The Li-Yau-Hamilton Harnack inequality

Proof. (Step 1. Decide your quantity.) One first sees that $$P:=\partial_t \log u- |\nabla \log u|^2=\frac{\Delta u}{u}-\frac{|\nabla u|^2}{u^2}=\Delta \log u$$ The last equality is a quick computation: $$\nabla^2 \log u=\nabla (\frac{\nabla u}{u})=\frac{1}{u}\nabla^2 u-\frac{\nabla u\otimes \nabla u}{u^2}$$ (Step 2. Compute its variation and get a favorable inequality.) The idea of making this transformation is that one can now compute its variation and we want to get an equation as $\partial_t P=\Delta P+$some other terms possibly involving $P$. Because time and space derivatives commute, we have $$\frac{\partial}{\partial t}(\Delta \log u) =\Delta\left(\frac{\partial}{\partial t} \log u\right)=\Delta(\frac{\Delta u}{u})=\Delta\left(\Delta \log u+|\nabla \log u|^2\right) $$ To take care of the second term, we need to commute $\Delta$ and $\nabla$. This is the so called "Bochner formulae" which follows form commutation of derivatives. $$\Delta(\nabla_i f)=\nabla_i (\Delta f)+R_{i}^j\nabla_j f$$ We obtain: $$\nabla^2 |\nabla f|^2 =2\langle \nabla(\nabla f),\nabla(\nabla f)\rangle+2\langle \nabla^2 (\nabla f),\nabla f\rangle$$ which shows $$\Delta |\nabla f|^2=2|\nabla^2 f|^2+2\langle \Delta(\nabla f),\nabla f\rangle$$ and using Bochner, $$\Delta|\nabla f|^2=2|\nabla^2 f|^2+2\langle \nabla (\Delta f),\nabla f\rangle+2\operatorname{Ric}(\nabla f,\nabla f)$$ Returning to $P$, we have $$\partial_t P=\Delta P+2|\nabla^2 \log u|^2+2\langle \nabla P,\nabla \log u\rangle +2\operatorname{Ric}(\nabla \log u,\nabla \log u)$$ We can drop the $\operatorname{Ric}$ term and use Cauchy-Schwars on the square term and we will still be going in the right direction to obtain $$\partial_t P\geq \Delta P+2\langle \nabla P,\nabla \log u\rangle +\frac{2}{n}P^2$$ (Step 3. Apply the WEAK Maximum principle.) We are in the right position to apply maximum principle! The final step $$\begin{aligned} \partial_t (P+\frac{n}{2t})&\geq \Delta (P+\frac{n}{2t})+2\langle \nabla (P+\frac{n}{2t}),\nabla \log u\rangle \\ &+\frac{2}{n}(P+\frac{n}{2t})^2-\frac{2}{t}(P+\frac{n}{2t}) \end{aligned}$$ We have obtained a parabolic inequality for our desired quantity. To obtain a contradiction assume that $\inf_{M\times (0,T]} (P+\frac{n}{2t})<0$. Because $M$ is compact and our quantity blows up as $t\to 0$, this minima is actually attained, say at $(x_0,t_0)$. Then at $(x_0,t_0)$, looking at the inequality above, we see $$0\geq -\frac{2}{t_0}(P(x_0,t_0)+\frac{n}{2t_0})$$ which is a contradiction. The theorem is proved. ∎

