With this in mind, the next question that naturally arises is:

Question 2: Is there an appropriate notion of curvature dimension inequality that applies to subelliptic operators ?

At this level of generality, this question certainly does not admit a definite satisfying answer. However with my co-author, N. Garofalo, we introduced a curvature dimension inequality that applies to a large class of subelliptic operators.

To introduce the relevant setting we consider a smooth, connected manifold \mathbb{M} endowed with a smooth measure \mu and a smooth, locally subelliptic, second order differential operator L which is symmetric with respect to \mu. As above, we associate with L the following symmetric, first-order, differential bilinear form:
\Gamma(f,g) =\frac{1}{2}(L(fg)-fLg-gLf), \quad f,g \in C^\infty(\mathbb{M}).
There is a genuine distance d canonically associated with L which is continuous and defines the topology of \mathbb{M}. It is given by
d(x,y)=\sup \left\{ |f(x) -f(y) | \mid f \in C^\infty(\mathbb{M}) , \| \Gamma(f) \|_\infty \le 1 \right\},\ \ \ \ x,y \in \mathbb{M}.

We always assume that this distance is finite and that the metric space (\mathbb{M},d) is complete. This implies that L is essentially self-adjoint.

We also suppose, and this is the main new ingredient, that we are given on \mathbb{M} a symmetric, first-order differential bilinear form \Gamma^Z, satisfying
\Gamma^Z(fg,h) = f\Gamma^Z(g,h) + g \Gamma^Z(f,h).
The following hypothesis on \Gamma^Z plays a pervasive role in the results that we will describe and is natural in several geometric situations: For every f \in C^\infty(\mathbb{M}), we have \Gamma(f, \Gamma^Z(f))=\Gamma^Z( f, \Gamma(f)).

Before we proceed with the discussion, we pause to stress that, in the generality in which we work the bilinear differential form \Gamma^Z, unlike \Gamma, is not a priori intrinsic. Whereas \Gamma is determined once L is assigned, the form \Gamma^Z in general is not intrinsically associated with L. However, in several geometric examples the choice of \Gamma^Z will be natural and even canonical, up to a constant. This is the case, for instance, for one of the important geometric examples covered by our analysis: The CR Sasakian manifolds. Roughly speaking, we can think of \Gamma^Z as an orthogonal complement of \Gamma: the bilinear form \Gamma represents the square of the length of the gradient in the horizontal directions, whereas \Gamma^Z represents the square of the length of the gradient along the vertical directions.

Given the subelliptic operator L and the first-order bilinear forms \Gamma and \Gamma^Z on \mathbb{M}, we introduce the following second-order differential bilinear forms:
\Gamma_{2}(f,g) = \frac{1}{2}\big[L\Gamma(f,g) - \Gamma(f, Lg)-\Gamma (g,Lf)\big],
and
\Gamma^Z_{2}(f,g) = \frac{1}{2}\big[L\Gamma^Z (f,g) - \Gamma^Z(f, Lg)-\Gamma^Z (g,Lf)\big].
Observe that if \Gamma^Z\equiv 0, then \Gamma^Z_2 \equiv 0 as well. As for \Gamma and \Gamma^Z, we will use the notations \Gamma_2(f) = \Gamma_2(f,f), \Gamma_2^Z(f) = \Gamma^Z_2(f,f). The main definition and tool we proposed is the following:

Definition: We say that the subelliptic operator L satisfies the generalized curvature-dimension inequality CD(\rho_1,\rho_2,\kappa,d) if there exist constants \rho_1 \in \mathbb{R}, \rho_2 > 0, \kappa \ge 0, and 0 < d \le +\infty such that the inequality
\Gamma_2(f) +\nu \Gamma_2^Z(f) \ge \frac{1}{d} (Lf)^2 +\left( \rho_1 -\frac{\kappa}{\nu} \right) \Gamma(f) +\rho_2 \Gamma^Z(f)
holds for every f\in C^\infty(\mathbb{M}) and every \nu > 0.

The motivation behind this definition comes from examples arising in geometry.

Theorem: Let (\mathbb{M},\theta) be a strictly pseudo convex CR manifold with real dimension 2n+1 and vanishing Tanaka-Webster pseudohermitian torsion, i.e., a Sasakian manifold. The Tanaka-Webster Ricci tensor satisfies the lower bound
\emph{Ric}_x(v,v)\ \ge \rho_1|v|^2,\ \ \ \text{for every horizontal vector}\ v\in \mathcal H_x=\mathbf{Ker} ( \theta_x),
if and only if the CR sub-Laplacian of \mathbb{M} satisfies the curvature-dimension inequality CD(\rho_1,\frac{d}{4},1,d) with d = 2n. Moreover, the hypothesis \Gamma(f, \Gamma^Z(f))=\Gamma^Z( f, \Gamma(f)) is satisfied.

With this examples in mind let us now come back to the general framework and give a good answer to Question 1 in many purely subelliptic examples. In the following theorem that I proved with N. Garofalo and M. Bonnefont, the hypothesis \Gamma(f, \Gamma^Z(f))=\Gamma^Z( f, \Gamma(f)) is supposed to be satisfied.

Theorem: Let L be a subelliptic second order operator that satisfies CD(\rho_1,\rho_2,\kappa,n), with \rho_1 \ge 0, \rho_2 > 0, \kappa \ge 0 and n > 0. Then, there exist constants C_d, C_p > 0, depending only on \rho_1 and n, for which one has for every x\in \mathbb{M} and every r > 0:
\mu(B(x,2r)) \le C_d\ \mu(B(x,r));
and
\int_{B(x,r)} |f - f_B|^2 d\mu \le C_p r^2 \int_{B(x,r)} \Gamma(f) d\mu,
for every f\in C^1( B(x,r)).

Besides this result, the generalized curvature dimension inequality was also used to prove a Bonnet-Myers type theorem, the first of its kind in sub-Riemannian geometry.

The study of this generalized curvature dimension condition and of some of its extensions has generated quite a lot of research with my colleagues Michel Bonnefont, Nicola Garofalo, Isidro Munive and my students Bumsik Kim and Jing Wang.

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Lecture 18. The Gaussian upper bound for the heat kernel

Posted on September 18, 2013 by Fabrice Baudoin

Let \mathbb{M} be a complete n-dimensional Riemannian manifold and, as usual, denote by L its Laplace-Beltrami operator. As in the previous lecture, we will assume that the Ricci curvature of \mathbb{M} is bounded from below by -K with K \ge 0. Our purpose in this lecture is to prove a Gaussian upper bound for the heat kernel. Our main tools are the parabolic Harnack inequality proved in the previous lecture and the following integrated maximum principle:

Proposition Let g :\mathbb{M}\times \mathbb{R}_{\ge 0} \to \mathbb{R} be a non positive continuous function such that, in the sense of distributions,
\frac{\partial g}{\partial t} +\frac{1}{2} \Gamma(g) \le 0,
then, for every f \in L^2_\mu(\mathbb{M}), we have
\int_{\mathbb{M}} e^{g(y,t)} ) (P_t f)^2 (y) d\mu(y) \le \int_{\mathbb{M}} e^{g(y,0)} ) f^2 (y) d\mu(y).

Proof: Since
\left(L-\frac{\partial }{\partial t}\right)(P_tf)^2 = 2 P_tf\left(L-\frac{\partial }{\partial t}\right)(P_t f) + 2 \Gamma(P_t f) = 2 \Gamma(P_tf), multiplying this identity by h_n^2(y) e^{g(y,t)}, where h_n is the usual localizing sequence , and integrating by parts, we obtain
0 = 2 \int_0^\tau \int_{\mathbb{M}} h_n^2 e^g \Gamma(P_tf) d\mu(y) dt - \int_0^\tau \int_{\mathbb{M}}h_n^2 e^g \left(L-\frac{\partial }{\partial t}\right)(P_tf)^2 d\mu(y) dt
= 2 \int_0^\tau \int_{\mathbb{M}} h_n^2 e^g \Gamma(P_tf) d\mu(y) dt + 4 \int_0^\tau \int_{\mathbb{M}} h_n e^g P_tf \Gamma(h_n,P_tf) d\mu(y) dt + 2 \int_0^\tau\int_{\mathbb{M}}h_n^2 e^g P_tf \Gamma(P_tf,g)d\mu(y) dt
- \int_0^\tau \int_{\mathbb{M}} h_n e^g (P_tf)^2 \frac{\partial g}{\partial t} d\mu(y) dt - \int_{\mathbb{M}} h_n e^g (P_tf)^2 d\mu(y)\bigg|_{t=0} + \int_{\mathbb{M}} h_n e^g (P_tf)^2 d\mu(y)\bigg|_{t=\tau}
\ge 2 \int_0^\tau \int_{\mathbb{M}} h_n^2 e^g \left(\Gamma(P_tf)+P_tf\Gamma(P_tf,g)+\frac{P_tf^2}{4} \Gamma(g)\right) d\mu(y) dt + 4 \int_0^\tau \int_{\mathbb{M}} h_n e^g P_tf \Gamma(h_n,P_tf) d\mu(y) dt + \int_{\mathbb{M}} h_n e^g (P_tf)^2 d\mu(y)\bigg|_{t=\tau} - \int_{\mathbb{M}} h_n e^g (P_tf)^2 d\mu(y)\bigg|_{t=0},
where in the last inequality we have made use of the assumption on g. From this we conclude
\int_{\mathbb{M}} h_n e^g (P_t f)^2 d\mu(y)\bigg|_{t=\tau} \le \int_{\mathbb{M}} h_n e^g (P_tf)^2 d\mu(y)\bigg|_{t=0} - 4 \int_0^\tau \int_{\mathbb{M}} h_n e^g P_tf \Gamma(h_n,P_tf) d\mu(y) dt.
We now claim that
\underset{n\to \infty}{\lim} \int_0^\tau \int_{\mathbb{M}} h_n e^g P_tf \Gamma(h_n,P_tf) d\mu(y) dt = 0.
To see this we apply Cauchy-Schwarz inequality which gives
\left|\int_0^\tau \int_{\mathbb{M}} h_n e^g P_tf \Gamma(h_n,P_tf) d\mu(y) dt\right|\le \left(\int_0^\tau \int_{\mathbb{M}} h_n^2 e^g (P_tf)^2 \Gamma(h_n) d\mu(y) dt\right)^{\frac{1}{2}} \left(\int_0^\tau \int_{\mathbb{M}} e^g \Gamma(P_tf) d\mu(y) dt\right)^{\frac{1}{2}}
\le \left(\int_0^\tau \int_{\mathbb{M}} e^g (P_tf)^2 \Gamma(h_n) d\mu(y) dt\right)^{\frac{1}{2}} \left(\int_0^\tau \int_{\mathbb{M}} e^g \Gamma(P_tf) d\mu(y) dt\right)^{\frac{1}{2}} \to 0,
as n\to \infty. With the claim in hands we now let n\to \infty in the above inequality
obtaining
\int_{\mathbb{M}} e^{g(y,t)} ) (P_t f)^2 (y) d\mu(y) \le \int_{\mathbb{M}} e^{g(y,0)} ) f^2 (y) d\mu(y)
\square

We are now ready for the main result of this lecture:

Theorem: For any 0 < \epsilon < 1 there exist positive constants C_1=C_1(\epsilon) and C_2=C_2 (n,\epsilon), such that for every x,y\in \mathbb{M} and t > 0 one has
p(x,y,t)\le \frac{C_1}{\mu(B(x,\sqrt t))^{\frac{1}{2}}\mu(B(y,\sqrt t))^{\frac{1}{2}}} \exp \left(C_2Kt-\frac{d(x,y)^2}{(4+\epsilon)t}\right).

