# Sharpening a gap theorem: nonnegative Ricci and small curvature
concentration

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

We sharpen a gap theorem of Chan & Lee [4] for nonnegative Ricci curvature
manifolds that have positive asymptotic volume ratio and small enough scale-
invariant integral curvature (so-called “curvature concentration”), by showing
that the curvature concentration need only depend linearly on the asymptotic
volume ratio. We prove the result by exhibiting a long-time Ricci flow
solution with faster than $1/t$ curvature decay, which allows us to shift the
limiting contradiction argument to time infinity and thus obtain an explicit
bound on the size of the gap.

##  1\. Introduction

The Ricci flow, introduced by Hamilton [13], deforms an initial Riemannian
metric $g(0)=g_{0}$ on a $n$-dimensional manifold $M^{n}$ according to the
equation

$\frac{\partial}{\partial t}g(t)=-2\text{Rc}_{g(t)}.$

A central aspect of Ricci flow analysis is the tendency for the flow to
roughly preserve the geometry of almost Euclidean regions, despite the
manifold’s global geometry being possibly not well understood. This is the so-
called “pseudolocality” of the Ricci flow. Although the sense in which a
region is “almost Euclidean” may differ, the general arguments of
pseudolocality theorems often follow this outline:

  1. 1.

Construct a sequence of Ricci flows which are becoming more “locally
Euclidean” but for which some undesirable behaviour occurs (often of the form
of large curvature).

  2. 2.

Normalize the metrics to focus on the local region where the undesirable
behavior occurs.

  3. 3.

Analyze the sequence of normalized metrics as they approach a limiting space,
wherein a contradiction is apparent.

We will refer to this method of proof (where a contradiction is derived in the
limiting space) as a “limiting contradiction argument”. A notable aspect of
such arguments is that they often involve constants that are not explicitly
computable due to the nature of the contradiction appearing in the limit space
instead of at a specific moment along the sequence.

One of the reasons why pseudolocality theorems are important is that they
relay geometric and topological information in the local region. In
particular, if the sense in which a manifold $M$ is “almost Euclidean” is
invariant under scaling and is satisfied everywhere on $M$, then one can apply
the result to a sequence of blow-downs of the initial metric, thus providing
information about the manifolds global geometric structure. Consider for
example the following notable result of Chan & Lee.

######  Theorem 1.1.

[4, Theorem 1.1] For any $n\geq 4$ and $v>0$ there exists some $\varepsilon$,
depending on $n$ and $v$, such that if $(M^{n},g)$ is complete with
nonnegative Ricci curvature $\text{Rc}_{g}\geq 0$, asymptotic volume ratio
satisfying

$\lim_{r\to\infty}\frac{\text{Vol}_{g}B_{g}(x_{0},r)}{\omega_{n}r^{n}}\geq v,$

and globally small scale-invariant integral curvature (curvature
concentration)

(1) $\int_{M}|\text{Rm}|_{g}^{\frac{n}{2}}\,dV_{g}\leq\varepsilon(n,v),$

then $(M^{n},g)$ is isometric to $({\mathbb{R}}^{n},g_{E})$.

The proof given in [4] is a limiting contradiction argument wherein they fix
some $v>0$ and then assume there is a sequence of nonnegative Ricci curved
$(M_{i}^{n},g_{i})$, each with asymptotic volume ratio at least $v$ and
integral curvatures satisfying (1) with $\varepsilon_{i}\to 0$, which results
in a contradiction in the limit. This raises two questions: 1. For any fixed
$v>0$, precisely how small does $\varepsilon(v)$ need to be to lie within the
desired gap? 2. Given two distinct $v_{1},v_{2}>0$, how does
$\varepsilon(v_{1})$ relate to $\varepsilon(v_{2})$? The purpose of this paper
is to provide an answer to both of these questions. We prove the following.

######  Theorem 1.2.

For $n\geq 4$, there exists a dimensional constant
$\varepsilon(n)\geq(10n)^{-10n^{3}}$ such that the following holds. Let
$(M^{n},g)$ be any complete Riemannian manifold (not necessarily bounded
curvature) with nonnegative Ricci curvature $\text{Rc}_{g}\geq 0$. Suppose
that $(M^{n},g)$ has positive asymptotic volume ratio

$\lim_{r\to\infty}\frac{\text{Vol}_{g}B_{g}(x_{0},r)}{\omega_{n}r^{n}}\geq
v>0,$

and curvature concentration satisfying

(2) $\int_{M}|\text{Rm}|_{g}^{\frac{n}{2}}\,dV_{g}\leq\varepsilon(n)v.$

Then $(M^{n},g)$ is isometric to $({\mathbb{R}}^{n},g_{E})$.

Even though Theorem 1.2 is stated for dimensions $n\geq 4$, there is no
generality lost with this assumption because it follows from work of Cheeger
[6, Theorem 6.10] and Colding’s Volume Continuity Theorem [12] that for any
odd $n$ (in particular, $n=3$), merely the assumption of
$\|\text{Rm}\|_{L^{\frac{n}{2}}(g)}<\infty$ is sufficient to guarantee that
any nonnegative Ricci curvature manifold with Euclidean volume growth is
necessarily isometric to $({\mathbb{R}}^{n},g_{E})$. When $n$ is even, the
picture is less clear. For instance, we have the example of the Eguchi-Hanson
metric which is a nonflat 4-dimensional Ricci flat manifold with
$\operatorname{AVR}\geq\frac{1}{2}$ and
$\|\text{Rm}\|_{L^{\frac{n}{2}}}<\infty$. For the dimensions $n$ in which such
examples exist, there is a well-defined function $f_{n}:(0,1]\to{\mathbb{R}}$
which defines the “size of the gap” for manifolds with at least the given
asymptotic volume ratio, i.e., $f_{n}(v)$ is the supremum of all $\rho>0$ for
which

$\left\\{(M^{n},g)\text{ with }\text{Rc}_{g}\geq 0,\;\operatorname{AVR}(g)\geq
v\text{ and
}\int_{M}|\text{Rm}|_{g}^{\frac{n}{2}}\,dV_{g}\leq\rho\right\\}=\\{({\mathbb{R}}^{n},g_{E})\\},$

where both of the sets of Riemannian manifolds above is understood in the
sense of equivalent classes modulo isomorphism. The general behaviour of
$f_{n}(v)$ is not currently well understood. For instance, it is clear that
$v\mapsto f_{n}(v)$ is nondecreasing, but it is not known (to the author’s
knowledge) if we can even say that $\lim_{v\to 0^{+}}f_{n}(v)=0$. In light of
Theorem 1.2 however, we have the following natural question.