Proof. (Step 1. Decide your quantity.) Firstly we may assume $u$ is continuous on $B_1\times [0,1]$ for otherwise, carry out the proof on $B_1\times [\varepsilon,1]$ and take $\varepsilon\to 0$. Recall the identity $$\Delta \log u=\frac{\Delta u}{u}-|\nabla \log u|^2$$ From here we derive that the Harnack quantity is: $$\partial_t v-|\nabla v|^2=\Delta v$$ $$\partial_t v-\alpha|\nabla v|^2+\frac{n}{2 \alpha t}=\Delta v +(1-\alpha)|\nabla v|^2+\frac{n}{2\alpha t}$$ which means $w:=\Delta v=\Delta \log u$ is the quantity to inverstigate now. One can compute in tensorial notation, $$\begin{aligned} \frac{\partial}{\partial t}(\Delta \log u) & =\Delta\left(\frac{\partial}{\partial t} \log u\right)=\Delta\left(\Delta \log u+|\nabla \log u|^2\right) \\ & =\Delta(\Delta \log u)+2 \nabla \log u \cdot \Delta \nabla \log u+2|\nabla \nabla \log u|^2 \\ & \underset{(*)}{=}\Delta(\Delta \log u)+2 \nabla \log u \cdot \nabla \Delta \log u+2|\nabla \nabla \log u|^2 \end{aligned}$$ So we have $$\partial_t w= \Delta w+2\langle\nabla v, \nabla w\rangle+2|\nabla^2 v|^2$$ Let us set $$S:=\Delta v+(1-\alpha)|\nabla v|^2$$ And we compute the variation of this quantity. Now we have $$\begin{aligned} &\partial_t |\nabla v|^2=2\langle\nabla (\partial_t v),\nabla v\rangle\\ &=2\langle \nabla (\Delta v)+\nabla |\nabla v|^2,\nabla v\rangle\\ &\underset{(*)}{=}2\langle \Delta (\nabla v),\nabla v\rangle+2\langle \nabla |\nabla v|^2,\nabla v\rangle\\ &=2[\frac{1}{2}\Delta |\nabla v|^2-|\nabla^2 v|^2]+2\langle\nabla |\nabla v|^2,\nabla v\rangle \end{aligned}$$ From this we obtain $$\partial_t S=\Delta S+2\alpha |\nabla^2 v|^2+2\langle \nabla v,\nabla S\rangle$$ In order to stay away from the boundary of $B_1$, we need to use a cut-off function. Let $\eta\in C_c^\infty(B_1)$ such that $\eta\equiv 1$ on $B_{1/2}$ and $\eta:\mathbb{R}^n\to [0,1]$. Now we set $$H=t\eta^2 S$$ and start our long computation $$\begin{aligned} \partial_t H&=\eta^2 S+\eta^2 t(\Delta S+2\alpha |\nabla^2 v|^2+2\langle \nabla v,\nabla S\rangle) \end{aligned}$$ while $$\Delta H=t(2\eta \Delta \eta S+2|\nabla \eta|^2S+\eta^2 \Delta S)$$ $$\nabla H=tS(2\eta\nabla \eta)+t\eta^2 \nabla S$$ giving us $$\begin{aligned} \partial_t H&=\eta^2 S+\Delta H-2\eta t\Delta \eta S-2|\nabla \eta|^2 S\\ &+\eta^2 t(2\alpha |\nabla^2 v|^2)+2\langle \nabla v,\nabla H-2tS\eta \nabla \eta\rangle \end{aligned}$$ Cleaning it up we end up at $$\begin{aligned} \partial_t H&=\Delta H+2\langle\nabla v,\nabla H\rangle+\frac{H}{t}-2H\frac{\Delta \eta }{\eta}\\ &-2|\nabla \eta|^2 \frac{H}{t\eta^2}-4\frac{H}{\eta} \langle \nabla v,\nabla \eta\rangle\\ &+\eta^2 t(2\alpha |\nabla^2 v|^2) \end{aligned}$$ It is still not very clear whether we are ready to apply maximum principle simply because there are many terms with mixed sign. Remember we need to bring more positive terms on the right hand side. We can arrive at $$\begin{aligned} \partial_t H&\geq \Delta H+2\langle\nabla v,\nabla H\rangle+\frac{H}{t}-2H\frac{\Delta \eta }{\eta}\\ &-\frac{2H}{t}(|\nabla \eta|^2 \frac{1}{\eta^2}+2\frac{t}{\eta} \langle \nabla v,\nabla \eta\rangle)\\ &+\eta^2 t(\frac{2\alpha}{n} w^2) \end{aligned}$$ One now sees after