If $2s\geq C|\bar{\nabla}s|$ at $z_{0}$, then

$\omega(z_{0})\leq 4s\leq 4R_{2}.$

If $2s\leq C|\bar{\nabla}s|$ at $z_{0}$, then

$\frac{|\bar{\nabla}s|}{R_{2}+\delta-s}\leq 2C,$

and

$\omega(z_{0})\leq(2C)^{2}(R_{2}+\delta-s(z_{0}))\leq 4C^{2}(R_{2}+\delta-
R_{1}).$

Hence, there exists a constant $C=C(Q,R_{1},R_{2})$ such that $\omega(z)\leq
C$ for all $z\in\Sigma$. This implies that

$|\bar{\nabla}s|^{2}=(R_{2}+\delta-s(z))\omega(z)\leq C(R_{2}+\delta-R_{1})$

for all $z\in\Sigma$. Since $\delta>0$ is arbitrary, the estimate (9.4)
follows. ∎

The estimate (9.1) depends on the origin $p_{t}$ we choose. The following
lemma shows the existence of a ball with a fixed center enclosed by our flow
hypersurfaces on a suitable fixed time interval.

######  Lemma 9.2.

For any time $t_{0}\in[0,T)$, let $p_{0}$ be the center such that the estimate
(9.1) holds on $M_{t_{0}}$. Then there exists some positive $\tau$ depending
only on $n,k,{\gamma},\alpha,\mathrm{Vol}(K_{0})$ and $V_{n+1-k}(K_{0},W)$
such that the anisotropic support function $s(z,t)$ of $M_{t}$ with respect to
the center $p_{0}$ satisfies

$s(z,t)\geq R_{1}/2,\qquad\forall~{}t\in[t_{0},\min\\{T,t_{0}+\tau\\}).$ (9.6)

###### Proof.

Let $\underline{s}(t)=\min_{\Sigma}s(\cdot,t)=s(z_{t},t)$. At the point
$z_{t}$, we have

$\tau_{ij}[s(z_{t},t)]\geq\underline{s}(t)\bar{g}_{ij}.$

Then the evolution equation (8.7) of $s(z,t)$ implies that

$\frac{d}{dt}\underline{s}(t)\geq-\underline{s}(t)^{-\alpha}+\phi(t)\geq-\underline{s}(t)^{-\alpha}.$

By the initial condition $\underline{s}(t_{0})\geq R_{1}$, we solve the above
inequality as follows

$\underline{s}(t)\geq\left(R_{1}^{1+\alpha}-(\alpha+1)(t-t_{0})\right)^{\frac{1}{1+\alpha}}\geq
R_{1}/2,$

provided that
$t-t_{0}\leq(1+\alpha)^{-1}(1-2^{-\alpha-1})R_{1}^{\alpha+1}=:\tau$. The
conclusion (9.6) follows immediately. ∎

We now estimate the upper bound on the speed of the flow.

######  Proposition 9.3.

Let $M_{t}$ be a smooth convex solution of the flow (1.9) on $[0,T)$ with the
global term $\phi(t)$ given by (1.10). Then

$\max_{M_{t}}{E}_{k}\leq~{}C\left(1+t^{-\frac{\alpha}{1+\alpha}}\right)$ (9.7)

for any $t\in[0,T)$, where $C$ depends on
$n,k,{\gamma},\alpha,\mathrm{Vol}(K_{0})$ and $V_{n+1-k}(K_{0},W)$.

###### Proof.

The proof is by applying the technique of Tso [38]. For any given time
$t_{0}\in[0,T)$, the estimate (9.6) implies that the anisotropic support
function $s(z,t)$ with respect to the center $p_{0}$ satisfies $s(z,t)\geq
R_{1}/2$ in the time interval $t\in[t_{0},\min\\{T,t_{0}+\tau\\})$. Then the
following function $Z$

$Z(z,t)=\frac{F_{*}^{-\alpha}}{s(z,t)-\frac{1}{4}R_{1}},\quad z\in\Sigma,$

is well defined for all time $t\in[t_{0},\min\\{T,t_{0}+\tau\\})$. The
estimate (9.6) together with (7.10) also implies that the anisotropic support
function $s(z,t)$ with respect to the center $p_{0}$ is bounded from above by
$2R_{2}$. Since ${E}_{k}^{\alpha/k}=F_{*}^{-\alpha}$, to prove (9.7) it
suffices to estimate the upper bound of $Z$. Combining (8.9) and (8.11), we
obtain

$\displaystyle\frac{\partial}{\partial t}Z-$
$\displaystyle\dot{\Psi}^{k\ell}\left(\bar{\nabla}_{k}\bar{\nabla}_{\ell}Z-\frac{1}{2}Q_{k\ell
p}\bar{\nabla}_{p}Z\right)$ $\displaystyle=$
$\displaystyle\frac{1}{s-\frac{1}{4}R_{1}}\left(\frac{\partial}{\partial
t}F_{*}^{-\alpha}-\dot{\Psi}^{k\ell}\left(\bar{\nabla}_{k}\bar{\nabla}_{\ell}F_{*}^{-\alpha}-\frac{1}{2}Q_{k\ell
p}\bar{\nabla}_{p}F_{*}^{-\alpha}\right)\right)$
$\displaystyle-\frac{Z}{s-\frac{1}{4}R_{1}}\left(\frac{\partial}{\partial
t}s-\dot{\Psi}^{k\ell}\left(\bar{\nabla}_{k}\bar{\nabla}_{\ell}s-\frac{1}{2}Q_{k\ell
p}\bar{\nabla}_{p}s\right)\right)+2\dot{\Psi}^{k\ell}\frac{\bar{\nabla}_{k}Z\bar{\nabla}_{\ell}s}{s-\frac{1}{4}R_{1}}$
$\displaystyle=$ $\displaystyle
2\dot{\Psi}^{k\ell}\frac{\bar{\nabla}_{k}Z\bar{\nabla}_{\ell}s}{s-\frac{1}{4}R_{1}}+\frac{1}{s-\frac{1}{4}R_{1}}(F_{*}^{-\alpha}-\phi(t))\dot{\Psi}^{k\ell}\bar{g}_{k\ell}$
$\displaystyle-\frac{Z}{s-\frac{1}{4}R_{1}}\left(\phi(t)+(1+\alpha)\Psi+s\dot{\Psi}^{k\ell}\bar{g}_{k\ell}\right)$
$\displaystyle=$ $\displaystyle
2\dot{\Psi}^{k\ell}\frac{\bar{\nabla}_{k}Z\bar{\nabla}_{\ell}s}{s-\frac{1}{4}R_{1}}-\frac{R_{1}Z}{4(s-\frac{1}{4}R_{1})}\dot{\Psi}^{k\ell}\bar{g}_{k\ell}+(1+\alpha)Z^{2}-\phi(t)\frac{\dot{\Psi}^{k\ell}\bar{g}_{k\ell}+Z}{s-\frac{1}{2}R_{1}}.$
(9.8)

The second term on the right-hand side of (9) can be estimated as follows:

$\displaystyle\frac{R_{1}Z}{4(s-\frac{1}{4}R_{1})}\dot{\Psi}^{k\ell}\bar{g}_{k\ell}=$
$\displaystyle\frac{\alpha
R_{1}Z}{4(s-\frac{1}{4}R_{1})}F_{*}^{-\alpha-1}\dot{F}_{*}^{k\ell}\bar{g}_{k\ell}$
$\displaystyle\geq$ $\displaystyle\frac{\alpha
R_{1}Z}{4(s-\frac{1}{4}R_{1})}F_{*}^{-\alpha-1}$ $\displaystyle=$
$\displaystyle\frac{\alpha
R_{1}}{4}Z^{2+\frac{1}{\alpha}}(s-\frac{1}{4}R_{1})^{1/{\alpha}}$
$\displaystyle\geq$
$\displaystyle\alpha\left(\frac{R_{1}}{4}\right)^{1+1/{\alpha}}Z^{2+\frac{1}{\alpha}},$

where we used the fact
$\dot{F}_{*}^{k\ell}\bar{g}_{k\ell}=\sum_{i}\dot{f}^{i}_{*}\geq 1$ due to the
concavity of $F_{*}$. Let $\tilde{Z}(t)=\sup Z(\cdot,t)$. Since $\phi(t)\geq
0$, Equation (9) implies that (in the form given in [20, Lemma 3.5])

$\displaystyle\frac{d}{dt}\tilde{Z}\leq$
$\displaystyle-\alpha\left(\frac{R_{1}}{4}\right)^{1+1/{\alpha}}\tilde{Z}^{2+\frac{1}{\alpha}}+(1+\alpha)\tilde{Z}^{2}.$

Let

$\tilde{Z}_{0}(\alpha,R_{1})=\left(\frac{4}{R_{1}}\right)^{\alpha+1}\left(\frac{2(1+\alpha)}{\alpha}\right)^{\alpha},$

which is a positive constant depending only on $\alpha$ and $R_{1}$. Whenever
$\tilde{Z}(t)\geq\tilde{Z}_{0}$, we have

$\displaystyle\frac{d}{dt}\tilde{Z}\leq$
$\displaystyle-(1+\alpha)\tilde{Z}^{2}.$

It follows that

$\displaystyle\tilde{Z}(t)\leq$
$\displaystyle\max\left\\{\left(\frac{4}{R_{1}}\right)^{\alpha+1}\left(\frac{2(1+\alpha)}{\alpha}\right)^{\alpha},\left(\tilde{Z}(t_{0})^{-\frac{1+\alpha}{\alpha}}+\frac{1+\alpha}{2}\left(\frac{R_{1}}{4}\right)^{\frac{1+\alpha}{\alpha}}(t-t_{0})\right)^{-\frac{\alpha}{1+\alpha}}\right\\}$
$\displaystyle\leq$
$\displaystyle~{}C\left(1+(t-t_{0})^{-\frac{\alpha}{1+\alpha}}\right)$

for all $t\in(t_{0},\min\\{T,t_{0}+\tau\\})$, where $C$ depends only on
$n,k,{\gamma},\alpha,\mathrm{Vol}(K_{0})$ and $V_{n+1-k}(K_{0},W)$. The upper
bound of ${E}_{k}^{\alpha/k}=F_{*}^{-\alpha}$ follows from the bound on $Z$
and the fact $s(z,t)\leq 2R_{2}$. Proposition 9.3 follows since $t_{0}$ is
arbitrary. ∎

Proposition 9.3 implies the following estimate on the global term $\phi(t)$.

######  Corollary 9.4.

Let $M_{t}$ be a smooth convex solution of the flow (1.9) on $[0,T)$ with the
global term $\phi(t)$ given by (1.10). Then for any $p>0$ we have

$0<C_{1}\leq\frac{1}{|M_{t}|_{\gamma}}\int_{M_{t}}{E}_{k}^{p}d\mu_{\gamma}\leq
C_{2}$ (9.9)

on $[0,T)$, where the constants $C_{1},C_{2}$ depend only on
$n,k,{\gamma},\alpha,p,\mathrm{Vol}(K_{0})$ and $V_{n+1-k}(K_{0},W)$. In
particular,

$C_{1}\leq~{}\phi(t)\leq C_{2}$ (9.10)

on $[0,T)$.

###### Proof.

The upper bound in (9.9) follows from the upper bound on $E_{k}$. For the
lower bound in (9.9), if $0<p\leq 1$, by the Alexandrov–Fenchel inequality
(2.16) and the upper bound on $E_{k}$,

$\displaystyle(n+1)^{n+1}\mathrm{Vol}(K_{t})^{\frac{n-k}{n+1}}\mathrm{Vol}(W)^{\frac{k+1}{n+1}}\leq$
$\displaystyle\int_{M_{t}}E_{k}d\mu_{\gamma}\leq\sup_{M_{t}}E_{k}^{1-p}\int_{M_{t}}E_{k}^{p}d\mu_{\gamma}.$

Since the enclosed $K_{t}$ has fixed volume, the lower bound in (9.9) follows
from the above inequality and the upper bound on $|M_{t}|_{\gamma}$ in (7.9).
If $p>1$, the lower bound follows similarly by using the following inequality

$\displaystyle(n+1)^{n+1}\mathrm{Vol}(K_{t})^{\frac{n-k}{n+1}}\mathrm{Vol}(W)^{\frac{k+1}{n+1}}\leq\int_{M_{t}}E_{k}d\mu_{\gamma}\leq$
$\displaystyle~{}\left(\int_{M_{t}}E_{k}^{p}d\mu_{\gamma}\right)^{1/p}|M_{t}|_{\gamma}^{1-\frac{1}{p}}.$

∎

Let $[0,T)$ be the maximal interval such that the smooth solution $M_{t}$ of
the flow (1.9) exists.

######  Proposition 9.5.

If $M_{0}$ is strictly convex, then the solution $M_{t}$ of the flow (1.9) is
strictly convex for all $t\in[0,T)$. If $T<\infty$, there exists a constant
$C$ depending on $M_{0},{\gamma}$ and $T$ such that the anisotropic principal
curvatures of $M_{t}$ satisfy $\kappa_{i}\geq C>0$ on $t\in[0,T)$.

###### Proof.

To prove this estimate, we use the anisotropic Gauss map parametrization of
the flow (1.9) as described in §8. The flow (1.9) is equivalent to the scalar
parabolic equation (8.7) on $\Sigma$ of the anisotropic support function
$s(z,t)$. Since the eigenvalues of $(\tau_{ij})$ are equal to the reciprocal
of the anisotropic principal curvatures $\kappa$, in order to estimate the
lower bound of $\kappa$, it suffices to estimate the upper bound of the
eigenvalues of $\tau_{ij}$.

