Lecture 17. The parabolic Harnack inequality

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 !

This entry was posted in Curvature dimension inequalities. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s