Proof: Given T > 0, and \alpha > 0 we fix 0 < \tau \le (1+\alpha)T. For a function \psi\in C^\infty_0(\mathbb{M}), with \psi \ge 0, in \mathbb{M} \times (0,\tau) we consider the function
f(y,t) = \int_{\mathbb{M}} p(y,z,t) p(x,z,T) \psi(z) d\mu(z),\ \ \ x\in \mathbb{M}.
Since f = P_t(p(x,\cdot,T)\psi), it satisfies the Cauchy problem
\begin{cases} Lf - f_t = 0 \ \ \ \ \text{in}\ \mathbb{M} \times (0,\tau), \\ f(z,0) = p(x,z,T)\psi(z),\ \ \ z\in \mathbb{M}. \end{cases}
Let g :\mathbb{M}\times [0,\tau] \to \mathbb{R} be a non positive continuous function such that, in the sense of distributions,
\frac{\partial g}{\partial t} +\frac{1}{2} \Gamma(g) \le 0.
From the previous lemma, we know that:
\int_{\mathbb{M}} e^{g(y,\tau)} f^2(y,\tau) d\mu(y) \le \int_{\mathbb{M}} e^{g(y,0)} f^2(y,0) d\mu(y).
At this point we fix x\in \mathbb{M} and for 0 < t \le\tau consider the indicator function \mathbf 1_{B(x,\sqrt t)} of the ball B(x,\sqrt t). Let \psi_k\in C^\infty_0(\mathbb{M}), \psi_k \ge 0, be a sequence such that \psi_k \to \mathbf 1_{B(x,\sqrt t)} in L^2(\mathbb{M}), with supp\ \psi_k\subset B(x,100\sqrt t). Slightly abusing the notation we now set f(y,s) = P_s(p(x,\cdot,T)\mathbf{1}_{B(x,\sqrt t)})(y) = \int_{B(x,\sqrt t)} p(y,z,s) p(x,z,T) d\mu(z). Thanks to the symmetry of p(x,y,s) = p(y,x,s), we have
f(x,T) = \int_{B(x,\sqrt t)} p(x,z,T)^2 d\mu(z).

Applying the integrated maximum principle to f_k(y,s) = P_s(p(x,\cdot,T)\psi_k)(y), we find
\int_{\mathbb{M}} e^{g(y,\tau)} f^2_k(y,\tau) d\mu(y) \le \int_{\mathbb{M}} e^{g(y,0)} f^2_k(y,0) d\mu(y).
At this point we observe that as k\to \infty
\left|\int_{\mathbb{M}} e^{g(y,\tau)} f^2_k(y,\tau) d\mu(y) - \int_{\mathbb{M}} e^{g(y,\tau)} f^2(y,\tau) d\mu(y)\right|
\le 2 ||e^{g(\cdot,\tau)}||_{L^\infty(\mathbb{M})} ||p(x,\cdot,T)||_{L^2(\mathbb{M})} ||p(x,\cdot,\tau)||_{L^\infty(B(x,110 \sqrt t))} ||\psi_k - \mathbf 1_{B(x,\sqrt t)}||_{L^2(\mathbb{M})} \to 0.
By similar considerations we find
\left|\int_{\mathbb{M}} e^{g(y,0)} f^2_k(y,0) d\mu(y) - \int_{\mathbb{M}} e^{g(y,0)} f^2(y,0) d\mu(y)\right|
\le 2 ||e^{g(\cdot,0)}||_{L^\infty(\mathbb{M})} ||p(x,\cdot,T)||_{L^\infty(B(x,110 \sqrt t))} ||\psi_k - \mathbf 1_{B(x,\sqrt t)}||_{L^2(\mathbb{M})} \to 0.
Letting k\to \infty we thus conclude that the same inequality holds with f_k replaced by f(y,s) = P_s(p(x,\cdot,T)1_{B(x,\sqrt t)})(y). This implies in particular the basic estimate
\underset{z\in B(x,\sqrt t)}{\inf}\ e^{g(z,\tau)} \int_{B(x,\sqrt t)} f^2(z,\tau) d\mu(z)
\le \int_{B(x,\sqrt t)} e^{g(z,\tau)} f^2(z,\tau) d\mu(z) \le \int_{\mathbb{M}} e^{g(z,\tau)} f^2(z,\tau) d\mu(z)
\le \int_{\mathbb{M}} e^{g(z,0)} f^2(z,0) d\mu(z) = \int_{B(y,\sqrt t)} e^{g(z,0)} p(x,z,T)^2 d\mu(z)
\le \underset{z\in B(y,\sqrt t)}{\sup}\ e^{g(z,0)} \int_{B(y,\sqrt t)} p(x,z,T)^2 d\mu(z).

At this point we choose g(y,t) = g_x(y,t) = - \frac{d(x,y)^2}{2((1+2\alpha) T - t)}. Using the fact that \Gamma(d)\le 1, one can easily check that g satisfies
\frac{\partial g}{\partial t} +\frac{1}{2} \Gamma(g) \le 0.
Taking into account that
\inf_{z\in B(x,\sqrt t)} e^{g_x(z,\tau)} = \inf_{z\in B(x,\sqrt t)} e^{-\frac{d(x,z)^2} {2((1+2\alpha)T- \tau)}} \ge e^{\frac{-t}{2((1+2\alpha)T- \tau)}},
if we now choose \tau = (1+\alpha)T, then from the previous inequality we conclude that
\int_{B(x,\sqrt t)} f^2(z,(1+\alpha)T) d\mu(z) \le \left(\underset{z\in B(y,\sqrt t)}{\sup}\ e^{-\frac{d(x,z)^2}{2(1+2\alpha)T} + \frac{t}{2\alpha T}}\right) \int_{B(y,\sqrt t)} p(x,z,T)^2 d\mu(z).
We now apply the Harnack inequality which gives for every z\in B(x,\sqrt t)
f(x,T)^2 \le f(z,(1+\alpha)T)^2(1+\alpha)^{n} e^{\frac{t}{2\alpha T}+\frac{Kt}{3} +\frac{nK\alpha T}{2 }}.
Integrating this inequality on B(x,\sqrt t) we find
\left(\int_{B(y,\sqrt t)} p(x,z,T)^2 d\mu(z)\right)^2 = f(x,T)^2
\le \frac{(1+\alpha)^{n} e^{\frac{t}{2\alpha T}+\frac{Kt}{3} +\frac{nK\alpha T}{2 }}}{\mu(B(x,\sqrt t)) } \int_{B(x,\sqrt t)} f^2(z,(1+\alpha)T) d\mu(z).
Thus, we obtain
\int_{B(y,\sqrt t)} p(x,z,T)^2 d\mu(z) \le \frac{(1+\alpha)^{n} e^{\frac{t}{2\alpha T}+\frac{Kt}{3} +\frac{nK\alpha T}{2 }}}{\mu(B(x,\sqrt t)) } \left(\underset{z\in B(y,\sqrt t)}{\sup}\ e^{-\frac{d(x,z)^2}{2(1+2\alpha)T} + \frac{t}{2\alpha T}}\right).
Choosing T = (1+\alpha)t in this inequality we find
\int_{B(y,\sqrt t)} p(x,z,(1+\alpha)t)^2 d\mu(z) \le \frac{(1+\alpha)^{n} e^{\frac{Kt}{3} +\frac{nK}{2} \alpha (1+\alpha)t+ \frac{1}{2\alpha (1+\alpha)}}}{\mu(B(x,\sqrt t))} \left(\underset{z\in B(y,\sqrt t)}{\sup}\ e^{-\frac{d(x,z)^2}{2(1+2\alpha)(1+\alpha)t} + \frac{1}{2\alpha (1+\alpha)}}\right).
We now apply again the Harnack inequality to obtain for every z\in B(y,\sqrt t)
p(x,y,t)^2 \le p(x,z,(1+\alpha)t)^2 (1+\alpha)^{n} \exp\left(\frac{1}{2\alpha }+\frac{Kt}{3}+\frac{nK\alpha t}{4} \right).
Integrating this inequality in z\in B(y,\sqrt t), we have
\mu(B(y,\sqrt t)) p(x,y,t)^2 \le (1+\alpha)^{n} \exp\left(\frac{1}{2\alpha }+\frac{Kt}{3}+\frac{nK\alpha t}{4} \right) \int_{B(y,\sqrt t)} p(x,z,(1+\alpha)t)^2 d\mu(z).
Combining this inequality with the above inequality we conclude
p(x,y,t) \le \frac{(1+\alpha)^{n} e^{\frac{3+\alpha}{4\alpha(1+\alpha)} +\frac{Kt}{3} +\frac{nKt}{4} \left( \alpha^2 +\frac{3}{2} \alpha \right) }}{\mu(B(x,\sqrt t))^{\frac{1}{2}}\mu(B(y,\sqrt t))^{\frac{1} {2}}}\left(\underset{z\in B(y,\sqrt t)}{\sup}\ e^{-\frac{d(x,z)^2}{4(1+2\alpha)(1+\alpha)t}}\right).
If now x\in B(y,\sqrt t), then
d(x,z)^2 \ge (d(x,y) - \sqrt t)^2 > d(x,y)^2 - t,
and therefore
\underset{z\in B(y,\sqrt t)}{\sup}\ e^{-\frac{d(x,z)^2}{4(1+2\alpha)(1+\alpha)t}} \le e^{\frac{1}{4(1+2\alpha)(1+\alpha)}} e^{-\frac{d(x,y)^2}{4(1+2\alpha)(1+\alpha)t}}.
If instead x\not\in B(y,\sqrt t), then for every \delta > 0 we have d(x,z)^2 \ge (1-\delta) d(x,y)^2 - (1+ \delta^{-1}) t. Choosing \delta = \alpha/(\alpha+1) we find d(x,z)^2 \ge \frac{d(x,y)^2}{1+\alpha} - (2 + \alpha^{-1}) t, and therefore
\underset{z\in B(y,\sqrt t)}{\sup}\ e^{-\frac{d(x,z)^2}{4(1+2\alpha)(1+\alpha)t}} \le e^{-\frac{d(x,y)^2} {4(1+2\alpha)(1+\alpha)^2 t} + \frac{2 + \alpha^{-1}}{4(1+2\alpha)(1+\alpha)}}
For any \epsilon > 0 we now choose \alpha > 0 such that 4(1+2\alpha)(1+\alpha)^2 = 4+\epsilon to reach the desired conclusion \square