For any given time $t_{0}\in[0,T)$, we choose the origin as $p_{0}$ such that
the estimate (9.1) holds. Then Lemma 9.2 implies that the anisotropic support
function $s(z,t)$ of $M_{t}$ with respect to the origin $p_{0}$ satisfies

$R_{1}/2\leq s(z,t)\leq
2R_{2},\qquad\forall~{}t\in[t_{0},\min\\{T,t_{0}+\tau\\})$ (9.11)

for some fixed constant $\tau$ depending only on $n,k,\alpha,\gamma,M_{0}$.
Suppose $e_{1}$ is the direction where the maximum eigenvalue of $(\tau_{ij})$
occurs. By the evolution equation (8.5) of $\tau_{ij}$, we have

$\displaystyle\frac{\partial}{\partial t}\tau_{11}\leq$
$\displaystyle~{}\dot{\Psi}^{k\ell}\left(\bar{\nabla}_{k}\bar{\nabla}_{\ell}\tau_{11}-\frac{1}{2}Q_{k\ell
p}\bar{\nabla}_{p}\tau_{11}\right)+\ddot{\Psi}^{k\ell,pq}\bar{\nabla}_{1}\tau_{k\ell}\bar{\nabla}_{1}\tau_{pq}$
$\displaystyle\quad+(1-\alpha)\Psi+\phi(t)+C_{1}\dot{\Psi}^{k\ell}\bar{g}_{k\ell}\tau_{11},$
(9.12)

where $C_{1}=C_{1}(Q,\bar{\nabla}Q)$ is a constant depending only on
$Q,\bar{\nabla}Q$. Applying Theorem 9.3 and Corollary 9.4, we have
$(1-\alpha)\Psi+\phi(t)\leq C$ for some uniform constant $C>0$. However,
$\dot{\Psi}^{k\ell}\bar{g}_{k\ell}=\alpha
F_{*}^{-\alpha-1}\dot{F}_{*}^{k\ell}\bar{g}_{k\ell}$, which has no control
from above. Therefore, the parabolic maximum principle for tensors can not be
applied directly to (9) to deduce the upper bound of $\tau_{11}$.

To overcome this problem, we define

$\zeta=\sup\\{\tau_{ij}\xi^{i}\xi^{j}:~{}|\xi|=1\\}$

and consider

$\omega(z,t)=\frac{\zeta(z,t)}{(s(z,t)-\frac{R_{1}}{4})^{a}},$ (9.13)

where the power $a>0$ will be determined later. The function $\omega$ is well
defined on the time interval $[t_{0},\min\\{T,t_{0}+\tau\\})$. We consider the
spatial maximum point $(z_{1},t_{1})$ where
$\omega(z_{1},t_{1})=\max_{\Sigma}\omega(\cdot,t_{1})$ for each
$t_{1}\in[t_{0},\min\\{T,t_{0}+\tau\\})$. By rotation of the local orthonormal
frame, we assume that $\xi=e_{1}$ and
$(\tau_{ij})=\mathrm{diag}(\tau_{1},\dots,\tau_{n})$ is diagonal at
$(z_{1},t_{1})$. Then we have $\zeta=\tau_{11}$ at $(z_{1},t_{1})$. Combining
(8.9) and (9) gives

$\displaystyle\frac{\partial}{\partial
t}\omega-\dot{\Psi}^{k\ell}\left(\bar{\nabla}_{k}\bar{\nabla}_{\ell}\omega-\frac{1}{2}Q_{k\ell
p}\bar{\nabla}_{p}\omega\right)-\frac{2}{(s-\frac{R_{1}}{4})^{a}}\dot{\Psi}^{k\ell}\bar{\nabla}_{k}\omega\bar{\nabla}_{\ell}(s-\frac{R_{1}}{4})^{a}$
$\displaystyle=$
$\displaystyle\frac{1}{(s-\frac{R_{1}}{4})^{a}}\left(\frac{\partial}{\partial
t}\tau_{11}-\dot{\Psi}^{k\ell}\left(\bar{\nabla}_{k}\bar{\nabla}_{\ell}\tau_{11}-\frac{1}{2}Q_{k\ell
p}\bar{\nabla}_{p}\tau_{11}\right)\right)$
$\displaystyle-\frac{a\omega}{s-\frac{R_{1}}{4}}\left(\frac{\partial}{\partial
t}s-\dot{\Psi}^{k\ell}\left(\bar{\nabla}_{k}\bar{\nabla}_{\ell}s-\frac{1}{2}Q_{k\ell
p}\bar{\nabla}_{p}s\right)\right)+\frac{a(a-1)}{(s-\frac{R_{1}}{4})^{2}}\omega\dot{\Psi}^{k\ell}\bar{\nabla}_{k}s\bar{\nabla}_{\ell}s$
$\displaystyle\leq$
$\displaystyle\frac{1}{(s-\frac{R_{1}}{4})^{a}}\left(\ddot{\Psi}^{k\ell,pq}\bar{\nabla}_{1}\tau_{k\ell}\bar{\nabla}_{1}\tau_{pq}+(1-\alpha)\Psi+\phi(t)+C_{1}\dot{\Psi}^{k\ell}\bar{g}_{k\ell}\tau_{11}\right)$
$\displaystyle-\frac{a\omega}{s-\frac{R_{1}}{4}}\left(\phi(t)+(1+\alpha)\Psi+s\dot{\Psi}^{k\ell}\bar{g}_{k\ell}\right)+\frac{a(a-1)}{(s-\frac{R_{1}}{4})^{2}}\omega\dot{\Psi}^{k\ell}\bar{\nabla}_{k}s\bar{\nabla}_{\ell}s$
$\displaystyle\leq$
$\displaystyle\frac{1}{(s-\frac{R_{1}}{4})^{a}}\ddot{\Psi}^{k\ell,pq}\bar{\nabla}_{1}\tau_{k\ell}\bar{\nabla}_{1}\tau_{pq}+\frac{a(a-1)}{(s-\frac{R_{1}}{4})^{2}}\omega\dot{\Psi}^{k\ell}\bar{\nabla}_{k}s\bar{\nabla}_{\ell}s$
$\displaystyle+\omega\sum_{k}\dot{\psi}^{k}\left(C_{1}-\frac{as}{s-\frac{R_{1}}{4}}\right)+\frac{1}{(s-\frac{R_{1}}{4})^{a}}((1-\alpha)\Psi+\phi(t))$
$\displaystyle-\frac{a\omega}{s-\frac{R_{1}}{4}}\left(\phi(t)+(1+\alpha)\Psi\right).$
(9.14)

We estimate gradient terms in (9). Since $\Psi$ is concave, we have
$\ddot{\psi}\leq 0$ and
$(\dot{\psi}^{k}-\dot{\psi}^{\ell})(\tau_{k}-\tau_{\ell})\leq 0$. Using the
equation (8.6) we estimate the first term on the right-hand side of (9)

$\displaystyle\ddot{\Psi}^{k\ell,pq}\bar{\nabla}_{1}\tau_{k\ell}\bar{\nabla}_{1}\tau_{pq}=$
$\displaystyle~{}\ddot{\psi}^{k\ell}\bar{\nabla}_{1}\tau_{kk}\bar{\nabla}_{1}\tau_{\ell\ell}+2\sum_{k>\ell}\frac{\dot{\psi}^{k}-\dot{\psi}^{\ell}}{\tau_{k}-\tau_{\ell}}(\bar{\nabla}_{1}\tau_{k\ell})^{2}$
$\displaystyle\leq$
$\displaystyle~{}-2\sum_{k>1}\tau_{1}^{-1}(\dot{\psi}^{k}-\dot{\psi}^{1})(\bar{\nabla}_{1}\tau_{k1})^{2}.$

Codazzi equation (8.4) implies that

$\displaystyle(\bar{\nabla}_{1}\tau_{k1})^{2}=$
$\displaystyle\left(\bar{\nabla}_{k}\tau_{11}+\frac{1}{2}Q_{11p}\tau_{kp}-\frac{1}{2}Q_{k1p}\tau_{1p}\right)^{2}\geq\frac{1}{2}(\bar{\nabla}_{k}\tau_{11})^{2}-C(\tau_{1})^{2}.$

Since $\bar{\nabla}_{k}\omega=0$ at $(z_{1},t_{1})$, we have

$\frac{\bar{\nabla}_{k}\tau_{11}}{\tau_{11}}=\frac{a\bar{\nabla}_{k}s}{s-\frac{R_{1}}{4}}.$
(9.15)

Then the gradient terms on the right-hand side of (9) satisfy

$\displaystyle\frac{1}{(s-\frac{R_{1}}{4})^{a}}\ddot{\Psi}^{k\ell,pq}\bar{\nabla}_{1}\tau_{k\ell}\bar{\nabla}_{1}\tau_{pq}+\frac{a(a-1)}{(s-\frac{R_{1}}{4})^{2}}\omega\dot{\Psi}^{k\ell}\bar{\nabla}_{k}s\bar{\nabla}_{\ell}s$
$\displaystyle=$
$\displaystyle\omega\biggl{(}\frac{1}{\tau_{11}}\ddot{\Psi}^{k\ell,pq}\bar{\nabla}_{1}\tau_{k\ell}\bar{\nabla}_{1}\tau_{pq}+\frac{a-1}{a}\frac{1}{(\tau_{11})^{2}}\sum_{k}\dot{\psi}^{k}(\bar{\nabla}_{k}\tau_{11})^{2}\biggr{)}$
$\displaystyle\leq$
$\displaystyle~{}\omega\biggl{(}-2\sum_{k>1}\tau_{11}^{-2}(\dot{\psi}^{k}-\dot{\psi}^{1})\left(\frac{1}{2}(\bar{\nabla}_{k}\tau_{11})^{2}-C(\tau_{1})^{2}\right)+\frac{a-1}{a}\frac{1}{(\tau_{11})^{2}}\sum_{k}\dot{\psi}^{k}(\bar{\nabla}_{k}\tau_{11})^{2}\biggr{)}$
$\displaystyle=$
$\displaystyle~{}\omega\biggl{(}\dot{\psi}^{1}\tau_{11}^{-2}\sum_{k}(\bar{\nabla}_{k}\tau_{11})^{2}+2C\sum_{k>1}(\dot{\psi}^{k}-\dot{\psi}^{1})\biggr{)}$
$\displaystyle\leq$
$\displaystyle~{}\omega\biggl{(}\dot{\psi}^{1}a^{2}\frac{|\bar{\nabla}s|^{2}}{(s-\frac{R_{1}}{4})^{2}}+C_{2}\sum_{k}\dot{\psi}^{k}\biggr{)}$
(9.16)

at $(z_{1},t_{1})$.

Substituting (9) into (9), we have

$\displaystyle\frac{\partial}{\partial t}\omega\leq~{}$
$\displaystyle\omega\sum_{k}\dot{\psi}^{k}\left(C_{1}+C_{2}-\frac{as}{s-\frac{R_{1}}{4}}\right)+\omega\dot{\psi}^{1}a^{2}\frac{|\bar{\nabla}s|^{2}}{(s-\frac{R_{1}}{4})^{2}}$
$\displaystyle+\frac{1}{(s-\frac{R_{1}}{4})^{a}}((1-\alpha)\Psi+\phi(t))-\frac{a\omega}{s-\frac{R_{1}}{4}}\left(\phi(t)+(1+\alpha)\Psi\right)$
(9.17)

at $(z_{1},t_{1})$. Since $s(z,t)$ satisfies the estimate (9.11) on the time
interval $[t_{0},\min\\{T,t_{0}+\tau\\})$, by choosing a suitable constant
$a>0$ depending on $R_{1},R_{2}$ and $C_{1},C_{2}$ we can ensure that the
first term on the right-hand side of (9) is negative. This kills the bad term
$\sum_{k}\dot{\psi}^{k}$. The second term on the right-hand side of (9) is
bounded, since $|\bar{\nabla}s|$ is bounded by (9.2) and

$\omega\dot{\psi}^{1}=\frac{\dot{\psi}^{1}\tau_{1}}{(s-\frac{R_{1}}{4})^{a}}\leq\frac{\sum_{i=1}^{n}\dot{\psi}^{i}\tau_{i}}{(s-\frac{R_{1}}{4})^{a}}=\frac{\alpha{E}_{k}^{\alpha/k}}{(s-\frac{R_{1}}{4})^{a}}\leq
C.$

Then at the spatial maximum point $(z_{1},t_{1})$, the function $\omega$
satisfies

$\frac{\partial}{\partial t}\omega\leq C_{3}+C_{4}\omega,$ (9.18)

where $C_{3}$, $C_{4}$ are constants depending on $a,R_{1},R_{2}$, the bound
on $\phi(t)$ in Corollary 9.4 and the upper bound on $-\Psi=E_{k}^{\alpha/k}$
in Proposition 9.3. Applying the parabolic maximum principle to (9.18), we see
that the function $\omega$ in the time interval
$[t_{0},\min\\{T,t_{0}+\tau\\})$ is bounded by a constant depending on its
value at the time $t_{0}$ and on the parameter $\tau$. Since $0<R_{1}/2\leq
s(z,t)\leq 2R_{2}$ for all $t\in[t_{0},\min\\{T,t_{0}+\tau\\})$, this also
implies that the largest anisotropic principal radius $\tau_{1}$ is bounded by
a constant depending on its value at the time $t_{0}$ and on the parameter
$\tau$. If the maximal existence time $T$ is finite, after a finite number of
iterations, we conclude that the largest anisotropic principal radius
$\tau_{1}$ is bounded from above by a positive constant $C$ depending on the
maximal existence time $T$. This completes the proof of Proposition 9.5. ∎

Now we can prove the long time existence of the flow (1.9), i.e., the maximum
existence time $T=\infty$.

######  Theorem 9.6.

Let $W$ be a Wulff shape in $\mathbb{R}^{n+1}$ with the smooth support
function $\gamma\in C^{\infty}(\mathbb{S}^{n})$, and
$X_{0}:M^{n}\to\mathbb{R}^{n+1}$ be a smooth embedding such that
$M_{0}=X_{0}(M)$ is a closed strictly convex hypersurface in
$\mathbb{R}^{n+1}$ enclosing a convex body $K_{0}$. Then for any
$k\in\\{1,\dots,n\\}$ and $\alpha>0$, the volume preserving flow (1.9) with
the global term $\phi(t)$ given by (1.10) has a smooth strictly convex
solution $M_{t}$ for all time $t\in[0,\infty)$.