######  Question 1.3.

Is the linear relationship between the required smallness of the curvature
concentration and the asymptotic volume ratio $v$ in Theorem 1.2 optimal? In
other words, is it true that

$\lim_{v\to 0^{+}}\frac{f_{n}(v)}{v}<\infty?$

Some evidence towards an affirmative answer to this question is the
relationship between between the curvature concentration and an
$L^{2}$-Sobolev inequality via Hölder’s inequality:

$\int_{M}|\text{Rm}|u^{2}\leq\left(\int_{M}|\text{Rm}|^{\frac{n}{2}}\right)^{\frac{2}{n}}\left(\int_{M}u^{\frac{2n}{n-2}}\right)^{\frac{n-2}{n}}\lesssim
v^{\frac{2}{n}}\left(\int_{M}u^{\frac{2n}{n-2}}\right)^{\frac{n-2}{n}}.$

This coefficient of $v^{\frac{2}{n}}$ on the RHS turns out to be the largest
one possible that still allows the proof to run because our translations
between volume ratios and Sobolev constants is sharp with respect to $v$ (see
Section 2) and the particular power of $\frac{2}{n}$ is precisely the one that
is necessary to eliminate the scalar curvature term in the $\nu$-functional as
well as the $|\text{Rm}|^{\frac{n}{2}+1}$ term in the evolution of
$|\text{Rm}|^{\frac{n}{2}}$ along the Ricci flow (see the proof of Theorem
1.4). This is suggestive that the linear relationship of curvature
concentration to $v$ as in (2) may be the largest one that the Ricci flow
approach can handle.

The way in which we are able to obtain an explicit bound on $\varepsilon(n)$
is to demonstrate the existence of a long-time solution to the Ricci flow with
faster than $1/t$ curvature decay, which allows us to shift the limiting
contradiction argument to time infinity. It is worth noting in the following
result that we obtain _no control on the rate_ of which
$t|\text{Rm}|_{g(t)}\to 0$. This is the sacrifice we make in exchange for
obtaining an explicit bound on the constant $\varepsilon(n)$ in the statement
of Theorem 1.2.

######  Theorem 1.4.

For any $n\geq 4$ and $v>0$, there exists $c(n)>0$, $\iota(n,v)>0$ such that
the following holds. Suppose that $(M^{n},g)$ satisfies the hypotheses of
Theorem 1.2 with this choice of $n$, $v$. Then there exists a long-time
solution $g(t)$, $t\in[0,\infty)$ to the Ricci flow with $g(0)=g$ which
satisfies:

  1. (i)

${\mathcal{R}}_{g(t)}\geq 0$ for every $t\in[0,\infty)$;

  2. (ii)

$|\text{Rm}|_{g(t)}\leq\frac{1}{t}$ for every $t\in(0,\infty)$;

  3. (iii)

$\operatorname{inj}_{g(t)}\geq\iota(n,v)\sqrt{t}$ for every $t\in(0,\infty)$;

  4. (iv)

$\text{Vol}_{g(t)}B_{g(t)}(x,r)\geq c(n)v\,r^{n}$ for every $x\in M$, $r>0$,
$t\in(0,\infty)$; and

  5. (v)

$\lim_{t\to\infty}t|\text{Rm}|_{g(t)}=0$.

###### Proof of Theorem 1.2.

Let $g(t)$, $t\in[0,\infty)$ be the long-time Ricci flow emanating from $g$
given by Theorem 1.4. Now take any sequence of times $t_{i}\to\infty$, and
define the sequence of parabolically rescaled flows
$g_{i}(t)=\frac{1}{t_{i}}g(t_{i}t)$, each defined on $[0,1]$. Note that the
curvature bound (ii) and the injectivity radius bound (iii) survive this
rescaling, and so by Hamilton’s Compactness Theorem [14] and a diagonal
argument, the $(M,g_{i}(t),x_{0})$ subconverges (in the pointed Cheeger-Gromov
sense) to some limiting complete flow
$(\widetilde{M},\tilde{g}(t),x_{\infty})$, $t\in(0,1]$ which, by nature of the
faster than $1/t$ curvature decay (v), must satisfy

$|\text{Rm}|_{\tilde{g}(t)}\equiv 0\;\text{ for each }t\in(0,1].$

Furthermore, each $\tilde{g}(t)$ must have a positive asymptotic volume ratio
because (iv) passes to the limit as well. Therefore, for each $t\in(0,1]$, we
must have that $(\widetilde{M},\tilde{g}(t))$ is isometric to
$({\mathbb{R}}^{n},g_{E})$. Thus if we fix any $\delta>0$ and $t\in(0,1]$
small enough (depending on $\delta$ and $n$), we can say that for all
sufficiently large $i\in{\mathbb{N}}$, there holds

$\displaystyle\text{Vol}_{g_{i}(0)}B_{g_{i}(0)}(x_{0},1)$
$\displaystyle\geq\text{Vol}_{g_{i}(t)}B_{g_{i}(0)}(x_{0},1)$
$\displaystyle\geq\text{Vol}_{g_{i}(t)}B_{g_{i}(t)}(x_{0},1-\beta\sqrt{t})$
$\displaystyle\geq(1-\delta)\omega_{n}(1-\beta\sqrt{t})^{n}$
$\displaystyle\geq(1-2\delta)\omega_{n}$

where we have used the nonnegative scalar curvature of the flow (i) in the
first inequality, the Shrinking Balls Lemma [19, Corollary 3.3] in the second
(here $\beta=\beta(n)$ is a dimensional constant), the convergence
$g_{i}(t)\to g_{E}$ in the third, and our smallness condition on $t$ in the
fourth. Scaling back tells us that

$\lim_{r\to\infty}\frac{\text{Vol}_{g}B_{g}(x_{0},r)}{r^{n}}\geq(1-2\delta)\omega_{n}.$

Since $\delta>0$ was arbitrary, we can use our assumption of
$\text{Rc}_{g}\geq 0$ again to in fact see that

$\lim_{r\to\infty}\frac{\text{Vol}_{g}B_{g}(x_{0},r)}{r^{n}}=\omega_{n}.$

Therefore $(M^{n},g)$ is isometric to $({\mathbb{R}}^{n},g_{E})$ by the
rigidity statement in the Bishop-Gromov Volume Comparison Theorem. ∎

If we knew in advance that $(M^{n},g)$ had _bounded curvature_ , we could
circumvent the need for Theorem 1.4 in the proof of Theorem 1.2 by directly
applying the results from [5]. In particular, by Corollary 2.11, we could
obtain the Sobolev inequality

$\left(\int_{M}|u|^{\frac{2n}{n-2}}\,dV_{g}\right)^{\frac{n-2}{n}}\leq
C(n)v^{-\frac{2}{n}}\int_{M}|\nabla^{g}u|^{2}_{g}\,dV_{g}$

for all compactly supported $u\in W^{1,2}(M)$. If we restrict our
$\varepsilon(n)$ small enough so that

$\varepsilon(n)\leq\left(\frac{\delta(n)}{C(n)}\right)^{\frac{n}{2}}$

where $\delta(n)$ is the dimensional constant in [5, Theorem 1.2], then by the
conclusion of that theorem, we know that the complete bounded curvature Ricci
flow $g(t)$ with initial condition $g(0)=g$, exists for all times
$t\in[0,\infty)$ and satisfies the curvature estimates

