# Long-time Ricci flow existence and topological rigidity from manifolds with
pinched scale-invariant integral curvature

Albert Chau1 Department of Mathematics, The University of British Columbia,
Room 121, 1984 Mathematics Road, Vancouver, B.C., Canada V6T 1Z2
<EMAIL_ADDRESS>and  Adam Martens  Department of Mathematics, The University
of British Columbia, 1984 Mathematics Road, Vancouver, B.C., Canada V6T 1Z2
<EMAIL_ADDRESS>
###### Abstract.

We prove long-time existence of the Ricci flow starting from complete
manifolds with bounded curvature and scale-invariant integral curvature
sufficiently pinched with respect to the inverse of its Sobolev constant.
Moreover, if the curvature is sub-critical $L^{p}$ integrable, this flow
converges locally smoothly to a limiting metric $g(\infty)$ on $M$ with
$(M,g(\infty))$ isometric to the standard flat $\mathbb{R}^{n}$, which implies
topological rigidity of $M$. This generalizes work of Chen [9], who proved
analogous results for asymptotically flat manifolds. We also prove a long-time
Ricci flow existence (and likewise topological rigidity) result for unbounded
curvature initial data, assuming the initial data is a locally smooth limit of
bounded curvature manifolds as described above.

1Research partially supported by NSERC grant no. #327637-06

##  1\. Introduction

The Ricci flow is the following evolution equation for an initial Riemannian
metric $g_{0}$ on a $n$-dimensional manifold $M^{n}$:

(1) $\left\\{\begin{aligned} \frac{{\partial}}{{\partial
t}}g(t)&=-2\text{Rc}_{g(t)},\\\ g(0)&=g_{0}.\end{aligned}\right.$

A classical result of Shi [20] says that if $g_{0}$ is complete with sectional
curvatures uniformly bounded by $K$, then (1) admits a complete bounded
curvature solution $g(t)$ on $M^{n}\times[0,T(n,K))$ for some time $T(n,K)>0$
depending only on $n$ and $K$. Extending Shi’s theorem to cases of unbounded
curvature, or by extending the solution for all times $t\in[0,\infty)$ under
suitable conditions are of fundamental importance in Ricci flow theory and
applications. One theme of recent prevalance is to consider manifolds
satisfying some Sobolev-type control coupled with a smallness of the scale-
invariant integral curvature

$\int_{\Omega}|\text{Rm}|^{\frac{n}{2}}\,dV_{g}.$

See for instance [2] or [3] where short time existence of (1) is proved
assuming local control of these quantities. If one replaces the local
assumptions in exchange for global ones, one may obtain stronger conclusions.
For example, in [4] a gap theorem is proved for manifolds which have
nonnegative Ricci curvature, Euclidean volume growth, and globally small
scale-invariant integral curvature as a consequence of rescalings of the
short-time Ricci flow existence results proved in [3]. Another example is the
following rigidity theorem of Chen for asymptotically flat manifolds.

######  Theorem 1.1.

[9, Theorem A] For $n\geq 3$, there exists a dimensional constant
$\delta(n)>0$ such that the following holds. Let $(M^{n},g)$ be an
asymptotically flat manifold of order $\tau>0$. Suppose that $(M^{n},g)$
satisfies the Sobolev inequality

$\left(\int u^{\frac{2n}{n-2}}\,dV_{g}\right)^{\frac{n-2}{n}}\leq
C_{g}\int|\nabla u|^{2}\,dV_{g}$

for all $u\in W^{1,2}$, and the scale-invariant curvature pinching

$\left(\int_{M^{n}}|\text{Rm}|_{g}^{\frac{n}{2}}\,dV_{g}\right)^{\frac{2}{n}}\leq\delta(n)\frac{1}{C_{g}}.$

Then the complete Ricci flow $g(t)$ with initial condition $g(0)=g$ exists for
all times $t\in[0,\infty)$ and satisfies

$\lim_{t\to\infty}t\sup_{x\in M}|\text{Rm}|(x,t)=0.$

In particular, $g(t)$ converges in $C^{\infty}_{-\tau^{\prime}}(M^{n})$ for
any $\tau^{\prime}\in(0,\min(\tau,n-2))$ to the $\mathbb{R}^{n}$ with the
Euclidean metric and thus $M^{n}$ is diffeomorphic to $\mathbb{R}^{n}$.

The purpose of this paper is to generalize Theorem 1.1 to complete noncompact
bounded curvature manifolds. We prove the following.

######  Theorem 1.2.

For $n\geq 3$, there are dimensional constants $\delta(n)>0$ and
$C_{0}(n)=(3e)^{\frac{n}{2}}$ such that the following holds. Let $(M^{n},g)$
be a complete bounded curvature manifold. Suppose that $(M^{n},g)$ satisfies
the Sobolev inequality

(2) $\left(\int u^{\frac{2n}{n-2}}\,dV_{g}\right)^{\frac{n-2}{n}}\leq
C_{g}\int|\nabla u|^{2}\,dV_{g}$

for all $u\in W^{1,2}$, and the scale-invariant curvature pinching

(3)
$\left(\int_{M}|\text{Rm}|_{g}^{\frac{n}{2}}\,dV_{g}\right)^{\frac{2}{n}}\leq\delta(n)\frac{1}{C_{g}}.$

Then the complete Ricci flow $g(t)$ with initial condition $g(0)=g$ exists for
all times $t\in[0,\infty)$ and satisfies the following curvature estimates:

  1. (a)

For all $t\in(0,\infty)$ there holds

$\sup_{x\in M}|\text{Rm}|(x,t)\leq\frac{C_{0}(n)}{t}.$

  2. (b)

The singularity at infinity satisfies

$\lim_{t\to\infty}t\sup_{x\in M}|\text{Rm}|(x,t)=0.$

Moreover, if $|\text{Rm}|_{g}\in L^{p}$ for some $p<\frac{n}{2}$, then $g(t)$
converges locally smoothly to a complete flat limit metric $g(\infty)$ on
$M^{n}$ with Euclidean volume growth, and thus $M^{n}$ is diffeomorphic to
$\mathbb{R}^{n}$.

######  Remark 1.3.

The constant $\delta(n)$ in Theorem 1.2 can be taken to be

$\delta(n)=\frac{1}{40nL(n)},$

where the dimensional constant $L(n)$ is from Theorem 3.4 and is estimated in
the appendix. This value of $\delta(n)$ is henceforth fixed in the remainder
of the paper.

######  Remark 1.4.

When $(M^{n},g)$ is asymptotically flat, the condition that $|\text{Rm}|\in
L^{p}$ for some $p<\frac{n}{2}$ is automatically satisfied whenever
$p>\frac{n}{2+\tau}$ where $\tau>0$ is the order of the asymptotic end. Our
proof of Proposition 4.10 establishes that for any $\sigma<\tau/2$, we have
the global decay estimate $|\nabla^{k}\text{Rm}|\lesssim t^{-1-\sigma-k/2}$.
Using this, it follows from [16, Theorem 5.1] that in fact $g(t)$ converges in
$C^{\infty}_{-\tau^{\prime}}(M^{n})$ for any
$\tau^{\prime}\in(0,\min(\tau,n-2))$ to $\mathbb{R}^{n}$ with the Euclidean
metric, in the same sense as Theorem 1.2. Thus, Theorem 1.2 generalizes
Theorem 1.1.

We also prove - as a corollary of our main theorem - the following long-time
Ricci flow existence and rigidity result for which the initial condition may
have unbounded curvature.

######  Theorem 1.5.

With $n\geq 3$ and $\delta(n)$, $C_{0}(n)$ as in Theorem 1.2, the following
holds. Suppose that $(M^{n},g)$ is a complete Riemannian manifold (not
necessarily bounded curvature) which is the pointed smooth local limit of a
sequence of complete bounded curvature Riemannian manifolds $(M_{i},g_{i})$
($i=1,2,\dots$) such that for each $i\in\mathbb{N}$, there holds:

  1. (i)

The Sobolev inequality

$\left(\int u^{\frac{2n}{n-2}}\,dV_{g_{i}}\right)^{\frac{n-2}{n}}\leq
C_{g_{i}}\int|\nabla u|^{2}\,dV_{g_{i}};\;\text{ and }$

  2. (ii)

Sufficiently pinched scale-invariant integral curvature conditions

$\sup_{i\in\mathbb{N}}\left(\int_{M}|\text{Rm}|_{g_{i}}^{\frac{n}{2}}\,dV_{g_{i}}\right)^{\frac{2}{n}}\leq\frac{\delta(n)}{\sup_{i\in\mathbb{N}}C_{g_{i}}}.$

Then there exists a complete solution $g(t)$ to the Ricci flow on
$M\times[0,\infty)$ with $g(0)=g$ that satisfies the curvature estimate

$\sup_{x\in M}|\text{Rm}|(x,t)\leq\frac{C_{0}(n)}{t}$

for all $t\in(0,\infty)$. Moreover if
$\sup_{i\in\mathbb{N}}\||\text{Rm}|_{g_{i}}\|_{L^{p}(g_{i})}<\infty$ for some
$p<\frac{n}{2}$, then $g(t)$ converges locally smoothly to a complete flat
limit metric $g(\infty)$ on $M^{n}$ with Euclidean volume growth, and thus
$M^{n}$ is diffeomorphic to $\mathbb{R}^{n}$.

######  Remark 1.6.

The hypotheses of Theorem 1.2 impose no restriction on the supremum of, or
decay/growth rate of the curvature. In fact, given any sequence of positive
numbers $\\{\alpha_{i}\\}\subset\mathbb{R}_{+}$, one can construct a complete
metric $g$ on $\mathbb{R}^{n}$ such that

$\sup_{B_{g_{E}}(i+1)\setminus B_{g_{E}}(i)}|\text{Rm}|_{g}=\alpha_{i}$

but for which $g$ satisfies the Sobolev inequality (2) and the integral
curvature pinching estimate (3) with $C_{g}$ arbitrarily close to $C_{g_{E}}$.
In particular, one can define a sequence of complete bounded curvature metrics
$g_{i}$ on $\mathbb{R}^{n}$ that converge locally smoothly to an unbounded
curvature metric $g$ on $\mathbb{R}^{n}$ such that the metrics $g_{i}$ satisfy
the hypotheses i and ii from Theorem 1.5. The described constructions can be
accomplished by gluing into the flat Euclidean space $(\mathbb{R}^{n},g_{E})$
countably many “wrinkled balls”, by which we mean a compact perturbation $h$
of the Euclidean metric on the open unit ball $B$ which satisfies: (a)
$(1+c_{1})^{-1}g_{E}\leq h\leq(1+c_{1})g_{E}$ on $B$, (b)
$\sup_{B}|\text{Rm}|_{h}=c_{2}$, and (c)
$\int_{B}|\text{Rm}|_{h}^{\frac{n}{2}}\,dV_{h}<c_{3}$, where
$c_{1},c_{2},c_{3}>0$ can be prescribed arbitrarily.

The organization of this paper is as follows. Section 2 is devoted to showing
that finiteness of $\int|\text{Rm}|^{p}\,dV_{t}$ ($p\geq 1$) is preserved
along complete bounded curvature Ricci flows. Though this is a standard
maximum principle argument, the conclusion is essential to the proof of
Theorem 1.2. In Section 3 we discuss why we have a uniform global Sobolev
inequality along the flow, which follows from work of Ye [24]. Section 4
contains the bulk of the proof of Theorem 1.2 following the outline of Chen
[9] very closely. In fact, the numbering of the theorems in that section are
specifically chosen to be the same as the analogous results in [9]. The proof
of Theorem 1.5 is carried out in Section 5. Finally in the appendix we obtain
a bound on the constant $\delta(n)$ by tracking through the constants from
[24].

## Acknowledgement

The authors would like to thank Eric Chen for the helpful comments, and Oliver
Díaz Espinosa for pointing out to us an improved bound in the Marcinkiewicz
Interpolation Thoerem A.1.

##  2\. Boundedness of $\int|\operatorname{Rm}|^{p}\,dV_{t}$ along the Ricci
flow

For this section we fix the notation that $(M,g(t))$, $t\in[0,T]$ is a
complete bounded curvature Ricci flow and we write

$K=\sup_{M\times[0,T]}|\text{Rm}|(x,t)<\infty.$

Also, given a function $f(x,t)$ on $M^{n}\times[0,T]$ we will often abbreviate
as

$\int fdV_{t}:=\int_{M}f(x,t)dV_{g(t)}(x).$

We will prove that $\int|\text{Rm}|^{p}\,dV_{t}$ remains finite for all
$t\in[0,T]$ and any $p\geq 1$, assuming it was finite initially. In fact we
show that this quantity is upper-continuous, which will be used in the proof
of our main theorem. If we were to restrict to $p\geq 2$, then the results in
this section would be much simpler. When $1\leq p<2$ however (for example if
$n=3$ and $p=\frac{n}{2}$), there is an added difficulty due to the fact that

$\partial_{t}|\text{Rm}|^{p}=\frac{p}{2}|\text{Rm}|^{p-2}\partial_{t}|\text{Rm}|^{2},$

and the curvature is certainly permitted to be zero in places.

We get around this as follows. First, choose a smooth positive function
$v_{0}:M\to\mathbb{R}$ which is integrable (with respect to $g_{0}$) and
satisfies $0<v_{0}\leq 1$. Now for any $\epsilon>0$, define $v_{\epsilon}$ to
be the solution to the initial value problem

$\begin{cases}&\partial_{t}v_{\epsilon}=\Delta_{g(t)}v_{\epsilon}\\\
&v_{\epsilon}(0,x)=\epsilon v_{0}(x).\end{cases}$

Coupling the well-known evolution

$\partial_{t}|\text{Rm}|^{2}\leq\Delta|\text{Rm}|^{2}-2|\nabla\text{Rm}|^{2}+16|\text{Rm}|^{3}$

with the evolution of $v_{\epsilon}^{2}$ which satisfies

$\partial_{t}v_{\epsilon}^{2}=\Delta v_{\epsilon}^{2}-2|\nabla
v_{\epsilon}|^{2},$

we can write

$\partial_{t}(|\text{Rm}|^{2}+v_{\epsilon}^{2})\leq\Delta(|\text{Rm}|^{2}+v_{\epsilon}^{2})-2|\nabla\text{Rm}|^{2}-2|\nabla
v_{\epsilon}|^{2}+16|\text{Rm}|^{3}.$

By Cauchy-Schwarz and Young’s inequality, there holds

$\displaystyle|\nabla(|\text{Rm}|^{2}+v_{\epsilon}^{2})|^{2}$
$\displaystyle=4|\text{Rm}|^{2}|\nabla|\text{Rm}||^{2}+4v_{\epsilon}^{2}|\nabla
v_{\epsilon}|^{2}+8|\text{Rm}|v_{\epsilon}\langle\nabla|\text{Rm}|,\nabla
v_{\epsilon}\rangle$ $\displaystyle\leq
4|\text{Rm}|^{2}|\nabla|\text{Rm}||^{2}+4v_{\epsilon}^{2}|\nabla
v_{\epsilon}|^{2}+8|\text{Rm}|v_{\epsilon}|\nabla|\text{Rm}||\,|\nabla
v_{\epsilon}|$ $\displaystyle\leq
4|\text{Rm}|^{2}|\nabla|\text{Rm}||^{2}+4v_{\epsilon}^{2}|\nabla
v_{\epsilon}|^{2}+4|\text{Rm}|^{2}|\nabla
v_{\epsilon}|^{2}+v_{\epsilon}^{2}|\nabla|\text{Rm}||^{2}$
$\displaystyle=4(|\text{Rm}|^{2}+v_{\epsilon}^{2})(|\nabla|\text{Rm}||^{2}+|\nabla
v_{\epsilon}|^{2}).$

So if we set $K_{\epsilon}^{2}=|\text{Rm}|^{2}+v_{\epsilon}^{2}$, we can see
that

$4K_{\epsilon}^{2}|\nabla K_{\epsilon}|^{2}\leq
4K_{\epsilon}^{2}(|\nabla|\text{Rm}||^{2}+|\nabla v_{\epsilon}|^{2}).$

Subsequently, by the Kato inequality $|\nabla|T||\leq|\nabla T|$, there holds

$\partial_{t}K_{\epsilon}^{2}\leq\Delta K_{\epsilon}^{2}-2|\nabla
K_{\epsilon}|^{2}+16K_{\epsilon}^{3}.$

We now use this evolution of $K_{\epsilon}^{2}$ to determine a useful
evolution inequality for $K_{\epsilon}^{p}$.

