Why $\frac{d}{dt}\overline\lambda(g_{ij}(t))\ge \frac{d}{dt}(\mathcal F(g_{ij}(t),f(t))\cdot V^{2/n}(g_{ij}(t)))$?

Question

$M$ is a Riemannian manifold,$g_{ij}(t)$ evolve under Ricci flow.

$\lambda (g_{ij}) = \inf \{\mathcal F(g_{ij}, f) \mid f \in C^\infty (M), \int \limits _M \Bbb e^{-f} \Bbb d V = 1 \}$.

$\overline\lambda(g_{ij}) = \lambda (g_{ij}) V^{\frac 2 n} (g_{ij})$.

$\mathcal F(g_{ij},f)=\int_M(R+|\nabla f|^2)e^{-f}dV$.

$V(g_{ij}(t))$ is the volume of $(M,g_{ij}(t))$.

$f$ is the solution of the backward equation $\partial_t f=-\Delta f+|\nabla f|^2-R$, $f=f_0$ when $t=t_0$, and $f_0$ is the minimizer of $\lambda(g_{ij}(t_0))$.

Why $\frac{d}{dt}\overline\lambda(g_{ij}(t))\ge \frac{d}{dt}(\mathcal F(g_{ij}(t),f(t))\cdot V^{2/n}(g_{ij}(t)))$ at $t=t_0$?

Answer 1

The question is more or less answered in other answer. Let me just fill in the details. You want to show that $\overline \lambda(t) := \overline \lambda (g_{ij}(t))$ is nondecreasing. Using

$$ \frac{\overline\lambda(t_0)-\overline\lambda(t)}{t_0-t}\ge \frac{\mathcal F(g_{ij}(t_0),f_0(t_0))\cdot V^{2/n}(g_{ij}(t_0))-\mathcal F(g_{ij}(t),f(t))\cdot V^{2/n}(g_{ij}(t))}{t_0-t} $$

we have that ((1.5.8) in here)

$$D_- \overline \lambda (t_0) \ge \frac{d}{dt} \mathcal F(g_{ij}(t),f(t))\cdot V^{2/n}(g_{ij}(t)) \bigg|_{t = t_0} \ge 0,$$

where $$D_- \overline \lambda (t_0) := \liminf_{t \to t_0^-} \frac{\overline \lambda (t_0) - \overline \lambda (t)}{t_0 - t}.$$

As $t_0 >0$ is arbitrary, we have $D_-\overline \lambda \ge 0$ on $[0,T)$. This suffices us to say that $\overline \lambda$ is nondecreasing, once we prove the following two lemma:

Lemma 1: Let $f : [a,b] \to \mathbb R$ satisfies $D_- f \ge 0$ on $[a, b]$ and $$\tag{1} \liminf_{t\to t_0^+} f(t) \ge f(t_0).$$

Then $f$ is nondecreasing.

Proof: Let $\epsilon >0$ and $f_\epsilon (t) : = f(t) + \epsilon t$. Then we have $D_- f \ge \epsilon >0$ on $[a,b]$. Now let $y \in [a,b]$ and consider $$A = \{ t\in [a,y] : f_\epsilon (s) \le f_\epsilon (y)\ \ \forall s\in [t,y]\}. $$ $A$ is nonempty as $y\in A$. Let $t_1 > t_2 > t_3 \cdots$ be a sequence in $A$ so that $\{t_n \}$ converges to $t$. Since $f$ satisfies $(1)$, so is $f_\epsilon$. Hence $$f_\epsilon (t) \le \liminf_{n\to \infty} f_\epsilon(t_n) \le f_\epsilon (y)\Rightarrow t\in A$$ and $A$ is closed. To show that $A$ is open, let $t_0\in A$. Then as $D_- f_\epsilon \ge \epsilon$, there is $\delta >0$ so that $$\frac{f_\epsilon (t_0) - f_\epsilon(t)}{t_0-t} \ge \frac{\epsilon}{2} \ \ \ \ \forall t\in (t_0-\delta, t_0)$$ In particular, $$f_\epsilon (t) < f_\epsilon (t_0) \le f(y)$$ and so $(t_0 - \delta, t_0]\in A$. Thus $A$ is open. As $[a,y]$ is connected, we have $A = [a,y]$ and thus $f_\epsilon (x) \le f_\epsilon (y)$ for all $x\le y$. By definition of $f_\epsilon$, $$f(x) + \epsilon x \le f(y) + \epsilon y$$ for all $x\le y$. Take $\epsilon \to 0$, we have $f(x) \le f(y)$ whenever $x\le y$. Thus $f$ is nondecreasing.

Lemma 2: $\overline \lambda(t)$ satisfies (1).

Proof: This is easy as $\lambda (t)$ is nondecreasing, so when $t_0< t$, $$\lambda (t_0) \le \lambda (t) \Rightarrow \lambda (t_0) V^{\frac{2}{n}}(t)\le \overline \lambda (t).$$ Taking liminf on both sides, as $t\mapsto V(t)$ is continuous, $$ \overline \lambda (t_0) \le \liminf_{t\to t_0^+} \overline \lambda (t)$$ and we are done.

Answer 2

Using the define of limit.

Firstly, $\overline\lambda(g_{ij}(t))\le \mathcal F(g_{ij}(t),f(t))\cdot V^{2/n}(g_{ij}(t))$

Secondly, $\overline\lambda(g_{ij}(t_0))= \mathcal F(g_{ij}(t_0),f_0(t_0))\cdot V^{2/n}(g_{ij}(t_0))$

Then $$ \frac{\overline\lambda(g_{ij}(t_0))-\overline\lambda(g_{ij}(t))}{t_0-t}\ge \frac{\mathcal F(g_{ij}(t_0),f_0(t_0))\cdot V^{2/n}(g_{ij}(t_0))-\mathcal F(g_{ij}(t),f(t))\cdot V^{2/n}(g_{ij}(t))}{t_0-t} $$

Then,let $t\rightarrow t_0$.

Besides,it should can be proved by elliptic theory, but I don't finish it. If there are any answer using the elliptic theory , I will be glad to choice it as right answer.