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

for any $t\in(0,\infty)$ and

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

This gives conditions akin to (ii) and (v). Moreover, by [5, Corollary 4.3]
and Lemma 2.1, we know that there exists a $v_{0}(n,v)>0$ such that

$\text{Vol}_{g(t)}B_{g(t)}(x,r)\geq v_{0}(n,v)r^{n}\;\text{ for any }\;t\geq
0,\;r>0,\;x\in M,$

which gives (iv). Penultimately, for any $t>0$, the conditions

$\text{Vol}_{g(t)}B_{g(t)}(x,\sqrt{t})\geq v_{0}t^{\frac{n}{2}}\;\text{ and
}\;|\text{Rm}|_{g(t)}\leq\frac{C_{0}}{t},$

together imply the injectivity radius bound

$\operatorname{inj}_{g(t)}(x)\geq\iota(n,v_{0})\sqrt{t}$

by [7] for some $\iota(n,v_{0})>0$ which does not depend on $t>0$. This gives
condition (iii). Finally, condition (i) follows immediately from the
assumption that $g(t)$ is a bounded curvature flow (so in particular, lower
scalar curvature bounds are preserved by the maximum principle). The proof of
Theorem 1.2 then follows synonymously as above.

The issue that arises when $(M,g)$ has unbounded curvature is _not_
establishing short-time existence of the flow. Indeed, by [3, Theorem 4.1], we
can say that if $(M,g)$ satisfies the hypotheses of Theorem 1.2 (for some
$v>0$), then for any $\alpha>0$ there is a Ricci flow $g(t)$ with $g(0)=g$,
defined on some short time interval $[0,T(\alpha,g)]$, which enjoys $\alpha/t$
curvature control. Rather, the issues that we primarily face in this case are:
1. Extending the flow for all times $t\in[0,\infty)$; and 2. Establishing the
faster than $1/t$ curvature decay as in (v) in Theorem 1.4. We get around both
of these issues by showing that there exists a111The question of uniqueness of
Ricci flows emanating from unbounded curvature complete metrics is still
unanswered in general. Ricci flow $g(t)$ which preserves the Sobolev
inequality and curvature concentration of the initial metric (up to
dimensional constants). Let us explain this in slightly more detail: Just as
mentioned in the bounded curvature case above, we obtain from an asymptotic
volume ratio at least $v>0$, the Sobolev inequality

$\left(\int_{M}|u|^{\frac{2n}{n-2}}\,dV_{g}\right)^{\frac{n-2}{n}}\leq
C(n)v^{-\frac{2}{n}}\int_{M}|\nabla^{g}u|^{2}_{g}\,dV_{g}.$

The Ricci flow that we construct will satisfy the properties

(3) $\begin{cases}&|\text{Rm}|_{g(t)}\leq\frac{1}{t},\\\
&\int_{M}|\text{Rm}|_{g(t)}^{\frac{n}{2}}\,dV_{g(t)}\leq
C_{1}(n)\varepsilon(n)v,\text{ and }\\\
&\left(\int_{M}|u|^{\frac{2n}{n-2}}\,dV_{g(t)}\right)^{\frac{n-2}{n}}\leq
C_{1}(n)C(n)v^{-\frac{2}{n}}\int_{M}|\nabla^{g(t)}u|^{2}_{g(t)}\,dV_{g(t)}.\end{cases}$

This preservation of the Sobolev inequality and smallness of the curvature
concentration allows us to apply [5, Theorem 1.2] with the bounded curvature
metric $g(1)$ replacing $g(0)$, from which we obtain the faster than $1/t$
curvature decay.

The organization of this paper is as follows. In section 2 we investigate the
relationship between volume ratios, $L^{2}$-Sobolev inequalities, and control
on the $\bar{\nu}$-functional in spaces with nonnegative Ricci curvature. In
fact, we prove that these three notions of Sobolev control are equivalent in
this setting. In section 3 we discuss how distances evolve under the Ricci
flow, proving in particular, a version of the so-called “Expanding Balls
Lemma” where the time interval depends only on the dimension, $\alpha/t$
curvature bound, and lower bound on scalar curvature. In section 4, we prove a
short-time Ricci flow existence result for _bounded curvature_ metrics with
the property that the bound on the local curvature concentration is preserved
(up to a dimensional constant). Finally in section 5, we prove Theorem 1.4 by
approximating blow-downs of $(M,g)$ by bounded curvature metrics, from which
our short-time bounded curvature Ricci flow existence result is readily
applied. This proves the existence of a long-time Ricci flow solution
emanating from $g$ with the desired properties given in (3), and thus also
with the faster than $1/t$ curvature decay.

## Acknowledgement

I would like to thank Albert Chau for the helpful comments on earlier drafts
of this paper, as well as Man-Chun Lee for the discussions of related problems
at CRM.

##  2\. Relationship between Sobolev quantities

In this section, we will investigate the relationship between three different
notions of “Sobolev control” in manifolds which enjoy lower Ricci bounds:
volume ratios, $L^{2}$-Sobolev inequality, and control on the
$\bar{\nu}$-functional. We will see that when $\text{Rc}_{g}\geq 0$, all three
of these notions of Sobolev control are equivalent (up to dimensional
constants).

For the sake of brevity, we introduce the following notation. For a connected
$\Omega\subset(M,g)$, we will write

$\operatorname{Sob}(\Omega)\leq[A,B]$

to mean that the $L^{2}$-Sobolev inequality

$\left(\int_{\Omega}|u|^{\frac{2n}{n-2}}\,dV_{g}\right)^{\frac{n-2}{n}}\leq
A\left(\int_{\Omega}|\nabla^{g}u|^{2}_{g}+Bu^{2}\,dV_{g}\right)$

holds for all $u\in W^{1,2}(\Omega)$ with
$\text{supp}\,u\subset\subset\Omega$. We may also write
$\operatorname{Sob}(\Omega,g)$ to emphasize which metric is being used.

The fact that an $L^{2}$-Sobolev inequality implies control on volume ratios
is well-known, but since we care about the exact relationship between these
quantities, we will provide a short proof. The following is only a slight
modification of [22, Theorem 6.1].