######  Lemma 2.1.

For any $p\geq 1$, there holds

(4) $\partial_{t}K_{\epsilon}^{p}\leq\Delta
K_{\epsilon}^{p}+8pK_{\epsilon}^{p+1}.$

###### Proof.

Denote $\alpha=\frac{p}{2}$. Since $K_{\epsilon}$ is a strictly positive
function on $M\times[0,T]$, we freely differentiate and see that

$\displaystyle\partial_{t}K_{\epsilon}^{2\alpha}$
$\displaystyle=\partial_{t}(K_{\epsilon}^{2})^{\alpha}=\alpha
K_{\epsilon}^{2(\alpha-1)}\partial_{t}K_{\epsilon}^{2}$
$\displaystyle\leq\alpha K_{\epsilon}^{2(\alpha-1)}(\Delta
K_{\epsilon}^{2}-2|\nabla K_{\epsilon}|^{2}+16K_{\epsilon}^{3})$
$\displaystyle=\alpha K_{\epsilon}^{2(\alpha-1)}\Delta
K_{\epsilon}^{2}-2\alpha K_{\epsilon}^{2(\alpha-1)}|\nabla
K_{\epsilon}|^{2}+16\alpha K_{\epsilon}^{2\alpha+1}$ $\displaystyle=\Delta
K_{\epsilon}^{2\alpha}-4\alpha(\alpha-1)K_{\epsilon}^{2(\alpha-1)}|\nabla
K_{\epsilon}|^{2}-2\alpha K_{\epsilon}^{2(\alpha-1)}|\nabla
K_{\epsilon}|^{2}+16\alpha K_{\epsilon}^{2\alpha+1}$ $\displaystyle\leq\Delta
K_{\epsilon}^{2\alpha}+16\alpha K_{\epsilon}^{2\alpha+1},$

where in the last line we have used the fact that $\alpha\geq\frac{1}{2}$
implies that $-4\alpha(\alpha-1)-2\alpha\leq 0$. ∎

We can use the heat-type evolution of $K_{\epsilon}^{p}$ as in (4), along with
the Maximum Principle for complete bounded curvature Ricci flows to obtain the
following.

######  Proposition 2.2.

Let $p\geq 1$ and suppose that $|\text{Rm}|_{g(0)}\in L^{p}_{g(0)}$. Then
there holds

$\int_{M}|\text{Rm}|_{g(t)}^{p}\,dV_{t}\leq
e^{(8p+n)Kt}\int_{M}|\text{Rm}|_{g(0)}^{p}\,dV_{0}.$

###### Proof.

First let $\Omega_{0}\subset\subset\Omega_{1}\subset\subset\dots\subset\subset
M$ be an exhausting sequence of relatively compact domains with smooth
boundaries in $M$, and for each $j\in\mathbb{N}$ let $\phi_{j}:M\to[0,1]$ be a
smooth cutoff function which is supported in $\Omega_{j}$ and $\phi_{j}\equiv
1$ in $\Omega_{j-1}$. Now for any $\epsilon>0$ and for each $j\in\mathbb{N}$,
we denote by $u_{j}:M\times[0,T]\to\mathbb{R}$ the solution to the initial
value Dirichlet boundary condition problem

$\begin{cases}&\partial_{t}u_{j}=\Delta_{g(t)}u_{j}\\\
&u_{j}(x,0)=\phi_{j}K_{\epsilon}^{p}(x,0)\\\ &u_{j}(x,t)=0\text{ if
}x\in\partial\Omega_{j}.\end{cases}$

Then by the Maximum Principle, we know that the $u_{j}$ are all nonnegative
functions which are uniformly bounded

(5) $\sup_{M\times[0,T]}u_{j}(x,t)\leq\sup_{x\in
M}K_{\epsilon}^{p}(x,0)\leq\sup_{x\in
M}(|\text{Rm}|^{2}(x,0)+\epsilon^{2})^{\frac{p}{2}}.$

Then by the Dominated Convergence Theorem and the evolution
$\partial_{t}dV_{t}=-R_{t}dV_{t}$, we have

(6)
$\frac{d}{dt}\int_{\Omega_{j}}u_{j}\,dV_{t}=\int_{\Omega_{j}}\partial_{t}u_{j}-R_{t}u_{j}\,dV_{t}=\int_{\Omega_{j}}\Delta_{g(t)}u_{j}-R_{t}u_{j}\,dV_{t}\leq\int_{\Omega_{j}}\Delta_{g(t)}u_{j}+nK\int_{\Omega_{j}}u_{j}\,dV_{t}.$

Here we have applied the bound $|R|\leq n|\text{Rm}|$, see [15, Proposition
7.28] for instance. Now Green’s Theorem says that

$\int_{\Omega_{j}}\Delta_{g(t)}u_{j}\,dV_{t}=\int_{\partial\Omega_{j}}\nabla_{v(t)}u_{j}\,dV_{t}$

where $v(t)$ is the outward unit normal on $\partial\Omega_{j}$ relative to
$g(t)$. Then by the nonnegativitiy of $u_{j}$ and the Dirichlet condition
$u_{j}=0$ on $\partial\Omega_{j}$, this boundary integral is nonpositive.
Subsequently, putting this back into (6) shows that

(7) $\int_{M}u_{j}\,dV_{t}=\int_{\Omega_{j}}u_{j}\,dV_{t}\leq
e^{nKt}\int_{M}u_{j}(x,0)\,dV_{0}\leq
e^{nKt}\int_{M}(|\text{Rm}|_{g(0)}^{2}+\epsilon^{2}v_{0}^{2})^{\frac{p}{2}}\,dV_{0}.$

Moreover for each fixed point in spacetime $(x,t)\in M\times[0,T]$, the
sequence $u_{j}(x,t)$ is monotone increasing (another consequence of the
Maximum Principle), and thus has a pointwise limit which we denote by
$u(x,t)$. By the uniform boundedness (5), Schauder estimates, the Arzelà-
Ascoli Theorem, and a diagnoal subsequence argument, we obtain that the limit
function $u$ is smooth on $M\times[0,T)$ and solves the initial value problem

$\begin{cases}&\partial_{t}u=\Delta_{g(t)}u\\\
&u(x,0)=K_{\epsilon}^{p}(x,0).\end{cases}$

Moreover, the integral bound (7) descends to $u$ as well via the Monotone
Convergence Theorem. That is to say

(8) $\int_{M}u\,dV_{t}\leq
e^{nKt}\int(|\text{Rm}|_{g(0)}^{2}+\epsilon^{2}v_{0}^{2})^{\frac{p}{2}}\,dV_{0}.$

We also define
$\tilde{u}(x,t)=e^{8pt(K^{2}+\epsilon^{2})^{\frac{1}{2}}}u(x,t)$ so that

$\partial_{t}\tilde{u}=8p(K^{2}+\epsilon^{2})^{\frac{1}{2}}\tilde{u}+e^{8pt(K^{2}+\epsilon^{2})^{\frac{1}{2}}}\partial_{t}u=8p(K^{2}+\epsilon^{2})^{\frac{1}{2}}\tilde{u}+e^{8pt(K^{2}+\epsilon^{2})^{\frac{1}{2}}}\Delta_{g(t)}u=8p(K^{2}+\epsilon^{2})^{\frac{1}{2}}\tilde{u}+\Delta_{g(t)}\tilde{u},$

and $\tilde{u}(x,0)=K_{\epsilon}^{p}(x,0)$. Comparing this with the evolution
inequality (4), another application of the Maximum Principle shows that

$K_{\epsilon}^{p}(x,t)\leq\tilde{u}(x,t)$

for all $(x,t)\in M\times[0,T]$. Integrating this and applying our integral
bound (8) yields

$\displaystyle\int_{M}|\text{Rm}|_{g(t)}^{p}\,dV_{t}$
$\displaystyle\leq\int_{M}K_{\epsilon}^{p}\,dV_{t}\leq\int_{M}\tilde{u}(x,t)\,dV_{t}(x)$
$\displaystyle=e^{8pt(K^{2}+\epsilon^{2})^{\frac{1}{2}}}\int_{M}u(x,t)\,dV_{t}(x)$
$\displaystyle\leq
e^{(8p+n)t(K^{2}+\epsilon^{2})^{\frac{1}{2}}}\int_{M}(|\text{Rm}|_{g(0)}^{2}+\epsilon^{2}v_{0}^{2})^{\frac{p}{2}}\,dV_{0}.$

Now because of the assumed integrability of $|\text{Rm}|_{g(0)}^{p}$ and
$v_{0}$ (with respect to $g_{0}$), we can take $\epsilon\to 0$ and once again
apply the Dominated Convergence Theorem to obtain the result. ∎

##  3\. Sobolev Inequality along the Ricci flow

The purpose of this section is to show that we have a uniform global Sobolev
inequality along a complete bounded curvature Ricci flow (Theorem 3.4). This
has been proved on compact manifolds in [24] and this result was verified and
applied in various complete noncompact cases such as [9] in the asymptotically
flat case. The steps to proving this are as follows: (1) Show a log-Sobolev
inequality for the initial metric $g_{0}$; (2) Show the log-Sobolev inequality
is preserved along the flow using monotonicity of Perelman’s $\mu$-functional;
and (3) At a given time $t\in[0,T]$ along the flow, use regularity properties
of the heat equation on $M$ with the _fixed_ metric $g(t)$ to get back to a
Sobolev inequality from the log-Sobolev inequality at time $t$. We briefly
outline these steps and refer the reader to the other works in which these
methods have been employed for more detail.

Step 1: log-Sobolev inequality for the initial metric $g_{0}$.

We first prove a log-Sobolev inequality for $g_{0}$. This has been done many
times in the literature (cf. [24], [9], etc.), but we will sketch the proof
here for convenience and for our future reference.

######  Lemma 3.1 (Log-Sobolev inequality for $g_{0}$).

Let $(M^{n},g)$, $n\geq 3$ be any Riemannian manifold satisfying

(9) $\left(\int|u|^{\frac{2n}{n-2}}\,dV_{g}\right)^{\frac{n-2}{n}}\leq
C_{g}\int|\nabla u|^{2}\,dV_{g}$

for all $u\in W^{1,2}(M,g)$. Then each of the following is true:

(1) For all $u\in W^{1,2}(M,g)$ with $\int u^{2}\,dV_{g}=1$ and any $\tau>0$
there holds

$\int u^{2}\log u^{2}\,dV_{g}\leq 4\tau\int|\nabla
u|^{2}\,dV_{g}-\frac{n}{2}\log\tau+\frac{n}{2}\left(\log
C_{g}+\log\frac{n}{8}-1\right).$

(2) Assume moreover that

$\left(\int|R_{g}|^{\frac{n}{2}}\,dV_{g}\right)^{\frac{2}{n}}\leq\frac{2}{C_{g}}.$

Then for all $u\in W^{1,2}(M,g)$ with $\int u^{2}\,dV_{g}=1$ and any $\tau>0$
there holds

(10) $\int u^{2}\log u^{2}\,dV_{g}\leq 4\tau\left(\int|\nabla
u|^{2}+\frac{R_{g}}{4}u^{2}\,dV_{g}\right)-\frac{n}{2}\log\tau+\frac{n}{2}\left(\log
C_{g}+\log\frac{n}{4}-1\right).$

###### Proof.

We denote $q=\frac{2n}{n-2}$ and apply Jensen’s inequality and (9) to see

$\displaystyle\int_{M}u^{2}\log u^{2}\,dV$ $\displaystyle=2\int_{M}u^{2}\log
u\,dV=\frac{2}{q-2}\int_{M}u^{2}\log u^{q-2}\,dV$
$\displaystyle\leq\frac{2}{q-2}\log\int_{M}u^{q}\,dV=\frac{q}{q-2}\log\|u\|_{q}^{2}\leq\frac{n}{2}\log\left(C_{g}\int|\nabla
u|^{2}\,dV_{g}\right).$

Then (1) is established after using the inequality

$\log\left(\int|\nabla u|^{2}\,dV_{g}\right)\leq a\int|\nabla
u|^{2}\,dV_{g}-1-\log a$

for $a=\frac{8\tau}{n}$. To prove (2) we use Hölder’s inequality and (9) to
see that

$2\tau\int
R_{g}u^{2}\,dV_{g}\geq-2\tau\left(\int|R_{g}|^{\frac{n}{2}}\,dV_{g}\right)^{\frac{2}{n}}\left(\int|u|^{\frac{2n}{n-2}}\,dV_{g}\right)^{\frac{n-2}{n}}\geq-4\tau\int|\nabla
u|^{2}\,dV_{g}.$

Then plugging this into (1) gives

$\int u^{2}\log u^{2}\,dV_{g}\leq 8\tau\int|\nabla u|^{2}\,dV_{g}+2\tau\int
R_{g}u^{2}\,dV_{g}-\frac{n}{2}\log\tau+\frac{n}{2}\left(\log
C_{g}+\log\frac{n}{8}-1\right).$

This is equivalent to (10) after relabeling. ∎

Step 2: log-Sobolev inequality is preserved along the flow.

The preservation of the log-Sobolev inequality obtained in Lemma 3.1 for
$g=g(0)$, to times $t>0$ along the Ricci flow is shown via the monotonicity of
Perelman’s $\mu$-functional. Let’s recall the definition of this.

######  Definition 3.1 (Perelman’s $\mu$-functional).

Given a complete bounded curvature Riemannian manifold $(M^{n},g)$, $n\geq 3$
and $\tau>0$, we define the $\mathcal{W}$-functional to be

$\mathcal{W}(g,u,\tau)=\int\left[\tau(4|\nabla u|^{2}+Ru^{2})-u^{2}\log
u^{2}\right]\,dV_{g}-n-\frac{n}{2}\log(4\pi\tau),$

where $u\in W^{1,2}(M,g)$ satisfies $\int u^{2}\,dV_{g}=1$. Then _Perelman’s
$\mu$-functional_ is given by

$\mu(g,\tau)=\inf\left\\{\mathcal{W}(g,u,\tau)\;:\;u\in W^{1,2}(M,g),\;\;\int
u^{2}\,dV_{g}=1\right\\}.$

It is well known that the Ricci flow corresponds to a gradient flow of the
$\mu$-functional and therefore satisfies a monotonicity formula. This
monotonicity was first shown to hold for compact flows by Perelman [18] and
has since recieved a lot of attention for the noncompact case (cf. [5], [25],
[9], etc.). The following lemma follows from the (much more general) almost-
monotonicity of the local $\mu$-functional [23, Theorem 5.4] and the lower-
continuity of the local $\mu$-functional [23, Proposition 2.3].

######  Lemma 3.2.

Suppose that $(M^{n},g(t))$, $t\in[0,T]$ is a complete bounded curvature Ricci
flow. Then for any $t\in[0,T]$ and $\tau>t$, there holds

$\mu(g(0),\tau)\leq\mu(g(t),\tau-t).$

So if we have any complete bounded curvature Ricci flow $(M,g(t))$,
$t\in[0,T]$ such that inequality (10) holds for all $L^{2}$ normalized
$W^{1,2}$ functions $u$, then this establishes the inequality

$\mu(g(0),\tau+t)\geq-\frac{n}{2}\left(\log
C_{g(0)}+\log\frac{n}{4}-1\right)-n-\frac{n}{4}\log(4\pi)$

for any $t\in[0,T]$ and any $\tau>0$. Now the monotonicity of the
$\mu$-functional implies that the above holds with $\mu(g(t),\tau)$ replacing
$\mu(g(0),\tau+t)$. This in turn is equivalent to a log-Sobolev inequality at
time $t$. More precisely, there holds

(11) $\int u^{2}\log u^{2}\,dV_{g(t)}\leq 4\tau\left(\int|\nabla
u|_{g(t)}^{2}+\frac{R_{g(t)}}{4}u^{2}\,dV_{g(t)}\right)-\frac{n}{2}\log\tau+\frac{n}{2}\left(\log
C_{g(0)}+\log\frac{n}{4}-1\right)$

for any $\tau>0$.

Step 3: log-Sobolev inequality implies Sobolev inequality.

The method of obtaining a Sobolev inequality from a log-Sobolev inequality is
well known in the literature: Davies [11, Chapters 1-2] describes the process
in the general context of abstract measure spaces $(\Omega,\text{d}x)$; Ye
[24, Theorems 5.3-5.5] works out this process for compact Riemannian manifolds
$(M,g)$, showing in particular that the Sobolev constant obtained does not
depend on the geometry of $g$; and Chen [10, Appendix B] extends this to
asymptotically flat manifolds $(M,g)$ by applying Ye’s result to obtain a
uniform Sobolev inequality for each member of an exhaustion by compact sets
$\Omega_{1}\subset\Omega_{2}\subset\dots\subset M$, then extending this to a
Sobolev inequality on $M$ by this uniformity in $j$, and the fact that
$W^{1,2}_{0}(M,g)=\bigcup_{j\in\mathbb{N}}W^{1,2}_{0}(\Omega_{j},g)$. We
observe that in this argument, the condition of asymptotic flatness can
actually be weakened to the condition of $g$ being complete with bounded
curvature as in our setting. In summary, we have the following.

######  Proposition 3.3 (cf. [11], [24], [10]).

For $n\geq 3$ and $C>0$, there exists a constant $\tilde{C}(n,C)>0$ such that
the following holds. Let $(M^{n},h)$, $n\geq 3$ be any complete bounded
curvature manifold such that the log-Sobolev inequality

$\int_{M}u^{2}\log u^{2}\,dV\leq\sigma\left(\int_{M}|\nabla u|^{2}+\Psi
u^{2}\,dV\right)-\frac{n}{2}\log\sigma+C$

holds for all $\sigma>0$ and all $L^{2}$ normalized $\|u\|_{2}=1$ functions
$u\in W^{1,2}_{0}(M)$. Here

$W^{1,2}_{0}(M):=\\{u\in W^{1,2}(M)\;:\;u\text{ has compact support}\\},$

and $\Psi$ is a nonnegative bounded function on $M$. Then the Sobolev
inequality

$\left(\int_{M}u^{\frac{2n}{n-2}}\,dV_{h}\right)^{\frac{n-2}{n}}\leq\tilde{C}(n,C)\left(\int|\nabla
u|^{2}+\Psi u^{2}\,dV_{h}\right)$

holds for all $u\in W_{0}^{1,2}(M,h)$.