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Lecture 17. The parabolic Harnack inequality

Posted on September 18, 2013 by Fabrice Baudoin

Let \mathbb{M} be a complete n-dimensional Riemannian manifold and, as usual, denote by L its Laplace-Beltrami operator. Throughout the Lecture, we will assume that the Ricci curvature of \mathbb{M} is bounded from below by -K with K \ge 0. Our purpose is to prove a first important consequence of the Li-Yau inequality: The parabolic Harnack inequality.

Theorem: Let f \in L^\infty(\mathbb{M}), f \ge 0. For every s \le t and x,y \in \mathbb{M},
P_s f(x) \le P_t f(y) \left(\frac{t}{s}\right)^{\frac{n}{2}} \exp\left(\frac{d(x,y)^2}{4(t-s)} +\frac{K d(x,y)^2}{6} +\frac{nK}{4}(t-s)\right).

Proof: We first assume that f \in C_0^\infty(\mathbb{M}). Let x,y \in \mathbb{M} and let \gamma:[s,t] \to \mathbb{M}, s < t be an absolutely continuous path such that \gamma(s)=x, \gamma(t)=y.
We write the Li-Yau inequality in the form
\Gamma( \ln P_u f (x) ) \le a(u) \frac{ L P_u f (x)}{P_u f (x)} +b(u),
where a(u)=1+\frac{2K}{3} u, and b(u)=\frac{n}{2} \left(\frac{1}{u}+\frac{K^2 u}{3}+K \right). Let us now consider \phi(u)=\ln P_u f(\gamma(u)). We compute \phi'(u)= ( \partial_u \ln P_u f) (\gamma(u))+\langle \nabla \ln P_u f (\gamma(u)),\gamma'(u) \rangle. Now, for every \lambda > 0, we have
\langle \nabla \ln P_u f (\gamma(u)),\gamma'(u) \rangle \ge -\frac{1}{2\lambda^2} \| \nabla \ln P_u f (x) \|^2 -\frac{\lambda^2}{2} \| \gamma'(u) \|^2.
Choosing \lambda=\sqrt{\frac{a(u)}{2} } and using then the Li-Yau inequality yields
\phi'(u) \ge -\frac{b(u)}{a(u)} -\frac{1}{4} a(u) \| \gamma'(u) \|^2.
By integrating this inequality from s to t we get as a result.
\ln P_tf(y)-\ln P_s f(x)\ge -\int_s^t \frac{b(u)}{a(u)} du -\frac{1}{4} \int_s^t a(u) \| \gamma'(u) \|^2 du.
We now minimize the quantity \int_s^t a(u) \| \gamma'(u) \|^2 du over the set of absolutely continuous paths such that \gamma(s)=x, \gamma(t)=y. By using reparametrization of paths, it is seen that
\int_s^t a(u) \| \gamma'(u) \|^2 du \ge \frac{d^2(x,y)}{\int_s^t \frac{dv}{a(v)}},
with equality achieved for \gamma(u)=\sigma\left( \frac{\int_s^u \frac{dv}{a(v)}}{\int_s^t \frac{dv}{a(v)}} \right) where \sigma:[0,1] \to \mathbb{M} is a unit geodesic joining x and y. As a conclusion,
P_sf(x) \le \exp\left( \int_s^t \frac{b(u)}{a(u)} du + \frac{d^2(x,y)}{4\int_s^t \frac{dv}{a(v)}} \right) P_tf(y).
Now, from Cauchy-Schwarz inequality we have
\int_s^t \frac{dv}{a(v)} \ge \frac{(t-s)^2}{\int_s^t a(v)dv}=\frac{(t-s)^2}{(t-s)+\frac{2K}{3}(t-s)^2 },
and also
\int_s^t \frac{b(u)}{a(u)} du=\frac{n}{2} \int_s^t \frac{ 1/u +K^2 u/3+K}{1+2Ku/3} du\le \frac{n}{2} \int_s^t \left( \frac{ 1}{u}+\frac{K}{2} \right) du.
This proves the inequality when f \in C_0^\infty(\mathbb{M}). We can then extend the result to f \in L^\infty(\mathbb{M}) by considering the approximations h_n P_\tau f \in C_0^\infty(\mathbb{M}) , where h_n \in C_0^\infty(\mathbb{M}), h_n \ge 0, h_n \to_{n \to \infty} 1 and let n \to \infty and \tau \to 0 \square

The following result represents an important consequence of the Harnack inequality.

Corollary: Let p(x,y,t) be the heat kernel on \mathbb{M}. For every x,y, z\in \mathbb{M} and every 0 < s < t < \infty one has
p(x,y,s) \le p(x,z,t) \left(\frac{t}{s}\right)^{\frac{n}{2}} \exp\left(\frac{d(y,z)^2}{4(t-s)} +\frac{K d(y,z)^2}{6} +\frac{nK}{4}(t-s)\right).

Proof: Let \tau > 0 and x\in \mathbb{M} be fixed. By the hypoellipticity of L, we know that p(x,\cdot,\cdot + \tau)\in C^\infty(\mathbb{M} \times (-\tau,\infty)). From the semigroup property we
have p (x,y,s+\tau)=P_s (p(x,\cdot,\tau))(y) and p (x,z,t+\tau)=P_t (p(x,\cdot,\tau))(z). Since we cannot apply the inequality directly to u(y,t) = P_t(p(x,\cdot,\tau))(y), we consider u_n(y,t) = P_t(h_n p(x,\cdot,\tau))(y), where h_n\in C^\infty_0(\mathbb{M}), 0\le h_n\le 1, and h_n\nearrow 1. From Harnack’s inequality we find
P_s (h_np(x,\cdot,\tau))(y) \le P_t (h_np(x,\cdot,\tau))(z) \left(\frac{t}{s}\right)^{\frac{n}{2}} \exp\left(\frac{d(y,z)^2}{4(t-s)} +\frac{K d(y,z)^2}{6} +\frac{nK}{4}(t-s)\right).
Letting n \to \infty, by Beppo Levi’s monotone convergence theorem we obtain
p (x,y,s+\tau) \le p (x,z,t+\tau) \left(\frac{t}{s}\right)^{\frac{n}{2}} \exp\left(\frac{d(y,z)^2}{4(t-s)} +\frac{K d(y,z)^2}{6} +\frac{nK}{4}(t-s)\right).
The desired conclusion follows by letting \tau \to 0 \square

A nice consequence of the parabolic Harnack inequality for the heat kernel is the following lower bound for the heat kernel:

Proposition: For x,z \in \mathbb{M} and t > 0,
p(x,z,t) \ge \frac{1}{(4\pi t)^{n/2}} \exp\left(-\frac{d(x,z)^2}{4t} -\frac{K d(x,z)^2}{6} -\frac{nK}{4}t \right).

Proof: We just need to use the above Harnack inequality with y=x and let s \to 0 using the asymptotics \lim_{s\to 0} s^{n/2} p_s(x,x)=\frac{1}{(4\pi )^{n/2}}. \square

Observe that when K=0, the inequality is sharp, since it is actually an equality on the Euclidean space !

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Lecture 16. The Li-Yau inequality

Posted on September 17, 2013 by Fabrice Baudoin

Let \mathbb{M} be a complete n-dimensional Riemannian manifold and, as usual, denote by L its Laplace-Beltrami operator. Throughout the Lecture, we will assume again that the Ricci curvature of \mathbb{M} is bounded from below by \rho \in \mathbb{R}. The Lecture is devoted to the proof of a beautiful inequality due to P. Li and S.T. Yau.
Henceforth, we will indicate C_b^\infty(\mathbb M) = C^\infty(\mathbb M)\cap L^\infty(\mathbb M).

Lemma: Let f \in C^\infty_b(\mathbb{M}), f > 0 and T > 0, and consider the function
\phi (x,t)=(P_{T-t} f) (x)\Gamma (\ln P_{T-t}f)(x),
which is defined on \mathbb{M} \times [0,T]. We have
L\phi+\frac{\partial \phi}{\partial t} =2 (P_{T-t} f) \Gamma_2 (\ln P_{T-t}f)

Proof: Let for simplicity g(x,t) = P_{T-t} f(x). A simple computation gives \frac{\partial \phi}{\partial t} = g_t \Gamma(\ln g) + 2 g \Gamma \left(\ln g,\frac{g_t}{g}\right).
On the other hand,
L\phi = Lg \Gamma(\ln g) + g L \Gamma(\ln g) + 2 \Gamma(g,\Gamma(\ln g)).
Combining these equations we obtain
L\phi + \frac{\partial \phi}{\partial t} = g L\Gamma(\ln g) + 2\Gamma(g,\Gamma(\ln g)) + 2 g \Gamma \left(\ln g,\frac{g_t}{g}\right).
From the above equation we see that
2 g \Gamma_2(\ln g) = g (L \Gamma(\ln g) - 2 \Gamma(\ln g,L(\ln g)))
= g L\Gamma(\ln g) - 2 g \Gamma(\ln g,L(\ln g)).
Observing that
L(\ln g) = - \frac{\Gamma(g)}{g^2} - \frac{g_t}{g},
we conclude that
L\phi+\frac{\partial \phi}{\partial t} =2 (P_{T-t} f) \Gamma_2 (\ln P_{T-t}f)
\square

We now turn to an important variational inequality that shall extensively be used throughout these lectures. Given a function f\in C^\infty_b(\mathbb{M}) and \varepsilon > 0, we let f_\varepsilon=f+\varepsilon.