######  Lemma 2.1 (Volume from Sobolev).

Let $(M^{n},g)$, $n\geq 3$ be a (possibly incomplete) Riemannian manifold with
$B_{g}(x,L)\subset\subset M$ for some $L>0$. Assume that
$\operatorname{Sob}(B_{g}(x,L))\leq[A,B]$ for some $A>0$, $B\geq 0$. Then
whenever $r\leq L$, there holds

$\text{Vol}_{g}B_{g}(x,r)\geq\left(\frac{1}{A(2^{n+3}+2BL^{2})}\right)^{\frac{n}{2}}r^{n}.$

###### Proof.

Assume for the sake of contradiction that there exists $(M^{n},g,x)$, $L>0$,
$A>0$, $B\geq 0$ for which the hypotheses hold, and some $r\in(0,L]$ such that

$\text{Vol}_{g}B_{g}(x,r)<\left(\frac{1}{A(2^{n+3}+2BL^{2})}\right)^{\frac{n}{2}}r^{n}=:\delta
r^{n}.$

Set $\tilde{g}=r^{-2}g$ so that
$\text{Vol}_{\tilde{g}}B_{\tilde{g}}(x,1)<\delta$. Then by the assumed Sobolev
inequality for the original metric $g$ and the fact that $r\leq L$, we obtain
$\operatorname{Sob}(B_{\tilde{g}}(x,1))\leq[A,BL^{2}]$. Now by Hölder’s
inequality, $\text{Vol}_{\tilde{g}}B_{\tilde{g}}(x,1)<\delta$, and the
definition of $\delta$, we have

$ABL^{2}\int_{B_{\tilde{g}}(x,1)}u^{2}\,dV_{\tilde{g}}\leq
ABL^{2}\delta^{\frac{2}{n}}\left(\int_{B_{\tilde{g}}(x,1)}|u|^{\frac{2n}{n-2}}\,dV_{\tilde{g}}\right)^{\frac{n-2}{n}}\leq\frac{1}{2}\left(\int_{B_{\tilde{g}}(x,1)}|u|^{\frac{2n}{n-2}}\,dV_{\tilde{g}}\right)^{\frac{n-2}{n}}.$

Therefore this term can be absorbed into the LHS of the Sobolev inequality for
$\tilde{g}$ to give $\operatorname{Sob}(B_{\tilde{g}}(x,1))\leq[2A,0]$. The
rest of the proof then is verbatim to the latter half of the proof of [22,
Theorem 6.1]. ∎

Now we recall the definition of local entropy formulated by Wang [20]: Let
$(M^{n},g)$ be a complete Riemannian manifold of dimension $n$, and
$\Omega\subset M$ a connected open subset with smooth boundary and define

$\displaystyle\mathscr{S}(\Omega):=\left\\{u\bigg{|}u\in
W^{1,2}_{0}(\Omega),\;\;u\geq 0,\;\;\int_{\Omega}u^{2}\,dV_{g}=1\right\\},$
$\displaystyle\overline{W}(\Omega,g,u,\tau):=\int_{\Omega}4\tau|\nabla^{g}u|_{g}^{2}-u^{2}\log
u^{2}\,dV_{g}-\frac{n}{2}\log(4\pi\tau)-n,$ $\displaystyle
W(\Omega,g,u,\tau):=\int_{\Omega}\tau(4|\nabla^{g}u|_{g}^{2}+{\mathcal{R}}_{g}u^{2})-u^{2}\log
u^{2}\,dV_{g}-\frac{n}{2}\log(4\pi\tau)-n,$
$\displaystyle\bar{\mu}(\Omega,g,\tau):=\inf_{u\in\mathscr{S}(\Omega)}\overline{W}(\Omega,g,u,\tau),\;\;\;\mu(\Omega,g,\tau):=\inf_{u\in\mathscr{S}(\Omega)}W(\Omega,g,u,\tau),$
$\displaystyle\bar{\nu}(\Omega,g,\tau):=\inf_{s\in(0,\tau]}\bar{\mu}(\Omega,g,s),\;\;\;\nu(\Omega,g,\tau):=\inf_{s\in(0,\tau]}\mu(\Omega,g,s),$
$\displaystyle\bar{\nu}(\Omega,g):=\inf_{s\in(0,\infty)}\bar{\mu}(\Omega,g,s),\;\;\;\nu(\Omega,g):=\inf_{s\in(0,\infty)}\mu(\Omega,g,s).$

In the above (and throughout the paper), ${\mathcal{R}}_{g}$ denotes the
scalar curvature of $g$. Although these definitions are given for domains
$\Omega$ with _smooth_ boundaries, we will often take the domain to be some
geodesic ball within $M$, which will not necessarily have smooth boundary.
This slight abuse of notation is validated for our purposes by [20, Lemma 2.6
and Remark 2.7].

The following shows the implication of $\bar{\nu}$-functional control given a
Sobolev inequality. Its proof is a simple consequence of Jensen’s inequality
(see for instance [22], [9], [5]).

######  Lemma 2.2 ($\bar{\nu}$-functional from Sobolev).

For any $n\geq 3$, there exists a dimensional constant $C(n)>0$ such that the
following holds. Let $(M^{n},g)$ be a (possibly incomplete) Riemannian
manifold with $\Omega\subset\subset M$ and
$\operatorname{Sob}(\Omega)\leq[A,\frac{a}{\tau}]$ for some $A\geq 1$, $a\geq
0$, $\tau>0$. Then $\bar{\nu}(\Omega,g,\tau)\geq-\frac{n}{2}\log(C(n)A)-4a$.
If in addition
$\int_{\Omega}|\text{Rm}|_{g}^{\frac{n}{2}}\,dV_{g}\leq\left(\frac{2}{nA}\right)^{\frac{n}{2}}$,
or more generally
$\int_{\Omega}{\mathcal{R}}_{g}^{\frac{n}{2}}\,dV_{g}\leq\left(\frac{2}{A}\right)^{\frac{n}{2}}$,
then $\nu(\Omega,g,\tau)\geq-\frac{n}{2}\log(C(n)A)-4a$.

###### Proof.

Just as in the proof of [5, Lemma 3.1], Jensen’s inequality and the fact that
$\log x\leq bx-1-\log b$ (for any $b>0$) implies

(4) $\begin{split}\int_{\Omega}u^{2}\log
u^{2}\,dV_{g}&\leq\frac{n}{2}\log\left(\int_{\Omega}|\nabla^{g}u|^{2}_{g}+\frac{a}{\tau}u^{2}\,dV_{g}\right)+\frac{n}{2}\log
A\\\ &\leq
4s\left(\int_{\Omega}|\nabla^{g}u|^{2}_{g}+\frac{a}{\tau}u^{2}\,dV_{g}\right)+\frac{n}{2}\log
A-1-\log\left(\frac{8s}{n}\right).\end{split}$

This shows that $\bar{\nu}(\Omega,g,\tau)\geq-\frac{n}{2}\log(C_{1}(n)A)-4a$
since $\frac{4sa}{\tau}\leq 4a$ when $s\leq\tau$. For the other implication,
we apply Hölder’s and our Sobolev inequality

$2s\int_{\Omega}{\mathcal{R}}_{g}u^{2}\,dV_{g}\geq-4s\left(\int_{\Omega}|\nabla^{g}u|^{2}_{g}+\frac{a}{\tau}u^{2}\,dV_{g}\right).$

Putting this back into (4) gives

$\int_{\Omega}u^{2}\log u^{2}\,dV_{g}\leq
8s\left(\int_{\Omega}|\nabla^{g}u|^{2}_{g}+\frac{{\mathcal{R}}_{g}}{4}u^{2}+\frac{a}{\tau}u^{2}\,dV_{g}\right)+\frac{n}{2}\log
A-1-\log\left(\frac{8s}{n}\right).$

Now relabel $\tilde{s}=2s$ and restrict $\tilde{s}\leq\tau$ to obtain
$\nu(\Omega,g,\tau)\geq-\frac{n}{2}\log(2C_{1}(n)A)-4a$. For simplicity, we
will assume $C(n)=2C_{1}(n)$ from which both implications clearly follow. ∎

It is also well-known that rough control on the $\bar{\nu}$-functional on a
set $\Omega$ is sufficient to obtain a Sobolev inequality on $\Omega$. We
include a precise statement for our ease of reference.