Now steps 1-3 above can be combined to give a uniform global Sobolev
inequality along the flow.

######  Theorem 3.4.

For $n\geq 3$, there exists a dimensional constant $L(n)>0$ such that the
following holds. Suppose that $(M^{n},g(t))$, $t\in[0,T]$ is a complete
bounded curvature Ricci flow and that the initial metric $g(0)=g_{0}$
satisfies the curvature condition

$\left(\int_{M}|R|_{g_{0}}^{\frac{n}{2}}\,dV_{g_{0}}\right)^{\frac{2}{n}}\leq\frac{2}{C_{g_{0}}}.$

Then with $A=LC_{g_{0}}$, the weighted Sobolev inequality

(12) $\left(\int|u|^{\frac{2n}{n-2}}\,dV_{g(t)}\right)^{\frac{n-2}{n}}\leq
A\left(\int|\nabla_{g(t)}u|^{2}+R_{g(t)}^{+}u^{2}\,dV_{g(t)}\right)$

holds for all $u\in W_{0}^{1,2}(M^{n},g(t))$ on $[0,T]$.

###### Proof.

Part (2) of Lemma 3.1 gives us the following log-Sobolev inequality for the
metric $g_{0}$:

$\int u^{2}\log u^{2}\,dV_{g_{0}}\leq 4\tau\left(\int|\nabla
u|^{2}+\frac{R}{4}u^{2}\,dV_{g_{0}}\right)-\frac{n}{2}\log\tau+\frac{n}{2}\left(\log
C_{g_{0}}+\log\frac{n}{4}-1\right).$

Monotonicity of the $\mu$-functional (Lemma 3.2) implies that the uniform log-
Sobolev inequality (11) holds for any $t\in[0,T]$ and any $\tau>0$. Now given
a $t\in[0,T]$, we replace $R_{g(t)}$ with $R_{g(t)}^{+}$ in (11) to obtain

(13) $\int u^{2}\log u^{2}\,dV_{g(t)}\leq 4\tau\left(\int|\nabla
u|_{g(t)}^{2}+\frac{R^{+}_{g(t)}}{4}u^{2}\,dV_{g(t)}\right)-\frac{n}{2}\log\tau+\frac{n}{2}\left(\log
C_{g_{0}}+\log\frac{n}{4}-1\right).$

We are then in a position to apply Proposition 3.3 which allows us to conclude
that (12) is true at any given time $t$. Since the RHS of (13) is independent
of $t\in[0,T]$, this establishes the result. The fact that $A$ depends
linearly on $C_{g_{0}}$ is realized by tracking the constants in [24] \- this
computation is carried out in the appendix wherein we also provide a bound on
$L(n)$. ∎

##  4\. Proof of Theorem 1.2

The proof of Theorem 1.2 will be broken down into 10 intermediate steps within
this section. The first eight of which (Lemma 4.1 to Proposition 4.8) follow
the same outline as done by Chen [9] and the precise naming and numbering of
those results are specifically chosen to align with the analogous statements
in [9, Section 4]. The proofs are (naturally) very similar the originals and,
in some cases, are identical. For the readers’ convenience, we will present
the proofs in full when there are notable differences, and will omit them
otherwise.

We now start the proof of Theorem 1.2, and let $(M^{n},g)$ satisfy the
hypotheses of the theorem for some constant $C_{g}\in(0,\infty]$. We also let
$g(t)$ on $M\times[0,T_{max})$ be the unique complete bounded curvature Ricci
flow with $g(0)=g$ with existence guaranteed from [20] and uniqueness from [8]
(here $T_{max}$ is the maximal existence time for the flow and may equal
$\infty$). Note that if $C_{g}=\infty$, then the integral curvature pinching
condition (3) implies $|\text{Rm}|_{g}\equiv 0$ on $M$ and subsequently the
conclusion is trivially satisfied by the stationary solution $g(t)\equiv g$.
So without loss of generality, we may assume $C_{g}<\infty$ in which case we
will prove the theorem assuming that the curvature pinching inequality (3) is
strict, albeit with the slightly larger constant

$\delta_{1}(n):=\frac{1}{39nL(n)}\frac{1}{C_{g}}$

where $L(n)$ is the dimensional constant from Theorem 3.4.

For ease of reference, we summarize our assumptions as follows:

(14) $\begin{cases}&(M^{n},g(t)),\;n\geq 3,\;t\in[0,T_{max}),\;g(0)=g,\text{
is a complete Ricci flow satisfying:}\\\ &\text{For any
}\;T<T_{max},\;\sup_{M\times[0,T]}|\text{Rm}|(x,t)<\infty\;\text{ and }\\\
&\left(\int|\text{Rm}|_{g}^{\frac{n}{2}}\,dV_{g}\right)^{\frac{2}{n}}<\delta_{1}(n)\frac{1}{C_{g}}\;\text{
where }\;C_{g}<\infty\;\text{ is such that }\\\ &\left(\int
u^{\frac{2n}{n-2}}\,dV_{g}\right)^{\frac{n-2}{n}}\leq C_{g}\int|\nabla
u|^{2}\,dV_{g}\text{ for all }u\in W^{1,2}.\end{cases}$

We are now ready to continue with the proof.

Our first step is to determine the evolution of $L^{p}$ norm of the curvature.
We choose to break up into two different cases based on the value of $p$
because the proof is simpler when $p/2\geq 1$ and may be of independent
interest.

######  Lemma 4.1.

Let $(M^{n},g(t))$, $t\in[0,T]$ be a complete bounded curvature Ricci flow and
fix some smooth function $\phi:M\to\mathbb{R}$ with compact support. Then the
following hold.

  1. (a)

If $\alpha\geq 1$, then there holds

$\frac{d}{dt}\int\phi^{2}|\text{Rm}|^{2\alpha}\,dV_{t}\leq-2\int|\nabla(\phi|\text{Rm}|^{\alpha})|^{2}\,dV_{t}+(16\alpha+n)\int\phi^{2}|\text{Rm}|^{2\alpha+1}\,dV_{t}+2\int|\nabla\phi|^{2}|\text{Rm}|^{2\alpha}\,dV_{t}.$

  2. (b)

If $\frac{1}{2}<\alpha<1$, then for any $\beta>0$ there holds

$\displaystyle\frac{d}{dt}\int\phi^{2}(|\text{Rm}|^{2}+\beta)^{\alpha}\,dV_{t}\leq-
C_{1}(\alpha)\int|\nabla(\phi(|\text{Rm}|^{2}+\beta)^{\frac{\alpha}{2}})|^{2}\,dV_{t}$
$\displaystyle\;\;\;+(16\alpha+n)\int\phi^{2}(|\text{Rm}|^{2}+\beta)^{\alpha+\frac{1}{2}}\,dV_{t}+C_{2}(\alpha)\int|\nabla\phi|^{2}(|\text{Rm}|^{2}+\beta)^{\alpha}\,dV_{t}.$

Moreover, $C_{1}(\alpha)\to 0$ as $\alpha\to\frac{1}{2}$ and if
$\frac{2}{3}<\alpha<1$, then one can take $C_{1}(\alpha)=\frac{1}{2}$.

###### Proof.

First recall the well-known evolution of $|\text{Rm}|^{2}$ under the Ricci
flow:

(15)
$\partial_{t}|\text{Rm}|^{2}\leq\Delta|\text{Rm}|^{2}-2|\nabla\text{Rm}|^{2}+16|\text{Rm}|^{3}.$

For part a, we combine this with $\partial_{t}dV_{t}=-R_{t}dV_{t}$, the Kato
inequality $|\nabla|T||\leq|\nabla T|$, $|R|\leq n|\text{Rm}|$ (from [15,
Proposition 7.28]), and $\alpha\geq 1$ which gives

$\displaystyle\frac{d}{dt}\int\phi^{2}|\text{Rm}|^{2\alpha}\,dV_{t}$
$\displaystyle=\int\phi^{2}\frac{\partial}{\partial
t}\left(|\text{Rm}|^{2\alpha}\,dV_{t}\right)=\int\phi^{2}\left(\alpha|\text{Rm}|^{2(\alpha-1)}\partial_{t}|\text{Rm}|^{2}-R|\text{Rm}|^{2\alpha}\right)\,dV_{t}$
$\displaystyle\leq\int\phi^{2}\alpha|\text{Rm}|^{2(\alpha-1)}\left(\Delta|\text{Rm}|^{2}-2|\nabla\text{Rm}|^{2}+16|\text{Rm}|^{3}\right)-\phi^{2}R|\text{Rm}|^{2\alpha}\,dV_{t}$
$\displaystyle\leq\int\phi^{2}\left[\alpha|\text{Rm}|^{2(\alpha-1)}\Delta|\text{Rm}|^{2}-2\alpha|\text{Rm}|^{2\alpha-2}|\nabla|\text{Rm}||^{2}+\left(16\alpha+n\right)|\text{Rm}|^{2\alpha+1}\right]\,dV_{t}$
$\displaystyle=\int\phi^{2}\alpha|\text{Rm}|^{2(\alpha-1)}\Delta|\text{Rm}|^{2}\,dV_{t}-\frac{2}{\alpha}\int\phi^{2}|\nabla|\text{Rm}|^{\alpha}|^{2}\,dV_{t}+\left(16\alpha+n\right)\int\phi^{2}|\text{Rm}|^{2\alpha+1}\,dV_{t}.$

Now we integrate by parts the Laplacian term using that the support of $\phi$
is compact (so that no boundary integrals are introduced) to obtain

$\displaystyle\int\phi^{2}\alpha|\text{Rm}|^{2(\alpha-1)}\Delta|\text{Rm}|^{2}\,dV_{t}$
$\displaystyle=-\int\alpha\nabla\left(\phi^{2}|\text{Rm}|^{2(\alpha-1)}\right)\nabla|\text{Rm}|^{2}\,dV_{t}$
$\displaystyle=-\int\alpha\nabla\phi^{2}|\text{Rm}|^{2(\alpha-1)}\nabla|\text{Rm}|^{2}\,dV_{t}-\int\alpha\phi^{2}\nabla|\text{Rm}|^{2(\alpha-1)}\nabla|\text{Rm}|^{2}\,dV_{t}$
$\displaystyle=-2\int\nabla\phi^{2}|\text{Rm}|^{\alpha}\nabla|\text{Rm}|^{\alpha}\,dV_{t}-\int\alpha\phi^{2}\nabla|\text{Rm}|^{2(\alpha-1)}\nabla|\text{Rm}|^{2}\,dV_{t}$
$\displaystyle=-4\int\phi\nabla\phi|\text{Rm}|^{\alpha}\nabla|\text{Rm}|^{\alpha}\,dV_{t}-\int\alpha\phi^{2}\nabla|\text{Rm}|^{2(\alpha-1)}\nabla|\text{Rm}|^{2}\,dV_{t}$
$\displaystyle=\textbf{I}-\alpha\,\textbf{II}.$

The first term above is simplified by expanding out the derivative

(16)
$\textbf{I}=-2\int|\nabla(\phi|\text{Rm}|^{\alpha})|^{2}\,dV_{t}+2\int|\nabla\phi|^{2}|\text{Rm}|^{2\alpha}\,dV_{t}+2\int\phi^{2}|\nabla|\text{Rm}|^{\alpha}|^{2}\,dV_{t}.$

For the second term, we have to be careful because
$\nabla|\text{Rm}|^{2(\alpha-1)}$ is singular at points where $|\text{Rm}|=0$
when $\alpha<3/2$ (though this singularity is countered by a contribution of
$|\text{Rm}|$ from the $\nabla|\text{Rm}|^{2}$ term). Since $\alpha\geq 1$,
for any $\epsilon>0$ there holds

$\displaystyle\nabla(|\text{Rm}|+\epsilon)^{2(\alpha-1)}\nabla(|\text{Rm}|+\epsilon)^{2}$
$\displaystyle=4(\alpha-1)(|\text{Rm}|+\epsilon)^{2\alpha-3}(|\text{Rm}|+\epsilon)|\nabla(|\text{Rm}|+\epsilon)|^{2}$
$\displaystyle=4(\alpha-1)(|\text{Rm}|+\epsilon)^{2\alpha-2}|\nabla(|\text{Rm}|+\epsilon)|^{2}$
$\displaystyle=\frac{4(\alpha-1)}{\alpha^{2}}|\nabla(|\text{Rm}|+\epsilon)^{\alpha}|^{2}.$

Of course the RHS converges to
$\frac{4(\alpha-1)}{\alpha^{2}}|\nabla|\text{Rm}|^{\alpha}|^{2}$ pointwise as
$\epsilon\to 0$, and since $\phi$ is compactly supported we can apply the
Dominated Convergence Theorem (twice) to see that

II $\displaystyle=\lim_{\epsilon\to
0}\int\phi^{2}\nabla(|\text{Rm}|+\epsilon)^{2(\alpha-1)}\nabla(|\text{Rm}|+\epsilon)^{2}\,dV_{t}$
(17) $\displaystyle=\lim_{\epsilon\to
0}\frac{4(\alpha-1)}{\alpha^{2}}\int\phi^{2}|\nabla(|\text{Rm}|+\epsilon)^{\alpha}|^{2}\,dV_{t}=\frac{4(\alpha-1)}{\alpha^{2}}\int\phi^{2}|\nabla|\text{Rm}|^{\alpha}|^{2}\,dV_{t}.$

So using (16) and (17), we have

$\int\phi^{2}\alpha|\text{Rm}|^{2(\alpha-1)}\Delta|\text{Rm}|^{2}\,dV_{t}=-4\int\phi\nabla\phi|\text{Rm}|^{\alpha}\nabla|\text{Rm}|^{\alpha}\,dV_{t}-\frac{4(\alpha-1)}{\alpha}\int\phi^{2}|\nabla|\text{Rm}|^{\alpha}|^{2}\,dV_{t}.$

Finally, we gather all the $\phi^{2}|\nabla|\text{Rm}|^{\alpha}|^{2}$ terms to
see that

(18)
$\left[-\frac{2}{\alpha}-\frac{4(\alpha-1)}{\alpha}+2\right]\int\phi^{2}|\nabla|\text{Rm}|^{\alpha}|^{2}\,dV_{t}=-\frac{2(\alpha-1)}{\alpha}\int\phi^{2}|\nabla|\text{Rm}|^{\alpha}|^{2}\,dV_{t}\leq
0,$

where we have again used that $\alpha\geq 1$. This proves part a.