###### Proof.

Suppose that the maximal existence time $T$ is finite. By Proposition 9.3 and
Proposition 9.5, there exists a constant $C>0$ depending on $M_{0},{\gamma}$
and $T$ such that the anisotropic principal curvatures satisfy

$0<\frac{1}{C}\leq\kappa_{i}\leq C,\qquad i=1,\cdots,n$

on $M_{t}$ for $t\in[0,T)$. This together with $C^{0}$ and $C^{1}$ estimates
implies the $C^{2}$ estimate of the evolving hypersurface $M_{t}$. To derive
the higher regularity estimate, we note that the flow (1.9) is equivalent to
the equation (8.7) for the anisotropic support function $s(z,t)$. Since
$F_{*}$ is concave and $\alpha>0$, the right-hand side of (8.7) is concave
with respect to the spatial second derivatives of $s(z,t)$. As the global term
$\phi(t)$ is bounded, Theorem 1.1 of [37] can be applied to derive the
$C^{2,\beta}$ estimate of $s$ for some $\beta\in(0,1)$. The higher order
regularity estimate of $M_{t}$ follows from the standard parabolic Schauder
theory [26]. Thus $M_{t}$ converges to a hypersurface $M_{T}$ smoothly as
$t\to T$. The short time existence theorem yields that the flow (1.9) can be
extended past $T$ for a short time interval, contradicting the maximality of
$T$. ∎

##  10\. Hausdorff convergence

We showed in the previous section that the smooth solution $M_{t}$ of the flow
(1.9) exists for all time $t\in[0,\infty)$. In this section, using Theorem 1.2
and Theorem 3.10 for the anisotropic curvature measures and applying the
argument as in [6, §6], we show that the solution $M_{t}$ converges to a
scaled Wulff shape as $t\to\infty$ in the Hausdorff sense. As applications, we
further improve the estimate on the global term $\phi(t)$ and obtain the
uniform positive lower bound on the $k$th anisotropic mean curvature $E_{k}$.

###  10.1. Hausdorff convergence to the Wulff shape

Let

$\bar{E}_{k}=~{}\frac{1}{|M_{t}|_{\gamma}}\int_{M_{t}}{E}_{k}d\mu_{\gamma}=~{}\frac{V_{n-k}(K_{t},W)}{V_{n}(K_{t},W)}$

denote the average integral of $E_{k}$. By Proposition 7.3, the mixed volume
$V_{n+1-k}(K_{t},W)$ is monotone non-increasing in time. Since
$V_{n+1-k}(K_{t},W)$ is bounded from below, by the long-time existence of the
flow, the equation (7.2) implies that

$\int_{0}^{\infty}\int_{M_{t}}\left(E_{k}^{\alpha/k}-\bar{E}_{k}^{\alpha/k}\right)\left(E_{k}-\bar{E}_{k}\right)d\mu_{\gamma}~{}\leq~{}V_{n+1-k}(K_{0},W)<\infty.$

Therefore there exists a sequence of times $t_{i}\to\infty$ such that

$\displaystyle\int_{M_{t_{i}}}\left(E_{k}^{\alpha/k}-\bar{E}_{k}^{\alpha/k}\right)\left(E_{k}-\bar{E}_{k}\right)d\mu_{\gamma}~{}\to
0.$ (10.1)

If $0<\alpha<k$, we have

$\displaystyle\left(E_{k}^{\alpha/k}-\bar{E}_{k}^{\alpha/k}\right)\left(E_{k}-\bar{E}_{k}\right)=$
$\displaystyle\frac{\alpha}{k}\int_{0}^{1}\left((1-s)\bar{E}_{k}+sE_{k}\right)^{\alpha/k-1}ds\cdot(E_{k}-\bar{E}_{k})^{2}$
$\displaystyle\geq$
$\displaystyle\frac{\alpha}{k}\left(\sup_{M_{t}}E_{k}\right)^{\alpha/k-1}(E_{k}-\bar{E}_{k})^{2}$
$\displaystyle\geq$ $\displaystyle C(E_{k}-\bar{E}_{k})^{2}$

by the upper bound on $E_{k}$ that proved in Proposition 9.3. On the other
hand, for $\alpha>k$ we have

$\displaystyle\left(E_{k}^{\alpha/k}-\bar{E}_{k}^{\alpha/k}\right)\left(E_{k}-\bar{E}_{k}\right)=$
$\displaystyle\frac{\alpha}{k}\int_{0}^{1}\left((1-s)\bar{E}_{k}+sE_{k}\right)^{\alpha/k-1}ds\cdot(E_{k}-\bar{E}_{k})^{2}$
$\displaystyle\geq$
$\displaystyle\frac{\alpha}{k}\int_{0}^{1}(1-s)^{\alpha/k-1}ds\bar{E}_{k}^{\alpha/k-1}(E_{k}-\bar{E}_{k})^{2}$
$\displaystyle\geq$ $\displaystyle C(E_{k}-\bar{E}_{k})^{2}$

by Corollary 9.4. Therefore, (10.1) implies that

$\displaystyle\int_{M_{t_{i}}}\left({E}_{k}-\bar{E}_{k}\right)^{2}d\mu_{\gamma}~{}\to$
$\displaystyle~{}0,\quad\mathrm{as}~{}i\to\infty$ (10.2)

for a sequence of times $t_{i}\to\infty$.

######  Proposition 10.1.

Let $M_{t}$ be the smooth solution to the flow (1.9) with the global term
$\phi(t)$ given by (1.10). Denote the closure of the enclosed domain of
$M_{t}$ by $K_{t}$. Then $K_{t}$ converges to a scaled Wulff shape $\bar{r}W$
in the Hausdorff sense as $t\to\infty$, where $\bar{r}$ is the radius such
that $\mathrm{Vol}(K_{0})=\bar{r}^{n+1}\mathrm{Vol}(W)$.

###### Proof.

The proof is similar to Lemma 6.3 in [6]. We include it here for completeness.
By the estimate (7.10) on the anisotropic outer radius of $K_{t}$, the
Blaschke selection theorem (see Theorem 1.8.7 of [33]) implies that there
exists a subsequence of times $t_{i}$ and a convex body $\hat{K}$ such that
$K_{t_{i}}$ converges to $\hat{K}$ in the Hausdorff sense as $t_{i}\to\infty$.
As each $K_{t_{i}}$ has the anisotropic inner radius $r(K_{t_{i}})\geq R_{1}$,
the limit convex body $\hat{K}$ has a positive anisotropic inner radius.
Without loss of generality, we may assume that the sequence $t_{i}$ is the
same sequence such that (10.2) holds.

We will show that the limit convex body $\hat{K}$ satisfies
$\Phi_{n-k}(\hat{K};.)=c\,\Phi_{n}(\hat{K};.)$ where
$c=V_{n-k}(\hat{K},W)/V_{n}(\hat{K},W)$. The weak continuity of the
anisotropic curvature measure $\Phi_{m}$ proved in Theorem 3.10 is equivalent
to the statement that $\int fd\Phi_{m}(K_{i})$ converges to $\int
fd\Phi_{m}(K)$ whenever $f$ is a bounded continuous function on
${\mathbb{R}}^{n+1}$ and $K_{i}$ is a sequence of convex sets converging to
$K$ in the Hausdorff distance. In particular we get that $\int
fd\Phi_{n-k}(K_{t_{i}})$ converges to $\int fd\Phi_{n-k}(\hat{K})$, and $\int
fd\Phi_{n}(K_{t_{i}})$ converges to $\int fd\Phi_{n}(\hat{K})$, as
$i\to\infty$. Since $K_{t_{i}}$ is smooth and uniformly convex, by (3.16) we
obtain for any bounded continuous $f$,

$\displaystyle\left|\int fd\Phi_{n-k}(K_{t_{i}})-c\int
fd\Phi_{n}(K_{t_{i}})\right|$
$\displaystyle=\left|\int_{M_{t_{i}}}fE_{k}\gamma(\nu)d{\mathcal{H}}^{n}-\int_{M_{t_{i}}}fc\gamma(\nu)d{\mathcal{H}}^{n}\right|$
$\displaystyle\leq\sup|f|\int_{M_{t_{i}}}|E_{k}-c|\gamma(\nu)d{\mathcal{H}}^{n}$
$\displaystyle\leq\sup|f|\int_{M_{t_{i}}}|E_{k}-\bar{E}_{k}|\gamma(\nu)d{\mathcal{H}}^{n}$
$\displaystyle\quad\hbox{}+\sup|f|V_{n}(K_{t_{i}},W)\left|\frac{V_{n-k}(K_{t_{i}},W)}{V_{n}(K_{t_{i}},W)}-\frac{V_{n-k}(\hat{K},W)}{V_{n}(\hat{K},W)}\right|.$

The left-hand side converges to $\left|\int fd\Phi_{n-k}(\hat{K})-c\int
fd\Phi_{n}(\hat{K})\right|$ by the weak continuity of the anisotropic
curvature measures, while the first term on the right-hand side converges to
zero by (10.2), and the second also does by the continuity of the mixed
volumes with respect to the Hausdorff distance. It follows that

$\int fd\Phi_{n-k}(\hat{K})=c\int fd\Phi_{n}(\hat{K})$

for all bounded continuous functions $f$, and therefore that
$\Phi_{n-k}(\hat{K};.)=c\,\Phi_{n}(\hat{K};.)$ as claimed. By Theorem 1.2, the
convex body $\hat{K}$ is a scaled Wulff shape.

By Proposition 7.3, ${\mathcal{I}}_{n+1-k}(K_{t},W)$ is non-increasing in
time. Since ${\mathcal{I}}_{n+1-k}(K_{t_{i}},W)\to 1$ as $i\to\infty$ for some
subsequence of times $t_{i}\to\infty$, we conclude that
${\mathcal{I}}_{n+1-k}(K_{t},W)\to 1$ as $t\to\infty$. It follows from the
stability estimate (7.124) in [33] that the whole family of $K_{t}$ converges
to a scaled Wulff shape $\bar{r}W$ as $t\to\infty$, where $\bar{r}$ is the
radius such that $\mathrm{Vol}(K_{0})=\bar{r}^{n+1}\mathrm{Vol}(W)$. This
completes the proof of Proposition 10.1. ∎

###  10.2. Improved estimate on the global term $\phi(t)$

The Hausdorff convergence of the solution has the following consequence on the
global term $\phi(t)$ in the flow equation (1.9).

######  Corollary 10.2.

If $\alpha\geq k$ in the flow equation (1.9), the global term $\phi(t)$ given
by (1.10) satisfies

$\liminf_{t\to\infty}\phi(t)\geq\bar{r}^{-\alpha},$ (10.3)

where $\bar{r}$ is the radius such that
$\mathrm{Vol}(K_{0})=\bar{r}^{n+1}\mathrm{Vol}(W)$. In general, for any
$\alpha>0$, the global term $\phi(t)$ satisfies

$\int_{t}^{t+d}\biggl{|}\phi(s)-\frac{1}{\bar{r}^{\alpha}}\biggr{|}ds~{}\to~{}0$
(10.4)

as $t\to\infty$ for any fixed constant $d>0$.

###### Proof.

If $\alpha\geq k$, then by the Hölder inequality we get

$\phi(t)=\frac{1}{|M_{t}|_{\gamma}}\int_{M_{t}}{E}_{k}^{\alpha/k}d\mu_{\gamma}\geq\left(\frac{\int_{M_{t}}{E}_{k}d\mu_{\gamma}}{|M_{t}|_{\gamma}}\right)^{\alpha/k}=\left(\frac{V_{n-k}(K_{t},W)}{V_{n}(K_{t},W)}\right)^{\alpha/k}.$

Since $K_{t}$ converges to a scaled Wulff shape $\bar{r}W$ in the Hausdorff
sense as $t\to\infty$, the continuity of the mixed volumes implies the
estimate (10.3).

To show (10.4) for a general $\alpha>0$, as at the beginning of this section
we note that

$\displaystyle\left(E_{k}^{\alpha/k}-\bar{E}_{k}^{\alpha/k}\right)^{2}=$
$\displaystyle\frac{\alpha}{k}\int_{0}^{1}\left((1-s)\bar{E}_{k}+sE_{k}\right)^{\alpha/k-1}ds\cdot(E_{k}-\bar{E}_{k})\left(E_{k}^{\alpha/k}-\bar{E}_{k}^{\alpha/k}\right).$

If $\alpha>k$,

$\displaystyle\frac{\alpha}{k}\int_{0}^{1}\left((1-s)\bar{E}_{k}+sE_{k}\right)^{\alpha/k-1}ds\leq$
$\displaystyle\frac{\alpha}{k}\left(\sup_{M_{t}}E_{k}\right)^{\alpha/k-1}.$

If $0<\alpha<k$,

$\displaystyle\frac{\alpha}{k}\int_{0}^{1}\left((1-s)\bar{E}_{k}+sE_{k}\right)^{\alpha/k-1}ds\leq$
$\displaystyle\bar{E}_{k}^{\alpha/k-1}\frac{\alpha}{k}\int_{0}^{1}(1-s)^{\alpha/k-1}ds.$

In both cases, we obtain

$\displaystyle\left(E_{k}^{\alpha/k}-\bar{E}_{k}^{\alpha/k}\right)^{2}\leq$
$\displaystyle
C(E_{k}-\bar{E}_{k})\left(E_{k}^{\alpha/k}-\bar{E}_{k}^{\alpha/k}\right),$
(10.5)

where $C=C(n,k,\alpha,\gamma,K_{0})$ comes from the bounds on
$E_{k},\bar{E}_{k}$ in Proposition 9.3 and Corollary 9.4. Since
$V_{n+1-k}(K_{t},W)$ is non-increasing and converges to the value
$V_{n+1-k}(\bar{r}W,W)$, for any small $\varepsilon>0$, there exists a large
time $N_{0}$ such that for all times $t_{1}>t_{2}>N_{0}$,

$V_{n+1-k}(K_{t_{2}},W)-V_{n+1-k}(K_{t_{1}},W)\leq\varepsilon.$

In particular, for any fixed constant $d>0$ the estimates (7.2) and (10.5)
imply that

$\displaystyle\int_{t}^{t+d}\int_{M_{s}}\left(E_{k}^{\alpha/k}-\bar{E}_{k}^{\alpha/k}\right)^{2}d\mu_{\gamma}ds\leq$
$\displaystyle C\varepsilon$ (10.6)

for all $t>N_{0}$, where $C$ depends on $n,k,\alpha,\gamma,K_{0}$. Applying
the Hölder inequality