######  Lemma 2.3 (Sobolev from $\bar{\nu}$-functional).

For any $n\geq 3$, there exists a dimensional constant $C(n)>0$ such that the
following holds. Suppose $(M^{n},g)$, is a (possibly incomplete) Riemannian
manifold, $\Omega\subset\subset M$ is connected, and $A,\tau>0$.

  1. (1)

If $\bar{\nu}(\Omega,g,\tau)\geq-\frac{n}{2}\log A$, then
$\operatorname{Sob}(\Omega)\leq[C(n)A,\frac{1}{\tau}]$.

  2. (2)

If $\nu(\Omega,g,\tau)\geq-\frac{n}{2}\log A$ and
$\int_{\Omega}|\text{Rm}|_{g}^{\frac{n}{2}}\,dV_{g}\leq\left(\frac{2}{nC(n)A}\right)^{\frac{n}{2}}$,
or more generally
$\int_{\Omega}{\mathcal{R}}_{g}^{\frac{n}{2}}\,dV_{g}\leq\left(\frac{1}{C(n)A}\right)^{\frac{n}{2}}$,
then $\operatorname{Sob}(\Omega)\leq[2C(n)A,\frac{1}{\tau}]$.

###### Proof.

We will only discuss the proof of part (2) since part (1) is simpler. Fix some
$\lambda>0$, a compact $K\subset\subset\Omega$, and observe that the condition
on the $\nu$-functional implies

$\displaystyle\int_{\Omega}u^{2}\log u^{2}\,dV_{g}$ $\displaystyle\leq
s\left(\int_{\Omega}|\nabla^{g}u|_{g}^{2}+\frac{{\mathcal{R}}_{g}}{4}u^{2}\,dV_{g}\right)-\frac{n}{2}\log
s+\frac{n}{2}\log A-\frac{n}{2}\log(\pi)-n$ $\displaystyle\leq
s\left(\int_{\Omega}|\nabla^{g}u|_{g}^{2}+\frac{{\mathcal{R}}_{g}^{+}}{4}u^{2}+\lambda
u^{2}\,dV_{g}\right)-\frac{n}{2}\log s+\frac{n}{2}\log
A-\frac{n}{2}\log(\pi)-n$

for all $s\in(0,4\tau]$ and $u\in W^{1,2}_{0}(K)$ with
$\int_{K}u^{2}\,dV_{g}=1$. Now by [22, Theorem 5.5], we may conclude a Sobolev
inequality

(5) $\left(\int_{K}u^{\frac{2n}{n-2}}\,dV_{g}\right)^{\frac{n-2}{n}}\leq
C(n,A)\left(\int_{K}|\nabla^{g}u|^{2}_{g}+\frac{{\mathcal{R}}_{g}^{+}}{4}u^{2}+\frac{1}{\tau}u^{2}+\lambda
u^{2}\,dV_{g}\right)$

for all $u\in W^{1,2}_{0}(K)$. Even though the results in [22] are stated for
closed manifolds, equivalent statements hold for compact manifolds with
boundary (such as our $K\subset\subset\Omega$). Now since $K$ is compact, we
may send $\lambda\to 0$. The fact that the constant $C(n,A)$ is of the precise
form $C(n)A$ follows by tracking through the constants in [22, Theorems
5.3-5.5] as was first noticed by Chen [9] (see also [5, Appendix]). So it
remains to deal with the remaining scalar curvature term. For this, we apply
Hölder’s inequality

$\frac{C(n)A}{4}\int_{K}{\mathcal{R}}_{g}^{+}u^{2}\,dV_{g}\leq\frac{C(n)A}{4}\left(\int_{K}|{\mathcal{R}}_{g}|^{\frac{n}{2}}\,dV_{g}\right)^{\frac{2}{n}}\left(\int_{K}u^{\frac{2n}{n-2}}\,dV_{g}\right)^{\frac{n-2}{n}}\leq\frac{1}{2}\left(\int_{K}u^{\frac{2n}{n-2}}\,dV_{g}\right)^{\frac{n-2}{n}}.$

So we can absorb this term into the LHS of (5) to give
$\operatorname{Sob}(K)\leq[2C(n)A,\frac{1}{\tau}]$. The result follows by
taking an exhaustion of compact $K_{1}\subset K_{2}\subset\dots\subset\Omega$.
∎

######  Remark 2.4.

In our application, we will have some Ricci flow $g(t)$, $t\in[0,T]$, an
initial Sobolov inequality $\operatorname{Sob}(\Omega,g(0))\leq[A,0]$, and
some assumed smallness of $\int|\text{Rm}|_{g(0)}^{\frac{n}{2}}\,dV_{g(0)}$.
Lemma 2.2 then will give us $\nu$-functional control
$\nu(\Omega,g(0),\tau+T)\geq-\frac{n}{2}\log(C_{1}(n)A)$, which is preserved
under the Ricci flow to give
$\nu(\Omega,g(t),\tau)\geq-\frac{n}{2}\log(C_{1}(n)A)$. Then Lemma 2.3 will
return us to a Sobolev inequality for $g(t)$:
$\operatorname{Sob}(\Omega,g(t))\leq[2C_{2}(n)C_{1}(n)A,\frac{1}{\tau}]$. It
follows from [5, Proof of Theorem 3.4 and Appendix] that $C_{1}(n)C_{2}(n)\leq
250n$. We will use this estimate in obtaining a bound for our constant
$\varepsilon(n)$.

In the presence of a (possibly negative) lower Ricci curvature bound
$\text{Rc}_{g}\geq-K$, Wang [20, Theorem 3.6] proved that one can obtain local
control on the $\bar{\nu}$-functional depending on the volume ratio:

(6)
$\bar{\nu}(B_{g}(x,r),g,r^{2})\geq-10n^{2}(1+Kr)+\log\left(\frac{\text{Vol}_{g}B_{g}(x,r)}{\omega_{n}r^{n}}\right)^{n+1}.$

Here $\omega_{n}$ is the volume of an $n$-dimensional Euclidean unit ball. In
particular, when $\text{Rc}_{g}\geq 0$, we can obtain global control on the
$\bar{\nu}$-functional (and subsequently a global Sobolev inequality) if we
know that _all the volume ratios_ of $(M,g)$ are uniformly bounded from below.
Such a manifold is said to have _positive asymptotic volume ratio_ , which we
will denote by

$\operatorname{AVR}(g):=\lim_{r\to\infty}\frac{\text{Vol}_{g}B_{g}(x_{0},r)}{\omega^{n}r^{n}}.$

It is well-known that when $\text{Rc}_{g}\geq 0$, $\operatorname{AVR}(g)$ does
not depend on the choice of fixed point $x_{0}$, is a well-defined number in
$[0,1]$, and that $\operatorname{AVR}(g)=1$ if and only if $(M^{n},g)$ is
isometric to $({\mathbb{R}}^{n},g_{E})$ where $g_{E}$ is the Euclidean metric.

