## Lecture 16. The Li-Yau 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 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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