Suppose that T > 0, and x\in \mathbb{M} be given. For a function f\in C^\infty_b(\mathbb M) with f \ge 0 we define for t\in [0,T], \Phi (t)=P_t \left( (P_{T-t} f_\varepsilon) \Gamma (\ln P_{T-t}f_\varepsilon) \right).

Theorem: Let a \in C^1([0,T],[0,\infty)) and \gamma \in C((0,T),\mathbb R). Given f \in C_0^\infty(\mathbb M), with f\ge 0, we have
a(T) P_T \left( f_\varepsilon \Gamma (\ln f_\varepsilon) \right) -a(0)(P_{T} f_\varepsilon) \Gamma (\ln P_{T}f_\varepsilon)
\ge \int_0^T \left(a'+2\rho a -\frac{4a\gamma}{n} \right)\Phi (s) ds +\left(\frac{4}{n}\int_0^T a\gamma ds\right)LP_{T} f_\varepsilon -\left(\frac{2 }{n}\int_0^T a\gamma^2ds\right)P_T f_\varepsilon.

Proof: Let f \in C^\infty(\mathbb{M}), f \ge 0. Consider the function
\phi (x,t)=a(t)(P_{T-t} f) (x)\Gamma (\ln P_{T-t}f)(x).
Applying the previous lemma and the curvature-dimension inequality, we obtain
L\phi+\frac{\partial \phi}{\partial t} =a' (P_{T-t} f) \Gamma (\ln P_{T-t}f)+2a (P_{T-t} f) \Gamma_2 (\ln P_{T-t}f)
\ge \left(a'+2\rho a \right)(P_{T-t} f) \Gamma (\ln P_{T-t}f)+\frac{2a}{n} (P_{T-t} f) (L(\ln P_{T-t} f))^2.
But, we have
(L(\ln P_{T-t} f))^2 \ge 2\gamma L(\ln P_{T-t}f) -\gamma^2,
and
L(\ln P_{T-t}f)=\frac{LP_{T-t}f}{P_{T-t}f} -\Gamma(\ln P_{T-t} f ).
Therefore we obtain,
L\phi+\frac{\partial \phi}{\partial t} \ge \left(a'+2\rho a -\frac{4a\gamma}{n} \right) (P_{T-t} f) \Gamma (\ln P_{T-t}f) +\frac{4a\gamma}{n} LP_{T-t} f - \frac{2a\gamma^2}{n} P_{T-t} f.
We then easily reach the conclusion by using the parabolic comparison theorem in L^\infty \square

As a first application the previous result, we derive a family of Li-Yau type inequalities. We choose the function \gamma in a such a way that
a' -\frac{4a\gamma}{n} +2\rho a=0.
That is
\gamma=\frac{n}{4} \left( \frac{a'}{a}+2\rho \right).
Integrating the inequality from 0 to T, and denoting V=\sqrt{a}, we obtain the following result.

Proposition: Let V:[0,T]\rightarrow \mathbb{R}^+ be a smooth function such that V(0)=1, V(T)=0. We have
\Gamma (\ln P_T f) \le \left( 1-2\rho\int_0^T V^2(s) ds\right) \frac{L P_Tf}{P_T f} +\frac{n}{2} \left( \int_0^T V'(s)^2 ds +\rho^2 \int_0^T V(s)^2 ds -\rho \right).

A first family of interesting inequalities may be obtained with the choice V(t)=\left( 1-\frac{t}{T}\right)^\alpha, \alpha>\frac{1}{2}. In this case we have \int_0^T V(s)^2 ds=\frac{T}{2\alpha+1} and \int_0^T V'(s)^2 ds=\frac{\alpha^2}{(2\alpha-1)T}. In particular, we therefore proved the celebrated Li-Yau inequality:

Theorem: If f \in C_0^\infty(\mathbb{M}), f \ge 0. For \alpha > \frac{1}{2} and T > 0, we have
\Gamma (\ln P_T f) \le \left( 1-\frac{2\rho T}{2\alpha+1}\right) \frac{L P_Tf}{P_T f} +\frac{n}{2} \left( \frac{\alpha^2}{(2\alpha-1)T}+\frac{\rho^2 T}{2\alpha+1} -\rho \right).

In the case, \rho=0 and \alpha=1, it reduces to the beautiful sharp inequality:
\Gamma(\ln P_t f) \le \frac{ L P_t f }{P_t f } + \frac{n}{2t}.

Although in the sequel, we shall first focus on the case \rho=0, let us presently briefly discuss the case \rho > 0.

Using the Li-Yau inequality with \alpha=3/2 leads to the Bakry-Qian inequality:
\frac{L P_tf}{P_t f}\le \frac{n \rho}{4},\ \ \ \ \ t \ge \frac{2}{\rho}.
Also, by using
V(t)=\frac{e^{-\frac{\rho t}{3}} (e^{-\frac{2\rho t}{3}}-e^{-\frac{2\rho T}{3}})}{1-e^{-\frac{2\rho T}{3}}},
we obtain the following inequality that shall be later used in the lectures:
\Gamma(\ln P_t f) \le e^{-\frac{2\rho t}{3}} \frac{ L P_t f }{P_t f } +\frac{n\rho}{3} \frac{e^{-\frac{4\rho t}{3}}}{ 1-e^{-\frac{2\rho t}{3}}}

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Lecture 15. Convergence of the semigroup, Poincare and log-Sobolev

Posted on September 12, 2013 by Fabrice Baudoin

Let \mathbb{M} be a complete n-dimensional Riemannian manifold and denote by L its Laplace-Beltrami operator. As usual, we denote by P_t the heat semigroup generated by L. Throughout the Lecture, we will assume that the Ricci curvature of \mathbb{M} is bounded from below by \rho > 0. We recall that this is equivalent to the fact that for every f \in C^\infty(\mathbb{M}),
\Gamma_2(f,f) \ge \frac{1}{n} (Lf)^2 + \rho \Gamma(f,f).
Readers knowing Riemannian geometry know that from Bonnet-Myers theorem, the manifold needs to be compact and we therefore expect the semigroup to converge to equilibrium. However for several lectures, our goal will be to not use the Bonnet-Myers theorem, because eventually we shall provide a proof of this fact using semigroup theory. Thus the results in this Lecture will not use the compactness of \mathbb{M}.

Lemma: The Riemannian measure \mu is finite, i.e. \mu(\mathbb{M}) < +\infty and for every f \in L_\mu^2(\mathbb{M}), the following convergence holds pointwise and in L^2_\mu(\mathbb{M}),
P_t f \to_{t \to +\infty} \frac{1}{\mu(\mathbb{M})} \int_\mathbb{M} f d\mu.

Proof: Let f,g \in C^\infty_0(\mathbb{M}), we have
\int_{\mathbb{M}} (P_t f -f) g d\mu = \int_0^t \int_{\mathbb{M}}\left( \frac{\partial}{\partial s} P_s f \right) g d\mu ds
= \int_0^t \int_{\mathbb{M}}\left(L P_s f \right) g d\mu ds=- \int_0^t \int_{\mathbb{M}} \Gamma ( P_s f , g) d\mu ds.
By means of Cauchy-Schwarz inequality, we
find
\left| \int_{\mathbb{M}} (P_t f -f) g d\mu \right| \le \left(\int_0^t e^{-2\rho s} ds\right) \sqrt{ \| \Gamma (f) \|_\infty } \int_{\mathbb{M}}\Gamma (g)^{\frac{1}{2}}d\mu.
Now it is seen from spectral theorem that in L^2(\mathbb{M}) we have a convergence P_t f \to P_\infty f, where P_\infty f belongs to the domain of L. Moreover LP_\infty f=0. By ellipticity of L we deduce that P_\infty f is a smooth function. Since LP_\infty f=0, we have \Gamma(P_\infty f)=0 and therefore P_\infty f is constant.

Let us now assume that \mu(\mathbb{M})=+\infty. This implies in particular that P_\infty f =0 because no constant besides 0 is in L^2(\mathbb{M}). Using then the previous inequality and letting t \to +\infty, we infer
\left| \int_{\mathbb{M}} f g d\mu \right| \le \left(\int_0^{+\infty} e^{-2\rho s} ds\right) \sqrt{ \| \Gamma (f) \|_\infty } \int_{\mathbb{M}}\Gamma (g)^{\frac{1}{2}}d\mu.
Let us assume g \ge 0, g \ne 0 and take for f the usual localizing sequence h_n. Letting n \to \infty, we deduce \int_\mathbb{M} g d\mu \le 0, which is clearly absurd. As a consequence \mu (\mathbb{M}) < +\infty.

The invariance of \mu implies then \int_\mathbb{M} P_\infty f d\mu =\int_\mathbb{M} f d\mu,
and thus P_\infty f =\frac{1}{\mu(\mathbb{M})} \int_\mathbb{M} f d\mu. Finally, using the Cauchy-Schwarz inequality, we find that for x \in \mathbb{M}, f \in L^2(\mathbb{M}), s,t,\tau \ge 0,
| P_{t+\tau} f (x)-P_{s+\tau} f (x) | = | P_\tau (P_t f -P_s f) (x) |
=\left| \int_\mathbb{M} p(\tau, x, y) (P_t f -P_s f) (y) \mu(dy) \right|
\le \int_\mathbb{M} p(\tau, x, y)^2 \mu(dy) \| P_t f -P_s f\|^2_2
\le p(2\tau,x,x) \| P_t f -P_s f\|^2_2.

Thus, we also have P_t f (x) \to_{t \to +\infty} \frac{1}{\mu(\mathbb{M})} \int_{\mathbb{M}} f d\mu \square

Proposition: The following Poincare inequality is satisfied: For f \in \mathcal{D}(L), \frac{1}{\mu(\mathbb{M})}\int_{\mathbb{M}} f^2 d\mu \le \left( \frac{1}{\mu(\mathbb{M})} \int_{\mathbb{M}} f d\mu \right)^2 +\frac{n-1}{n\rho} \frac{1}{\mu(\mathbb{M})} \int_{\mathbb{M}} \Gamma(f,f) d\mu.