(b) The proof is similar to [2, Lemma 2.1], though in that proof it is assumed
that $\alpha\geq\frac{n}{4}$ and that $\phi$ is a function of space _and
time_. Since our condition on $\alpha$ is different and our $\phi$ is a
function of space only, we provide the details for the readers’ convenience
and to keep track of the constants. We begin just as before:

$\displaystyle\frac{d}{dt}\int\phi^{2}(|\text{Rm}|^{2}+\beta)^{\alpha}\,dV_{t}$
$\displaystyle=\int\phi^{2}\left(\alpha(|\text{Rm}|^{2}+\beta)^{\alpha-1}\partial_{t}|\text{Rm}|^{2}-R(|\text{Rm}|^{2}+\beta)^{\alpha}\right)\,dV_{t}$
$\displaystyle\leq\int\phi^{2}\alpha(|\text{Rm}|^{2}+\beta)^{\alpha-1}\left(\Delta|\text{Rm}|^{2}-2|\nabla|\text{Rm}||^{2}\right)+(n+16\alpha)\phi^{2}(|\text{Rm}|^{2}+\beta)^{\alpha+\frac{1}{2}}\,dV_{t}.$

Now we integrate by parts the Laplacian term which gives

$\displaystyle\int\phi^{2}\alpha(|\text{Rm}|^{2}+\beta)^{\alpha-1}\Delta|\text{Rm}|^{2}\,dV_{t}=-\int\alpha\nabla\left(\phi^{2}(|\text{Rm}|^{2}+\beta)^{\alpha-1}\right)\nabla|\text{Rm}|^{2}\,dV_{t}$
$\displaystyle\;\;\;=-4\alpha\int\phi\langle\nabla\phi,\nabla|\text{Rm}|\rangle(|\text{Rm}|^{2}+\beta)^{\alpha-1}|\text{Rm}|\,dV_{t}-4\alpha(\alpha-1)\int\phi^{2}(|\text{Rm}|^{2}+\beta)^{\alpha-2}|\text{Rm}|^{2}|\nabla|\text{Rm}||^{2}\,dV_{t}.$

Note that the coefficient in the latter term is actually positive because
$\alpha<1$, so we have

$-4\alpha(\alpha-1)\int\phi^{2}(|\text{Rm}|^{2}+\beta)^{\alpha-2}|\text{Rm}|^{2}|\nabla|\text{Rm}||^{2}\,dV_{t}\leq-4\alpha(\alpha-1)\int\phi^{2}(|\text{Rm}|^{2}+\beta)^{\alpha-1}|\nabla|\text{Rm}||^{2}\,dV_{t}.$

For the former term, we apply Cauchy-Schwarz, Hölder’s, and Young’s
inequalities

$\displaystyle-4\alpha\int\phi\langle\nabla\phi,\nabla|\text{Rm}|\rangle(|\text{Rm}|^{2}+\beta)^{\alpha-1}|\text{Rm}|\,dV_{t}$
$\displaystyle\leq
4\alpha\int\phi|\nabla\phi|\,|\nabla|\text{Rm}||\,(|\text{Rm}|^{2}+\beta)^{\alpha-1}|\text{Rm}|\,dV_{t}$
$\displaystyle\leq
4\alpha\int\phi|\nabla\phi|\,|\nabla|\text{Rm}||\,(|\text{Rm}|^{2}+\beta)^{\alpha-\frac{1}{2}}\,dV_{t}$
$\displaystyle\leq
4\alpha\left(\int\phi^{2}(|\text{Rm}|^{2}+\beta)^{\alpha-1}|\nabla|\text{Rm}||^{2}\,dV_{t}\right)^{\frac{1}{2}}\left(\int|\nabla\phi|^{2}(|\text{Rm}|^{2}+\beta)^{\alpha}\,dV_{t}\right)^{\frac{1}{2}}$
$\displaystyle\leq
2\alpha\epsilon\int\phi^{2}(|\text{Rm}|^{2}+\beta)^{\alpha-1}|\nabla|\text{Rm}||^{2}\,dV_{t}+\frac{2\alpha}{\epsilon}\int|\nabla\phi|^{2}(|\text{Rm}|^{2}+\beta)^{\alpha}\,dV_{t}=\textbf{I}+\textbf{II}.$

Now we can gather up the terms in the form of I and restrict $\epsilon>0$
small enough so that the coefficient is negative. Then we can use Cauchy-
Schwarz again (but in reverse) to see that

$\displaystyle\left[-2\alpha+4\alpha(1-\alpha)+2\alpha\epsilon\right]\int\phi^{2}(|\text{Rm}|^{2}+\beta)^{\alpha-1}|\nabla|\text{Rm}||^{2}\,dV_{t}$
$\displaystyle\leq\left[-2\alpha+4\alpha(1-\alpha)+2\alpha\epsilon\right]\int\phi^{2}(|\text{Rm}|^{2}+\beta)^{\alpha-2}|\text{Rm}|^{2}|\nabla|\text{Rm}||^{2}\,dV_{t}$
$\displaystyle=\left[-2\alpha+4\alpha(1-\alpha)+2\alpha\epsilon\right]\frac{1}{\alpha^{2}}\int\phi^{2}|\nabla(|\text{Rm}|^{2}+\beta)^{\frac{\alpha}{2}}|^{2}\,dV_{t}$
$\displaystyle\leq\left[-2\alpha+4\alpha(1-\alpha)+2\alpha\epsilon\right]\frac{1}{\alpha^{2}}\left(\frac{1}{2}\int|\nabla\phi(|\text{Rm}|^{2}+\beta)^{\frac{\alpha}{2}}|^{2}\,dV_{t}-\int|\nabla\phi|^{2}(|\text{Rm}|^{2}+\beta)^{\alpha}\,dV_{t}\right).$

Now whenever $\alpha>\frac{1}{2}$, we can restrict $\epsilon$ small enough so
that

$-C_{1}(\alpha):=\left[-2\alpha+4\alpha(1-\alpha)+2\alpha\epsilon\right]\frac{1}{2\alpha^{2}}<0,$

and if $\alpha>\frac{2}{3}$, $\epsilon$ can be taken small enough so that
$C_{1}(\alpha)=\frac{1}{2}$. This proves the result. ∎

We now prove that our hypotheses of (14) are enough to ensure that the scale-
invariant integral curvature quantity $\int|\text{Rm}|^{\frac{n}{2}}\,dV_{t}$
is nonincreasing in time. The following Lemma 4.2 is similar to [4, Lemma
2.2]. In fact the necessary tools needed in both proofs are the same: (a) a
weighted Sobolev inequality with scalar curvature term along the flow; and (b)
smallness of the scale-invariant integral curvature
$\int|\text{Rm}|^{\frac{n}{2}}\,dV_{t}$ _uniformly along the flow_ relative to
the constant in the Sobolev inequality. In [4] when $n\geq 4$, both (a) and
(b) are guaranteed by a rescaling argument from the short-time existence
result in [3]. For us however, we obtain (a) from Theorem 3.4 while (b) is
guaranteed for some short time interval by the upper-continuity of
$t\mapsto\int|\text{Rm}|^{\frac{n}{2}}\,dV_{t}$ from Proposition 2.2, which
will be sufficient for our proof below. Since our hypotheses and some lines of
argument differ from those in [4], we will include the proof for completeness.
We will also refer to these steps in establishing the case for $n=3$, and as
well as Lemma 4.4 which extends this monotonicity result to other powers of
$|\text{Rm}|$.

######  Lemma 4.2.

Let $(M^{n},g(t))$ satisfy the conditions in (14). Then
$t\mapsto\int|\text{Rm}|^{\frac{n}{2}}\,dV_{t}$ is nonincreasing on
$[0,T_{max})$.

###### Proof.

We first prove the result for dimensions $n\geq 4$ and comment on the
necessary modification for dimension $n=3$ at the end. Fix some $T<T_{max}$.
It clearly suffices to show that
$t\mapsto\int|\text{Rm}|^{\frac{n}{2}}\,dV_{t}$ is nonincreasing on $[0,T]$.

Let $\phi:M\to\mathbb{R}$ be some smooth compactly supported function to be
chosen later. Lemma 4.1 part a (applied with $\alpha=\frac{n}{4}$) gives

$\displaystyle\frac{d}{dt}\int\phi^{2}|\text{Rm}|^{\frac{n}{2}}\,dV_{t}$
$\displaystyle\leq-2\int|\nabla(\phi|\text{Rm}|^{\frac{n}{4}})|^{2}\,dV_{t}+(4n+n)\int\phi^{2}|\text{Rm}|^{\frac{n}{2}+1}\,dV_{t}+2\int|\nabla\phi|^{2}|\text{Rm}|^{\frac{n}{2}}\,dV_{t}.$

Note that $\delta_{1}(n)<\frac{2}{n}$ so that the conclusion of Theorem 3.4
holds with $A=L(n)C_{g}$, thus giving

$\displaystyle-2\int|\nabla(\phi|\text{Rm}|^{\frac{n}{4}})|^{2}\,dV_{t}$
$\displaystyle\leq-\frac{2}{A}\left(\int(\phi|\text{Rm}|^{\frac{n}{4}})^{\frac{2n}{n-2}}\,dV_{t}\right)^{\frac{n-2}{n}}+2\int
R^{+}\phi^{2}|\text{Rm}|^{\frac{n}{2}}\,dV_{t}$
$\displaystyle\leq-\frac{2}{A}\left(\int(\phi|\text{Rm}|^{\frac{n}{4}})^{\frac{2n}{n-2}}\,dV_{t}\right)^{\frac{n-2}{n}}+2n\int\phi^{2}|\text{Rm}|^{\frac{n}{2}+1}\,dV_{t}.$

Now we gather the terms with $|\text{Rm}|^{\frac{n}{2}+1}$ in the integrand
and apply Hölder’s inequality to obtain

$7n\int\phi^{2}|\text{Rm}|^{\frac{n}{2}+1}\,dV_{t}\leq
7n\left(\int|\text{Rm}|^{\frac{n}{2}}\,dV_{t}\right)^{\frac{2}{n}}\left(\int\phi^{\frac{2n}{n-2}}|\text{Rm}|^{\frac{n}{4}\frac{2n}{n-2}}\,dV_{t}\right)^{\frac{n-2}{n}}.$

To summarize the above, for any compactly supported smooth $\phi$ we have

(19)
$\frac{d}{dt}\int\phi^{2}|\text{Rm}|^{\frac{n}{2}}\,dV_{t}\leq\left(\int\phi^{\frac{2n}{n-2}}|\text{Rm}|^{\frac{n}{4}\frac{2n}{n-2}}\,dV_{t}\right)^{\frac{n-2}{n}}\left[-\frac{2}{A}+7n\left(\int|\text{Rm}|^{\frac{n}{2}}\,dV_{t}\right)^{\frac{2}{n}}\right]+2\int|\nabla\phi|^{2}|\text{Rm}|^{\frac{n}{2}}\,dV_{t}.$

The choice of $\phi$ we will be employing is as follows. Let
$\psi:\mathbb{R}\to[0,1]$ be a smooth decreasing cutoff function which
satisfies $\psi\equiv 1$ on $(-\infty,1]$ and $\psi\equiv 0$ on $[2,\infty)$.
Fix some $x_{0}\in M$ and define, for $r>0$,

$\phi_{r}(x)=\psi\left(\frac{\tilde{d}_{0}(x)}{r}\right).$

Here $\tilde{d}_{0}$ is smooth with uniformly bounded gradient $\sup_{x\in
M}|\nabla\tilde{d}_{0}|_{g}\leq c$, and uniformly bounded from below by the
distance function from $x_{0}\in M$ relative to $g_{0}$ in the sense that
$c^{-1}d_{g_{0}}(x_{0},x)\leq\tilde{d}_{0}(x)$ for all $x\in M$ for some $c>0$
depending on $g_{0}$ (in particular on a lower bound of $\text{Rc}_{g_{0}}$,
see [19, Theorem 4.2]). From this, assumption (14), and Proposition 2.2, we
can define the following constants which may depend on $T<T_{max}$ :

$C:=\sup_{r\geq
1}\sup_{M\times[0,T]}r^{2}|\nabla^{g(t)}\phi_{r}|_{g(t)}^{2}<\infty;\,\,\,\,K:=\sup_{M\times[0,T]}|\text{Rm}|<\infty;\,\,\,\,\beta:=\sup_{t\in[0,T]}\int|\text{Rm}|^{\frac{n}{2}}\,dV_{t}<\infty.$

Now we define the function

$F(r,t):=\int\phi_{r}^{2}|\text{Rm}|^{\frac{n}{2}}\,dV_{t}$

where $r\geq 1$ and $t\in[0,T]$ and define

$T^{\prime}:=\sup\\{t\in[0,T]:F(\infty,t)\leq(\delta_{1}/C_{g})^{\frac{n}{2}}\\}>0$

where the non-emptyness of the set on the RHS follows from assumption (14)
while the positivity of $T^{\prime}$ follows from Proposition 2.2 (in
particular, the upper-continuity of
$t\mapsto\int|\text{Rm}|^{\frac{n}{2}}\,dV_{t}$).

Now using the fact that $[-\frac{2}{A}+7n\frac{\delta_{1}}{C_{g(0)}}]<0$ which
follows from $A=L(n)C_{g(0)}$ and our choice of $\delta_{1}$ in assumption
(14), and our definition of $C,\beta$ above, the evolution (19) then gives

$\frac{d}{dt}F(r,t)\leq 2\frac{C}{\beta r^{2}}$

for all $r\geq 1$ and all $t\in[0,T^{\prime}]$. In particular, for all
$0<s_{1}<s_{2}<T^{\prime}$ this gives

$\int|\text{Rm}|^{\frac{n}{2}}\,dV_{s_{2}}=\lim_{r\to\infty}F(r,s_{2})\leq\lim_{r\to\infty}\left(F(r,s_{1})+2\frac{C(s_{2}-s_{1})}{\beta
r^{2}}\right)=F(\infty,s_{1})=\int|\text{Rm}|^{\frac{n}{2}}dV_{s_{1}}$

so that $\int|\text{Rm}|^{\frac{n}{2}}\,dV_{t}$ is non-increasing for
$t\in[0,T^{\prime}]$. Here we have used that $\tilde{d}_{0}(x)\geq
c^{-1}d_{g}(x,x_{0})$ and thus $\\{\phi_{r}=1\\}$ exhausts $M$ as
$r\to\infty$. But since
$\int|\text{Rm}|^{\frac{n}{2}}\,dV_{0}<(\delta_{1}/C_{g(0)})^{\frac{n}{2}}$,
this just imples that $T^{\prime}=T$ so that
$\int|\text{Rm}|^{\frac{n}{2}}\,dV_{t}$ is nonincreasing for $t\in[0,T]$ which
concludes the proof of the lemma for $n\geq 4$.

For the dimension $n=3$ case, we proceed just as we did above with part b of
Lemma 4.1 instead of part a. The analogue of the evolution inequality (19)
that we obtain in this case is

$\displaystyle\frac{d}{dt}\int\phi^{2}(|\text{Rm}|^{2}+\beta)^{\alpha}\,dV_{t}\leq
C_{2}(3/4)\int|\nabla\phi|^{2}(|\text{Rm}|^{2}+\beta)^{\alpha}\,dV_{t}$
$\displaystyle\;\;\;+\left(\int\phi^{6}(|\text{Rm}|^{2}+\beta)^{\frac{9}{4}}\,dV_{t}\right)^{\frac{1}{3}}\left[-\frac{1}{2A}+\left(15+\frac{3}{2}\right)\left(\int_{\text{supp}\phi}(|\text{Rm}|^{2}+\beta)^{\frac{3}{4}}\,dV_{t}\right)^{\frac{2}{3}}\right].$

Now for fixed $r\gg 0$ and $\beta>0$, we can integrate the above inequality on
the interval $[t_{0},t_{1}]$, then send $\beta\to 0$ (which is validated by
the precense of the cutoff function $\phi$) to obtain

$\displaystyle\int\phi^{2}|\text{Rm}|^{\frac{3}{2}}\,dV_{t_{1}}-\int\phi^{2}|\text{Rm}|^{\frac{3}{2}}\,dV_{t_{0}}\leq
C_{2}(3/4)\int_{t_{0}}^{t_{1}}\int|\nabla\phi|^{2}|\text{Rm}|^{\frac{3}{2}}\,dV_{t}\,dt$
(20)
$\displaystyle\;\;\;+\int_{t_{0}}^{t_{1}}\left(\int\phi^{6}|\text{Rm}|^{\frac{9}{2}}\,dV_{t}\right)^{\frac{1}{3}}\left[-\frac{1}{2A}+\left(15+\frac{3}{2}\right)\left(\int_{\text{supp}\phi}|\text{Rm}|^{\frac{3}{2}}\,dV_{t}\right)^{\frac{2}{3}}\right]\,dt.$

From here, one can show $t\mapsto\int|\text{Rm}|^{\frac{3}{2}}\,dV_{t}$ is
nonincreasing just as in the higher dimensional case above. In the process, we
use the fact that

$\delta_{1}(3)<\frac{1}{33L(3)},$

so that the bracked term in (20) is nonpositive. ∎

The following corollary is not essential to our proof (any time it is used we
could use Theorem 3.4 and Lemma 4.2 instead) but we include its statement for
ease of reference. The proof is identical to the proof of [9, Corollary 4.3].

######  Corollary 4.3.

Let $(M^{n},g(t))$ satisfy the conditions in (14). Then the uniform Sobolev
inequality

(21) $\left(\int u^{\frac{2n}{n-2}}\,dV_{t}\right)^{\frac{n-2}{n}}\leq
A_{0}\int|\nabla u|^{2}\,dV_{t}$

holds for all $t\in[0,T_{max})$ and $u\in W_{0}^{1,2}(M,g(t))$. Moreover we
can take $A_{0}=2L(n)C_{g_{0}}$.

######  Lemma 4.4.

Let $(M^{n},g(t))$ satisfy the conditions in (14). Then
$t\mapsto\int|\text{Rm}|^{q}\,dV_{t}$ is nonincreasing on $[0,T_{max})$ for
any $q$ satisfying $\max\\{2,\frac{n}{2}\\}\leq
q\leq\frac{n}{2}\frac{n}{n-2}$.