Now suppose that $(M,g)$ were some nonnegative Ricci curvature manifold with a
known volume bound

$\text{Vol}_{g}B_{g}(x,1)\geq v>0.$

Then by (6) we may obtain
$\bar{\nu}(B_{g}(x,1),g,1)\geq-C+\log(v^{\frac{n+1}{n}})$, which in turn gives
$\operatorname{Sob}(B_{g}(x,1))\leq[C^{\prime}v^{-\frac{2(n+1)}{n}},1]$ by
Lemma 2.3. Now if we return to a bound on the volume via Lemma 2.1, we obtain

$\text{Vol}_{g}B_{g}(x,1)\geq cv^{n+1},$

which is a much worse volume estimate than we started with as $v\to 0$.
Therefore one step along this process is not sharp with respect to $v$. It
turns out that this bluntness occurs in going from volume to
$\bar{\nu}$-functional in (6), and is rectified in the following lemma. Put
another way, Lemma 2.5 completes the circle of equivalences between volume
ratios, Sobolev inequality, and control on the $\bar{\nu}$-functional (up to
dimensional constants).

######  Lemma 2.5 (Sobolev from volume).

For any $n\geq 3$, there exists a dimensional constant $C(n)>0$ such that the
following holds. Let $(M^{n},g)$ be a complete Riemannian manifold (not
assumed bounded curvature) such that $\text{Rc}_{g}\geq 0$ and
$\operatorname{AVR}(g)>0$. If for some $r>0$ and $v>0$ there holds

(7) $\frac{\text{Vol}_{g}B_{g}(x,r)}{r^{n}}=v,$

then $\operatorname{Sob}(B_{g}(x,r))\leq[C(n)v^{-\frac{2}{n}},0]$.

######  Remark 2.6.

The constant $v$ in the volume ratio hypothesis (7) is permitted to be much
larger than the value of $\omega_{n}\cdot\operatorname{AVR}(g)$. In fact, the
assumption that $\operatorname{AVR}(g)>0$ in Lemma 2.5 is likely superfluous
because the proof of [1, Theorem 1.1] (which we will be modifying) is local in
nature until the end when they let $r\to\infty$ (instead we will take a
determined value of $r$). Regardless, the stated version will be sufficient
for our purposes.

###### Proof of Lemma 2.5.

First note that by volume comparison, we have

$\frac{v}{2^{n}}=\frac{\text{Vol}_{g}B_{g}(x,r)}{(2r)^{n}}\leq\frac{\text{Vol}_{g}B_{g}(x,2r)}{(2r)^{n}}\leq\frac{\text{Vol}_{g}B_{g}(x,r)}{r^{n}}=v.$

Therefore, it suffices to prove the lemma with $v$ replaced by

$v_{0}:=\frac{\text{Vol}_{g}B_{g}(x,2r)}{(2r)^{n}},$

at the expense of replacing the dimensional constant $C(n)$ with (at worst)
$2^{n}C(n)$.

Fix some smooth nonnegative function $u\in C^{\infty}(B_{g}(x,r))$ with
$u\equiv 0$ on $\partial B_{g}(x,r)$. Since the hypotheses and conclusion of
the lemma are invariant under scaling, we may assume that

(8)
$\int_{B_{g}(x,r)}|\nabla^{g}u|_{g}\,dV_{g}=n\int_{B_{g}(x,r)}u^{\frac{n}{n-1}}\,dV_{g}.$

If this seems like a somewhat mysterious condition, we refer the reader to the
proof of [1, Theorem 1.1], which we will be modifying (note that the
coefficient of $n$ in (8) is only there to match the convention in [1]). We
start with a couple elementary claims.

######  Claim 2.7.

$\text{Vol}_{g}B_{g}(x,r)\leq(2r)^{n}(1-(\frac{1}{7})^{n})v_{0}$

###### Proof of Claim.

Let $\gamma:[0,\infty)\to M$ be a geodesic ray with $\gamma(0)=x$. Setting
$y=\gamma(3r/2)$, we therefore have

$B_{g}(y,r/2)\subset B_{g}(x,2r)\setminus B_{g}(x,r).$

Then by volume comparison and the fact that $B_{g}(y,\frac{7}{2}r)\supset
B_{g}(x,2r)$, we can see that

$\frac{\text{Vol}_{g}B_{g}(y,r/2)}{(r/2)^{n}}\geq\frac{\text{Vol}_{g}B_{g}(y,7r/2)}{(7r/2)^{n}}\geq\frac{\text{Vol}_{g}B_{g}(x,2r)}{(7r/2)^{n}}\geq\frac{\text{Vol}_{g}B_{g}(x,2r)}{(2r)^{n}}(4/7)^{n}=v_{0}(4/7)^{n}.$

Then we obtain the upper bound

$\text{Vol}_{g}B_{g}(x,r)\leq\text{Vol}_{g}B_{g}(x,2r)-\text{Vol}_{g}B_{g}(y,r/2)\leq(2r)^{n}v_{0}-(2r/7)^{n}v_{0}=(2r)^{n}(1-(1/7)^{n})v_{0}.$

∎

######  Claim 2.8.

There exists dimensional constants $\delta(n)>0$, $C_{1}(n)\geq 1$ such that
for any measurable nonnegative function $f:B_{g}(x,r)\to{\mathbb{R}}$, there
holds

$\frac{1}{(2r)^{n}}\int_{B_{g}(x,r)}(1+f(x))^{n}\,dV_{g}\leq(1-\delta)v_{0}+\frac{C_{1}}{(2r)^{n}}\int_{B_{g}(x,r)}f(x)^{n}\,dV_{g}.$

###### Proof of Claim.

Applying the Binomial Theorem gives

$\int_{B_{g}(x,r)}(1+f(x))^{n}\,dV_{g}\leq\text{Vol}_{g}B_{g}(x,r)+\sum_{k=1}^{n}{n\choose
k}\int_{B_{g}(x,r)}f(x)^{k}\,dV_{g}.$

For the $1\leq k\leq n-1$ terms, we write $p_{k}=\frac{n}{k}$,
$q_{k}=\frac{n}{n-k}$ and apply Hölder’s then Young’s inequalities

$\displaystyle\int_{B_{g}(x,r)}f(x)^{k}\,dV_{g}$
$\displaystyle\leq\left(\int_{B_{g}(x,r)}f(x)^{n}\,dV_{g}\right)^{\frac{1}{p_{k}}}(\text{Vol}_{g}B_{g}(x,r))^{\frac{1}{q_{k}}}$
$\displaystyle\leq\frac{\varepsilon_{k}^{q_{k}}}{q_{k}}\text{Vol}_{g}B_{g}(x,r)+\frac{1}{\varepsilon_{k}^{p_{k}}p_{k}}\int_{B_{g}(x,r)}f(x)^{n}\,dV_{g}.$