$\displaystyle\biggl{|}\phi(t)-\bar{E}_{k}^{\alpha/k}\biggr{|}=$
$\displaystyle\frac{1}{|M_{t}|_{\gamma}}\biggl{|}\int_{M_{t}}\left(E_{k}^{\alpha/k}-\bar{E}_{k}^{\alpha/k}\right)d\mu_{\gamma}\biggr{|}$
$\displaystyle\leq$
$\displaystyle\left(\frac{1}{|M_{t}|_{\gamma}}\int_{M_{t}}\left(E_{k}^{\alpha/k}-\bar{E}_{k}^{\alpha/k}\right)^{2}d\mu_{\gamma}\right)^{1/2}$

and the uniform bounds on $|M_{t}|_{\gamma}$, we have

$\displaystyle\int_{t}^{t+d}\biggl{|}\phi(s)-\bar{E}_{k}^{\alpha/k}\biggr{|}ds\leq$
$\displaystyle C\varepsilon^{1/2}.$ (10.7)

By Proposition 10.1, $K_{t}$ converges to a scaled Wulff shape $\bar{r}W$ as
$t\to\infty$. The continuity of the mixed volume $V_{n-k}(\cdot,W)$ with
respect to the Hausdorff distance implies that

$\displaystyle\bar{E}_{k}=$
$\displaystyle~{}\frac{V_{n-k}(K_{t},W)}{V_{n}(K_{t},W)}\to~{}\frac{V_{n-k}(\bar{r}W,W)}{V_{n}(\bar{r}W,W)}=\bar{r}^{-k},\quad\mathrm{as}~{}t\to\infty.$

Then we conclude from (10.7) that for any fixed constant $d>0$ and any small
$\varepsilon>0$, there exists a large $N_{0}$ such that

$\displaystyle\int_{t}^{t+d}\biggl{|}\phi(s)-\bar{r}^{-\alpha}\biggr{|}ds\leq$
$\displaystyle C\varepsilon^{1/2}$ (10.8)

for all $t>N_{0}$, where $C$ is a uniform constant depending only on
$n,k,\alpha,\gamma$ and $K_{0}$. ∎

###  10.3. Lower bound on ${E}_{k}$

In this subsection, we prove a uniform positive lower bound on the $k$th
anisotropic mean curvature $E_{k}$. The key is the Hausdorff convergence of
$K_{t}$ to a scaled Wulff shape $\bar{r}W$ as $t\to\infty$. We also use the
uniform two-sided positive bounds on the global term $\phi(t)$.

######  Proposition 10.3.

Let $M_{t}$, $t\in[0,\infty),$ be a smooth solution of the flow (1.9). Then
there exists a positive constant $C=C(n,k,\alpha,\gamma,K_{0})$ such that
${E}_{k}\geq C$ on $M_{t}$ for all $t\geq 0$.

###### Proof.

It suffices to control ${E}_{k}$ from below for sufficiently large times. By
Proposition 10.1, $K_{t}$ converges to a scaled Wulff shape $\bar{r}W$ as
$t\to\infty$ in the Hausdorff sense. Then for any small $\epsilon>0$, there
exists a sufficiently large time $N_{0}>0$ such that for all $t\geq N_{0}$,
the anisotropic outer radius $R(t)$ and anisotropic inner radius $r(t)$ of
$K_{t}$ relative to $W$ satisfy

$(1-\epsilon)\bar{r}~{}\leq~{}r(t)\leq~{}R(t)\leq~{}(1+\epsilon)\bar{r}.$
(10.9)

In the following, we only consider times $t\geq N_{0}$.

Let $t_{1}>N_{0}$ and set

$\varphi_{t_{1}}(t)=\int_{t_{1}}^{t}\phi(s)ds.$

We use the anisotropic Gauss map parametrization described in §8. Choose the
origin to be the center of the inner Wulff shape such that $r(t_{1})W\subset
K_{t_{1}}$. Then

$s(z,t_{1})\geq~{}s(z_{1},t_{1})=r(t_{1})\geq~{}(1-\epsilon)\bar{r},$

where $z_{1}$ is the minimum point of $s(z,t_{1})$. Define a function $f(z,t)$
on $\Sigma=\partial W$ by

$f(z,t)~{}=~{}-(1+\alpha)(t-t_{1})\Psi(z,t)-(1+\alpha)\varphi_{t_{1}}(t)+s(z,t)$

for $t\geq t_{1}$. Combining (8.7) and (8.11) gives the evolution equation of
$f(z,t)$

$\displaystyle\frac{\partial}{\partial t}f(z,t)=$
$\displaystyle~{}\dot{\Psi}^{k\ell}(z,t)\bar{\nabla}_{k}\bar{\nabla}_{\ell}f(z,t)-\frac{1}{2}\dot{\Psi}^{k\ell}(z,t)Q_{k\ell
p}\bar{\nabla}_{p}f-\alpha\phi(t)$
$\displaystyle\quad+\biggl{(}s(z,t)-(1+\alpha)(t-t_{1})(\Psi+\phi(t))\biggr{)}\dot{\Psi}^{k\ell}\bar{g}_{k\ell}.$
(10.10)

We now show that the last term on the right-hand side of (10.3) is nonnegative
for a short time after time $t_{1}$. By Theorem 9.3,
$\Psi=-{E}_{k}^{\alpha/k}$ is bounded below by a uniform negative constant
$-c_{+}$. On the other hand, by Corollary 9.4 the function $\phi(t)$ is
bounded from above and below by positive constants:

$0<\phi_{-}\leq\phi(t)\leq~{}\phi_{+},$

where $\phi_{-},\phi_{+}$ depend only on $n,k,{\gamma},\alpha$ and $K_{0}$.
Then integrating the evolution equation (8.7) of $s(z,t)$ gives that

$\displaystyle s(z,t)\geq$
$\displaystyle~{}s(z,t_{1})+(\phi_{-}-c_{+})(t-t_{1})\geq(1-\epsilon)\bar{r}+(\phi_{-}-c_{+})(t-t_{1}).$

Let $t_{2}>t_{1}$ be the time such that

$t_{2}-t_{1}=~{}\frac{(1-\epsilon)\bar{r}}{c_{+}-\phi_{-}+(1+\alpha)\phi_{+}},$
(10.11)

which is positive and independent of the time $t_{1}$. Then for any
$t\in[t_{1},t_{2}]$, we have

$\displaystyle s(z,t)-(1+\alpha)(t-t_{1})(\Psi(z,t)+\phi(t))$
$\displaystyle\geq(1-\epsilon)\bar{r}+(\phi_{-}-c_{+})(t-t_{1})-(1+\alpha)(t-t_{1})\phi_{+}\geq~{}0.$
(10.12)

Since $\dot{\Psi}^{k\ell}\bar{g}_{k\ell}\geq 0$, combining (10.3) with (10.3)
gives that

$\displaystyle\frac{\partial}{\partial t}f(z,t)\geq$
$\displaystyle~{}\dot{\Psi}^{k\ell}(z,t)\bar{\nabla}_{k}\bar{\nabla}_{\ell}f(z,t)-\frac{1}{2}\dot{\Psi}^{k\ell}(z,t)Q_{k\ell
p}\bar{\nabla}_{p}f-\alpha\phi(t)$

for all time $t\in[t_{1},t_{2}]$. The maximum principle implies that

$\displaystyle f(z,t)\geq$
$\displaystyle\min_{M_{t_{1}}}f(z,t_{1})-\alpha\varphi_{t_{1}}(t)=~{}s(z_{1},t_{1})-\alpha\varphi_{t_{1}}(t)$

for all time $t\in[t_{1},t_{2}]$, which implies that

$-(1+\alpha)\Psi(z,t)\geq~{}\frac{\varphi_{t_{1}}(t)-s(z,t)+s(z_{1},t_{1})}{t-t_{1}}$
(10.13)

for all $t\in(t_{1},t_{2}]$ with $t_{2}$ defined in (10.11).

To estimate the lower bound of $-\Psi$, it remains to estimate the right-hand
side of (10.13). Recall that the origin is chosen such that $r(t_{1})W\subset
K_{t_{1}}$. Then by [2, Lem. 7.2], we have

$s(z,t_{1})\leq
R(t_{1})+C(R(t_{1})-r(t_{1}))\leq(1+(1+2C)\epsilon)\bar{r},\quad\forall~{}z\in\Sigma,$
(10.14)

where $C$ is a constant only depending on $\gamma$.

From the flow equation (8.7), we deduce that the spatial maximum
$\bar{s}(t)=\max_{\Sigma}s(\cdot,t)$ of the anisotropic support function $s$
satisfies

$\frac{d}{dt}\bar{s}(t)=~{}-F_{*}({\tau}_{ij})^{-\alpha}+\phi(t)\leq~{}\phi_{+},$
(10.15)

which implies that

$\displaystyle\bar{s}(t)\leq$
$\displaystyle~{}\bar{s}(t_{1})+\phi_{+}(t-t_{1})~{}\leq~{}2\bar{s}(t_{1}),\quad\forall~{}t\in[t_{1},t_{3}],$

where $t_{3}$ is given by $t_{3}-t_{1}=\phi_{+}^{-1}{\bar{s}}(t_{1})$. By the
definition of $\tau_{ij}$, at the maximum point of $s(\cdot,t)$, we have
${\tau}_{ij}\leq\bar{g}_{ij}s$. Therefore,

$\displaystyle\frac{d}{dt}\big{(}\varphi_{t_{1}}(t)-\bar{s}(t)+s(z_{1},t_{1})\big{)}=$
$\displaystyle F_{*}({\tau}_{ij})^{-\alpha}$ $\displaystyle\geq$
$\displaystyle~{}\bar{s}(t)^{-\alpha}$ $\displaystyle\geq$
$\displaystyle~{}2^{-\alpha}\bar{s}(t_{1})^{-\alpha},\quad\forall~{}t\in[t_{1},t_{3}],$

which implies that

$\displaystyle\varphi_{t_{1}}(t)-\bar{s}(t)+s(z_{1},t_{1})\geq$
$\displaystyle~{}-\bar{s}(t_{1})+s(z_{1},t_{1})+2^{-\alpha}\bar{s}(t_{1})^{-\alpha}(t-t_{1})$
$\displaystyle\geq$
$\displaystyle~{}2^{-1-\alpha}\bar{s}(t_{1})^{-\alpha}(t-t_{1}),$ (10.16)

provided that

$t\geq~{}t_{1}+2^{1+\alpha}r_{+}^{\alpha}(\bar{s}(t_{1})-s(z_{1},t_{1}))~{}=:t_{4}.$

By (10.9) and (10.14), we know that

$\bar{s}(t_{1})-s(z_{1},t_{1})\leq~{}2(1+C)\epsilon\bar{r}$

can be arbitrarily small by assuming $N_{0}$ is sufficiently large. Then the
waiting time $t_{4}-t_{1}$ can be made small such that
$t_{4}<\min\\{t_{2},t_{3}\\}$ by choosing $N_{0}$ sufficiently large.
Combining (10.13) and (10.3) yields that

$\displaystyle-(1+\alpha)\Psi(z,t)\geq$
$\displaystyle~{}\frac{\varphi_{t_{1}}(t)-s(z,t)+s(z_{1},t_{1})}{t-t_{1}}$
$\displaystyle\geq$ $\displaystyle~{}2^{-1-\alpha}\bar{s}(t_{1})^{-\alpha}$
$\displaystyle\geq$
$\displaystyle~{}2^{-1-\alpha}(1+(1+2C)\epsilon)^{-\alpha}\bar{r}^{-\alpha}$

for all time $t\in[t_{4},\min\\{t_{2},t_{3}\\}]$. This gives the uniform
positive lower bound on ${E}_{k}$. ∎

##  11\. Smooth convergence

In this section, we prove the smooth convergence of the flow (1.9) and
complete the proof of Theorem 1.3. We will consider the three cases (i) $k=1$,
$\alpha>0$; (ii) $k=n$, $\alpha>0$ and (iii) $\alpha\geq k=1,\dots,n$
separately.

###  11.1. Smooth convergence for $k=1$, $\alpha>0$

For $k=1$, from Proposition 9.3 we see that
$nE_{1}=\sum_{i=1}^{n}\kappa_{i}\leq C$ for some constant $C>0$. Since $M_{t}$
is convex for each $t>0$ by Proposition 9.5, we get that the anisotropic
principal curvatures are bounded

$0<\kappa_{i}\leq C,\quad i=1,\cdots,n$

along the flow (1.9). This is equivalent to the $C^{2}$ estimate of $M_{t}$.
In the following lemma, we derive the higher regularity of $M_{t}$.

######  Lemma 11.1.

Let $M_{t}$ be a smooth solution to the flow (1.9) with $k=1$, $\alpha>0$.
Then for any integer $\ell\geq 2$, there exists a constant $C_{\ell}$
depending only on $n,\alpha,\ell,\gamma$ such that

$|\hat{\nabla}^{\ell}\mathcal{W}_{\gamma}|\leq C_{\ell}$

for all $t>0$, where $\mathcal{W}_{\gamma}$ is the anisotropic Weingarten map
of $M_{t}$.