###### Proof.

We proceed in a similar fashion as Lemma 4.2. Fix $T<T_{max}$ and note that
Lemma 4.2 combined with our bounded curvature assumption implies

$\sup_{t\in[0,T]}\|\text{Rm}\|_{L^{q}_{g(t)}}<\infty.$

We apply Lemma 4.1 part (a) with $\alpha=q/2$ (note that $\alpha\geq 1$) and
Hölder’s inequality to see

$\displaystyle\frac{d}{dt}\int\phi^{2}|\text{Rm}|^{2\alpha}\,dV_{t}$
$\displaystyle\leq-2\int|\nabla(\phi|\text{Rm}|^{\alpha})|^{2}\,dV_{t}+2\int|\nabla\phi|^{2}|\text{Rm}|^{2\alpha}\,dV_{t}+(16\alpha+n)\int\phi^{2}|\text{Rm}|^{2\alpha+1}\,dV_{t}$
$\displaystyle\leq-2\int|\nabla(\phi|\text{Rm}|^{\alpha})|^{2}\,dV_{t}+2\int|\nabla\phi|^{2}|\text{Rm}|^{2\alpha}\,dV_{t}$
$\displaystyle\;\;\;\;+(16\alpha+n)\left(\int|\text{Rm}|^{\frac{n}{2}}\,dV_{t}\right)^{\frac{2}{n}}\left(\int\phi^{\frac{2n}{n-2}}|\text{Rm}|^{\alpha\frac{2n}{n-2}}\,dV_{t}\right)^{\frac{n-2}{n}}.$

Now Corollary 4.3 gives

$-2\int|\nabla(\phi|\text{Rm}|^{\alpha})|^{2}\,dV_{t}\leq-\frac{2}{A_{0}}\left(\int\phi^{\frac{2n}{n-2}}|\text{Rm}|^{\alpha\frac{2n}{n-2}}\,dV_{t}\right)^{\frac{n-2}{n}}.$

We will also retain from Lemma 4.2 the notation of $\phi_{r}$ and the constant
$C$ for which there holds

$|\nabla^{g(t)}\phi_{r}|_{g(t)}^{2}\leq\frac{C}{r^{2}}.$

Of course, the constant $C$ here will depend on the bound of the curvature on
$M\times[0,T]$. Then plugging in $A_{0}=2LC_{g_{0}}$, applying Lemma 4.2, and
using our definition of $\delta_{1}(n)$ yields

$\displaystyle\frac{d}{dt}\int\phi_{r}^{2}|\text{Rm}|^{q}\,dV_{t}$
$\displaystyle\leq\left(\int\phi_{r}^{\frac{2n}{n-2}}|\text{Rm}|^{q\frac{n}{n-2}}\,dV_{t}\right)^{\frac{n-2}{n}}\left[-\frac{2}{2LC_{g_{0}}}+(8q+n)\left(\int|\text{Rm}|^{\frac{n}{2}}\,dV_{t}\right)^{\frac{2}{n}}\right]+2\int|\nabla\phi_{r}|^{2}|\text{Rm}|^{2\alpha}\,dV_{t}$
$\displaystyle\leq\left(\int|\text{Rm}|^{q\frac{n}{n-2}}\,dV_{t}\right)^{\frac{n-2}{n}}\left[-\frac{1}{LC_{g_{0}}}+\left(\frac{4n^{2}}{n-2}+n\right)\frac{\delta_{1}(n)}{C_{g_{0}}}\right]+2\frac{C}{r^{2}}\int|\text{Rm}|^{2\alpha}\,dV_{t}$
$\displaystyle\leq
2\frac{C}{r^{2}}\sup_{t\in[0,T]}\int|\text{Rm}|^{q}\,dV_{t},$

where we have used the fact that $\frac{4n^{2}}{n-2}+n<39n$. Thus for fixed
$0\leq t_{0}<t_{1}\leq T$, we may integrate the above from $t_{0}$ to $t_{1}$
to see

$\displaystyle\int\phi_{r}^{2}|\text{Rm}|^{q}\,dV_{t_{1}}$
$\displaystyle\leq\int\phi_{r}^{2}|\text{Rm}|^{q}\,dV_{t_{0}}+2(t_{1}-t_{0})\frac{C}{r^{2}}\sup_{t\in[0,T]}\int|\text{Rm}|^{q}\,dV_{t}$
$\displaystyle\leq\int|\text{Rm}|^{q}\,dV_{t_{0}}+2(t_{1}-t_{0})\frac{C}{r^{2}}\sup_{t\in[0,T]}\int|\text{Rm}|^{q}\,dV_{t}.$

Taking $r\to\infty$ completes the proof. ∎

######  Proposition 4.5.

Let $(M^{n},g(t))$ satisfy the conditions in (14). Then the curvature estimate

$\sup_{x\in
M}|\text{Rm}|^{2}(x,t)\leq\max(c_{1}(g_{0})t^{-\frac{2(n-2)}{n}},c_{2}(g_{0})t^{\frac{4(n-2)}{n^{2}}})$

holds for all $t\in[0,T_{max})$. Here $c_{1}(g_{0})$ and $c_{2}(g_{0})$ are
positive constants depending only on the initial metric $g_{0}$. In
particular, they depend only on $C_{g_{0}}$ and the initial curvature
quanitity $\int|\text{Rm}|^{\frac{n}{2}\frac{n}{n-2}}\,dV_{0}$.

###### Proof.

Like the previous lemmas, we will fix some $T<T_{max}$ and prove the estimate
on $[0,T]$. For convenience of notation, let $f(x,t)=|\text{Rm}|^{2}(x,t)$ and
write $\phi$ to be some $\phi_{r}$ just as in the notation used in Lemma 4.2
(we’ll specify $r$ later). By Lemma 4.1, for any
$p\geq\frac{n}{4}\frac{n}{n-2}$ (note that this condition forces $p\geq 1$),
there holds

(22) $\begin{split}\frac{d}{dt}\int\phi^{2}f^{p}\,dV_{t}+\int|\nabla(\phi
f^{\frac{p}{2}})|^{2}\,dV_{t}&\leq-\int|\nabla(\phi
f^{\frac{p}{2}})|^{2}\,dV_{t}+(8p+n)\int\phi^{2}|\text{Rm}|f^{p}\,dV_{t}+2\int|\nabla\phi|^{2}f^{p}\,dV_{t}.\\\
&=:\textbf{I}+(8p+n)\textbf{II}+\textbf{III}.\end{split}$

We will estimate the above terms $\textbf{I},\textbf{II},\textbf{III}$
seperately.

Corollary 4.3 immediately yields

(23)
$\textbf{I}\leq-\frac{1}{A_{0}}\left(\int\phi^{\frac{2n}{n-2}}f^{p\frac{n}{n-2}}\,dV_{t}\right)^{\frac{n-2}{n}}$

where $A=2L(n)C_{g_{0}}$. To estimate II, we recall the generalized Hölder’s
inequality:

$\left\|\prod_{j=1}^{k}f_{j}\right\|_{L^{1}}\leq\prod_{j=1}^{k}\|f_{j}\|_{L^{r_{j}}}\;\text{
where }\;\sum_{j=1}^{k}\frac{1}{r_{i}}=1.$

We will apply this with

$\begin{cases}&f_{1}=|\text{Rm}|,\;\;f_{2}=\eta^{-\left(1-\frac{n}{2q}\right)\frac{n}{2q}}\phi^{2\left(1-\frac{n}{2q}\right)}f^{p\left(1-\frac{n}{2q}\right)},\;\;f_{3}=\eta^{\left(1-\frac{n}{2q}\right)\frac{n}{2q}}\phi^{\frac{n}{q}}f^{p\frac{n}{2q}},\\\
&r_{1}=q,\;\;r_{2}=\frac{1}{1-\frac{n}{2q}},\;\;r_{3}=\frac{2q}{n}\frac{n}{n-2},\end{cases}$

where $q=\frac{n}{2}\frac{n}{n-2}$ and $\eta>0$ will be chosen later. This
gives

$\textbf{II}\leq\left(\int|\text{Rm}|^{q}\,dV_{t}\right)^{\frac{1}{q}}\left(\eta^{-\frac{n}{2q}}\int\phi^{2}f^{p}\,dV_{t}\right)^{1-\frac{n}{2q}}\left(\eta^{\left(1-\frac{n}{2q}\right)\frac{n}{n-2}}\int\phi^{\frac{2n}{n-2}}f^{p\frac{n}{n-2}}\,dV_{t}\right)^{\frac{n-2}{n}\frac{n}{2q}}.$

Now we apply Lemma 4.4, which states that
$t\mapsto\|\text{Rm}\|_{L^{q}_{g(t)}}$ is nonincreasing along the flow. So
writing $\beta=\|\text{Rm}\|_{L^{q}_{g(0)}}$, we have that
$\|\text{Rm}\|_{L^{q}_{g(t)}}\leq\beta$ for all $t\geq 0$. Now to the
remaining terms we apply Young’s inequality

$ab\leq\frac{a^{r_{1}}}{r_{1}}+\frac{b^{r_{2}}}{r_{2}},\;\;\text{ where
}\frac{1}{r_{1}}+\frac{1}{r_{2}}=1,$

with $r_{1}=\frac{1}{1-\frac{n}{2q}}$, $r_{2}=\frac{2q}{n}$ to see that

II
$\displaystyle\leq\beta\left(\eta^{-\frac{n}{2q}}\int\phi^{2}f^{p}\,dV_{t}\right)^{1-\frac{n}{2q}}\left(\eta^{\left(1-\frac{n}{2q}\right)\frac{n}{n-2}}\int\phi^{\frac{2n}{n-2}}f^{\frac{p}{2}\frac{2n}{n-2}}\,dV_{t}\right)^{\frac{n-2}{n}\frac{n}{2q}}$
(24)
$\displaystyle\leq\beta\left(1-\frac{n}{2q}\right)\eta^{-\frac{n}{2q}}\int\phi^{2}f^{p}\,dV_{t}+\beta\frac{n}{2q}\eta^{1-\frac{n}{2q}}\left(\int\phi^{\frac{2n}{n-2}}f^{p\frac{n}{n-2}}\,dV_{t}\right)^{\frac{n-2}{n}}.$

Finally, to estimate III, we can use Lemma 4.2 combined with the fact that
$(M,g(t))$, $t\in[0,T]$ is a _bounded curvature_ flow to obtain

(25) $\sup_{t\in[0,T]}\int f^{p}\,dV_{t}<\infty.$

Then following just as in the proof of the previous lemmas, for any
$\epsilon>0$, there exists $r_{\epsilon}\gg 0$ such that whenever $r\geq
r_{\epsilon}$ and $t\in[0,T]$ there holds

(26) $\textbf{III}<\epsilon.$

Now puttting (23), (24) and (26) back into (22) we have

$\displaystyle\frac{d}{dt}\int\phi^{2}f^{p}\,dV_{t}$
$\displaystyle+\int|\nabla(\phi
f^{\frac{p}{2}})|^{2}\,dV_{t}\leq\beta(8p+n)\left(1-\frac{n}{2q}\right)\eta^{-\frac{n}{2q}}\int\phi^{2}f^{p}\,dV_{t}$
$\displaystyle\;\;\;+\left[\frac{\beta
n(8p+n)}{2q}\eta^{1-\frac{n}{2q}}-\frac{1}{A_{0}}\right]\left(\int\phi^{\frac{2n}{n-2}}f^{p\frac{n}{n-2}}\,dV_{t}\right)^{\frac{n-2}{n}}+\epsilon.$

Now we choose $\eta>0$ so that the bracketed quanitity above is zero.
Precisely, we take

$\eta^{1-\frac{n}{2q}}=\frac{2q}{\beta n(8p+n)A_{0}},$

so that plugging this value of $\eta$ back into the above, and simplifying
gives

$\displaystyle\frac{d}{dt}\int\phi^{2}f^{p}\,dV_{t}+\int|\nabla(\phi
f^{\frac{p}{2}})|^{2}\,dV_{t}$
$\displaystyle\leq\beta(8p+n)\left(1-\frac{n}{2q}\right)\left(\frac{2q}{\beta
n(8p+n)A_{0}}\right)^{-\frac{2q}{2q-n}\frac{n}{2q}}\int\phi^{2}f^{p}\,dV_{t}+\epsilon$
$\displaystyle=\beta(8p+n)\frac{2}{n}\left(\frac{A_{0}(n-2)\beta(8p+n)}{n}\right)^{\frac{n-2}{2}}\int\phi^{2}f^{p}\,dV_{t}+\epsilon$
(27) $\displaystyle\leq
12p\beta^{\frac{n}{2}}A_{0}^{\frac{n-2}{2}}\int\phi^{2}f^{p}\,dV_{t}+\epsilon=:K_{p}\int\phi^{2}f^{p}\,dV_{t}+\epsilon,$

where in the second line we used our value of $q=\frac{n}{2}\frac{n}{n-2}$,
and in the last line we used the estimates $8p+n\leq 12p$ and $n\geq 3$. Now
for $0<\tau<\tau^{\prime}\leq T$, define $\psi:[0,T]\to[0,1]$ by

$\psi(t)=\begin{cases}0&0\leq t\leq\tau\\\
\frac{t-\tau}{\tau^{\prime}-\tau}&\tau\leq t\leq\tau^{\prime}\\\
1&\tau^{\prime}\leq t\leq T\end{cases}.$

Now multiply (27) by $\psi$ and find that

$\frac{d}{dt}\left(\psi\int\phi^{2}f^{p}\,dV_{t}\right)+\psi\int|\nabla(\phi
f^{\frac{p}{2}})|^{2}\,dV_{t}\leq(K_{p}\psi+\psi^{\prime})\int\phi^{2}f^{p}\,dV_{t}+\epsilon,$

so that after integrating from $\tau$ to $\tilde{t}\in[\tau^{\prime},T]$, we
obtain

$\displaystyle\int\phi^{2}f^{p}\,dV_{\tilde{t}}+\int_{\tau^{\prime}}^{\tilde{t}}\int|\nabla(\phi
f^{\frac{p}{2}})|^{2}\,dV_{t}\,dt$
$\displaystyle\leq\psi(\tilde{t})\int\phi^{2}f^{p}\,dV_{\tilde{t}}-\psi(\tau)\int\phi^{2}f^{p}\,dV_{\tau}+\int_{\tau}^{\tilde{t}}\psi\int|\nabla(\phi
f^{\frac{p}{2}})|^{2}\,dV_{t}\,dt$
$\displaystyle=\int_{\tau}^{\tilde{t}}\left[\frac{d}{dt}\left(\psi\int\phi^{2}f^{p}\,dV_{t}\right)+\psi\int|\nabla(\phi
f^{\frac{p}{2}})|^{2}\,dV_{t}\right]\,dt$
$\displaystyle\leq\int_{\tau}^{\tilde{t}}(K_{p}\psi+\psi^{\prime})\int\phi^{2}f^{p}\,dV_{t}\,dt+(\tilde{t}-\tau)\epsilon$
$\displaystyle\leq\left(K_{p}+\frac{1}{\tau^{\prime}-\tau}\right)\int_{\tau}^{T}\int
f^{p}\,dV_{t}\,dt+T\epsilon.$

In particular, because both terms on the LHS are nonnegative, the above yields
two inequalities which will be used later. The first is

(28) $\int_{\tau^{\prime}}^{T}\int|\nabla(\phi
f^{\frac{p}{2}})|^{2}\,dV_{t}\,dt\leq\left(K_{p}+\frac{1}{\tau^{\prime}-\tau}\right)\int_{\tau}^{T}\int
f^{p}\,dV_{t}\,dt+T\epsilon\;\text{ for any }\epsilon>0,r\geq r_{\epsilon}.$

Likewise, considering the other term on the LHS gives

$\int\phi^{2}f^{p}\,dV_{\tilde{t}}\leq\left(K_{p}+\frac{1}{\tau^{\prime}-\tau}\right)\int_{\tau}^{T}\int
f^{p}\,dV_{t}\,dt+T\epsilon$

for any $\tilde{t}\in[\tau^{\prime},T]$. Since the integrand $\phi^{2}f^{p}$
is nonnegative however, we can take $r\to\infty$, followed by taking
$\epsilon\to 0$ which gives the second useful inequality:

(29) $\int
f^{p}\,dV_{\tilde{t}}\leq\left(K_{p}+\frac{1}{\tau^{\prime}-\tau}\right)\int_{\tau}^{T}\int
f^{p}\,dV_{t}\,dt.$

For $p\geq 1$ and $\tau\in[0,T]$, we define

$H(p,\tau)=\int_{\tau}^{T}\int f^{p}\,dV_{t}\,dt,$

and let $\nu=1+\frac{2}{n}$. We need to prove the following claim:

Claim: For $0\leq\tau<\tau^{\prime}\leq T$, there holds

(30) $H(\nu p,\tau^{\prime})\leq
A_{0}\left(K_{p}+\frac{1}{\tau^{\prime}-\tau}\right)^{\nu}H(p,\tau)^{\nu}.$

###### Proof of Claim:.

We will first show the desired inequality albeit with a $\phi^{2}$ term in the
integrand on the LHS. With this in mind, we apply Hölder’s inequality followed
by Corollary 4.3 to see that

$\displaystyle\int_{\tau^{\prime}}^{T}\int\phi^{2}f^{p\left(1+\frac{2}{n}\right)}\,dV_{t}\,dt$
$\displaystyle\leq\int_{\tau^{\prime}}^{T}\left(\int_{\text{supp}\phi}f^{p}\,dV_{t}\right)^{\frac{2}{n}}\left(\int(\phi
f^{\frac{p}{2}})^{\frac{2n}{n-2}}\,dV_{t}\right)^{\frac{n-2}{n}}\,dt$
$\displaystyle\leq A_{0}\left(\sup_{\tau^{\prime}\leq t\leq
T}\int_{\text{supp}\phi}f^{p}\,dV_{t}\right)^{\frac{2}{n}}\int_{\tau^{\prime}}^{T}\int|\nabla(\phi
f^{\frac{p}{2}})|^{2}\,dV_{t}\,dt.$

Applying (28), and noting that the supremum above is achieved at some
$\tilde{t}\in[\tau^{\prime},T]$ (of course, the time $\tilde{t}$ possibly
depends on $\phi$), we can continue our estimate above to

