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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