Let f \in C_0^\infty(\mathbb{M}). We have by assumption \Gamma_2(f,f) \ge \frac{1}{n} (Lf)^2 + \rho \Gamma (f,f). Therefore, by integrating the latter inequality we obtain
\int_{\mathbb{M}} \Gamma_2(f,f) d\mu \ge \frac{1}{n} (Lf)^2+ \rho \int_{\mathbb{M}} \Gamma (f,f) d\mu.
But we have
\int_{\mathbb{M}} \Gamma_2(f,f) d\mu=-\int_{\mathbb{M}} \Gamma(f,Lf) d\mu=\int_{\mathbb{M}} (Lf)^2 d\mu.
Therefore we obtain
\int_{\mathbb{M}} (Lf)^2 d\mu \ge \rho \int_{\mathbb{M}} \Gamma (f,f) d\mu=-\frac{n\rho}{n-1} \int_{\mathbb{M}} fLf d\mu.
By density, this last inequality is seen to hold for every function f\in \mathcal{D}(L). It means that the L^2 spectrum of -L lies in \{0\} \cup \left[\frac{n\rho}{n-1} ,+\infty\right). Since from the previous proof the projection of f onto the 0-eigenspace is given by \frac{1}{\mu(\mathbb{M})} \int_{\mathbb{M}} f d\mu, we deduce that
\int_\mathbb{M} \left( f-\frac{1}{\mu(\mathbb{M})} \int_\mathbb{M} f d\mu\right)^2d\mu \le \frac{n-1}{n\rho} \int_{\mathbb{M}} \Gamma(f,f) d\mu
which is exactly the inequality we wanted to prove \square

As observed in the proof, the Poincare inequality
\int_{\mathbb{M}} f^2 d\mu \le \left( \int_{\mathbb{M}} f d\mu \right)^2 +\frac{n-1}{n\rho} \int_{\mathbb{M}} \Gamma(f,f) d\mu.
is equivalent to the fact that the L^2 spectrum of -L lies in \{0\} \cup \left[\frac{n\rho}{n-1} ,+\infty\right), or in other words that -L has a spectral gap of size at least \frac{n\rho}{n-1}. This is Lichnerowicz estimate. It is sharp, because on the n-dimensional sphere it is known that \rho=n-1 and that the first non zero eigenvalue is exactly equal to n.

As a basic consequence of the spectral theorem and of the above spectral gap estimate, we also get the rate convergence to equilibrium in L^2_\mu(\mathbb{M}) for P_t.

Proposition: Let f \in L^2_\mu(\mathbb{M}), then for t \ge 0,
\left\| P_tf -\frac{1}{\mu(\mathbb{M})}\int_\mathbb{M} f d\mu \right\|^2_2 \le e^{-\frac{2n\rho}{n-1} t} \left\| f -\frac{1}{\mu(\mathbb{M})}\int_\mathbb{M} f d\mu \right\|^2_2 .

Exercise: By using the Riesz-Thorin interpolation theorem, show that for p \in [2,+\infty), and f \in L^p_\mu(\mathbb{M}),
\left\| P_tf -\frac{1}{\mu(\mathbb{M})}\int_\mathbb{M} f d\mu \right\|^p_p \le 2^{p-2} e^{-\frac{2n\rho}{n-1} t} \left\| f -\frac{1}{\mu(\mathbb{M})}\int_\mathbb{M} f d\mu \right\|^p_p.
By using duality, prove a corresponding statement when p \in (1,2].

As we have just seen, the convergence in L^2 of P_t is connected and actually equivalent to the Poincare inequality.

We now turn to the so-called log-Sobolev inequality which is connected to the convergence in entropy for P_t. This inequality is much stronger (and more useful) than the Poincare inequality. To simplify a little the expressions, we assume in the sequel that \mu(\mathbb{M})=1 (Otherwise, just replace \mu by \frac{\mu}{\mu(\mathbb{M})} in the following results).

Proposition: For f \in \mathcal{D}(L), f \ge 0,
\int_{\mathbb{M}} f^2 \ln f^2 d\mu \le \int_{\mathbb{M}} f^2 d\mu \ln \left( \int_{\mathbb{M}} f^2 d\mu \right) +\frac{2}{ \rho} \int_{\mathbb{M}} \Gamma(f,f) d\mu.

Proof: By considering \sqrt{f} instead of f, it is enough to show that if f is positive,
\int_{\mathbb{M}} f \ln f d\mu \le \int_{\mathbb{M}} f d\mu \ln \left( \int_{\mathbb{M}} f d\mu \right) +\frac{1}{2 \rho} \int_{\mathbb{M}} \frac{\Gamma(f,f)}{f} d\mu.
We now have
\int_{\mathbb{M}} f \ln f d\mu - \int_{\mathbb{M}} f d\mu \ln \left( \int_{\mathbb{M}} f d\mu \right)= - \int_0^{+\infty} \frac{d}{dt} \int_{\mathbb{M}} P_t f \ln P_t f d\mu dt
=- \int_0^{+\infty} \int_{\mathbb{M}} L P_t f \ln P_t f d\mu dt
= \int_0^{+\infty} \int_{\mathbb{M}} \Gamma (P_t f, \ln P_t f) d\mu dt
=\int_0^{+\infty} \int_{\mathbb{M}} \frac{\Gamma (P_t f, P_t f) }{P_t f} d\mu dt
Now, we know that
\Gamma (P_t f, P_t f) \le e^{-2\rho t} \left( P_t \sqrt{\Gamma(f,f)}\right)^2. And, from Cauchy-Schwarz inequality, (P_t \sqrt{\Gamma(f,f)})^2 \le P_t \frac{\Gamma(f,f)}{f} P_t f. Therefore,
\int_{\mathbb{M}} f \ln f d\mu - \int_{\mathbb{M}} f d\mu \ln \left( \int_{\mathbb{M}} f d\mu \right) \le \int_0^{+\infty} e^{-2\rho t} dt \int_{\mathbb{M}} \frac{\Gamma(f,f)}{f} d\mu,
which is the inequality we claimed \square

We finally prove the entropic convergence of P_t.

Theorem: Let f \in L^2_\mu(\mathbb{M}), f \ge 0. For t \ge 0,
\int_{\mathbb{M}} P_t f \ln P_t f d\mu - \int_{\mathbb{M}} P_t f d\mu \ln \left( \int_{\mathbb{M}} P_t f d\mu\right) \le e^{-2\rho t} \left( \int_{\mathbb{M}} f \ln f d\mu- \int_{\mathbb{M}} f d\mu \ln \left( \int_{\mathbb{M}} f d\mu\right)\right).

Proof: Let us assume \int_{\mathbb{M}} f d\mu=1, otherwise we use the following argument with \frac{f}{\int_\mathbb{R}^n f d\mu} and consider the functional
\Phi(t)= \int_{\mathbb{M}} P_t f \ln P_t f d\mu,
which by differentiation gives
\Phi'(t)= \int_{\mathbb{M}} LP_t f \ln P_t f d\mu=-\int_{\mathbb{M}} \frac{ \Gamma(P_t f)}{P_t f} d\mu.
Using now the log-Sobolev inequality, we obtain
\Phi'(t)\le -2\rho \Phi(t).
The Gronwall’s differential inequality implies then:
\Phi(t) \le e^{-2\rho t} \Phi(0),
that is
\int_{\mathbb{M}} P_t f \ln P_t f d\mu \le e^{-2\rho t} \int_{\mathbb{M}} f \ln f d\mu
\square

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Lecture 14. Stochastic completeness of the heat semigroup

Posted on September 10, 2013 by Fabrice Baudoin

In this Lecture, we will prove a first interesting consequence of the Bochner’s identity: We will prove that if, on a complete Riemannian manifold \mathbb{M}, the Ricci curvature is bounded from below, then the heat semigroup is stochastically complete, that is P_t 1=1. This result is due to S.T. Yau, and we will see this property is also equivalent to the uniqueness in L^\infty for solutions of the heat equation. The proof we give is due to D. Bakry.

Let \mathbb{M} be a complete Riemannian manifold and denote by L its Laplace-Beltrami operator. As usual, we denote by P_t the heat semigroup generated by L. Throughout the Lecture, we will assume that the Ricci curvature of \mathbb{M} is bounded from below by \rho \in \mathbb{R}. As seen in the previous Lecture, this is equivalent to the fact that for every f \in C^\infty(\mathbb{M}),
\Gamma_2(f,f) \ge \frac{1}{n} (Lf)^2 + \rho \Gamma(f,f).

We start with a technical lemma:

Lemma: If f \in L^2_\mu(\mathbb{M}), then for every t > 0, the functions \Gamma (P_t f), L\Gamma(P_t f), \Gamma(P_tf, LP_t f) and \Gamma_2 (P_t f) are in L^1_\mu(\mathbb{M}).

Proof: It is straightforward to see from the spectral theorem that \Gamma (P_t f) \in L^1_\mu(\mathbb{M}). Similarly, | \Gamma(P_tf, LP_t f) | \le \sqrt{\Gamma(P_tf) \Gamma(LP_tf) } \in L^1_\mu(\mathbb{M}). Since, \Gamma_2 (P_t f) =\frac{1}{2} \left( L\Gamma(P_t f) -2 \Gamma(P_tf, LP_t f)\right), we are let with the problem of proving that \Gamma_2 (P_t f) \in L^1_\mu(\mathbb{M}). If g \in C_0^\infty(\mathbb{M}), then an integration by parts easily yields \int_\mathbb{M} \Gamma_2 (g) d\mu=\int_\mathbb{M} (Lg)^2 d\mu. As a consequence,
\int_\mathbb{M} \Gamma_2 (g)-\rho \Gamma(g) d\mu=\int_\mathbb{M} (Lg)^2 +\rho g Lg d\mu,
and we obtain
\int_\mathbb{M} | \Gamma_2 (g)-\rho \Gamma(g) | d\mu \le\left(1 +\frac{1}{2}| \rho | \right) \int_\mathbb{M} (Lg)^2 d\mu +\frac{1}{2}| \rho | \int_\mathbb{M} g^2 d\mu.
Using a density argument, it is then easily proved that for g \in \mathcal{D}(L)\cap C^\infty(\mathbb{M}) we have
\int_\mathbb{M} | \Gamma_2 (g)-\rho \Gamma(g) | d\mu \le\left(1 +\frac{1}{2}| \rho | \right) \int_\mathbb{M} (Lg)^2 d\mu +\frac{1}{2}| \rho | \int_\mathbb{M} g^2 d\mu.
In particular, we deduce that if g \in \mathcal{D}(L)\cap C^\infty(\mathbb{M}), then \Gamma_2(g) \in L^1_\mu(\mathbb{M}) \square

We will also need the following fundamental parabolic comparison theorem that shall be extensively used throughout these lectures.