$\displaystyle\int_{\tau^{\prime}}^{T}\int\phi^{2}f^{p\left(1+\frac{2}{n}\right)}\,dV_{t}\,dt$
$\displaystyle\leq A_{0}\left(\int
f^{p}\,dV_{\tilde{t}}\right)^{\frac{2}{n}}\left[\left(K_{p}+\frac{1}{\tau^{\prime}-\tau}\right)\int_{\tau}^{T}\int
f^{p}\,dV_{t}\,dt+T\epsilon\right]$ $\displaystyle\leq A_{0}\left(\int
f^{p}\,dV_{\tilde{t}}\right)^{\frac{2}{n}}\left(K_{p}+\frac{1}{\tau^{\prime}-\tau}\right)H(p,\tau)+A_{0}T\epsilon\left(\int
f^{p}\,dV_{\tilde{t}}\right)^{\frac{2}{n}}$ $\displaystyle\leq
A_{0}\left(K_{p}+\frac{1}{\tau^{\prime}-\tau}\right)^{\nu}H(p,\tau)^{\nu}+A_{0}T\epsilon\left(\sup_{t\in[0,T]}\int
f^{p}\,dV_{t}\right)^{\frac{2}{n}}.$

Now the RHS is independent of $r\geq r_{\epsilon}$, so we can take
$r\to\infty$, apply the Monotone Convergence Theorem, then take $\epsilon\to
0$. Note that the quantity $\sup_{t\in[0,T]}\int f^{p}\,dV_{t}$ is finite by
(25) above. This proves the claim. ∎

At this point, the remainder of the proof is verbatim to [9, Proposition 4.5].
We will however provide an outline to compute the constants involved and for
our future reference. Let $p_{0}=\frac{q}{2}=\frac{n}{4}\frac{n}{n-2}$, recall
$\nu=1+\frac{2}{n}$ and define

$\eta=\nu^{\frac{n}{2}},\;\;p_{k}=\nu^{k}p_{0},\;\;\tau_{k}=(1-\eta^{-k})T,\;\;\Phi_{k}=H(p_{k},\tau_{k})^{\frac{1}{p_{k}}}.$

So applying (30), we get

$\Phi_{k+1}\leq A_{0}^{\frac{1}{\nu
p_{k}}}\left(\beta^{\frac{n}{2}}A_{0}^{\frac{n-2}{2}}(12p_{0})^{\frac{n}{2}}+\frac{1}{(1-\eta^{-1})T}\right)^{\frac{1}{p_{k}}}\eta^{\frac{k}{p_{k}}}\Phi_{k}$

and iterating the above gives

(31) $\lim_{k\to\infty}\Phi_{k}\leq
A_{0}^{\frac{1}{\nu}\sum_{k=0}^{\infty}\frac{1}{p_{k}}}\left(\beta^{\frac{n}{2}}A_{0}^{\frac{n-2}{2}}(12p_{0})^{\frac{n}{2}}+\frac{1}{(1-\eta^{-1})T}\right)^{\sum_{k=0}^{\infty}\frac{1}{p_{k}}}\eta^{\sum_{k=0}^{\infty}\frac{k}{p_{k}}}\Phi_{0}.$

We can also calculate

$\sum_{k=0}^{\infty}\frac{1}{p_{k}}=\frac{2(n^{2}-4)}{n^{2}},\;\;\frac{1}{\nu}\sum_{k=0}^{\infty}\frac{1}{p_{k}}=\frac{2(n-2)}{n},\;\;\sum_{k=0}^{\infty}\frac{k}{p_{k}}=\frac{n^{2}-4}{n}.$

Finally,

$\eta^{\frac{n^{2}-4}{n}}=\nu^{\frac{n}{2}\frac{n^{2}-4}{n}}=\left(1+\frac{2}{n}\right)^{\frac{n^{2}-4}{2}}\leq
e^{n-1}$

because the function $\left(1+\frac{2}{x}\right)^{\frac{x^{2}-4}{2}}e^{1-x}$
is increasing on $[0,\infty)$ and bounded by 1. Plugging these identities and
$p_{0}=\frac{n}{4}\frac{n}{n-2}\leq n$ back into (31) gives

(32) $\lim_{k\to\infty}\Phi_{k}\leq
e^{n-1}A_{0}^{\frac{2(n-2)}{n}}\left(\beta^{\frac{n}{2}}A_{0}^{\frac{n-2}{2}}(12n)^{\frac{n}{2}}+\frac{1}{(1-\eta^{-1})T}\right)^{\frac{2(n^{2}-4)}{n^{2}}}\Phi_{0}$

Just as in [9], there holds

$\Phi_{0}\leq T^{\frac{4(n-2)}{n^{2}}}\beta^{2}\;\text{ and
}\;\lim_{k\to\infty}\Phi_{k}\geq\|f\|_{L^{\infty}_{g(T)}}.$

Using this in (32) gives

$\displaystyle\sup_{x\in M}|\text{Rm}|^{2}_{g(T)}$
$\displaystyle=\|f\|_{L^{\infty}_{g(T)}}\leq
e^{n-1}A_{0}^{\frac{2(n-2)}{n}}\left(\beta^{\frac{n}{2}}A_{0}^{\frac{n-2}{2}}(12n)^{\frac{n}{2}}+\frac{1}{(1-\eta^{-1})T}\right)^{\frac{2(n^{2}-4)}{n^{2}}}T^{\frac{4(n-2)}{n^{2}}}\beta^{2}$
$\displaystyle\leq
2^{\frac{2(n^{2}-4)}{n^{2}}}e^{n-1}A_{0}^{\frac{2(n-2)}{n}}\beta^{2}\max\left\\{\left(\beta^{\frac{n}{2}}A_{0}^{\frac{n-2}{2}}(12n)^{\frac{n}{2}}\right)^{\frac{2(n^{2}-4)}{n^{2}}},\left(\frac{1}{(1-\eta^{-1})T}\right)^{\frac{2(n^{2}-4)}{n^{2}}}\right\\}T^{\frac{4(n-2)}{n^{2}}}.$

Because $T>0$ was arbitrary, this completes the proof.

∎

######  Corollary 4.6.

Let $(M^{n},g(t))$ satisfy the conditions in (14). Then $T_{max}=\infty$ and

$\sup_{M\times[0,\infty)}|\text{Rm}|(x,t)<\infty.$

###### Proof.

The proof is identical to [9, Corollary 4.6]. ∎

######  Lemma 4.7.

Let $(M^{n},g(t))$ satisfy the conditions in (14). Then for any $\gamma>0$,
there exists $T_{\gamma}>0$ such that for all $t\geq T_{\gamma}$, there holds

$\left(\int|\text{Rm}|^{\frac{n}{2}\frac{n}{n-2}}dV_{t}\right)^{\frac{n-2}{n}}\leq\gamma
t^{-1}.$

###### Proof.

Start by fixing $T>0$. Recall that in Lemma 4.2, we proved that for any
$\epsilon>0$, there exists $r_{\epsilon}\gg 0$ such that whenever $r\geq
r_{\epsilon}$, there holds

$\int\phi_{r}^{2}|\text{Rm}|^{\frac{n}{2}}\,dV_{T}-\int\phi_{r}^{2}|\text{Rm}|^{\frac{n}{2}}\,dV_{0}\leq-\frac{1}{10A}\int_{0}^{T}\left(\int\phi_{r}^{\frac{2n}{n-2}}|\text{Rm}|^{\frac{n}{4}\frac{2n}{n-2}}\,dV_{t}\right)^{\frac{n-2}{n}}\,dV_{t}+T\epsilon.$

Rearranging this in turn gives

$\displaystyle\int_{0}^{T}\left(\int\phi_{r}^{\frac{2n}{n-2}}|\text{Rm}|^{\frac{n}{2}\frac{n}{n-2}}dV_{t}\right)^{\frac{n-2}{n}}\,dt$
$\displaystyle\leq
10A\left[\int\phi_{r}^{2}|\text{Rm}|^{\frac{n}{2}}\,dV_{0}-\int\phi_{r}^{2}|\text{Rm}|^{\frac{n}{2}}\,dV_{T}\right]+10TA\epsilon$
(33) $\displaystyle\leq
10A\int\phi_{r}^{2}|\text{Rm}|^{\frac{n}{2}}\,dV_{0}+10TA\epsilon\leq
10A\int|\text{Rm}|^{\frac{n}{2}}\,dV_{0}+10TA\epsilon.$

Now take $r\to\infty$, then $\epsilon\to 0$, then $T\to\infty$ to see that

$\int_{0}^{\infty}\left(\int_{M}|\text{Rm}|^{\frac{n}{2}\frac{n}{n-2}}dV_{t}\right)^{\frac{n-2}{n}}\,dt\leq
10A\int|\text{Rm}|^{\frac{n}{2}}\,dV_{0}<\infty.$

This combined with the fact that
$t\mapsto\int|\text{Rm}|^{\frac{n}{2}\frac{n}{n-2}}\,dt$ is nonincreasing
(Lemma 4.4) gives

$\lim_{t\to\infty}t\left(\int|\text{Rm}|^{\frac{n}{2}\frac{n}{n-2}}\,dV_{t}\right)^{\frac{n-2}{n}}=0.$

∎

######  Proposition 4.8.

Let $(M^{n},g(t))$ satisfy the conditions in (14). Then the curvature estimate
b holds.

###### Proof.

The proof is the same as [9, Proposition 4.8]. ∎

At this point, we diverge from the numbering from [9] to prove the other
curvature estimate and the convergence statement.

######  Proposition 4.9.

Let $(M^{n},g(t))$ satisfy the conditions in (14). Then the curvature estimate
a holds.

###### Proof.

By slightly modifying the proof of Proposition 4.5 (see also [9, Proposition
4.8]) with the only changes being

$\tau_{k}=\frac{T}{2}+(1-\eta^{-k})\frac{T}{2},\;\;\text{ and
}\;\;\beta=\sup_{t\in[T/2,T]}\||\text{Rm}|\|_{L^{q}_{g(t)}},$

we obtain the estimate

(34) $\sup_{x\in M}|\text{Rm}|_{g(t)}\leq
2^{\frac{n^{2}-4}{n^{2}}}e^{\frac{1}{2}(n-1)}A_{0}^{\frac{n-2}{n}}\beta\max\left\\{\left(\beta^{\frac{n}{2}}A_{0}^{\frac{n-2}{2}}(12n)^{\frac{n}{2}}\right)^{\frac{n^{2}-4}{n^{2}}},\left(\frac{2}{(1-\eta^{-1})t}\right)^{\frac{n^{2}-4}{n^{2}}}\right\\}t^{\frac{2(n-2)}{n^{2}}}.$

Note that Lemma 4.4 implies that $t\mapsto\int|\text{Rm}|^{q}\,dV_{t}$ is
nonincreasing, so we may take $\beta=\||\text{Rm}|\|_{L^{q}_{g(T/2)}}$. Now by
slightly modifying the proof of Lemma 4.7 (specificially taking $r\to\infty$
then $\epsilon\to 0$ in inequality (33)), we have

$\frac{t}{2}\beta^{\frac{n}{2}}\leq\int_{0}^{t/2}\left(\int|\text{Rm}|^{\frac{n}{2}\frac{n}{n-2}}\,dV_{s}\right)^{\frac{n-2}{n}}\,ds\leq
10A\int|\text{Rm}|^{\frac{n}{2}}\,dV_{0}\leq
10L(n)C_{g}\left(\frac{\delta_{1}(n)}{C_{g}}\right)^{\frac{n}{2}}.$

Then since $A_{0}=2L(n)C_{g}$, we can see that

$A_{0}^{\frac{n-2}{n}}\beta\leq(2LC_{g})^{\frac{n-2}{n}}(10LC_{g})^{\frac{2}{n}}\frac{2^{\frac{2}{n}}\delta_{1}(n)}{C_{g}t^{\frac{2}{n}}}\leq\frac{1}{4nt^{\frac{2}{n}}}.$

Plugging this back into our curvature bound (34) we have

$\displaystyle\sup_{x\in M}|\text{Rm}|_{g(t)}$ $\displaystyle\leq
2^{\frac{n^{2}-4}{n^{2}}}e^{\frac{1}{2}(n-1)}\frac{1}{4nt^{\frac{2}{n}}}\max\left\\{\left(\frac{(12n)^{\frac{n}{2}}}{(4n)^{\frac{n}{2}}t}\right)^{\frac{n^{2}-4}{n^{2}}},\left(\frac{2}{(1-\eta^{-1})t}\right)^{\frac{n^{2}-4}{n^{2}}}\right\\}t^{\frac{2(n-2)}{n^{2}}}$
$\displaystyle<(3e)^{\frac{n}{2}}t^{-\frac{2}{n}-1+\frac{4}{n^{2}}+\frac{2}{n}-\frac{4}{n^{2}}}=\frac{(3e)^{\frac{n}{2}}}{t}=\frac{C_{0}(n)}{t},$

where in the latter inequality we used the fact that

$\frac{2}{(1-\eta^{-1})}\leq 4\leq 3^{\frac{n}{2}}.$

∎

######  Proposition 4.10.

Let $(M^{n},g(t))$ satisfy the conditions in (14) and furthermore assume that
$|\text{Rm}|_{g}\in L^{p}$ for some $p<\frac{n}{2}$. Then $g(t)$ converges
locally smoothly to a complete flat limit metric $g(\infty)$ on $M^{n}$ with
Euclidean volume growth, and thus $M^{n}$ is diffeomorphic to
$\mathbb{R}^{n}$.

###### Proof.

In this proof, we will use $C(\cdot,\cdot,\cdots)$ to denote a constant which
depends on its arguments, and may differ line-by-line. First note that by
interpolation, we may assume that

(35) $\begin{cases}\frac{2}{3}<\frac{p}{2}<\frac{n}{4}&n=3,4\\\
\frac{n-2}{4}<\frac{p}{2}<\frac{n}{4}&n\geq 5.\end{cases}$

Given this (fixed) value of $p$, write $q=\frac{n}{n-2}p$. Now we note that in
establishing (27) in the proof of Proposition 4.5, the only assumptions on
$\tilde{p},\tilde{q}$ there were $\tilde{p}\geq 1$ and
$\max\\{2,n/2\\}\leq\tilde{q}\leq\frac{n}{2}\frac{n}{n-2}$ so that Lemma 4.4
in turn guaranteed that $t\mapsto\||\text{Rm}|\|_{L^{\tilde{q}}_{g(t)}}$ is
nonincreasing. Thus for any $\tilde{p}\geq 1$ and our choice of $q$ above, we
have

$\displaystyle\frac{d}{dt}\int\phi^{2}f^{\tilde{p}}\,dV_{t}+\int|\nabla(\phi
f^{\frac{\tilde{p}}{2}})|^{2}\,dV_{t}$
$\displaystyle\leq\beta(8\tilde{p}+n)\left(1-\frac{n}{2q}\right)\left(\frac{\beta
n(8\tilde{p}+n)A_{0}}{2q}\right)^{\frac{n}{2q-n}}\int\phi^{2}f^{\tilde{p}}\,dV_{t}+\epsilon$
$\displaystyle=(\tilde{p}\beta)^{\frac{2q}{2q-n}}C(n,q,C_{g})\int\phi^{2}f^{\tilde{p}}\,dV_{t}+\epsilon.$

Now we continue on in identical fashion to the proof of Proposition 4.5 (in
going from (27) to (31)) with the only modifications being that we take

$\tilde{p}_{0}=\frac{q}{2},\;\;\eta=\nu^{\frac{2q}{2q-n}},\;\;\tau_{k}=\frac{T}{2}+(1-\eta^{-k})\frac{T}{2},\;\;\text{
and }\;\;\beta=\sup_{t\in[T/2,T]}\||\text{Rm}|\|_{L^{q}_{g(t)}}.$

In light of our choice of $p,q$ and Lemma 4.4, we in fact have
$\beta=\||\text{Rm}|\|_{L^{q}_{g(T/2)}}$. Also, by our requirements on $p$, we
have that $\tilde{p}_{0}>1$. The result after iterating is

$\sup_{x\in M}|\text{Rm}|(x,T)^{2}\leq
C(n,p,C_{g})\left(\beta^{\frac{2q}{2q-n}}+T^{-1}\right)^{\sum_{k=0}^{\infty}\frac{1}{\tilde{p}_{k}}}\Phi_{0}.$

Also, in this case we have

$\sum_{k=0}^{\infty}\frac{1}{\tilde{p}_{k}}=\frac{1}{\tilde{p}_{0}}\sum_{k=0}^{\infty}\frac{1}{(1+\frac{2}{n})^{k}}=\frac{n+2}{q}.$

Now recalling the definition of $\Phi_{0}$ and the fact that
$\int_{M}|\text{Rm}|^{q}\,dV_{t}$ is nonincreasing in $t$, we have

$\Phi_{0}=H(q/2,\tau_{0})^{\frac{2}{q}}=\left(\int_{T/2}^{T}\int_{M}|\text{Rm}|^{q}\,dV_{t}\,dt\right)^{\frac{2}{q}}=C(q)T^{\frac{2}{q}}\||\text{Rm}|\|_{L^{q}_{g(T/2)}}^{2}.$

We need to control the value of $\||\text{Rm}|\|_{L^{q}_{g(T/2)}}$. For this,
we proceed just as in Lemma 4.2 and 4.4 with $\alpha=\frac{p}{2}$ (here we use
the assumption that $\frac{p}{2}>\frac{2}{3}$ so that we can apply part b of
Lemma 4.1) to obtain