###### Proof.

As in [2, §8], we write the evolving hypersurface locally as a graph
$x(y^{1},\dots,y^{n},t)=y^{i}e_{i}+u(y^{1},\dots,y^{n},t)e_{0}$, where
$\\{e_{i}\\}_{i=0}^{n}$ is the standard basis for $\mathbb{R}^{n+1}$. The
tangent space of the graph is spanned by $\partial_{i}=e_{i}+u_{i}e_{0}$,
where $u_{i}=\partial u/\partial y^{i}$, $i=1,\dots,n$. The isotropic normal
vector field is

$\nu=(1+|Du|^{2})^{-1/2}(Du,-1).$

Let $\gamma$ be the $1$-homogenous extension of the support function of the
Wulff shape $\Sigma$. Then $\gamma\in
C^{\infty}(\mathbb{R}^{n+1}\setminus\\{0\\})$. Define a function
$\hat{\gamma}$ on $\mathbb{R}^{n}$ by

$\hat{\gamma}(p_{1},\dots,p_{n})={\gamma}(p_{1},\dots,p_{n},-1).$ (11.1)

Then $\hat{\gamma}(Du)=\sqrt{1+|Du|^{2}}{\gamma}(\nu)$. Let
$\omega=(Du,-1)\in\mathbb{R}^{n+1}$. By the $1$-homogeneity of $\gamma$,

$\displaystyle\hat{\gamma}(Du)=$
$\displaystyle\sum_{i=1}^{n}D^{i}\gamma\big{|}_{\omega}D_{i}u-D^{0}\gamma\big{|}_{\omega}=\sum_{i=1}^{n}D^{i}\hat{\gamma}\big{|}_{Du}D_{i}u-D^{0}\gamma\big{|}_{\omega}.$

Then the anisotropic unit normal $\nu_{\gamma}=D\gamma(\nu)$ can be expressed
as

$\displaystyle\nu_{\gamma}=$
$\displaystyle\sum_{i=1}^{n}D^{i}\gamma\big{|}_{\omega}e_{i}+D^{0}\gamma\big{|}_{\omega}e_{0}$
$\displaystyle=$
$\displaystyle\sum_{i=1}^{n}D^{i}\hat{\gamma}|_{Du}e_{i}+\left(\sum_{i=1}^{n}D_{i}uD^{i}\hat{\gamma}|_{Du}-\hat{\gamma}(Du)\right)e_{0}.$
(11.2)

Differentiate (11.1) with respect to $y^{\ell}$:

$\displaystyle\partial_{\ell}\nu_{\gamma}=$ $\displaystyle u_{\ell
i}D^{i}D^{k}\hat{\gamma}\big{|}_{Du}e_{k}+\left(u_{\ell
i}D^{i}\hat{\gamma}|_{Du}+u_{k}u_{\ell i}D^{i}D^{k}\hat{\gamma}|_{Du}-u_{\ell
i}D^{i}\hat{\gamma}\big{|}_{Du}\right)e_{0}$ $\displaystyle=$ $\displaystyle
u_{\ell i}D^{i}D^{k}\hat{\gamma}\big{|}_{Du}(e_{k}+u_{k}e_{0})$
$\displaystyle=$ $\displaystyle u_{\ell
i}D^{i}D^{k}\hat{\gamma}\big{|}_{Du}\partial_{k}x.$

By the anisotropic Weingarten formula (2.8), we obtain

$\hat{h}_{\ell}^{k}=u_{\ell i}D^{i}D^{k}\hat{\gamma}\big{|}_{Du}.$ (11.3)

Moreover, by (2.7) and (11.1), the anisotropic second fundamental form
$\hat{h}$ is given by

$\displaystyle\hat{h}_{ij}=$
$\displaystyle-G(\nu_{\gamma})(\nu_{\gamma},\partial_{i}\partial_{j}x)$
$\displaystyle=$ $\displaystyle-G(\nu_{\gamma})(\nu_{\gamma},u_{ij}e_{0})$
$\displaystyle=$ $\displaystyle-
u_{ij}G(\nu_{\gamma})\left(\nu_{\gamma},-\frac{\nu_{\gamma}}{\hat{\gamma}(Du)}+\frac{D^{i}\hat{\gamma}\big{|}_{Du}}{\hat{\gamma}(Du)}\partial_{i}x\right)$
$\displaystyle=$ $\displaystyle\frac{u_{ij}}{\hat{\gamma}(Du)}.$

Then the metric $\hat{g}$ satisfies

$\displaystyle\hat{g}^{ij}=\hat{\gamma}(Du)D^{i}D^{j}\hat{\gamma}|_{Du}.$

The equation (11.3) also implies that the normalized anisotropic mean
curvature is given by

$E_{1}(\kappa)=\frac{1}{n}\hat{g}^{ij}\hat{h}_{ij}=\frac{1}{n}u_{ij}D^{i}D^{j}\hat{\gamma}|_{Du}.$

Now we derive the evolution of the function $u$ along the flow (1.9). Write
the position vector field of the evolving hypersurface near a point
$(x_{0},t_{0})$ as

$X(p,t)~{}=~{}y^{i}(p,t)e_{i}+u(y(p,t),t)e_{0},$

where $y(p,t)$ gives a time-dependent diffeomorphism of $\mathbb{R}^{n}$ such
that the graphical representation is preserved along the flow. Then

$\frac{\partial}{\partial t}X(p,t)~{}=~{}\frac{\partial}{\partial
t}y^{i}(p,t)e_{i}+\left(\frac{\partial}{\partial
t}u(y(p,t),t)+D_{i}u\frac{\partial}{\partial t}y^{i}(p,t)\right)e_{0}.$

Taking the expression (11.1) of $\nu_{\gamma}$ into account, from the flow
equation (1.9) we have

$\displaystyle\frac{\partial}{\partial t}y^{i}(p,t)$
$\displaystyle=(\phi(t)-E_{1}^{\alpha})D^{i}\hat{\gamma}|_{Du},$
$\displaystyle\frac{\partial}{\partial
t}u(y(p,t),t)+D_{i}u\frac{\partial}{\partial t}y^{i}(p,t)$
$\displaystyle=(\phi(t)-E_{1}^{\alpha})\left(D_{i}uD^{i}\hat{\gamma}|_{Du}-\hat{\gamma}(Du)\right).$

Combining the above two equations, we see that the flow (1.9) is equivalent to
the following scalar parabolic equation

$\displaystyle\frac{\partial}{\partial t}u=$
$\displaystyle~{}\hat{\gamma}(Du)(E_{1}^{\alpha}-\phi(t))$ $\displaystyle=$
$\displaystyle~{}\hat{\gamma}(Du)\left(\frac{1}{n}D^{i}D^{j}\hat{\gamma}|_{Du}u_{ij}\right)^{\alpha}-\hat{\gamma}(Du)\phi(t)$
(11.4)

of the function $u$ locally on $\mathbb{R}^{n}$.

The bound on the anisotropic principal curvatures and the convexity of $M_{t}$
implies that $|Du|+|D^{2}u|\leq C$ for some uniform constant on a uniform
space-time neighbourhood of any $(y_{0},t_{0})$ chosen such that
$u(y_{0},t_{0})=0$ and $Du|_{(y_{0},t_{0})}=0$ (see, e.g., the argument in [2,
§9]). Since $E_{1}$ is linear in $D^{2}u$, by the uniform bound

$0<1/C\leq{E}_{1}\leq C$

on $E_{1}$ and the bound on $|Du|$, the equation (11.4) is uniformly
parabolic. Since $\alpha>0$ and $E_{1}$ is linear with respect to $D^{2}u$, we
can apply Theorem 6 of [3] to derive the uniform $C^{2,\beta}$ estimate of $u$
for some $\beta\in(0,1)$ in a uniform space-time neighbourhood of
$(y_{0},t_{0})$. The parabolic Schauder theory [26] then implies the
$C^{\ell,\beta}$ estimate of $u$ for all $\ell\geq 2$. This completes the
proof of the lemma. ∎

By the Hausdorff convergence of $M_{t}$ to a scaled Wulff shape $W$, we
conclude that $M_{t}$ converges smoothly to a scaled Wulff shape
$M_{\infty}=\bar{r}\Sigma$ up to a translation.

###  11.2. Smooth convergence for $k=n$, $\alpha>0$

As mentioned in §2.2, in the case $k=n$, we have
$E_{n}(\kappa)=\det(A_{\gamma}(\nu))E_{n}(\lambda)$, where
$\lambda=(\lambda_{1},\dots,\lambda_{n})$ denote the isotropic principal
curvatures of $M_{t}$. Since the anisotropic normal vector field is given by
$\nu_{\gamma}=\gamma(\nu)\nu+\nabla^{\mathbb{S}}\nu$, the flow (1.9) is
equivalent to

$\frac{\partial}{\partial
t}X=\left(\phi(t)-\det(A_{\gamma}(\nu))^{\alpha/n}E_{n}(\lambda)^{\alpha/n}\right)\gamma(\nu)\nu$
(11.5)

up to a tangential diffeomorphism. If we define the isotropic support function
$h(z,t):\mathbb{S}^{n}\to\mathbb{R}$ of $M_{t}=\partial K_{t}$ by

$h(z,t):=\sup\\{\langle x,z\rangle:x\in K_{t}\\},$

we can recover $M_{t}$ via an embedding $\bar{X}(z,t)$:
$\mathbb{S}^{n}\to\mathbb{R}^{n+1}$:

$\bar{X}(z,t)=h(z,t)z+\nabla^{\mathbb{S}}_{i}h(z,t)\sigma^{ij}\nabla^{\mathbb{S}}_{j}z,$

where we set $g_{\mathbb{S}^{n}}=(\sigma_{ij})$. Denote by
$\mathfrak{r}=(\mathfrak{r}_{ij})$ the matrix

$\mathfrak{r}_{ij}=\nabla^{\mathbb{S}}_{i}\nabla^{\mathbb{S}}_{j}h+h\sigma_{ij},$

whose eigenvalues are the principal radii of curvature
$\mathfrak{r}_{i}=1/{\lambda_{i}}$, $i=1,\dots,n$ of $M_{t}$. Then it is well
known that we can rewrite (11.5) as a scalar parabolic PDE of the isotropic
support function $h(z,t)$ on the unit sphere:

$\displaystyle\frac{\partial}{\partial t}h(z,t)=$
$\displaystyle\left(\phi(t)-\det(A_{\gamma}(z))^{\alpha/n}E_{n}(\mathfrak{r})^{-\alpha/n}\right)\gamma(z)$
$\displaystyle=$
$\displaystyle\phi(t)\gamma(z)-\rho(z)E_{n}(\mathfrak{r})^{-\alpha/n},$ (11.6)

where we set $\rho(z)=\det(A_{\gamma}(z))^{\alpha/n}\gamma(z)$ for simplicity
of the notation. Now the equation (11.2) differs from [1, Equ. (13)] only by
the term $\phi(t)\gamma(z)$.

Since the matrix $A_{\gamma}(z)$ is uniformly positive definite, the two-sided
positive bound on $E_{n}(\kappa)$ is equivalent to

$0<C_{1}\leq E_{n}(\mathfrak{r})\leq C_{2}$ (11.7)

for some positive constants $C_{1}$ and $C_{2}$. Then we can apply the
argument in Theorem 10 of [1] to conclude that the principal radii
$\mathfrak{r}_{i}$ are uniformly bounded from above. In fact, noting that the
term $\phi(t)\gamma(z)$ in (11.2) only contributes some lower order terms in
the evolution of $\mathfrak{r}_{ij}$, which are uniformly controlled, we can
apply the calculation in Theorem 10 of [1] to obtain that

$\displaystyle\frac{\partial}{\partial t}\mathfrak{r}_{ij}\leq$
$\displaystyle\frac{\alpha}{n}\rho(z)E_{n}(\mathfrak{r})^{-(1+\frac{\alpha}{n})}\dot{E}_{n}^{k\ell}\nabla^{\mathbb{S}}_{k}\nabla^{\mathbb{S}}_{\ell}\mathfrak{r}_{ij}-\frac{\alpha}{n}\rho(z)E_{n}(\mathfrak{r})^{-(1+\frac{\alpha}{n})}\dot{E}_{n}^{k\ell}(\mathfrak{r})\sigma_{k\ell}\mathfrak{r}_{ij}$
$\displaystyle+CE_{n}(\mathfrak{r})^{-\alpha/n}\sigma_{ij}+\phi(t)A_{\gamma}(z)_{ij},$

where $C$ depends on $\gamma$ and $\alpha$. The second term here is crucial.
By the generalized Newton’s inequality, we have

$\displaystyle\dot{E}_{n}^{k\ell}(\mathfrak{r})\sigma_{k\ell}=$ $\displaystyle
nE_{n-1}(\mathfrak{r})\geq
nE_{n}(\mathfrak{r})^{\frac{n-2}{n-1}}E_{1}(\mathfrak{r})^{\frac{1}{n-1}}\geq
CE_{n}(\mathfrak{r})^{\frac{n-2}{n-1}}\mathfrak{r}_{\max}^{\frac{1}{n-1}}.$

Now work at a space-time point where a maximum eigenvalue of
$\mathfrak{r}_{ij}$ is attained. We have

$\displaystyle\frac{\partial}{\partial t}\mathfrak{r}_{\max}\leq$
$\displaystyle-\frac{\alpha}{n}\rho(z)E_{n}(\mathfrak{r})^{-(\frac{1}{n-1}+\frac{\alpha}{n})}\mathfrak{r}_{\max}^{\frac{n}{n-1}}+C\left(E_{n}(\mathfrak{r})^{-\alpha/n}+\phi(t)\right)$

for a constant $C$ depending only on $\alpha$ and $\gamma$. Given the bound
(11.7) on $E_{n}(\mathfrak{r})$ and Corollary 9.4 on $\phi(t)$, the parabolic
maximum principle implies that

$\mathfrak{r}_{\max}\leq C\left(1+\frac{1}{t^{n-1}}\right)$

for some positive constant $C$ depending only on $C_{1},C_{2}$ and
$n,\alpha,k,\gamma,\mathrm{Vol}(K_{0})$ and $V_{n+1-k}(K_{0})$.

The upper bound on $\mathfrak{r}_{i}$ together with $E_{n}(\mathfrak{r})\geq
C_{1}$ also implies the uniform positive lower bound of $\mathfrak{r}_{i}$.
Therefore the equation (11.2) is uniformly parabolic. The estimate on
$\mathfrak{r}_{i}$ also implies the $C^{2}$ estimate of the isotropic support
function $h(z,t)$. Since (11.2) is concave with respect to the second spatial
derivatives of $h$, we apply Theorem 1.1 of [37] to derive the $C^{2,\beta}$
estimate of $h(z,t)$. By the parabolic Schauder theory [26], we conclude that
the $C^{\ell,\beta}$ norm of $h$ is uniformly bounded for any $\ell\geq 2$.
This is equivalent to the uniform $C^{\ell,\beta}$ regularity estimate of the
evolving hypersurfaces $M_{t}$. This together with the Hausdorff convergence
proved in Proposition 10.1 implies that $M_{t}$ converges to a translated
scaled Wulff shape in the $C^{\infty}$ topology as $t\to\infty$.

###  11.3. Smooth convergence for $\alpha\geq k=1,\dots,n$

We first prove the improved $C^{0}$ and $C^{1}$ estimates.