We may choose the $\varepsilon_{k}$ small enough so that

$(1-(1/7)^{n})\left(1+\sum_{k=1}^{n-1}{n\choose
k}\frac{\varepsilon_{k}^{q_{k}}}{q_{k}}\right)\leq 1-\frac{1}{2}(1/7)^{n}.$

Therefore, by applying Claim 2.7, we can see that with
$\delta(n)=\frac{1}{2}(1/7)^{n}$ and

$C_{1}(n)=1+\sum_{k=1}^{n-1}{n\choose
k}\frac{1}{\varepsilon_{k}^{p_{k}}p_{k}},$

the claim holds. ∎

Now we recall some of the details from [1]. The following is taken out of the
proof of their main theorem (specifically right up until the end of their
Corollary 2.6 and the subsequent few lines), which we state as an independent
result for clarity.

######  Proposition 2.9 (Proof of [1, Theorem 1.1]).

Suppose that $(M,g)$ satisfies $\text{Rc}_{g}\geq 0$ and
$\operatorname{AVR}(g)>0$. Let $D\subset M$ be some compact connected region
with smooth boundary and $u\in C^{\infty}(D)$ is nonnegative with $u\equiv 0$
on $\partial D$ and

$\int_{D}|\nabla^{g}u|\,dV_{g}=n\int_{D}u^{\frac{n}{n-1}}\,dV_{g}.$

Then for all $\rho>0$, there holds

$\text{Vol}_{g}\\{y\in M:d_{g}(x,y)<\rho\;\text{ for all }x\in
D\\}\leq\int_{D}\left(1+\rho\,u^{\frac{1}{n-1}}\right)^{n}\,dV_{g}.$

We apply Proposition 2.9 with222The smooth boundary condition can be
accomplished by padding $B_{g}(x,r)$ slightly by taking $B_{g}(x,r)\subset
D_{\varepsilon}$, where $D_{\varepsilon}$ has smooth boundary and is contained
in the $\varepsilon$-ball of $B_{g}(x,r)$ (see [20, Lemma 2.6]), then take
$\varepsilon\to 0$. $D=B_{g}(x,r)$ and $\rho=3r$ to obtain

$\text{Vol}_{g}B_{g}(x,2r)=\text{Vol}_{g}\\{y\in M:d_{g}(p,y)<3r\;\text{ for
all }p\in
B_{g}(x,r)\\}\leq\int_{B_{g}(x,r)}\left(1+3r\,u^{\frac{1}{n-1}}\right)^{n}\,dV_{g}.$

Now divide both sides by $(2r)^{n}$ and apply Claim 2.8 with
$f=3ru^{\frac{1}{n-1}}$ to obtain

$v_{0}\leq(1-\delta(n))v_{0}+\frac{C_{1}}{(2r)^{n}}\int_{B_{g}(x,r)}\left(3r\,u^{\frac{1}{n-1}}\right)^{n}\,dV_{g}.$

Rearranging and writing $C_{2}(n)=\frac{C_{1}3^{n}}{\delta 2^{n}}$ gives

$v_{0}\leq C_{2}\int_{B_{g}(x,r)}u^{\frac{n}{n-1}}\,dV_{g}.$

Therefore, by applying our scaling condition (8), we immediately see

(9)
$\int_{B_{g}(x,r)}|\nabla^{g}u|_{g}\,dV_{g}=n\int_{B_{g}(x,r)}u^{\frac{n}{n-1}}\,dV_{g}\geq
nv_{0}^{\frac{1}{n}}C_{2}^{-\frac{1}{n}}\left(\int_{B_{g}(x,r)}u^{\frac{n}{n-1}}\,dV_{g}\right)^{\frac{n-1}{n}}.$

Thus we obtain an $L^{1}$-Sobolev inequality for the given function $u$, which
is scale-invariant. So we may scale back to our orinal value of $r$ and
therefore may assume that (9) holds for all nonnegative $u\in
C^{\infty}(B_{g}(x,r))$ with $u\equiv 0$ on $\partial B_{g}(x,r)$. Now it is
well-known that an $L^{1}$-Sobolev inequality implies an $L^{2}$-Sobolev
inequality (cf. [16]). We provide a brief proof for the reader’s convenience
and to verify the proper dependence on $v_{0}$. By the $L^{1}$-Sobolev
inequality (9) and Cauchy-Schwarz, we have

$\displaystyle\left(\int_{B_{g}(x,r)}u^{\frac{2n}{n-2}}\,dV_{g}\right)^{\frac{2(n-1)}{n}}$
$\displaystyle=\left(\int_{B_{g}(x,r)}\left(u^{\frac{2(n-1)}{n-2}}\right)^{\frac{n}{n-1}}\,dV_{g}\right)^{\frac{2(n-1)}{n}}$
$\displaystyle\leq\left(nv_{0}^{\frac{1}{n}}C_{2}^{-\frac{1}{n}}\right)^{-2}\left(\int_{B_{g}(x,r)}|\nabla^{g}u^{\frac{2(n-1)}{n-2}}|_{g}\,dV_{g}\right)^{2}$
$\displaystyle=\left(\left(nv_{0}^{\frac{1}{n}}C_{2}^{-\frac{1}{n}}\right)^{-1}\frac{2(n-1)}{n-2}\right)^{2}\left(\int_{B_{g}(x,r)}u^{\frac{n}{n-2}}|\nabla^{g}u|_{g}\,dV_{g}\right)^{2}$
$\displaystyle\leq
C(n)v_{0}^{-\frac{2}{n}}\left(\int_{B_{g}(x,r)}u^{\frac{2n}{n-2}}\,dV_{g}\right)\left(\int_{B_{g}(x,r)}|\nabla^{g}u|_{g}^{2}\,dV_{g}\right).$

We obtain the desired $L^{2}$-Sobolev inequality by rearranging the above and
approximation. This completes the proof of the lemma. ∎

######  Remark 2.10.

By taking the $\varepsilon_{k}=(2\cdot 7^{n}n^{n+1})^{-1}$ in the proof of
Claim 2.8, we can calculate a naive bound on the dimensional constant in Lemma
2.5 when $n\geq 4$ of

$C(n)\leq 2^{n}\underbrace{\left(\frac{2(n-1)}{n(n-2)}\right)^{2}}_{\leq
2}C_{2}^{\frac{2}{n}}\leq
2^{n+1}\left[\left(\frac{3}{2}\right)^{n}\underbrace{2\cdot
7^{n}}_{=1/\delta}\underbrace{n^{n+1}(2\cdot 7^{n}n^{n+1})^{n}}_{\leq
C_{1}}\right]^{\frac{2}{n}}\leq(10n)^{3n}.$

As previously mentioned, when $\text{Rc}_{g}\geq 0$, a positive lower bound on
the asymptotic volume ratio $\operatorname{AVR}(g)$ implies a lower bound on
_all_ volume ratios by volume comparison. Therefore Lemma 2.5 immediately
yields the following corollary.