Proposition: Let T > 0. Let u,v: \mathbb{M}\times [0,T] \to \mathbb{R} be smooth functions such that:

  • For every t \in [0,T], u(\cdot,t) \in L^2(\mathbb{M}) and \int_0^T \| u(\cdot,t)\|_2 dt <\infty;
  • \int_0^T \| \sqrt{\Gamma(u) (\cdot,t)} \|_p dt <\infty for some 1 \le p \le \infty;
  • For every t \in [0,T], v(\cdot,t) \in L^q(\mathbb{M}) and \int_ 0^T \| v(\cdot,t ) \|_q dt <\infty for some 1 \le q \le \infty.

If the inequality
Lu+\frac{\partial u}{\partial t} \ge v,
holds on \mathbb{M}\times [0,T], then we have
P_T u(\cdot,T)(x) \ge u(x,0) +\int_0^T P_s v(\cdot,s)(x) ds.

Proof: Let f,g \in C_0^\infty (\mathbb{M}), f,g \ge 0. We claim that we must have
\int_\mathbb{M} g P_T(fu(\cdot,T)) d\mu - \int_\mathbb{M} g f u(x,0) d\mu
\ge - \|\sqrt{\Gamma(f)}\|_\infty \int_0^T \int_\mathbb{M} (P_t g) \sqrt{\Gamma(u)}d\mu dt- \| \sqrt{\Gamma(f)} \|_\infty \int_0^T \| \sqrt{\Gamma(P_t g) }\|_2 \| u(\cdot,t) \|_2 dt + \int_\mathbb{M} g \int_0^T P_t( f v(\cdot,t)) d\mu dt

where for every 1\le p \le \infty and a measurable F, we have let ||F||_p = ||F||_{L^p(\mathbb{M})}. To establish this, we consider the function
\phi(t)=\int_\mathbb{M} g P_t (fu(\cdot,t)) d\mu.
Differentiating \phi we find
\phi'(t) =\int_\mathbb{M} g P_t \left(L( fu) + f\frac{\partial u}{\partial t} \right) d\mu
= \int_\mathbb{M} g P_t \left((L f) u+2 \Gamma (f,u) +f Lu + f\frac{\partial u}{\partial t} \right) d\mu
\ge \int_\mathbb{M} g P_t \left((L f) u+2 \Gamma (f,u) \right) d\mu+\int_\mathbb{M} g P_t( f v) d\mu.
Since
\int_\mathbb{M} g P_t \left((L f) u\right) d\mu = \int_\mathbb{M} (P_t g) (L f) u d\mu
= -\int_\mathbb{M} \Gamma( f, u(P_t g)) d\mu
=-\left( \int_\mathbb{M} P_t g \Gamma( f, u)+ u \Gamma(f,P_t g) d\mu\right),
we obtain
\phi'(t) \ge \int_\mathbb{M} P_t g \Gamma(f,u) d\mu - \int_\mathbb{M} u \Gamma(f,P_t g) d\mu + \int_\mathbb{M} g P_t(fv) d\mu.
Now, we can bound
\left| \int_\mathbb{M} (P_t g) \Gamma( f, u) d\mu\right| \le \| \sqrt{\Gamma(f) } \|_\infty \int_\mathbb{M} (P_t g) \sqrt{\Gamma( u)} d\mu,
and for a.e. t\in [0,T] the integral in the right-hand side is finite. We have thus obtained
\phi'(t) \ge - \| \sqrt{\Gamma(f) } \|_\infty \int_\mathbb{M} (P_t g) \sqrt{\Gamma(u)} d\mu- \int_\mathbb{M} u \Gamma(f , P_t g) d\mu+ \int_\mathbb{M} g P_t( f v(\cdot,t)) d\mu.
As a consequence, we find
\int_\mathbb{M} g P_T (fu(\cdot,T)) d\mu-\int_\mathbb{M} gfu(x,0)d\mu
\ge - \| \sqrt{\Gamma(f) } \|_\infty \int_0^T \int_\mathbb{M} (P_t g) \sqrt{\Gamma(u)} d\mu dt - \int_0^T \int_\mathbb{M} u \Gamma\left(f,P_t g\right) d\mu dt + \int_0^T \int_\mathbb{M} g P_t(f v(\cdot,t)) d\mu dt
\ge - \| \sqrt{\Gamma(f) } \|_\infty \int_0^T \int_\mathbb{M} (P_t g) \sqrt{\Gamma(u)} d\mu dt - \int_0^T \| u(\cdot,t) \|_2 \| \Gamma(f, P_t g) \|_2 dt + \int_\mathbb{M} g \int_0^T P_t( f v(\cdot,t)) dt d\mu
\ge - \| \sqrt{\Gamma(f) } \|_\infty \int_0^T \int_\mathbb{M} (P_t g) \sqrt{\Gamma(u)} d\mu dt - \| \sqrt{\Gamma(f)} \|_\infty \int_0^T \| u(\cdot,t) \|_2 \| \sqrt{\Gamma(P_t g) }\|_2 dt + \int_\mathbb{M} g \int_0^T P_t(f v(\cdot,t)) dt d\mu,
which proves what we claimed.
Let now h_k\in C^\infty_0(\mathbb{M}) be a sequence such that 0 \le h_k \le 1, \| \Gamma(h_k) \|_\infty \to 0 and h_k increases to 1.
Using h_k in place of f and letting k \to \infty, gives
\int_\mathbb{M} g P_T (u(\cdot,T)) d\mu - \int_\mathbb{M} g u(x,0)d\mu \ge \int_\mathbb{M} g \int_0^T P_t(v(\cdot,t)) dt d\mu.
We observe that the assumption on v and Minkowski’s integral inequality guarantee that the function x\to \int_0^T P_t(v(\cdot,t))(x) dt belongs to L^q(\mathbb{M}). We have in fact
\left(\int_\mathbb{M} \left|\int_0^T P_t(v(\cdot,t)) dt\right|^q d\mu\right)^{\frac 1q} \le \int_0^T \left| \int_\mathbb{M} \left|P_t(v(\cdot,t))\right|^q d\mu\right|^{\frac 1q} dt \le \int_0^T \left| \int_\mathbb{M} \left|v(\cdot,t)\right|^q d\mu\right|^{\frac 1q} dt
\le T^{\frac{1}{q'}} \left(\int_0^T \int_\mathbb{M} \left|v(\cdot,t)\right|^q d\mu dt \right)^{\frac 1q} < \infty.

Since this must hold for every non negative g \in C_0^\infty (\mathbb{M}), we conclude that
P_T(u(\cdot,T))(x) \ge u(x,0) +\int_0^T P_s (v(\cdot,s))(x) ds,
which completes the proof \square

We are in position to prove the first gradient bound for the semigroup P_t.

Proposition: If f is a smooth function in \mathcal{D}(L), then for every t \ge 0 and x \in \mathbb{M},
\sqrt{\Gamma(P_t f)}(x) \le e^{-\rho t} P_t \sqrt{\Gamma(f)} (x).

Proof: We fix T > 0 and consider the functional
\Phi(x,t)=e^{-\rho t} \sqrt{ \Gamma(P_{T-t} f)}(x).
We first assume that (x,t)\to \Gamma(P_t f)(x) > 0 on \mathbb{M} \times [0,T]. From the previous lemma, we have \Phi(t) \in L^2(\mathbb{M}). Moreover \Gamma(\Phi)(t)=e^{-2\rho t}\frac{\Gamma(\Gamma(P_{T-t} f))}{4 \Gamma(P_{T-t} f)}. So, we have \Gamma( \Phi)(t) \le e^{-2\rho t}( \Gamma_2(P_{T-t} f)-\rho\Gamma(P_tf)). Therefore, again from the previous proposition , we deduce that \Gamma( \Phi)(t) \in L^1(\mathbb{M}). Next, we easily compute that
\frac{\partial \Phi}{\partial t}+ L\Phi =e^{-\rho t} \left( \frac{\Gamma_2(P_{T-t} f)}{\sqrt{\Gamma(P_{T-t} f)}}-\frac{\Gamma(\Gamma(P_{T-t} f))}{4 \Gamma(P_{T-t} f)^{3/2} } -\rho \sqrt{\Gamma(P_{T-t} f)} \right).
Thus,
\frac{\partial \Phi}{\partial t}+ L\Phi \ge 0.
We can then use the parabolic comparison theorem to infer that
\sqrt{ \Gamma(P_{T} f)} \le e^{-\rho T} P_T \left(\sqrt{\Gamma (f)} \right).
If (x,t) \to \Gamma(P_t f)(x) vanishes on \mathbb{M} \times [0,T], we consider the functional
\Phi(t)=e^{-\rho t} g_\varepsilon (\Gamma(P_{T-t} f) ),
where, for 0 < \varepsilon < 1,
g_\varepsilon (y)=\sqrt{ y+\varepsilon^2}-\varepsilon.
Since \Phi(t) \in L^2(\mathbb{M}), an argument similar to that above (details are let to the reader) shows that
g_\varepsilon (\Gamma(P_{T} f) )\le e^{-\rho T} P_T \left( g_\varepsilon( \Gamma (f)) \right).
Letting \varepsilon \to 0, we conclude that
\sqrt{ \Gamma(P_{T} f)} \le e^{-\rho T} P_T \left(\sqrt{\Gamma (f)} \right) \square

We now prove the promised stochastic completeness result:

Theorem: For t \ge 0, one has P_t 1 =1.