######  Lemma 11.2.

Let $M_{t}$, $t\in[0,\infty)$ be a smooth solution to the flow (1.9) with
$k=1,\dots,n$ and $\alpha>0$. For any small $\varepsilon>0$, there exists a
large time $N_{0}$ and a fixed constant $d>0$ such that for all $t_{0}>N_{0}$,
if the origin $o$ is chosen with $|s(z,t_{0})-\bar{r}|\leq\varepsilon\bar{r}$,
then

$|s(z,t)-\bar{r}|\leq C\varepsilon\bar{r},\quad\forall~{}z\in\Sigma$ (11.8)

for all $t\in[t_{0},t_{0}+d]$, where $s(z,t)$ is the anisotropic support
function and $C$ depends on $d,\bar{r}$ but not on $t_{0}$. This combined with
the $C^{1}$ estimate (9.2) implies that

$|\bar{\nabla}s(z,t)|^{2}\leq C\varepsilon\bar{r}$ (11.9)

for all $t\in[t_{0},t_{0}+d]$.

###### Proof.

By Proposition 10.1 and Corollary 10.2, for any given $0<\varepsilon<1/4$
there exists $N_{0}$ such that for all time $t>N_{0}$, the anisotropic
Hausdorff distance of $M_{t}$ to a scaled Wulff shape $\bar{r}\Sigma$ is less
than $\varepsilon$, and

$\int_{t}^{t+d}\bigg{|}\phi(x)-\frac{1}{\bar{r}^{\alpha}}\bigg{|}dx\leq\frac{\varepsilon}{\bar{r}^{\alpha}}$

for any fixed constant $d>0$. For each $t_{0}>N_{0}$, we choose the origin $o$
such that the anisotropic support function of $M_{t_{0}}$ with respect to $o$
satisfies

$|s(z,t_{0})-\bar{r}|\leq\varepsilon\bar{r}.$

Since $s(z,t)$ satisfies the parabolic equation

$\frac{\partial}{\partial t}s=-F_{*}(\tau_{ij})^{-\alpha}+\phi(t),$

by adjusting $d$ if necessary, the uniform bounds on $F_{*}$ and $\phi(t)$
imply that

$|s(z,t)-\bar{r}|\leq\frac{1}{2}\bar{r}$ (11.10)

for all time $t\in[t_{0},t_{0}+d]$, where $d$ is independent of $t_{0}$. Let
$\underline{s}(t)=\min_{z\in\Sigma}s(z,t)$. We have

$\displaystyle\frac{d}{dt}\underline{s}(t)\geq$
$\displaystyle-\underline{s}(t)^{-\alpha}+\phi(t).$ (11.11)

We claim that $\underline{s}(t)-\bar{r}\geq-C\varepsilon\bar{r}$ on
$[t_{0},t_{0}+d]$ for a suitable $d$. Without loss of generality, we assume
$\underline{s}(t)<\bar{r}$. Integrating (11.11) implies that

$\displaystyle\underline{s}(t)-\underline{s}(t_{0})\geq$
$\displaystyle\int_{t_{0}}^{t}(\phi(x)-\underline{s}(x)^{-\alpha})dx$
$\displaystyle\geq$
$\displaystyle\int_{t_{0}}^{t}\left((1-\varepsilon)\bar{r}^{-\alpha}-\underline{s}(x)^{-\alpha}\right)dx$
$\displaystyle=$
$\displaystyle\int_{t_{0}}^{t}\frac{1}{\bar{r}^{\alpha}\underline{s}(x)^{\alpha}}\left(\underline{s}(x)^{\alpha}-\bar{r}^{\alpha}\right)dx-\varepsilon\bar{r}^{-\alpha}(t-t_{0}).$
(11.12)

Note that

$\displaystyle\underline{s}(t)^{\alpha}-\bar{r}^{\alpha}=$
$\displaystyle\alpha\int_{0}^{1}\left((1-x)\bar{r}+x\underline{s}(t)\right)^{\alpha-1}dx~{}(\underline{s}(t)-\bar{r})$
$\displaystyle\geq$ $\displaystyle
C\bar{r}^{\alpha-1}(\underline{s}(t)-\bar{r}),$

where we used (11.10). We have

$\displaystyle\underline{s}(t)-\bar{r}=$
$\displaystyle~{}\underline{s}(t)-\underline{s}(t_{0})+\underline{s}(t_{0})-\bar{r}$
$\displaystyle\geq$
$\displaystyle~{}C\bar{r}^{-\alpha-1}\int_{t_{0}}^{t}\left(\underline{s}(x)-\bar{r}\right)dx-\varepsilon\bar{r}^{-\alpha}(t-t_{0})+\underline{s}(t_{0})-\bar{r}$
$\displaystyle\geq$
$\displaystyle~{}C\bar{r}^{-\alpha-1}\int_{t_{0}}^{t}\left(\underline{s}(x)-\bar{r}\right)dx-\varepsilon\bar{r}^{-\alpha}(t-t_{0})-\varepsilon\bar{r},$
(11.13)

where we used the fact that the origin $o$ is chosen such that
$\underline{s}(t_{0})-\bar{r}\geq-\varepsilon\bar{r}$. Then the integral form
of Gronwall’s inequality implies that

$\displaystyle\underline{s}(t)-\bar{r}\geq$
$\displaystyle~{}-\varepsilon\left(\bar{r}^{-\alpha}(t-t_{0})+\bar{r}\right)\left(1+C\bar{r}^{-\alpha-1}(t-t_{0})e^{C\bar{r}^{-\alpha-1}(t-t_{0})}\right)$
$\displaystyle\geq$ $\displaystyle-C\varepsilon\bar{r},$

where $C$ depends only on $\bar{r}$ and $t-t_{0}$. So we obtain the claim.

A similar argument gives the upper bound on
$\bar{s}(t)=\max_{\Sigma}s(\cdot,t)$:

$\displaystyle\bar{s}(t)-\bar{r}\leq$ $\displaystyle~{}C\varepsilon\bar{r}.$

So we have the improved $C^{0}$ estimate (11.8). The improved $C^{1}$ estimate
(11.9) follows from (11.8) and (9.2). ∎

In the following, we focus on the case $\alpha\geq k=1,\dots,n$ and show that
the anisotropic principal curvatures $\kappa_{i}$ of $M_{t}$ have a uniform
positive lower bound.

######  Lemma 11.3.

Along the flow (1.9) with $\alpha\geq k=1,\dots,n$, the anisotropic principal
curvatures $\kappa_{i}$ of the solution $M_{t}$ have a uniform positive lower
bound for $t>0$.

###### Proof.

We have proved in Proposition 9.5 that $\kappa_{i}$ are positive on any finite
time interval. However, the lower bound on $\kappa_{i}$ may degenerate as the
time approaches the infinity. To prove that $\kappa_{i}$ has a uniform
positive lower bound for all time, it suffices to show this for sufficiently
large time.

Recall that in the case $\alpha\geq k=1,\dots,n$, the global term $\phi(t)$
satisfies (10.3). We employ the anisotropic Gauss map parametrization in §8.
We assume that $N_{0}>0$ is sufficiently large such that the anisotropic
Hausdorff distance of $M_{t}$ to $\bar{r}\Sigma$ is less than $\varepsilon$,
and

$\phi(t)\geq(1-\varepsilon)\bar{r}^{-\alpha}$ (11.14)

for all time $t\in[N_{0},\infty)$. By choosing $N_{0}$ sufficiently large, the
constant $\varepsilon$ in (11.14) can be any small positive constant. For each
fixed $t_{0}>N_{0}$, we choose the origin $o$ such that the anisotropic
support function of $M_{t_{0}}$ with respect to $o$ satisfies

$|s(z,t_{0})-\bar{r}|\leq\bar{r}\varepsilon,\qquad\forall~{}z\in\Sigma.$

By Lemma 11.2, there exists a constant $d>0$ and $C=C(d,\bar{r})$ independent
of $t_{0}$ such that (11.8) and (11.9) hold for all $t\in[t_{0},t_{0}+d]$.

In the following, we only focus on the time $t\geq N_{0}$. To estimate the
lower bound on the anisotropic principal curvatures $\kappa_{i}$, we can
equivalently estimate the upper bound on the anisotropic principal radii of
curvature $\tau_{i}$, which are eigenvalues of $(\tau_{ij})$ defined in (8.3).
We define

$\zeta=\sup\\{\tau_{ij}\xi^{i}\xi^{j}:~{}|\xi|=1\\}$

and for any $t_{0}>N_{0}$ we consider the function

$\omega(z,t)=\log\zeta(z,t)-\mu_{1}s(z,t)+\mu_{2}|\bar{\nabla}s|^{2}(z,t)$
(11.15)

on the interval $[t_{0},t_{0}+d]$, where $\mu_{1},\mu_{2}>0$ are positive
constants to be determined later. We consider a point $(z_{1},t_{1})$ where a
new maximum of the function $\omega$ is achieved. By rotation of the local
orthonormal frame, we assume that $\xi=e_{1}$ and
$(\tau_{ij})=\mathrm{diag}(\tau_{1},\dots,\tau_{n})$ is diagonal at
$(z_{1},t_{1})$. First, at $(z_{1},t_{1})$ we have $\zeta=\tau_{11}$ and

$\displaystyle\frac{\partial}{\partial t}\omega=$
$\displaystyle\frac{1}{\tau_{11}}\frac{\partial}{\partial
t}\tau_{11}-\mu_{1}\frac{\partial}{\partial
t}s+\mu_{2}\frac{\partial}{\partial t}|\bar{\nabla}s|^{2},$
$\displaystyle\bar{\nabla}_{\ell}\omega=$
$\displaystyle\frac{1}{\tau_{11}}\bar{\nabla}_{\ell}\tau_{11}-\mu_{1}\bar{\nabla}_{\ell}s+\mu_{2}\bar{\nabla}_{\ell}|\bar{\nabla}s|^{2},$
$\displaystyle\bar{\nabla}_{k}\bar{\nabla}_{\ell}\omega=$
$\displaystyle\frac{1}{\tau_{11}}\bar{\nabla}_{k}\bar{\nabla}_{\ell}\tau_{11}-\mu_{1}\bar{\nabla}_{k}\bar{\nabla}_{\ell}s+\mu_{2}\bar{\nabla}_{k}\bar{\nabla}_{\ell}|\bar{\nabla}s|^{2}-\frac{1}{(\tau_{11})^{2}}\bar{\nabla}_{k}\tau_{11}\bar{\nabla}_{\ell}\tau_{11}.$

Combining these equations with (8.9), (8.4) and (8.5) gives

$\displaystyle\frac{\partial}{\partial
t}\omega-\dot{\Psi}^{k\ell}\left(\bar{\nabla}_{k}\bar{\nabla}_{\ell}\omega-\frac{1}{2}Q_{k\ell
p}\bar{\nabla}_{p}\omega\right)$ $\displaystyle=$
$\displaystyle\frac{1}{\tau_{11}}\left(\frac{\partial}{\partial
t}\tau_{11}-\dot{\Psi}^{k\ell}\left(\bar{\nabla}_{k}\bar{\nabla}_{\ell}\tau_{11}-\frac{1}{2}Q_{k\ell
p}\bar{\nabla}_{p}\tau_{11}\right)\right)$
$\displaystyle-\mu_{1}\left(\frac{\partial}{\partial
t}s-\dot{\Psi}^{k\ell}\left(\bar{\nabla}_{k}\bar{\nabla}_{\ell}s-\frac{1}{2}Q_{k\ell
p}\bar{\nabla}_{p}s\right)\right)$
$\displaystyle+\mu_{2}\left(\frac{\partial}{\partial
t}|\bar{\nabla}s|^{2}-\dot{\Psi}^{k\ell}\left(\bar{\nabla}_{k}\bar{\nabla}_{\ell}|\bar{\nabla}s|^{2}-\frac{1}{2}Q_{k\ell
p}\bar{\nabla}_{p}|\bar{\nabla}s|^{2}\right)\right)$
$\displaystyle+\frac{1}{(\tau_{11})^{2}}\dot{\Psi}^{k\ell}\bar{\nabla}_{k}\tau_{11}\bar{\nabla}_{\ell}\tau_{11}$
$\displaystyle\leq$
$\displaystyle\frac{1}{\tau_{11}}\left(\ddot{\Psi}^{k\ell,pq}\bar{\nabla}_{1}\tau_{k\ell}\bar{\nabla}_{1}\tau_{pq}+(1-\alpha)\Psi+\phi(t)+C\dot{\Psi}^{k\ell}\bar{g}_{k\ell}\tau_{11}\right)$
$\displaystyle-\mu_{1}\left(\phi(t)+(1+\alpha)\Psi+s\dot{\Psi}^{k\ell}\bar{g}_{k\ell}\right)+\frac{1}{(\tau_{11})^{2}}\dot{\Psi}^{k\ell}\bar{\nabla}_{k}\tau_{11}\bar{\nabla}_{\ell}\tau_{11}$
$\displaystyle+\mu_{2}\dot{\Psi}^{k\ell}\biggl{(}-2\tau_{ik}\tau_{i\ell}-2s^{2}\bar{g}_{k\ell}+4s\tau_{k\ell}-2\tau_{ik}Q_{i\ell
p}s_{p}+2sQ_{k\ell p}s_{p}\biggr{)}$
$\displaystyle+\mu_{2}\dot{\Psi}^{k\ell}\biggl{(}2s_{k}s_{\ell}-\frac{1}{2}(2Q_{i\ell
m}Q_{mkp}-Q_{ipm}Q_{mk\ell})s_{i}s_{p}-\bar{\nabla}_{i}Q_{k\ell
p}s_{p}s_{i}\biggr{)}$ $\displaystyle\leq$
$\displaystyle\frac{1}{\tau_{11}}\ddot{\Psi}^{k\ell,pq}\bar{\nabla}_{1}\tau_{k\ell}\bar{\nabla}_{1}\tau_{pq}+\frac{1}{(\tau_{11})^{2}}\dot{\Psi}^{k\ell}\bar{\nabla}_{k}\tau_{11}\bar{\nabla}_{\ell}\tau_{11}$
$\displaystyle-\mu_{1}\left(\phi(t)+(1+\alpha)\Psi+s\sum_{k}\dot{\psi}^{k}\right)-2\mu_{2}\sum_{k}\dot{\psi}^{k}\tau_{k}^{2}$
$\displaystyle+(1-\alpha)\frac{1}{\tau_{11}}\Psi+\frac{1}{\tau_{11}}\phi(t)-\alpha\mu_{2}C\Psi+C(1+\mu_{2}\varepsilon)\sum_{k}\dot{\psi}^{k},$
(11.16)

where in the last inequality we used the $C^{0}$, $C^{1}$ estimates of $s$ and
the homogeneity of $\Psi$

$\displaystyle\dot{\Psi}^{k\ell}\tau_{k\ell}=$ $\displaystyle-\alpha\Psi$
$\displaystyle-2\dot{\Psi}^{k\ell}\tau_{ik}Q_{i\ell p}s_{p}\leq$
$\displaystyle C\sum_{k}\dot{\psi}^{k}\tau_{k}=-C\alpha\Psi.$

The constants $C$ in the last line of (11.3) depend on $\gamma$ and the
initial hypersurface $M_{0}$. We next apply the maximum principle to (11.3) to
obtain a uniform upper bound on $\tau_{11}$. Note that $\tau_{11}=\tau_{1}$ is
the largest eigenvalue of $(\tau_{ij})$ at the point $(z_{1},t_{1})$.

