## 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 !

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