Proof: Let f,g \in C^\infty_0(\mathbb M), we have
\int_{\mathbb{M}} (P_t f -f) g d\mu = \int_0^t \int_{\mathbb{M}}\left( \frac{\partial}{\partial s} P_s f \right) g d\mu ds= \int_0^t\int_{\mathbb{M}}\left(L P_s f \right) g d\mu ds=- \int_0^t \int_{\mathbb{M}}\Gamma ( P_s f , g) d\mu ds.
By means of the previous Proposition and Cauchy-Schwarz inequality, we
find
\left| \int_{\mathbb{M}} (P_t f -f) g d\mu \right| \le \left(\int_0^t e^{-\rho s} ds\right) \sqrt{ \| \Gamma (f) \|_\infty } \int_{\mathbb{M}}\Gamma (g)^{\frac{1}{2}}d\mu.

We now apply the previous inequality with f = h_n, and then let n\to \infty.
Since by Beppo Levi’s monotone convergence theorem we have P_t h_n(x)\nearrow P_t 1(x) for every x\in \mathbb{M}, we see that the left-hand side converges to \int_{\mathbb{M}} (P_t 1 -1) g d\mu. We thus reach the conclusion
\int_{\mathbb{M}} (P_t 1 -1) g d\mu=0,\ \ \ g\in C^\infty_0(\mathbb{M}).
It follows that P_t 1 =1 \square

A consequence of the stochastic completeness is the uniqueness in L^\infty of solutions of the heat equation. More precisely, the following L^\infty parabolic comparison theorem holds.

Proposition: Let T > 0. Let u,v: \mathbb{M}\times [0,T] \to \mathbb{R} be smooth functions such that for every T > 0, \sup_{t \in [0,T]} \| u(\cdot,t)\|_\infty < \infty, \sup_{t \in [0,T]} \| v(\cdot,t)\|_\infty <\infty; If the inequality

Lu+\frac{\partial u}{\partial t} \ge v

holds on \mathbb{M}\times [0,T], then we have

P_T(u(\cdot,T))(x) \ge u(x,0) +\int_0^T P_s(v(\cdot,s))(x) ds.

Proof: Let (X^x_t)_{t \ge 0} be the diffusion Markov process with semigroup (P_t)_{t \ge 0} and started at x \in \mathbb{M}. From P_t1=1, we deduce that (X^x_t)_{t\ge 0} has an infinite lifetime. We have then for t \ge 0,
u\left( X^x_t, t \right)=u\left( x,0\right)+\int_0^t \left( Lu+\frac{\partial u}{\partial t}\right)(X^x_s,s) ds +M_t,
where (M_t)_{t \ge0 } is a local martingale. From the assumption one obtains
u\left( X^x_t, t \right) \ge u\left( x,0\right)+\int_0^t v(X^x_s,s) ds +M_t.
Let now (T_n)_{n \in \mathbb{N}} be an increasing sequence of stopping times such that almost surely T_n \to +\infty and (M_{t\wedge T_n})_{t \ge 0} is a martingale.
From the previous inequality, we find
\mathbb{E}\left( u\left( X^x_{t\wedge T_n}, t\wedge T_n \right) \right) \ge u\left( x,0\right)+\mathbb{E}\left( \int_0^{t\wedge T_n} v(X^x_s,s) ds\right).
By using the dominated convergence theorem, we conclude
\mathbb{E}\left( u\left( X^x_{t}, t \right) \right) \ge u\left( x,0\right)+\mathbb{E}\left( \int_0^{t} v(X^x_s,s) ds\right),
which yields the conclusion \square

Posted in Curvature dimension inequalities | 1 Comment

Lecture 13. The Bochner’s formula

Posted on September 10, 2013 by Fabrice Baudoin

The goal of this lecture is to prove the Bochner formula: A fundamental formula that relates the so-called Ricci curvature of the underlying Riemannian structure to the analysis of the Laplace–Beltrami operator. The Bochner’s formula is a local formula, we therefore only need to prove it on \mathbb{R}^n.

Let (V_1,\cdots,V_n) be an elliptic system of smooth vector fields on \mathbb{R}^n. As usual, we introduce the structure constants of the underlying Riemannian metric:
[V_i,V_j]=\sum_{k=1}^n \omega_{ij}^k V_k.

We know that the Laplace-Beltrami operator is given by L=\sum_{i=1}^n V_i^2 +V_0, where
V_0= -\sum_{i,k}^n \omega_{ik}^k V_i.
We first introduce the Ricci curvature, which is seen in this lecture as a first order differential bilinear form.

If f is a smooth and compactly supported function on \mathbb{R}^n, we define
\mathcal R(f,f) = \sum_{k,l=1}^n \mathcal{R}_{k,l} V_k f V_l f
where
\mathcal{R}_{k,l} = \sum_{j=1}^n (V_l\omega^j_{kj} - V_j\omega^k_{l j}) + \sum_{i,j=1}^n \omega_{ji}^i \omega^l_{k j} - \sum_{i=1}^n\omega_{k i}^i \omega_{l i}^i
+ \frac{1}{2} \sum_{1\le i<j\le n} \bigg(\omega^l_{ij} \omega^k_{ij} - (\omega_{l j}^i +\omega_{li}^j)(\omega^i_{kj} + \omega^j_{ki})\bigg)

Though, it is not apparent, it is actually an intrinsic Riemannian invariant. That is, \mathcal{R} only depends on the Riemannian metric g induced by the vector fields.

In the sequel, we will use the following differential bilinear form that already has been widely used throughout these lectures:
\Gamma(f,g) =\frac{1}{2}(L(fg)-fLg-gLf)=\sum_{i=1}^n V_i f V_i g,
and we now introduce its iteration
\Gamma_{2}(f,g) = \frac{1}{2}\left(L\Gamma(f,g) - \Gamma(f, Lg)-\Gamma (g,Lf)\right).

Henceforth, we adopt the notation
f_{,ij} = \frac{V_i V_j f + V_j V_i f}{2}
for the entries of the symmetrized Hessian of f with respect to the vector fields V_1,...,V_d. Noting that V_iV_j f\ =\ f_{,ij}\ +\ \frac{1}{2}\ [V_i,V_j] f and using the structure constants we obtain the useful formula
V_iV_jf = f_{,ij} + \frac{1}{2} \sum_{l=1}^n \omega^l_{ij} V_l f

Our principal result of this lecture is the following:

Theorem: (Bochner’s identity) For every smooth function f :\mathbb{R}^n \rightarrow \mathbb{R},
\Gamma_{2}(f,f)
= \sum_{l=1}^n \left(f_{,ll} -\sum_{i=1}^n \omega_{il}^l V_i f \right)^2 +2 \sum_{1 \le l<j \le n} \left( f_{,l j}-\sum_{i=1}^n \frac{\omega_{il}^j +\omega_{ij}^l}{2} V_i f \right)^2 +\mathcal{R}(f,f).
where \mathcal R(f,f) is the quadratic form defined above.

Proof: We begin by observing that for any smooth function F on \mathbb{R}^n L(F^2) = 2 F LF + 2 \Gamma(F,F).
This and the definition of \Gamma gives
L \Gamma(f,f) = \sum_{i=1}^n L((V_i f)^2) = 2 \sum_{i=1}^n V_i f L(V_i f) + 2 \sum_{i=1}^n \Gamma(V_i f,V_if) .
We now have
L(V_i f) = V_0 V_i f + \sum_{j=1}^n V_j^2 V_i f = V_i V_0 f + [V_0,V_i]f + \sum_{j=1}^n V_j (V_iV_j f) + V_j[V_j,V_i]f
= V_i V_0 f + [V_0,V_i]f + \sum_{j=1}^n \big\{V_i (V_jV_j f) + [V_j,V_i]V_j f + V_j[V_j,V_i]f\big\}
= V_i(Lf) + [V_0,V_i]f + \sum_{j=1}^n \big\{[V_j,V_i]V_j f + V_j[V_j,V_i]f\big\}
= V_i(Lf) + [V_0,V_i]f + 2 \sum_{j=1}^n [V_j,V_i]V_j f +\sum_{j=1}^n [V_j,[V_j,V_i]]f.
Using this identity we find
L \Gamma(f,f) = 2 \sum_{i=1}^n V_i f \left\{V_i(Lf) + [V_0,V_i]f + 2 \sum_{j=1}^n [V_j,V_i]X_j f + \sum_{j=1}^n [V_j,[V_j,V_i]]f\right\} + 2 \sum_{i,j=1}^n (V_jV_i f)^2
= 2 \Gamma(f,Lf) + 2 \sum_{i=1}^n V_i f [V_0,V_i]f + 4 \sum_{i,j=1}^n V_if [V_j,V_i]V_j f + 2 \sum_{i,j=1}^n V_i f [V_j,[V_j,V_i]]f + 2 \sum_{i,j=1}^n (V_jV_i f)^2.
Since, thanks to the skew-symmetry of the matrix \{[V_i,V_j]f\}_{i,j=1,...,n}, we have \sum_{i,j=1}^n f_{,ij} [V_i,V_j] f = 0, we find
\sum_{i,j=1}^n (V_jV_i f)^2 = \sum_{i,j=1}^n f_{,ij}^2 + \frac{1}{4} \sum_{i,j=1}^n ([V_i,V_j] f)^2 +\sum_{i,j=1}^n f_{,ij} [V_i,V_j] f
= \sum_{i,j=1}^n f_{,ij}^2 + \frac{1}{4} \sum_{i,j=1}^n ([V_i,V_j] f)^2.
We thus obtain

\frac{1}{2} \big[L \Gamma(f,f) - 2\Gamma(f,Lf)\big] = \sum_{i=1}^n V_i f [V_0,V_i]f + 2 \sum_{i,j=1}^n V_i f [V_j,V_i]V_j f + \sum_{i,j=1}^n V_i f [V_j,[V_j,V_i]]f + \sum_{i,j=1}^n f_{,ij}^2 +\frac{1}{4} \sum_{i,j=1}^n ([V_i,V_j] f)^2.

Since we have

\Gamma_{2}(f,f) = \frac{1}{2}\left(L\Gamma(f,f) - 2 \Gamma(f, Lf)\right), we conclude \Gamma_{2}(f,f) = \sum_{i,j=1}^n f_{,ij}^2 - 2 \sum_{i,j=1}^n V_i f [V_i,V_j]V_j f + \frac{1}{4} \sum_{i,j=1}^n ([V_i,V_j] f)^2 + \sum_{i=1}^n V_i f [V_0,V_i]f + \sum_{i,j=1}^n V_i f [[V_i,V_j],V_j]f

To complete the proof we need to recognize that the right-hand side coincides with that in the statement of our result.