$\displaystyle-\int|\text{Rm}|^{p}\,dV_{0}$
$\displaystyle\leq\int_{0}^{t}\left(\int_{M}|\text{Rm}|^{q}dV_{s}\right)^{\frac{n-2}{n}}\,ds\left[-\frac{1}{4A}+\left(16\alpha+n\right)\frac{\delta_{1}(n)}{C_{g}}\right]$
$\displaystyle\leq\int_{0}^{t}\left(\int_{M}|\text{Rm}|^{q}dV_{s}\right)^{\frac{n-2}{n}}\,ds\left[-\frac{1}{4A}+\frac{5n}{39nA}\right]$
$\displaystyle\leq-\frac{1}{10A}\int_{0}^{t}\left(\int_{M}|\text{Rm}|^{q}dV_{s}\right)^{\frac{n-2}{n}}\,ds\leq-\frac{t}{10A}\||\text{Rm}|\|_{L^{q}_{g(t)}}^{q\frac{n-2}{n}}.$

Of course, this argument relies on the fact that
$t\mapsto\int|\text{Rm}|^{p}_{g(t)}\,dV_{t}$ is uniformly bounded on finite
time intervals which is true by Proposition 2.2 because our assumption (35) in
particular implies $p>1$. So rearranging the above gives

$\||\text{Rm}|\|_{L^{q}_{g(t)}}^{q\frac{n-2}{n}}\leq
C(n,p,C_{g},\||\text{Rm}|\|_{L^{p}_{g_{0}}})\frac{1}{t}$

which implies

$\Phi_{0}\leq
C(n,p,C_{g},\||\text{Rm}|\|_{L^{p}_{g_{0}}})T^{\left(-\frac{n}{n-2}+1\right)\frac{2}{q}}=C(n,p,C_{g},\||\text{Rm}|\|_{L^{p}_{g_{0}}})T^{-\frac{4}{q(n-2)}},$

and

$\beta^{\frac{2q}{2q-n}}\leq
C(n,p,C_{g},\||\text{Rm}|\|_{L^{p}_{g_{0}}})T^{-\frac{2n}{(n-2)(2q-n)}}.$

Note that

$\frac{2n}{(n-2)(2q-n)}\geq 1\;\iff\;q\leq\frac{n^{2}}{2(n-2)}$

and the latter is true by our choice of $q$. Therefore for $T$ large, we have

(36) $\sup_{x\in M}|\text{Rm}|^{2}(x,T)\leq
C(n,p,C_{g},\||\text{Rm}|\|_{L^{p}_{g_{0}}})T^{-\frac{n^{2}}{q(n-2)}}$

But $q<\frac{n}{2}\frac{n}{n-2}$ and so we have that $\sup_{x\in
M}|\text{Rm}|_{g(t)}\lesssim t^{-(1+\sigma)}$ for some $\sigma>0$. Then by
Shi’s estimates [21] (see also [16, Theorem 4.7]), we have that for all $k\geq
0$, there holds

(37) $\sup_{x\in M}|\nabla^{k}\text{Rm}|_{g(t)}\leq C_{k}t^{-1-\sigma-k/2}$

for all $t\in(0,\infty)$ and some
$C_{k}(n,p,C_{g},\||\text{Rm}|\|_{L^{p}_{g_{0}}})>0$.

We now show that the integrability in $t$ of the RHS of (37) implies that
$g(t)$ converges in $C^{\infty}_{loc}(M)$ as $t\to\infty$ to a complete flat
metric $g(\infty)$ on $M$ which is equivalent to $g(0)$. This is by a standard
Ricci flow argument which we now sketch. In the proceeding paragraphs, the
constants $C,C_{i},c_{i}$ can be taken to depend only on
$n,p,C_{g},\||\text{Rm}|\|_{L^{p}_{g_{0}}}$.

Let $V\in T_{p}M$ denote an arbitrary unit vector on $M$ with respect to the
metric $g(1)$. Then for any $t\geq 1$ we have

$\log\|V\|_{g(t)}^{2}=\log\|V\|_{g(t)}^{2}-\log\|V\|_{g(1)}^{2}=\int_{1}^{t}\frac{d}{dt}\log\|V\|_{g(t)}^{2}\,dt=2\int_{1}^{t}\frac{\text{Rc}_{g(t)}(V,V)}{\|V\|^{2}_{g(t)}}\,dt$

while the integrand on the RHS is bounded in absolute value by
$n\sup_{M}|Rm(x,t)|$ and thus $C/t^{1+\sigma}$ for some $\sigma>0$ by (37)
with $k=0$. It follows that $\log\|V\|_{g(t)}^{2}$ converges - uniformly in
space - as $t\to\infty$ to a bounded function on $M$. From this we conclude
that $g(t)$ converges in $L^{\infty}(M)$ as $t\to\infty$ to a metric
$g(\infty)$ which is uniformly equivalent to $g(0)$, and thus also complete.

We show that in fact $g(t)\to g(\infty)$ in $C^{\infty}_{loc}(M)$ and thus
$g(\infty)$ is a smooth flat metric on $M$ equivalent to $g(0)$. Fix a local
coordinate domain $U\subset\subset M$ with coordinates $x^{i}$, and use
$\tilde{T}$ to denote the local components in $U$ of a global tensor $T$ on
$M$. Recall that if $\tilde{\Gamma}(t)$ are the Christoffel symbols of
$\tilde{g}(t)$, then $\frac{\partial}{\partial t}\tilde{\Gamma}(t)$ are the
components of a global tensor $g^{-1}*\nabla Rc$ on $M$ for each $t$.
Combining the uniform equivalence of $g(t)$ shown above with (37) for $k=1$ we
obtain

(38) $\frac{\partial}{\partial t}\tilde{\Gamma}(t)\leq C_{1}/t^{1+c_{1}}.$

Thus $\tilde{\Gamma}(t)$ remains uniformly bounded in $U$ for all $t$. Now if
we let $\partial$ denote partial derivatives in the coordinates $x^{i}$, then
in $U$ we have

(39) $\frac{\partial}{\partial
t}\partial\tilde{g}=-2\partial\tilde{Rc}=-2(\widetilde{\nabla
Rc}+\tilde{\Gamma}*\tilde{Rc})\leq C_{2}/t^{1+c_{2}}$

by the boundedness of $\tilde{\Gamma}(t)$ and again the uniform equivalence of
$g(t)$ shown above and (37) for $k=0,1$. From this we conclude that
$\partial\tilde{g}(t)$ converges to a bounded limit on $U$ as $t\to\infty$.

By applying a partial derivative $\partial$ to (39), we may express
$\frac{\partial^{2}}{\partial t^{2}}\partial\tilde{g}$ in terms of
$\widetilde{\nabla^{m}Rc}$ for $m=0,1,2$ and $\frac{\partial^{l}}{\partial
t^{l}}\tilde{\Gamma}(t)$ for $l=0,1$ and we can similarly show these terms are
bounded by $C_{3}/t^{1+c_{3}}$ from which we conclude that
$\partial^{2}\tilde{g}(t)$ converges to a bounded limit in $U$ as
$t\to\infty$. Continuing inductively in this way, we can show that
$\partial^{m}\tilde{g}(t)$ converges to a bounded limit in $U$ as $t\to\infty$
for each $m=3,4,...$. As $U$ was an arbitrary coordinate chart, this
establishes that $g(t)$ converges in $C^{\infty}_{loc}(M)$ as $t\to\infty$ to
a smooth limit $g(\infty)$ which is equivalent to $g(0)$ (and thus is
necessarily complete) and must also be flat by (37).

We have so far shown that $(M,g(\infty))$ is a complete flat manifold. In
particular, its universal cover is $\mathbb{R}^{n}$ with the Euclidean metric
$g_{E}$ and therefore

$(M,g(\infty))\cong(\mathbb{R}^{n},g_{E})/\Gamma$

where $\Gamma$ is a discrete group of isometries which acts freely on
$\mathbb{R}^{n}$. On the other hand, we know that $g(\infty)$ satisfies the
Sobolev inequality

$\left(\int_{M}|u|^{\frac{2n}{n-2}}\,dV_{g(\infty)}\right)^{\frac{n-2}{n}}\leq
A_{0}\int_{M}|\nabla u|^{2}\,dV_{g(\infty)}$

for all $u\in W^{1,2}_{0}$ (this follows because the above inequality holds
for all $t>0$ and $g(t)\to g(\infty)$ uniformly in $L^{\infty}(M)$). So for
any fixed domain $\Omega\subset M$ and any $u\in W^{1,2}_{0}(\Omega)$, we
apply Hölder’s inequality to see

$\int_{\Omega}|u|^{2}\leq\text{Vol}(\Omega)^{\frac{2}{n}}\left(\int_{\Omega}|u|^{\frac{2n}{n-2}}\,dV\right)^{\frac{n-2}{n}}\leq
A_{0}\text{Vol}(\Omega)^{\frac{2}{n}}\int_{\Omega}|\nabla u|^{2}$

and subsequently

$\text{Vol}(\Omega)^{\frac{2}{n}}\inf_{u\in
W^{1,2}_{0}(\Omega)}\frac{\int_{\Omega}|\nabla
u|^{2}}{\int_{\Omega}u^{2}}\geq\frac{1}{A_{0}}.$

Now by [1, Proposition 2.4] (or for a proof in English, see [24, Lemma 6.1]),
we have

$\text{Vol}_{g(\infty)}(B_{g(\infty)}(x_{0},\rho))\geq\left(\frac{1}{2^{n+2}A_{0}}\right)^{\frac{n}{2}}\rho^{n}$

for any $\rho>0$. That is to say, $(M,g(\infty))$ has a positive asymptotic
volume ratio which implies that the group of isometries $\Gamma$ is trivial.
This completes the proof. ∎

This completes the proof of Theorem 1.2.

##  5\. Proof of Theorem 1.5

In this section we will prove Theorem 1.5, which follows from Theorem 1.2 and
a diagonal argument using Hamilton’s Compactness Theorem for Ricci flows.

###### Proof of Theorem 1.5.

Let $(M_{i},g_{i})$ and $(M,g)$ be as in the Theorem. Thus there is an
exhaustion $\\{U_{i}\\}_{i=1}^{\infty}$ of $M$ by nested open sets, and maps
$\phi_{i}:U_{i}\to M_{i}$ which are diffeomorphisms onto their images, such
that we have the convergence

$\phi_{i}^{*}(g_{i})\xrightarrow[C^{\infty}_{loc}(M)]{}g.$

The hypothesis that

$\sup_{i\in\mathbb{N}}\left(\int_{M}|\text{Rm}|_{g_{i}}^{\frac{n}{2}}\,dV_{g_{i}}\right)^{\frac{2}{n}}\leq\frac{\delta(n)}{\sup_{i\in\mathbb{N}}C_{g_{i}}}$

implies either that (a) $|\text{Rm}|_{g_{i}}\equiv 0$ for all $i\in\mathbb{N}$
or else (b) that $\sup_{i\in\mathbb{N}}C_{g_{i}}<\infty$. In the former case,
we must necessarily have that $|\text{Rm}|_{g}\equiv 0$ since $g_{i}\to g$
locally smoothly and the result follows. For the latter case, we know by
Theorem 1.2 that for each $i\in\mathbb{N}$, there exists a long-time complete
Ricci flow $g_{i}(t)$ with $g_{i}(0)=g_{i}$. Moreover, the $\\{g_{i}(t)\\}$
uniformly satisfy the curvature estimate

(40) $\sup_{x\in M}|\text{Rm}|_{g_{i}(t)}\leq\frac{C_{0}(n)}{t}$

for all $t\in(0,\infty)$. Now to apply Hamilton’s Compactness to obtain a
limit solution on $M$ from the sequence $\\{g_{i}(t)\\}$, we need to obtain a
positive lower bound on the injectivity radius
$\text{inj}(g_{i}(t_{0}),x_{i})\geq C$ where $x_{i}=\phi_{i}(x)$ for some
fixed point $x\in M$ and $C,t_{0}>0$ and are independent of $i$. By (40),
there exists $r_{0}>0$ small enough so that

(41) $\sup_{i\in\mathbb{N}}\sup_{x\in
B_{g_{i}(1)}(x_{i},r_{0})}|\text{Rm}|_{g_{i}(1)}(x_{i})\leq r_{0}^{-2}.$

Now by Lemma 3.1 part (2), each $g_{i}(0)$ satisfies a log-Sobolev inequality
- equivalently a bound on the $\mu$-functional - with the constants on the RHS
depending only on a bound of $C_{g_{i}}$, which can be taken independently of
$i\in\mathbb{N}$ since we are assuming
$\sup_{i\in\mathbb{N}}C_{g_{i}}<\infty$. Now we can apply monotonicity of the
$\mu$-functional Lemma 3.2 for each $\tau\in(1,1+r_{0}^{2}]$ to conclude that

$\inf_{\tilde{\tau}\in(0,r_{0}^{2}]}\mu(B_{g_{i}(1)}(x_{i},r_{0}),g_{i}(1),\tilde{\tau})\geq\inf_{\tilde{\tau}\in(0,r_{0}^{2}]}\mu(M_{i},g_{i}(1),\tilde{\tau})\geq-C\left(\sup_{i\in\mathbb{N}}C_{g_{i}}\right)>-\infty.$

Now our curvature bound (41) and [23, Theorem 3.3] provides a uniform lower
bound of the volume ratio at $x_{i}$

(42) $\text{Vol}B_{g_{i}(1)}(x_{i},r_{0})\geq v_{0}r_{0}^{n}$

for some $v_{0}>0$ depending on $\sup_{i\in\mathbb{N}}C_{g_{i}}$ and $r_{0}$
depending on $C_{0}$. Then by (41), (42) and results of Cheeger-Gromov-Taylor
[6], there holds the uniform estimate

$\text{inj}_{g_{i}(1)}(x_{i})\geq c(n,r_{0},v_{0})>0$

which is independent from $i\in\mathbb{N}$, as was desired.

Now armed with the above, we can apply Hamilton’s Compactness Theorem [14],
coupled with a diagonal argument to obtain the convergence

$\phi_{i_{k}}^{*}(g_{i_{k}}(t))\xrightarrow[C^{\infty}_{loc}(M\times(0,\infty)]{}g(t)$

for some subsequence $\\{i_{k}\\}$, where $g(t)$ satisfies - by virtue of the
smooth convergence - the same curvature estimate (40).

To obtain the convergence of $g(t)$ to $g$ as $t\to 0$ we use (40) to apply
[7, Corollary 3.2] which states that

$\sup_{B_{g_{i}(0)}\left(x_{i},\frac{r_{x,i}}{2}\right)}|\text{Rm}|_{g_{i}(t)}\leq
Cr_{x,i}^{-2}.$

Here $C=C(n)>0$ is a dimensional constant and $r_{x,i}>0$ is such that

$\sup_{B_{g_{i}(0)}(x_{i},r_{x,i})}|\text{Rm}|_{g_{i}(0)}\leq r_{x,i}^{-2}.$

Moreover for any fixed point $x\in M$, the infimum
$r_{x}:=\inf_{i\in\mathbb{N}}r_{x,i}$ is strictly positive because of the
local smooth convergence $g_{i}(0)\to g(0)$. Therefore we have the locally
uniform curvature estimate

$\sup\left\\{|\text{Rm}|_{g_{i}(t)}(y):i\in\mathbb{N},\;t\in[0,1],\;y\in
B_{g_{i}(0)}\left(x_{i},\frac{r_{x}}{2}\right)\right\\}\leq C(n)r_{x}^{-2}.$

So because the flows $\\{g_{i}(t)\\}$ converge for all positive times to
$g(t)$ the convergence must occur at $t=0$ as well. In other words, we have
the smooth local convergence up to $t=0$ as

$\phi_{i_{k}}^{*}(g_{i_{k}}(t))\xrightarrow[C^{\infty}_{loc}(M\times[0,\infty)]{}g(t).$

The completeness of the limiting flow $g(t)$ follows from the Shrinking Balls
Lemma [22, Corollary 3.3] because the $C_{0}/t$ curvature bound (40) descends
to the limit flow $g(t)$, and $g(0)=g$ is assumed to be complete.

The convergence statement follows from the proof of Proposition 4.10 because
the estimates obtained there only depend on $n,p,C_{g_{i}}$ and
$\||\text{Rm}|\|_{L^{p}_{g_{i}}}$ which are all assumed to be uniform with
respect to $i\in\mathbb{N}$. Thus the estimates (36) and (37) pass down to the
limit $g(t)$. This gives the uniform convergence of $g(t)$ to a flat metric
$g(\infty)$ on $M$. The completeness of the limiting metric, and subsequently
the topological statement about $M$, are then identical to the proof of
Proposition 4.10. ∎

##  Appendix A Obtaining a bound for $L(n)$

Here we will be concerned with calculating the constants $A$ and $L(n)$ which
appear in Theorem 3.4. In the proof of that theorem, we establish that for any
$t\in[0,T]$ and any $\sigma>0$, there holds the following log-Sobolev
inequality:

(43) $\int u^{2}\log u^{2}\,dV_{t}\leq\sigma\left(\int|\nabla
u|^{2}+\frac{R^{+}_{t}}{4}\,dV_{t}\right)-\frac{n}{2}\log\sigma+\frac{n}{2}\left(\log
C_{g}+\log n-1\right).$

As described in Section 3, we can now fix a time $t\in[0,T]$ in this log-
Sobolev inequality and use regularity properties of the (fixed metric) heat
operator $H=-\Delta_{g(t)}+R_{g(t)}^{+}/4$ to obtain a Sobolev inequality at
time $t$. We will track through the proof presented by Ye [24, Appendix C] to
calculate the constants in this process. Although Ye restricts to compact
manifolds, the proof is verbatim (with the same constants) for bounded domains
$\Omega\subset M$ and Dirichlet boundary conditions (see [10, Appendix B]).