We first estimate the gradient terms in (11.3). By (8.6) and the fact that
$\Psi=-F_{*}^{-\alpha}$ is concave, we have

$\displaystyle\ddot{\Psi}^{k\ell,pq}\bar{\nabla}_{1}\tau_{k\ell}\bar{\nabla}_{1}\tau_{pq}=$
$\displaystyle~{}\ddot{\psi}^{k\ell}\bar{\nabla}_{1}\tau_{kk}\bar{\nabla}_{1}\tau_{\ell\ell}+2\sum_{k>\ell}\frac{\dot{\psi}^{k}-\dot{\psi}^{\ell}}{\tau_{k}-\tau_{\ell}}(\bar{\nabla}_{1}\tau_{k\ell})^{2}$
$\displaystyle\leq$
$\displaystyle~{}-2\sum_{k>1}\frac{1}{\tau_{1}}(\dot{\psi}^{k}-\dot{\psi}^{1})(\bar{\nabla}_{1}\tau_{k1})^{2}.$
(11.17)

Codazzi equation (8.4) implies that

$\displaystyle(\bar{\nabla}_{1}\tau_{k1})^{2}=$
$\displaystyle\left(\bar{\nabla}_{k}\tau_{11}+\frac{1}{2}Q_{11p}\tau_{kp}-\frac{1}{2}Q_{k1p}\tau_{1p}\right)^{2}\geq\frac{1}{2}(\bar{\nabla}_{k}\tau_{11})^{2}-C(\tau_{11})^{2}.$
(11.18)

Substituting (11.18) into (11.3), and noting that

$\frac{1}{\tau_{11}}\bar{\nabla}_{k}\tau_{11}=\mu_{1}\bar{\nabla}_{k}s-\mu_{2}\bar{\nabla}_{k}|\bar{\nabla}s|^{2}$

holds at $(z_{1},t_{1})$, we have

$\displaystyle\frac{1}{\tau_{11}}\ddot{\Psi}^{k\ell,pq}\bar{\nabla}_{1}\tau_{k\ell}\bar{\nabla}_{1}\tau_{pq}+\frac{1}{(\tau_{11})^{2}}\sum_{k}\dot{\psi}^{k}(\bar{\nabla}_{k}\tau_{11})^{2}$
$\displaystyle\leq$
$\displaystyle~{}-2\sum_{k>1}\frac{1}{\tau_{11}^{2}}(\dot{\psi}^{k}-\dot{\psi}^{1})\left(\frac{1}{2}(\bar{\nabla}_{k}\tau_{11})^{2}-C(\tau_{11})^{2}\right)+\frac{1}{(\tau_{11})^{2}}\sum_{k}\dot{\psi}^{k}(\bar{\nabla}_{k}\tau_{11})^{2}$
$\displaystyle=$
$\displaystyle~{}\dot{\psi}^{1}\tau_{11}^{-2}\sum_{k}(\bar{\nabla}_{k}\tau_{11})^{2}+2C\sum_{k>1}(\dot{\psi}^{k}-\dot{\psi}^{1})$
$\displaystyle\leq$
$\displaystyle~{}\dot{\psi}^{1}\sum_{k}\left(\mu_{1}\bar{\nabla}_{k}s-\mu_{2}\bar{\nabla}_{k}|\bar{\nabla}s|^{2}\right)^{2}+2C\sum_{k>1}(\dot{\psi}^{k}-\dot{\psi}^{1})$

at $(z_{1},t_{1})$. By the definition (8.3) of $\tau_{ij}$ and the improved
$C^{0}$, $C^{1}$ estimates of $s$ in Lemma 11.2, we have

$\displaystyle\left(\mu_{1}\bar{\nabla}_{k}s-\mu_{2}\bar{\nabla}_{k}|\bar{\nabla}s|^{2}\right)^{2}=$
$\displaystyle\left(\mu_{1}\bar{\nabla}_{k}s-2\mu_{2}\bar{\nabla}_{k}\bar{\nabla}_{\ell}s\bar{\nabla}_{\ell}s\right)^{2}$
$\displaystyle=$
$\displaystyle\left(\mu_{1}\bar{\nabla}_{k}s-2\mu_{2}\bar{\nabla}_{\ell}s(\tau_{k\ell}-s\bar{g}_{k\ell}+\frac{1}{2}Q_{k\ell
p}\bar{\nabla}_{p}s)\right)^{2}$ $\displaystyle\leq$ $\displaystyle
C\varepsilon\left(\mu_{1}^{2}+\mu_{2}^{2}(\tau_{1}^{2}+1)\right),\quad
k=1,\dots,n$

for some constant $C$ depending only on $M_{0}$ and $\gamma$. In summary, the
gradient terms on the right-hand side of (11.3) satisfies

$\displaystyle\frac{1}{\tau_{11}}\ddot{\Psi}^{k\ell,pq}\bar{\nabla}_{1}\tau_{k\ell}\bar{\nabla}_{1}\tau_{pq}+\frac{1}{(\tau_{11})^{2}}\dot{\Psi}^{k\ell}\bar{\nabla}_{k}\tau_{11}\bar{\nabla}_{\ell}\tau_{11}$
$\displaystyle\leq$ $\displaystyle
C\varepsilon\dot{\psi}^{1}\left(\mu_{1}^{2}+\mu_{2}^{2}(\tau_{1}^{2}+1)\right)+C\sum_{k}\dot{\psi}^{k}$
(11.19)

at $(z_{1},t_{1})$.

Since the left hand side of (11.3) is nonnegative at $(z_{1},t_{1})$,
rearranging the terms in (11.3) and applying (11.3), we have

$\displaystyle\mu_{1}\left(\phi(t)+(1+\alpha)\Psi+s\sum_{k}\dot{\psi}^{k}\right)+2\mu_{2}\sum_{k}\dot{\psi}^{k}\tau_{k}^{2}$
$\displaystyle\leq$ $\displaystyle
C\varepsilon\dot{\psi}^{1}\left(\mu_{1}^{2}+\mu_{2}^{2}(\tau_{1}^{2}+1)\right)+C(\mu_{2}\varepsilon+1)\sum_{k}\dot{\psi}^{k}$
$\displaystyle+(1-\alpha)\frac{1}{\tau_{11}}\Psi+\frac{1}{\tau_{11}}\phi(t)-\alpha\mu_{2}C\Psi$
(11.20)

at $(z_{1},t_{1})$. Note that

$\Psi=-F_{*}^{-\alpha},\quad\psi^{k}=\alpha F_{*}^{-\alpha-1}\dot{f}^{k}_{*},$

where $F_{*}(\tau_{ij})=f_{*}(\tau)$. Rewriting the terms in (11.3) and
multiplying by $F_{*}^{\alpha+1}$ both sides give that

$\displaystyle\mu_{1}\left(\phi(t)F_{*}^{\alpha+1}-(1+\alpha)F_{*}+\alpha
s\right)+\mu_{1}\alpha
s\left(\sum_{k}\dot{f}^{k}_{*}-1\right)+2\mu_{2}\alpha\sum_{k}\dot{f}_{*}^{k}\tau_{k}^{2}$
$\displaystyle\leq$ $\displaystyle
C\varepsilon\alpha\dot{f}_{*}^{1}\left(\mu_{1}^{2}+\mu_{2}^{2}(\tau_{1}^{2}+1)\right)+C(\mu_{2}\varepsilon+1)\alpha\sum_{k}\dot{f}^{k}_{*}$
$\displaystyle+(\alpha-1)\frac{1}{\tau_{11}}F_{*}+\frac{1}{\tau_{11}}\phi(t)F_{*}^{\alpha+1}+\alpha\mu_{2}CF_{*}$
(11.21)

at $(z_{1},t_{1})$. Using the estimate (11.14) and Lemma 11.2, the first term
on the left side of (11.3) satisfies

$\displaystyle\mu_{1}\left(\phi(t)F_{*}^{\alpha+1}-(1+\alpha)F_{*}+\alpha
s\right)$ $\displaystyle\geq$
$\displaystyle\mu_{1}\bar{r}\left((1-\varepsilon)\left(\frac{F_{*}}{\bar{r}}\right)^{\alpha+1}-(1+\alpha)\frac{F_{*}}{\bar{r}}+\alpha\right)+\alpha\mu_{1}(s-\bar{r})$
$\displaystyle\geq$
$\displaystyle\mu_{1}\bar{r}\left(\left(\frac{F_{*}}{\bar{r}}\right)^{\alpha+1}-(1+\alpha)\frac{F_{*}}{\bar{r}}+\alpha\right)-\mu_{1}\varepsilon\bar{r}^{-\alpha}F_{*}^{\alpha+1}-C\varepsilon\alpha\mu_{1}\bar{r}$
$\displaystyle\geq$
$\displaystyle-\mu_{1}\varepsilon\bar{r}^{-\alpha}F_{*}^{\alpha+1}-C\varepsilon\alpha\mu_{1}\bar{r},$
(11.22)

where $C$ is the constant in Lemma 11.2 depending only on $d$ and $\bar{r}$.
Substituting (11.3) into (11.3), we have the following estimate

$\displaystyle 2\mu_{2}\alpha\sum_{k}\dot{f}_{*}^{k}\tau_{k}^{2}+\mu_{1}\alpha
s\left(\sum_{k}\dot{f}^{k}_{*}-1\right)$ $\displaystyle\leq$ $\displaystyle
C\varepsilon\alpha\dot{f}_{*}^{1}\mu_{2}^{2}\tau_{1}^{2}+C\varepsilon\alpha\dot{f}_{*}^{1}\left(\mu_{1}^{2}+\mu_{2}^{2}\right)+C(\mu_{2}\varepsilon+1)\alpha\sum_{k}\dot{f}^{k}_{*}$
$\displaystyle+C(\mu_{2}+1)+C\mu_{1}\varepsilon+C\varepsilon\alpha\mu_{1}\bar{r}$
(11.23)

holding at $(z_{1},t_{1})$, where the constants $C$ are uniform and
independent of the time $t_{0}$ and the length $d$ of the time interval.

Now, we apply an observation in [17] and [44] which relies on the property
that $f_{*}$ is increasing and concave in $\Gamma_{+}$ and vanishes on
$\partial\Gamma_{+}$. Precisely, for
$f_{*}=\left(\frac{E_{n}}{E_{n-k}}\right)^{1/k}$, we have the following
property.

######  Proposition 11.4 (Lemma 2.2 in [17]).

Let $K\subset\Gamma_{+}$ be a compact set and $\varrho>0$ a constant. Then
there exists a constant $\theta>0$ depending only on $K$ and $\varrho$ such
that for $\tau\in\Gamma_{+}$ and $\mu\in K$ satisfying
$|\nu_{\tau}-\nu_{\mu}|\geq\varrho$, we have

$\sum_{k}\dot{f}_{*}^{k}(\tau)\mu_{k}-f_{*}(\mu)\geq\theta\left(\sum_{k}\dot{f}_{*}^{k}(\tau)+1\right),$

where $\nu_{\tau}$ denotes the unit normal vector of the level set
$\\{x\in\Gamma_{+}:~{}f_{*}(x)=f_{*}(\tau)\\}$ at the point $\tau$, i.e.,
$\nu_{\tau}=Df_{*}(\tau)/|Df_{*}(\tau)|$.