With this objective in mind, using the structure constants we obtain
\sum_{i,j=1}^n f_{,ij}^2 - 2 \sum_{i,j=1}^n V_i f [V_i,V_j]V_j f
= \sum_{l=1}^n f_{,ll}^2 + 2 \sum_{1 \le l<j \le n} f_{,jl}^2 - 2 \sum_{i,j=1}^n V_i f \left(\sum_{l=1}^d \omega_{ij}^l V_l \right)V_j f
= \sum_{l=1}^n f_{,ll}^2 + 2 \sum_{1 \le l<j \le n} f_{,jl}^2 - 2 \sum_{i,j=1}^n \sum_{l=1}^n \omega_{ij}^l V_l V_j f\ V_i f
= \sum_{l=1}^n f_{,ll}^n + 2 \sum_{1 \le l<j \le n} f_{,jl}^2 - 2 \sum_{i,j=1}^n \sum_{l=1}^n \omega_{ij}^l f_{,l j}V_i f - \sum_{i,j=1}^n \sum_{l, k=1}^n \omega_{ij}^l \omega^k_{l j} V_k f V_i f
= \sum_{l=1}^n f_{,ll}^2 + 2 \sum_{1 \le l<j \le n} f_{,jl}^2 - 2 \sum_{l,j=1}^n \left(\sum_{i=1}^n \omega_{ij}^l V_i f\right) f_{,l j} - \sum_{i,j=1}^n \sum_{l, k=1}^n \omega_{ij}^l \omega^k_{l j} V_k f V_i f
= \sum_{l=1}^n \left(f_{,ll}^2 - 2 \left(\sum_{i=1}^n \omega^l_{il} V_i f\right) f_{,l l}\right) + 2 \sum_{1 \le l<j \le n}\left( f_{,jl}^2 - 2 \sum_{1\le l<j\le n} \left(\sum_{i=1}^n \frac{\omega_{ij}^l + \omega_{il}^j}{2} V_i f\right) f_{,l j} \right)-\sum_{i,j=1}^n \sum_{l, k=1}^n \omega_{ij}^l \omega^k_{l j} V_k f V_i f.
If we now complete the squares we obtain

\sum_{i,j=1}^n f_{,ij}^2 - 2 \sum_{i,j=1}^n V_i f [V_i,V_j]V_j f
= \sum_{l=1}^n \left( f_{,ll} -\sum_{i=1}^n \omega_{il}^l V_i f \right)^2 + 2 \sum_{1 \le l<j \le n} \left( f_{,jl} -\sum_{i=1}^n \frac{\omega_{ij}^l +\omega_{il}^j}{2} V_i f \right)^2
- \sum_{l=1}^n \left(\sum_{i=1}^n \omega_{il}^l V_i f \right)^2 - 2 \sum_{1 \le l<j \le n} \left(\sum_{i=1}^n \frac{\omega_{ij}^l +\omega_{il}^j}{2} V_i f \right)^2-\sum_{i,j,k,l=1}^n \omega_{ij}^l \omega^k_{l j} V_k f V_i f.

Next, we have
\sum_{i=1}^n V_i f [V_0,V_i]f = \sum_{i,j,k,l=1}^n \omega^k_{jk} \omega^l_{ij} V_l f V_i f + \sum_{i=1}^n \sum_{j,k=1}^n (V_i \omega^k_{jk}) V_if V_j f,
and also
\sum_{i,j=1}^n V_i f [[V_i,V_j],V_j]f = \sum_{i,j=1}^n \sum_{l= 1}^n [\omega^l_{ij} V_l,V_j]f V_if
= \sum_{i,j=1}^n \sum_{l=1}^n \omega^l_{ij} V_i f [V_l,V_j] f - \sum_{i,j=1}^n \sum_{l=1}^n (V_j\omega^l_{ij}) V_if V_l f.
Using the structure constants we find
\sum_{i,j=1}^n V_i f [[V_i,V_j],V_j]f = \sum_{i,j=1}^n \sum_{l,k=1}^n \omega^l_{ij} \omega^k_{l j} V_i f V_kf - \sum_{i,j=1}^d \sum_{l=1}^d (X_j\omega^l_{ij}) X_if X_l f.
Next we have
\frac{1}{4} \sum_{i,j=1}^n ([V_i,V_j]f)^2 = \frac{1}{2} \sum_{1\le i<j\le n}\left(\sum_{l=1}^n \omega^l_{ij} V_l f\right)^2 .
We obtain therefore
\Gamma_2(f,f) = \sum_{l=1}^n \left( f_{,ll} -\sum_{i=1}^n \omega_{il}^l V_i f \right)^2 + 2 \sum_{1 \le l<j \le n} \left( f_{,jl} -\sum_{i=1}^n \frac{\omega_{ij}^l +\omega_{il}^j}{2} V_i f \right)^2 + \mathcal{M}onster
where we have let

\mathcal{M}onster = - \sum_{l=1}^n \left(\sum_{i=1}^d \omega_{il}^l V_i f \right)^2 - 2 \sum_{1 \le l<j \le n} \left(\sum_{i=1}^n \frac{\omega_{ij}^l +\omega_{il}^j}{2} V_i f \right)^2
+ \sum_{i,j,k,l=1}^n \omega^k_{jk} \omega^l_{ij} V_l f V_i f - \sum_{i,j,k,l=1}^n \omega_{ij}^k \omega^l_{k j} V_l f V_i f + \sum_{i=1}^n \sum_{j,k=1}^n (V_i \omega^k_{jk}) V_if V_j f + \sum_{i,j=1}^n \sum_{l,k=1}^n \omega^l_{ij} \omega^k_{l j} V_i f V_kf
- \sum_{i,j=1}^n \sum_{l=1}^n (V_j\omega^l_{ij}) V_if V_l f+ \frac{1}{2} \sum_{1\le i<j\le n}\left(\sum_{l=1}^n \omega^l_{ij} V_l f\right)^2.

Simplifying the expression we obtain
\mathcal{M}onster = - \sum_{k,l=1}^n \sum_{i=1}^n \omega_{k i}^i \omega_{l i}^i V_k f V_l f - \frac{1}{2} \sum_{k,l=1}^n \sum_{1 \le i<j \le n} (\omega_{l j}^i +\omega_{l i}^j)(\omega^i_{kj} + \omega^j_{ki}) V_k f V_l f + \sum_{k,l=1}^n \sum_{j=1}^n (V_l\omega^j_{kj} - V_j\omega^k_{l j}) V_kf V_l f
+ \sum_{i,j,k,l=1}^n \omega_{ji}^i \omega^l_{k j} V_k f V_l f+ \frac{1}{2} \sum_{k,l=1}^n \sum_{1\le i<j\le n} \omega^l_{ij} \omega^k_{ij} V_k f V_l f.

To complete the proof we need to recognize that the monster coincides with \mathcal R(f,f). This simple computation is let to the reader \square

Let f:\mathbb{R}^n \rightarrow \mathbb{R} be a smooth function. The matrix with coefficient (l,j) given by f_{,l j} -\sum_{i=1}^n \frac{\omega_{il}^j +\omega_{ij}^l}{2} V_i f is a Riemannian invariant. This matrix is called the Riemannian Hessian of f and denoted by \mathbf{Hess} f or \nabla^2 f. As a consequence of this and of the Bochner's identity, \mathcal{R} is seen to be a Riemannian invariant.

Definition: Let \mathbb{M} be a Riemannian manifold. The bilinear form ( a (0,2) tensor) locally defined by \mathbf{Ricc} (\nabla f, \nabla f)= \mathcal{R}(f,f) is called the Ricci curvature of \mathbb{M}.

On a Riemannian manifold, the Bochner’s formula can therefore synthetically be written

\Gamma_2 (f,f)=\| \mathbf{Hess} f \|^2_{HS}+ \mathbf{Ric} (\nabla f, \nabla f).

As a consequence, it should come as no surprise that a lower bound on \mathbf{Ric} translates into a lower bound on \Gamma_2.

Theorem: Let \mathbb{M} be a Riemannian manifold. We have, in the sense of bilinear forms, \mathbf{Ric} \ge \rho if and only if for every f \in C^\infty(\mathbb{M}),
\Gamma_2(f,f) \ge \frac{1}{n} (Lf)^2 + \rho \Gamma(f,f).

Proof: Let us assume that \mathbf{Ric} \ge \rho. In that case, from Bochner’s formula we deduce that \Gamma_2 (f,f) \ge \| \mathbf{Hess} f \|^2_{HS}+ \rho \Gamma(f,f). From Cauchy-Schwartz inequality, we have the bound
\| \mathbf{Hess} f \|^2_{HS} \ge \frac{1}{n} \mathbf{Tr} \left(\mathbf{Hess} f \right)^2.
Since \mathbf{Tr} \left(\mathbf{Hess} f \right)=Lf, we conclude that
\Gamma_2(f,f) \ge \frac{1}{n} (Lf)^2 + \rho \Gamma(f,f).

Conversely, let us now assume that for every f \in C^\infty(\mathbb{M}),
\Gamma_2(f,f) \ge \frac{1}{n} (Lf)^2 + \rho \Gamma(f,f).

Let x \in \mathbb{M} and v \in \mathbf{T}_x \mathbb{M}. It is possible to find a function f \in C^\infty(\mathbb{M}) such that, at x, \mathbf{Hess} f =0 and \nabla f=v. We have then, by using Bochner’s identity at x, \mathbf{Ric} (v,v) \ge \rho \| v \|^2 \square

The inequality \Gamma_2(f,f) \ge \frac{1}{n} (Lf)^2 + \rho \Gamma(f,f) is called the curvature-dimension inequality. It is an intrinsic property of the operator L.

We finally mention another consequence of Bochner’s identity which shall be later used.

Lemma: Let \mathbb{M} be a Riemannian manifold such that \mathbf{Ric} \ge \rho. For every f \in C^\infty(\mathbb{M}),
\Gamma(\Gamma(f)) \le 4 \Gamma (f) \left( \Gamma_2(f)-\rho\Gamma(f)\right).

Proof: It follows from the fact that \Gamma_2 (f,f) \ge \| \mathbf{Hess} f \|^2_{HS}+ \rho \Gamma(f,f) and Cauchy-Schwartz inequality implies that \Gamma(\Gamma(f)) \le 4 \| \mathbf{Hess} f \|^2_{HS} \Gamma(f). Details are let to the reader \square

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