For fixed $t\in[0,T]$, we use (43) as the hypothesis for [24, Theorem 5.3]
with

$\beta(\sigma)=-\frac{n}{2}\log\sigma+\frac{n}{2}\left(\log C_{g}+\log
n-1\right).$

The output of this theorem is that

$\|e^{-sH}u\|_{L^{\infty}}\leq e^{\tau(s)}\|u\|_{L^{2}}$

where

$\displaystyle\tau(s)=\frac{1}{2s}\int_{0}^{s}\beta(\sigma)\,d\sigma$
$\displaystyle=-\frac{n}{2}\frac{1}{2s}(s\log s-s)+\frac{n}{4}\left(\log
C_{g}+\log n-1\right)$ $\displaystyle=-\frac{n}{4}\log s+\frac{n}{4}\left(\log
C_{g}+\log n\right)=-\frac{n}{4}\log s+\frac{n}{4}\log(nC_{g}).$

Then applying [24, Theorem 5.4] with $c=(nC_{g})^{\frac{n}{4}}$, $\mu=n$, and
tracking through the constants in the proof, we obtain

$\|u\|_{L_{g(t)}^{\frac{2n}{n-2}}}^{2}\leq
K^{2}C_{p_{0}}C_{p_{1}}\left(\int|\nabla
u|^{2}+\frac{R^{+}_{t}}{4}u^{2}\,dV_{t}\right).$

So we can take $A=K^{2}C_{p_{0}}C_{p_{1}}$ where $K$ and $C_{p_{i}}$ are as
follows. Here $K$ is the constant in the Marcinkiewicz Interpolation Theorem,
which we will obtain a bound for later (see Theorem A.1). We will take
$p_{0}\in(1,2)$ to be variable (to be optimized later), with $p_{1}$, $q_{0}$,
$q_{1}$ defined by

$\frac{1}{p_{0}}+\frac{1}{p_{1}}=1,\;\;\text{ and
}\;\;q_{i}=\frac{np_{i}}{n-p_{i}}.$

The constants $C_{p_{i}}$ are then given generally as

$C_{p}=\left(2^{\frac{p(2n-p)}{n-p}}c_{p}^{\frac{p^{2}}{n-p}}\Gamma\left(\frac{1}{2}\right)^{-p}\right)^{\frac{1}{q}},$

where

$c_{p}=\Gamma\left(\frac{1}{2}\right)^{-1}\left(2^{\frac{n}{2}}c^{2}\right)^{\frac{1}{p}}\frac{2p}{n-p}.$

Calculating each of the terms involved in $C_{p}$, we have

$\left(2^{\frac{p(2n-p)}{n-p}}2^{\frac{n}{2p}\frac{p^{2}}{n-p}}\right)^{\frac{n-p}{np}}=2^{\frac{2n-p}{n}}2^{\frac{1}{2}}=2^{\frac{5n-2p}{2n}},\;\;\text{
and
}\;\;\left(\Gamma\left(\frac{1}{2}\right)^{-\frac{p^{2}}{n-p}}\Gamma\left(\frac{1}{2}\right)^{-p}\right)^{\frac{n-p}{np}}=\sqrt{\pi}^{-1}.$

Recall $c=(nC_{g})^{\frac{n}{4}}$, so that

$c^{\frac{2}{p}\frac{p^{2}}{n-p}\frac{n-p}{np}}=(nC_{g})^{\frac{n}{4}\frac{2}{p}\frac{p^{2}}{n-p}\frac{n-p}{np}}=(nC_{g})^{\frac{1}{2}}.$

Therefore

$\displaystyle A$ $\displaystyle=K^{2}C_{p_{0}}C_{p_{1}}$
$\displaystyle=K^{2}\pi^{-1}2^{\frac{5n-2p_{0}}{2n}+\frac{5n-2p_{1}}{2n}}(nC_{g})\left(\frac{2p_{0}}{n-p_{0}}\right)^{\frac{p_{0}}{n}}\left(\frac{2p_{1}}{n-p_{1}}\right)^{\frac{p_{1}}{n}}$
(44)
$\displaystyle=C_{g}K^{2}\left[\frac{n}{\pi}2^{5}\left(\frac{p_{0}}{n-p_{0}}\right)^{\frac{p_{0}}{n}}\left(\frac{p_{1}}{n-p_{1}}\right)^{\frac{p_{1}}{n}}\right].$

Now we need to compute a bound for the constant $K$ which, as noted above,
comes from an application of the off-diagonal Markinciwicz Interpolation
Theorem (specifically, we require $C_{0}^{1-t}C_{1}^{t}$ dependance in the
strong-type estimate where $C_{0},C_{1}$ are the weak-type constants). There
are many citations for this particular result, for instance a proof in a more
general setting is given in [13, Theorem 1.4.19] and the upper bound that is
obtained after plugging in our constants is something like $K\approx
1000*2^{12}n$. Through an argument communicated to the authors by Espinoza
[17], one can obtain much sharper bounds by combining the ideas in [12] and
[13]. We duplicate this argument here for the readers’ convenience.

######  Theorem A.1 ([17] Off-Diagonal Markinciwicz Interpolation Theorem with
Improved Constant).

Suppose that $(X,\mu)$ and $(Y,\nu)$ are $\sigma$-finite measure spaces and
that $p_{0},p_{1},q_{0},q_{1}\in[1,\infty]$ such that $p_{0}<p_{1}$,
$p_{0}<q_{0}$, $p_{1}<q_{1}$, $q_{0}<q_{1}$, and define for $t\in(0,1)$,
$p_{t}$ and $q_{t}$ by

(45)
$\frac{1}{p_{t}}=\frac{1-t}{p_{0}}+\frac{t}{p_{1}},\;\;\frac{1}{q_{t}}=\frac{1-t}{q_{0}}+\frac{t}{q_{1}}.$

If $T:L^{p_{0}}(X,\mu)+L^{p_{1}}(X,\mu)\to L_{0}(\nu)$ is a sub-additive
operator which is weak type $(p_{i},q_{i})$ with constant $C_{i}$, then $T$ is
strong type $(p_{t},q_{t})$ with constant

$C_{t}=2\left(\frac{q_{t}}{q_{t}-q_{0}}+\frac{q_{t}}{q_{1}-q_{t}}\right)^{\frac{1}{q_{t}}}C_{0}^{1-t}C_{1}^{t}.$

###### Proof (Duplicated):.

Fix some $t\in(0,1)$ and, for ease of notation, denote $p=p_{t}$ and
$q=q_{t}$. Let $f\in L^{p}(X,\mu)$ be normalized so that $\|f\|_{p}=1$. We
will show that

(46) $\|Tf\|_{q}\leq
2\left(\frac{q}{q-q_{0}}+\frac{q}{q_{1}-q}\right)^{\frac{1}{q}}C^{1-t}_{0}C^{t}_{1}.$

Equations (45) imply that

(47)
$\frac{p_{0}(q-q_{0})}{q_{0}(p-p_{0})}=\frac{\frac{1}{p}\big{(}\frac{1}{q_{0}}-\frac{1}{q}\big{)}}{\frac{1}{q}\big{(}\frac{1}{p_{0}}-\frac{1}{p}\big{)}}=\frac{\frac{1}{p}\big{(}\frac{1}{q}-\frac{1}{q_{1}}\big{)}}{\frac{1}{q}\big{(}\frac{1}{p}-\frac{1}{p_{1}}\big{)}}=\frac{p_{1}(q_{1}-q)}{q_{1}(p_{1}-p)}.$

Let us denote by $\sigma$ this common value. For arbitrary $\alpha>0$ and a
selected $\epsilon>0$ (to be determined later) we have

$f=f\chi_{\\{|f|>\alpha^{\sigma}\epsilon\\}}+f\chi_{\\{|f|\leq\alpha^{\sigma}\epsilon\\}}=:g+h.$

Notably, $g\in L_{p_{0}}(\mu)$ and $h\in L_{p_{1}}(\mu)$. By the
$(p_{0},q_{0})$ and $(p_{1},q_{1})$ weak-type estimates, we have

$\displaystyle\nu(|Tf|>\alpha)$
$\displaystyle\leq\nu(|Tg|>\alpha/2)+\nu(|Th|>\alpha/2)$
$\displaystyle\leq\left(\frac{2C_{0}}{\alpha}\right)^{q_{0}}\left(\int_{\\{|f|>\alpha^{\sigma}\epsilon\\}}|f|^{p_{0}}\,d\mu\right)^{\tfrac{q_{0}}{p_{0}}}+\left(\frac{2C_{1}}{\alpha}\right)^{q_{1}}\left(\int_{\\{|f|\leq\alpha^{\sigma}\epsilon\\}}|f|^{p_{1}}\,d\mu\right)^{\frac{q_{1}}{p_{1}}}.$

Consequently, there holds

$\displaystyle\|Tf\|^{q}_{q}$
$\displaystyle=q\int^{\infty}_{0}\alpha^{q-1}\nu(|Tf|>\alpha)\,d\alpha$
$\displaystyle\leq
q(2C_{0})^{q_{0}}\int^{\infty}_{0}\alpha^{q-q_{0}-1}\Big{(}\int_{\\{|f|>\alpha^{\sigma}\epsilon\\}}|f|^{p_{0}}\,d\mu\Big{)}^{\tfrac{q_{0}}{p_{0}}}\,d\alpha\quad$
$\displaystyle\;\;\;\;+q(2C_{1})^{q_{1}}\int^{\infty}_{0}\alpha^{q-q_{1}-1}\Big{(}\int_{\\{|f|\leq\alpha^{\sigma}\epsilon\\}}|f|^{p_{1}}\,d\mu\Big{)}^{\tfrac{q_{1}}{p_{1}}}\,d\alpha.$

The assumptions of $p_{j}<q_{j}$ and the $\sigma$-finiteness of the spaces
involved allows for us to apply Minkowski’s inequality to each of the two
integrals on the RHS. This gives

$\displaystyle\left(\int^{\infty}_{0}\alpha^{q-q_{0}-1}\left(\int_{\\{|f|>\alpha^{\sigma}\epsilon\\}}|f|^{p_{0}}\,d\mu\right)^{\tfrac{q_{0}}{p_{0}}}\,d\alpha\right)^{\frac{p_{0}}{q_{0}}}$
$\displaystyle\leq\int_{X}|f|^{p_{0}}\left(\int^{\frac{|f|^{1/\sigma}}{\epsilon^{1/\sigma}}}_{0}\alpha^{q-q_{0}-1}\,d\alpha\right)^{\frac{p_{0}}{q_{0}}}\,d\mu$
$\displaystyle=\frac{\epsilon^{-\frac{q-q_{0}}{\sigma}\frac{p_{0}}{q_{0}}}}{(q-q_{0})^{\frac{p_{0}}{q_{0}}}}\int_{X}|f|^{p_{0}}|f|^{\frac{q-q_{0}}{\sigma}\frac{p_{0}}{q_{0}}}\,\mu$
$\displaystyle=\frac{\epsilon^{-(p-p_{0})}}{(q-q_{0})^{\frac{p_{0}}{q_{0}}}}\int_{X}|f|^{p}\,d\mu$

where the last equality follows from (47). Similarly

$\displaystyle\left(\int^{\infty}_{0}\alpha^{q-q_{1}-1}\left(\int_{\\{|f|\leq\alpha^{\sigma}\epsilon\\}}|f|^{p_{1}}\,d\mu\right)^{\tfrac{q_{1}}{p_{1}}}\,d\alpha\right)^{\frac{p_{1}}{q_{1}}}\leq\frac{\epsilon^{p_{1}-p}}{(q_{1}-q)^{\frac{p_{1}}{q_{1}}}}\int_{X}|f|^{p}\,d\mu.$

Putting things together, we obtain that for all $f\in L_{p}(\mu)$ with
$\|f\|_{p}=1$, there holds

$\|Tf\|^{q}_{q}\leq
q\left(\frac{(2C_{0})^{q_{0}}\epsilon^{-\frac{q_{0}}{p_{0}}(p-p_{0})}}{q-q_{0}}+\frac{(2C_{1})^{q_{1}}\epsilon^{\frac{q_{1}}{p_{1}}(p_{1}-p)}}{q_{1}-q}\right).$

Now we choose $\epsilon>0$ so that

(48)
$(2C_{0})^{q_{0}}\epsilon^{-\frac{q_{0}}{p_{0}}(p-p_{0})}=(2C_{1})^{q_{1}}\epsilon^{\frac{q_{1}}{p_{1}}(p_{1}-p)}$

which subsequently yields

(49)
$\|Tf\|^{q}_{q}\leq\left(\frac{q}{q-q_{0}}+\frac{q}{q_{1}-q}\right)(2C_{1})^{q_{1}}\epsilon^{\frac{q_{1}}{p_{1}}(p_{1}-p)}.$

Equation (48), along with (47), implies that

$\frac{(2C_{0})^{q_{0}}}{(2C_{1})^{q_{1}}}=\epsilon^{\frac{q_{1}}{p_{1}}(p_{1}-p)+\frac{q_{0}}{p_{0}}(p-p_{0})}=\epsilon^{\frac{q_{1}-q}{\sigma}+\frac{q-q_{0}}{\sigma}}=\epsilon^{\frac{q_{1}-q_{0}}{\sigma}}.$

Using this value of $\epsilon$, our definition of $\sigma$ from (47), and our
assumption on $p_{i},q_{i}$ as in (45) yields

$\displaystyle(2C_{1})^{q_{1}}\epsilon^{\frac{q_{1}}{p_{1}}(p_{1}-p)}=(2C_{1})^{q_{1}}\left(\frac{(2C_{0})^{q_{0}}}{(2C_{1})^{q_{1}}}\right)^{\sigma\frac{q_{1}(p_{1}-p)}{p_{1}(q_{1}-q_{0})}}=(2C_{1})^{q_{1}}\left(\frac{(2C_{0})^{q_{0}}}{(2C_{1})^{q_{1}}}\right)^{\frac{q_{1}-q}{q_{1}-q_{0}}}$
$\displaystyle\;\;\;\;=(2C_{0})^{q\frac{\frac{1}{q}-\frac{1}{q_{1}}}{\frac{1}{q_{0}}-\frac{1}{q_{1}}}}(2C_{1})^{q_{1}-q_{1}\frac{q_{1}-q}{q_{1}-q_{0}}}=(2C_{0})^{q(1-t)}(2C_{1})^{qq_{1}\left(\frac{1}{q}-\frac{(1-t)}{q_{0}}\right)}=2^{q}C_{0}^{q(1-t)}C_{1}^{qt}.$

Putting this back into (49) and taking $q$’th roots on both sides gives (46)
and completes the proof. ∎

In our setting, we are applying Theorem A.1 with our given $p_{i}$, $q_{i}$
and with $t=\frac{1}{2}$. Thus we have

$K=2\left(\frac{q}{q-q_{0}}+\frac{q}{q_{1}-q}\right)^{\frac{1}{q}}.$

Plugging this back into (44), we have

$A\leq
C_{g}\left[2\left(\frac{q}{q-q_{0}}+\frac{q}{q_{1}-q}\right)^{\frac{1}{q}}\right]^{2}\left[\frac{n}{\pi}2^{5}\left(\frac{p_{0}}{n-p_{0}}\right)^{\frac{p_{0}}{n}}\left(\frac{p_{1}}{n-p_{1}}\right)^{\frac{p_{1}}{n}}\right].$

Recall that we did all this with a variable $p_{0}\in(1,2)$. We can optimize
this bound by finding its infimum over this interval. Thus we take

$L(n)=\inf_{p_{0}\in(1,2)}\left\\{n\frac{2^{5}}{\pi}\left[2\left(\frac{q}{q-q_{0}}+\frac{q}{q_{1}-q}\right)^{\frac{1}{q}}\right]^{2}\left(\frac{p_{0}}{n-p_{0}}\right)^{\frac{p_{0}}{n}}\left(\frac{p_{1}}{n-p_{1}}\right)^{\frac{p_{1}}{n}}\right\\}.$

For any $n\geq 3$, this infimum is achieved for some $p_{0}\in(1,2)$. We can
calculate numerical bounds for $L(n)$ using a computer:

$L(n)<\begin{cases}646&n=3\\\ 434&n=4\\\ 430&n=5\\\ 78n&n\geq 6.\end{cases}.$

######  Remark A.2.

In [24, Theorem 5.4], Ye gives precise values of $p_{0},p_{1}$ of

$p_{0}=\frac{n+2}{n}\;\;\text{ and }\;\;p_{1}=\frac{n+2}{2}.$

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