Let $\mu=2^{-1}\bar{r}(1,\dots,1)\in\Gamma_{+}$ and $K=\\{\mu\\}$ in
Proposition 11.4. Then there exists a constant $\varrho=\bar{r}/8>0$ such that
$\nu_{\mu}-2\varrho(1,\dots,1)\in\Gamma_{+}$. We have two cases. First, if the
anisotropic principal radii of curvature $\tau$ at the point $(z_{1},t_{1})$
satisfy $|\nu_{\tau}-\nu_{\mu}|\geq\varrho$, Proposition 11.4 implies that

$\sum_{k}\dot{f}_{*}^{k}(\tau)-1\geq\frac{2\theta}{\bar{r}}\left(\sum_{k}\dot{f}_{*}^{k}(\tau)+1\right).$
(11.24)

Substituting (11.24) into (11.3), we have

$\displaystyle 2\mu_{2}\alpha\sum_{k}\dot{f}_{*}^{k}\tau_{k}^{2}+\mu_{1}\alpha
s\frac{2\theta}{\bar{r}}\left(\sum_{k}\dot{f}^{k}_{*}+1\right)$
$\displaystyle\leq$ $\displaystyle
C\varepsilon\alpha\dot{f}_{*}^{1}\mu_{2}^{2}\tau_{1}^{2}+C\varepsilon\alpha\dot{f}_{*}^{1}\left(\mu_{1}^{2}+\mu_{2}^{2}\right)+C(\mu_{2}\varepsilon+1)\alpha\sum_{k}\dot{f}^{k}_{*}$
$\displaystyle+C(\mu_{2}+1)+C\mu_{1}\varepsilon+C\varepsilon\alpha\mu_{1}\bar{r}$
(11.25)

at $(z_{1},t_{1})$. Based on the above inequality, we can assume that
$\varepsilon$ is chosen with

$C\varepsilon+C\varepsilon\alpha\bar{r}\leq\frac{1}{2}\alpha\theta.$

Note that $\varepsilon$ can be chosen to depend only on $M_{0},\gamma$ and
$\theta$. Now we first choose
$\mu_{2}=\mu_{2}(\varepsilon,\alpha,\gamma,M_{0})$ small enough such that

$C\varepsilon\alpha\dot{f}_{*}^{1}\mu_{2}^{2}\tau_{1}^{2}\leq\mu_{2}\alpha\sum_{k}\dot{f}_{*}^{k}\tau_{k}^{2}$

and then choose $\mu_{1}=\mu_{1}(\alpha,\gamma,M_{0})$ large enough such that

$C(\mu_{2}\varepsilon+1)\alpha\sum_{k}\dot{f}^{k}_{*}+C(\mu_{2}+1)\leq\frac{1}{2}\mu_{1}\alpha
s\frac{2\theta}{\bar{r}}\left(\sum_{k}\dot{f}^{k}_{*}+1\right).$

Then (11.3) becomes

$\displaystyle\mu_{2}\alpha\dot{f}_{*}^{1}\tau_{1}^{2}\leq$ $\displaystyle
C\varepsilon\alpha\dot{f}_{*}^{1}\left(\mu_{1}^{2}+\mu_{2}^{2}\right),$

which gives a uniform upper bound on $\tau_{1}$ at the point $(z_{1},t_{1})$:

$\tau_{1}\leq\left(\frac{1}{\mu_{2}}C\left(\mu_{1}^{2}+\mu_{2}^{2}\right)\right)^{1/2}.$

Second, if the anisotropic principal radii of curvature $\tau$ at the point
$(z_{1},t_{1})$ satisfy $|\nu_{\tau}-\nu_{\mu}|\leq\varrho$, it follows from
$\nu_{\mu}-2\varrho(1,\dots,1)\in\Gamma_{+}$ that
$\nu_{\tau}-\varrho(1,\dots,1)\in\Gamma_{+}$ and $\dot{f}_{*}^{i}\geq
C\sum_{k}\dot{f}_{*}^{k}$ for some constant $C$ and all $i=1,\dots,n$. Then

$C\leq
F_{*}=\sum_{k}\dot{f}_{*}^{k}\tau_{k}\leq(\sum_{k}\dot{f}_{*}^{k})\tau_{1}\leq
C\dot{f}_{*}^{1}\tau_{1}.$ (11.26)

We can still choose $\mu_{2}=\mu_{2}(\varepsilon,\alpha,\gamma,M_{0})$ small
enough such that

$C\varepsilon\alpha\dot{f}_{*}^{1}\mu_{2}^{2}\tau_{1}^{2}\leq\mu_{2}\alpha\sum_{k}\dot{f}_{*}^{k}\tau_{k}^{2}.$
(11.27)

The function $f_{*}$ is concave, we always have $\sum_{k}\dot{f}^{k}_{*}\geq
1$. Substituting (11.26) and (11.27) into (11.3) and discarding the second
term on the left-hand side of (11.3), we have

$\displaystyle\mu_{2}\alpha\dot{f}_{*}^{1}\tau_{1}^{2}\leq$ $\displaystyle
C\varepsilon\alpha\dot{f}_{*}^{1}\left(\mu_{1}^{2}+\mu_{2}^{2}\right)+C(\mu_{2}\varepsilon+1)\alpha\dot{f}^{1}_{*}$
$\displaystyle+C(\mu_{2}+1)+C\mu_{1}\varepsilon+C\varepsilon\alpha\mu_{1}\bar{r}$
$\displaystyle\leq$ $\displaystyle
C\varepsilon\alpha\dot{f}_{*}^{1}\left(\mu_{1}^{2}+\mu_{2}^{2}\right)+C(\mu_{2}\varepsilon+1)\alpha\dot{f}^{1}_{*}$
$\displaystyle+C\biggl{(}(\mu_{2}+1)+\mu_{1}\varepsilon+C\varepsilon\alpha\mu_{1}\bar{r}\biggr{)}\dot{f}_{*}^{1}\tau_{1}$
(11.28)

at $(z_{1},t_{1})$. Multiplying by $1/{\dot{f}_{*}^{1}}$ both sides of (11.3)
we have

$\displaystyle\mu_{2}\alpha\tau_{1}^{2}\leq$ $\displaystyle
C\varepsilon\alpha\left(\mu_{1}^{2}+\mu_{2}^{2}\right)+C(\mu_{2}\varepsilon+1)\alpha$
$\displaystyle+C\biggl{(}(\mu_{2}+1)+\mu_{1}\varepsilon+C\varepsilon\alpha\mu_{1}\bar{r}\biggr{)}\tau_{1},$

which gives a uniform upper bound on $\tau_{1}$ at $(z_{1},t_{1})$.

Therefore, we can choose a small constant $\mu_{2}$ and a large constant
$\mu_{1}$ in the definition (11.15) of $\omega(z,t)$ such that in both the
above two cases, the largest anisotropic principal radius of curvature
$\tau_{1}$ has a uniform upper bound at $(z_{1},t_{1})$. This in turn implies
that $\omega(z_{1},t_{1})\leq C$ for some uniform positive constant $C$
depending only on the initial data $M_{0}$ and $\gamma,\alpha,n$. By the
definition of $\omega(z,t)$ and the $C^{0}$, $C^{1}$ estimates, we conclude
that $\tau_{1}$ is bounded from above uniformly on the time interval
$[t_{0},t_{0}+d]$. Note that the constants $\mu_{1}$, $\mu_{2}$ are chosen
uniformly and depend only on $M_{0}$, $\alpha$, $\gamma$. Since $t_{0}>N_{0}$
is arbitrary and $d>0$ is a fixed constant, repeating the above argument
completes the proof of Proposition 11.3. ∎

Combining Proposition 9.3 and Proposition 11.3, we obtain the uniform
curvature estimate of the solution $M_{t}$ to the flow (1.9). In fact, since

$\displaystyle E_{k}\geq$
$\displaystyle{\binom{n}{k}}^{-1}\kappa_{n}\cdots\kappa_{n-k+1}\geq{\binom{n}{k}}^{-1}\kappa_{n}\kappa_{1}^{k-1},$

the uniform upper bound on $E_{k}(\kappa)$ and the uniform positive lower
bound on the anisotropic principal curvature $\kappa_{i}$ imply that

$0<\frac{1}{C}\leq\kappa_{i}\leq C,\qquad i=1,\dots,n$

for some constant $C$ depending only on $\gamma$, $\alpha$ and $M_{0}$. Since
the dual function $F_{*}$ is concave, we apply Theorem 1 of [37] to the
equation (8.7) of $s$ and derive the $C^{2,\beta}$ estimate of $s(z,t)$ for
some $\beta\in(0,1)$. Then the parabolic Schauder theory [26] implies that the
$C^{\ell,\beta}$ norm of $s(z,t)$ is bounded for any $\ell\geq 2$. This means
that the solution $M_{t}$ of the flow (1.9) has uniform regularity estimates
for all time $t\in[0,\infty)$. Taking into account the Hausdorff convergence
of $M_{t}$, we conclude that $M_{t}$ converges smoothly to a scaled Wulff
shape as $t\to\infty$.

###  11.4. Exponential convergence

The stronger convergence results can be obtained by studying the linearization
of the flow around the Wulff shape (cf. [2, §12]). The goal is to prove that
the linearization of the flow about the Wulff shape $\Sigma$ is strictly
stable. Without loss of generality, we assume that the hypersurfaces near
$\Sigma$ can be written as graphs of smooth functions over $\Sigma$

$M_{t}=\\{r(z,t)z:z\in\Sigma\\},$

where $r(z,t)=1+\varepsilon\eta(z,t)$. By adding a smooth reparametrization to
preserve this graphical representation, the flow (1.9) is equivalent to the
following parabolic equation for the function $r$:

$\frac{\partial r}{\partial
t}=(\phi(t)-E_{k}^{\alpha/k}(\kappa))\left(G(\nu_{\gamma},z)-\frac{1}{r}G^{ij}G(\nu_{\gamma},\partial_{i}{z})\frac{\partial
r}{\partial z^{j}}\right),$ (11.29)

where $G$ is the metric at $z$ on $\mathbb{R}^{n+1}$ described in (2.4) and
$G^{ij}$ stands for the inverse of the matrix
$(G_{ij})_{i,j=1}^{n}=(G(\partial_{i}z,\partial_{j}z))_{i,j=1}^{n}$. Note that
on $\Sigma$ we have $G(\nu_{\gamma},z)=G(z,z)=1$,
$G(\nu_{\gamma},\partial_{i}{z})=G(z,\partial_{i}z)=0$ and
$\phi(t)-E_{k}^{\alpha/k}(\kappa)=0$. Differentiating (11.29) with respect to
$\epsilon$ and letting $\epsilon=0$, we have

$\frac{\partial}{\partial
t}\eta=\frac{\partial}{\partial\varepsilon}\bigg{|}_{\varepsilon=0}(\phi(t)-E_{k}^{\alpha/k}(\kappa)).$
(11.30)

On $\Sigma$ we have $\hat{h}_{i}^{j}=\delta_{i}^{j}$. The variation equations
(7.2) and (7.5) imply that

$\displaystyle\frac{\partial}{\partial\varepsilon}\bigg{|}_{\varepsilon=0}E_{k}^{\alpha/k}(\kappa)$
$\displaystyle=\frac{\alpha}{k}E_{k}^{\alpha/k-1}(I)E_{k}^{ij}(I)\left(-\bar{\nabla}^{i}\bar{\nabla}_{j}\eta-{A}_{j}{}^{pi}\partial_{p}\eta-\eta\hat{h}^{p}_{j}\hat{h}_{p}^{i}\right)$
$\displaystyle=$
$\displaystyle\frac{\alpha}{n}\left(-\bar{\Delta}\eta-{A}_{pi}{}^{i}\bar{\nabla}^{p}\eta-n\eta\right),$
(11.31)

and

$\displaystyle\frac{d}{d\varepsilon}\bigg{|}_{\varepsilon=0}\phi(t)=$
$\displaystyle\frac{1}{|\Sigma|_{\gamma}}\int_{\Sigma}\left(\frac{\partial}{\partial\varepsilon}\bigg{|}_{\varepsilon=0}E_{k}^{\alpha/k}(\kappa)d\mu_{\gamma}+E_{k}^{\alpha/k}(I)\frac{\partial}{\partial\varepsilon}\bigg{|}_{\varepsilon=0}d\mu_{\gamma}\right)$
$\displaystyle\quad-\frac{1}{|\Sigma|^{2}_{\gamma}}\int_{\Sigma}E_{k}^{\alpha/k}(I)d\mu_{\gamma}\frac{\partial}{\partial\varepsilon}\bigg{|}_{\varepsilon=0}|M_{t}|_{\gamma}$
$\displaystyle=$
$\displaystyle-\frac{\alpha}{n}\frac{1}{|\Sigma|_{\gamma}}\int_{\Sigma}n\eta
d\mu_{\gamma},$ (11.32)

where $I=(1,\dots,1)$ and we used
$\int_{\Sigma}(\bar{\Delta}\eta+{A}_{pi}{}^{i}\bar{\nabla}^{p}\eta)d\mu_{\gamma}=0$
by virtue of Lemma 2.8 in [43]. Substituting (11.31) and (11.32) into (11.30)
we get

$\frac{\partial}{\partial
t}\eta~{}=~{}\frac{\alpha}{n}\left(\bar{\Delta}\eta+{A}_{pi}{}^{i}\bar{\nabla}^{p}\eta+n\eta-\frac{1}{|\Sigma|_{\gamma}}\int_{\Sigma}n\eta
d\mu_{\gamma}\right)~{}=:\mathcal{L}\eta,$

which is the linearized equation of the flow (1.9) about $\Sigma$. As in
Proposition 12.1 of [2], the operator $\mathcal{L}$ maps the space
$W^{2,2}({\Sigma})\cap\\{f:\int_{\Sigma}fd\mu_{\gamma}=0\\}$ to the space
$L^{2}({\Sigma})\cap\\{f:\int_{\Sigma}fd\mu_{\gamma}=0\\}$. Moreover,
$\mathcal{L}$ is non-positive and self-adjoint with respect to the inner
product $\langle f_{1},f_{2}\rangle=\int_{\Sigma}f_{1}f_{2}d\mu_{\gamma}$, and
has kernel consisting precisely of the generators of translations. In other
words, the linearization of the flow (1.9) about the Wulff shape is strictly
stable modulo the action of the translation group. Then Theorem 9.1.2 of [27]
applies to conclude that initial hypersurfaces sufficiently close to the Wulff
shape converge modulo translations to the Wulff shape, exponentially fast in
the $C^{\infty}$ topology. Since the smooth convergence of $M_{t}$ is already
established, we conclude that the whole family of the solution $M_{t}$ of the
flow (1.9) converge modulo translations to a scaled Wulff shape
$M_{\infty}=\bar{r}\Sigma$ exponentially fast in $C^{\infty}$ topology. In
particular, the speed of the flow satisfies

$|\phi(t)-E_{k}^{\alpha/k}(\kappa)|\leq Ce^{-\delta t}$

for some constants $C$ and $\delta$, and for all $t>0$. Then we have for any
$t_{2}>t_{1}>0$,

$\displaystyle|X(\cdot,t_{2})-X(\cdot,t_{1})|\leq$
$\displaystyle\int_{t_{1}}^{t_{2}}\max_{M}\biggl{|}\frac{\partial X}{\partial
t}\biggr{|}dt$ $\displaystyle\leq$
$\displaystyle\int_{t_{1}}^{t_{2}}\max_{M}|\phi(t)-E_{k}^{\alpha/k}(\kappa)|dt$
$\displaystyle\leq$ $\displaystyle\frac{C}{\delta}(e^{-\delta
t_{1}}-e^{-\delta t_{2}}).$

By the Cauchy criterion, $X(\cdot,t)$ converges to a limit embedding
$X_{\infty}(\cdot)$ of $M_{\infty}$ as $t\to\infty$. Hence, $M_{t}$ converges
exponentially and smoothly to a scaled Wulff shape $M_{\infty}=\bar{r}\Sigma$
without any correction by translations. This completes the proof of the
convergence stated in Theorem 1.3.

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