a long computation that the above inequality is equivalnet to the following \begin{equation} \tag{\dagger}\label{HarVar} \begin{aligned} \partial_t H&\geq \Delta H+2\langle\nabla v,\nabla H\rangle\\ &-\frac{2H}{\eta^2}\left|\sqrt{\frac{2\alpha(1-\alpha)}{n}}\eta \nabla v+\frac{1}{\sqrt{\frac{2\alpha(1-\alpha)}{n}}}\nabla \eta\right|^2\\ &+\frac{H}{t\eta^2}[\eta^2-2\Delta \eta-2|\nabla \eta|^2+\frac{nt|\nabla \eta|^2}{2\alpha(1-\alpha)}+\frac{2\alpha}{n}H]\\ &-2|\nabla v|^2H\frac{2\alpha(1-\alpha)}{n}+2|\nabla v|^4t\eta^2\frac{(1-\alpha)^2\alpha}{n} \end{aligned} \end{equation} Notice that we have the right signs for everything! (Step 2. Carry out the maximum principle.) There exists some large $C_{\alpha,n}$ such that $$\frac{n|\nabla \eta|^2}{2\alpha(1-\alpha)}\leq C_{\alpha,n}$$ Thus we have on $B_{1/2}\times (0,T]$ where $H$ is nonpositive, $$\begin{aligned} \partial_t H&\geq \Delta H+2\langle\nabla v,\nabla H\rangle\\ &+\frac{H}{t\eta^2}[1+C_{\alpha,n}t+\frac{2\alpha}{n}H]\\ \end{aligned}$$ Let $$h(x,t)=1+C_{\alpha,n}t+\frac{2\alpha}{n}H(x,t)\qquad (x,t)\in B_1\times [0,T]$$ We claim that $h\geq 0$ on $B_{1/2}\times [0,T]$. Suppose not. Since $h(x,t)\geq 0$ for $x\in B_{1/2}$ and $t$ close to $0$, we see that there exists $(x_0,t_0)\in B_{1/2}\times (0,T]$ such that $$h(x_0,t_0)=\min_{(x,t)\in B_{1/2}\times [0,T]}h(x,t)$$ and so, we have at $(x_0,t_0)$ that $\partial_th\leq 0,\Delta h\geq 0,\nabla h=0$ which means at $(x_0,t_0)$, $H\leq 0, \Delta H\geq 0$ and $\nabla H=0$. But, at $(x_0,t_0)$, we have $$0\geq \frac{n}{2\alpha}(\partial_t h-C_{\alpha,n})=\partial_t H\geq 0$$ which is a contradiction! Thus, we have $$1+C_{\alpha,n}t+\frac{2\alpha}{n}H(x,t)\geq 0\qquad (x,t)\in B_{1/2}\times [0,T]$$ which gives $$\frac{n}{2\alpha t}+\frac{n}{2\alpha}C_{\alpha,n}+\frac{1}{t}H(x,t)\geq 0\qquad (x,t)\in B_{1/2}\times (0,T]$$ which finally arrives at $$\frac{n}{2\alpha t}+\frac{n}{2\alpha}C_{\alpha,n}+\Delta v+(1-\alpha)|\nabla v|^2\geq 0\qquad (x,t)\in B_{1/2}\times (0,T]$$ and that is the required Harnack! The proof of the theorem is complete. ∎

Proof. We have noted $(*)$ where we exchanged derivates in the previous proof. Set $w:=\Delta v$, $S=\Delta v+(1-\alpha)|\nabla v|^2$. Let $\eta\in C_c^\infty(B_\rho (p_0))$ such that $\eta\equiv 1$ on $B_{\rho/2}(p_0)$ and $\eta:M\to [0,1]$ such that on $M$, $$|\nabla \eta|^2\leq C/\rho^2$$ Now we set $$H=t\eta^2 S$$ We get after a look at previous proof that $$\partial_t S=\Delta S+2\alpha |\nabla^2 v|^2+2\langle \nabla v,\nabla S\rangle+2\alpha \operatorname{Ric}(\nabla v,\nabla v)$$ and \begin{equation} \tag{\dagger}\label{HarVar-2} \begin{aligned} \partial_t H&\geq \Delta H+2\langle\nabla v,\nabla H\rangle\\ &-\frac{2H}{\eta^2}\left|\sqrt{\frac{2\alpha(1-\alpha)}{n}}\eta \nabla v+\frac{1}{\sqrt{\frac{2\alpha(1-\alpha)}{n}}}\nabla \eta\right|^2\\ &+\frac{H}{t\eta^2}[\eta^2-2\Delta \eta-2|\nabla \eta|^2+\frac{nt|\nabla \eta|^2}{2\alpha(1-\alpha)}+\frac{2\alpha}{n}H]\\ &-2|\nabla v|^2H\frac{2\alpha(1-\alpha)}{n}+2|\nabla v|^4t\eta^2\frac{(1-\alpha)^2\alpha}{n} \end{aligned} \end{equation} There exists some large $C$ such that $$\frac{n|\nabla \eta|^2}{2\alpha(1-\alpha)}\leq \frac{C}{\rho^2}$$ Thus we have on $B_{\rho/2}(p_0)\times (0,T]$ where $H$ is nonpositive, $$\begin{aligned} \partial_t H&\geq \Delta H+2\langle\nabla v,\nabla H\rangle\\ &+\frac{H}{t\eta^2}[1+\frac{C}{\rho^2}t+\frac{2\alpha}{n}H]\\ \end{aligned}$$ The rest of the proof goes throughnd as before one gets at $$\frac{n}{2\alpha t}+\frac{C}{\rho^2}+\Delta v+(1-\alpha)|\nabla v|^2\geq 0\qquad (x,t)\in B_{1/2}\times (0,T]$$ proving the theorem. ∎

Proof. Let $u_R(x,t):=u(Rx,R^2t)$ and $v_R:=\log u_R$ and apply the Harnack on the unit ball to obtain $$\partial_t v_R-\alpha |\nabla v_R|^2+\frac{n}{2\alpha t}\geq -C\qquad B_{1/2}\times (0,T]$$ Unwinding, $$\partial_t v(Rx,R^2t)-\alpha |\nabla v|^2(Rx,R^2t)+\frac{n}{2\alpha t R^2}\geq -\frac{C}{R^2}\qquad B_{1/2}\times (0,T]$$ which means $$\partial_t v(x,t)-\alpha |\nabla v|^2(x,t)+\frac{n}{2\alpha t}\geq -\frac{C}{R^2}\qquad B_{R/2}\times (0,TR^2]$$ Take $R\to \infty$ to obtain the theorem! ∎

Proof. Recall that $v=\log u$ satisfies $$\partial_t v- |\nabla v|^2+\frac{n}{2 t}\geq 0$$ Let us consider a path $\gamma:[t_1,t_2]\to M,\gamma(t_1)=x_1,\gamma(t_2)=x_2$. We integrate: $$\begin{aligned} \log(\frac{u(x_1,t_1)}{u(x_2,t_2)})&=\int_{t_1}^{t_2}\partial_s [\log(u(\gamma(s),s))]\,ds\\ &=\int_{t_1}^{t_2}\frac{(\partial_t u)(\gamma(s),s)+\nabla u(\gamma(s),s)\cdot \gamma'(s)}{u(\gamma(s),s)}\\ &\geq \int_{t_1}^{t_2}|\nabla v|^2-\frac{n}{2s}+\nabla v\cdot \gamma'(s)\,ds\\ &=\int_{t_1}^{t_2}|\nabla v+\frac{1}{2}\gamma'(s)|^2-\frac{1}{4}|\gamma'(s)|^2-\frac{n}{2s}\,ds\\ &\geq -\frac{n}{2}\log(\frac{t_2}{t_1})-A \end{aligned}$$ where we have written $v(s)=v(\gamma(s),s)$. We have $$\frac{u(x_2,t_2)}{u(x_1,t_1)}\geq \frac{t_1^{n/2}}{t_2^{n/2}}e^{-A}$$ where $$A(x_1,x_2,t_1,t_2):=\inf \frac{1}{4}\int_{t_1}^{t_2}|\gamma'(s)|^2$$ where the infimum is taken over all paths $\gamma:[t_1,t_2]\to M$, with $\gamma(t_1)=x_1,\gamma(t_2)=x_2$. We need to bound $A$ by above and that can be done by simply taking $\sigma:[0,b]\to M$ to be the unit speed geodesic and taking $$\gamma(s):=\sigma(\varphi(t))\qquad \varphi(t)=\frac{b}{t_2-t_1}(t-t_1)$$ Then we get $$A\leq \frac{1}{4}\int_{0}^{t_2-t_1} \frac{b^2}{(t_2-t_1)^2}\,ds=\frac{1}{4(t_2-t_1)}d_g(x_1,x_2)$$ This yields us the result. ∎

Proof. We derived for any $\left(x_1, t_1\right),\left(x_2, t_2\right) \in M \times(0, T]$ with $t_2>t_1>0$, we have $$\frac{u(x_2,t_2)}{u(x_1,t_1)}\geq \frac{t_1^{n/2}}{t_2^{n/2}}e^{-\frac{1}{4(t_2-t_1)}d_g(x_1,x_2)}$$ One can easily bound the quantity on right by below and obtain the theorem. ∎