######  Corollary 2.11.

For any $n\geq 3$ and $C(n)$ as in Lemma 2.5, the following holds. Let
$(M^{n},g)$, $n\geq 3$ be a complete Riemannian manifold (not assumed bounded
curvature) such that $\text{Rc}_{g}\geq 0$ and $\operatorname{AVR}(g)>0$. Then
$\operatorname{Sob}(M,g)\leq[C(n)(\omega_{n}\operatorname{AVR}(g))^{-\frac{2}{n}},0]$.

##  3\. Evolution of distances and curvature concentration along the Ricci
flow

In this section, we first give two results which describe how the distance
function behaves under the Ricci flow. In particular, these results will allow
us to conclude that the radius of geodesic balls stay uniformly controlled
both forward and backward in time. The first is the well-known “Shrinking
Balls Lemma” [19, Corollary 3.3] which controlls the rate at which distances
can contract.

######  Lemma 3.1 (Shrinking Balls [19]).

For any $n\in{\mathbb{N}}$ and $\beta(n)=8\sqrt{2/3}(n-1)\geq 1$, the
following holds. Suppose $(M^{n},g(t))$, $t\in[0,T]$ is a (not necessarily
complete) $n$-dimensional Ricci flow such that
$B_{g(0)}(x_{0},r)\subset\subset M$ and $|\text{Rm}|_{g(t)}\leq c_{0}/t$ on
$B_{g(0)}(x_{0},r)\cap B_{g(t)}(x_{0},r-\beta\sqrt{c_{0}t})$ for each
$t\in(0,T]$ and some $c_{0}>0$. Then for any $t\in[0,T]$, there holds

$B_{g(t)}(x_{0},r-\beta\sqrt{c_{0}t})\subset B_{g(0)}(x_{0},r).$

Obtaining control in the other direction (i.e., the rate at which distances
can expand) is historically more challening; If it is known that the flow
enjoys a lower Ricci bound _for all times_ , then Simon & Topping have proved
an analogous “Expanding Balls Lemma” [19, Lemma 3.1]. The issue is however
that lower Ricci curvature bounds are not respected under Ricci flow when
$n\geq 4$. Therefore it is desireable to replace the lower Ricci bound
assumption with a lower _scalar curvature_ bound (which is preserved) along
with some volume estimates (see [15, Lemma 3.7] and [18, Lemma 2.2] for such
results of this flavour). The improvement that we offer here is uniform
control on the rate of the expansion depending only on the quotient of volume
ratios, albeit for a short amount of time. Importantly, the time interval does
_not_ depend on the given volume bounds.

######  Lemma 3.2 (Expanding Balls).

For any $n\in\mathbb{N}$ and any $\alpha,\sigma>0$, there exists
$\widehat{T}(n,\alpha,\sigma)>0$ such that the following holds. Let
$(M^{n},g(t))$, $t\in[0,T]$ be a complete bounded curvature Ricci flow such
that for some $x_{0}\in M$, $b_{0}>0$, and $c_{0}>0$ there holds:

  1. (a)

$\inf_{x\in M}{\mathcal{R}}_{g(0)}(x)\geq-\sigma$,

  2. (b)

$\sup_{x\in M}|\text{Rm}|_{g(t)}(x)\leq\alpha t^{-1}$ for all $t\in(0,T]$,

  3. (c)

$\text{Vol}_{g(0)}B_{g(0)}(x_{0},2)\leq b_{0}$, and

  4. (d)

$\text{Vol}_{g(t)}B_{g(t)}(x,r)\geq c_{0}r^{n}$ whenever $x\in
B_{g(0)}(x_{0},1)$, $r\in(0,\frac{1}{2}]$, and $t\in(0,T]$.

Then for all $t\in[0,T\land\widehat{T}]$, we have $B_{g(0)}(x_{0},1)\subset
B_{g(t)}(x_{0},2^{n+2}b_{0}c_{0}^{-1})$.

###### Proof.

We begin by setting our maximal time interval:

$\widehat{T}=\min\left\\{\frac{\log
2}{\sigma},\frac{1}{4\beta^{2}\alpha}\right\\}.$

Here $\beta=\beta(n)$ is from Lemma 3.1. Fix $\tau\in[0,T\land\widehat{T}]$
and define

$R=\inf\\{r>0:B_{g(0)}(x_{0},1)\subset B_{g(\tau)}(x_{0},r)\\}.$

We will show that $R\leq 2^{n+2}b_{0}c_{0}^{-1}$ which will prove the result.
By definition, there exists $y\in M$ such that $d_{g(0)}(x_{0},y)=1$ and
$d_{g(\tau)}(x_{0},y)=R$. Let $\gamma:[0,1]\to M$ be a minimizing unit-speed
$g(0)$-geodesic with $\gamma(0)=x_{0}$ and $\gamma(1)=y$. Notice that because
$\widehat{T}\leq\frac{1}{4\beta^{2}\alpha}$, we have that
$B_{g(\tau)}(x,1/2)\subset B_{g(0)}(x,1)$ for all $x\in M$ by Lemma 3.1. Thus
by taking $z_{1}=x_{0}$, we can inductively choose some set of points
$\\{z_{j}\\}_{j=1}^{k}\subset B_{g(0)}(x_{0},1)$ such that
$\\{B_{g(\tau)}(z_{j},\frac{1}{2})\\}_{j=1}^{k}$ are mutually disjoint and

$\bigsqcup_{j=1}^{k}B_{g(\tau)}(z_{j},1/2)\subset B_{g(0)}(x_{0},2)\;\;\text{
and }\;\;B_{g(0)}(x_{0},1)\subset\bigcup_{j=1}^{k}B_{g(\tau)}(z_{j},1).$

Then since $\gamma\subset\overline{B_{g(0)}(x_{0},1)}$, we can obtain a lower
bound on $d_{g(\tau)}(x_{0},y)$ by estimating the length of $\gamma$ with
respect to the metric $g(\tau)$ as

(10) $R=d_{g(\tau)}(x_{0},y)\leq\text{Length}_{g(\tau)}(\gamma)\leq 2k.$

By our assumptions on the bounds of volumes and the assumed disjointness of
$\\{B_{g(\tau)}(z_{j},\frac{1}{2})\\}_{j=1}^{k}$, we can obtain an upper bound
on $k$:

(11)
$\begin{split}kc_{0}(1/2)^{n}&\leq\sum_{j=1}^{k}\text{Vol}_{g(\tau)}B_{g(\tau)}(z_{j},1/2)\\\
&\leq\text{Vol}_{g(\tau)}(B_{g(0)}(x_{0},2))\\\ &\leq
2\text{Vol}_{g(0)}(B_{g(0)}(x_{0},2))\leq 2b_{0}.\end{split}$

In the third inequality, we have used the evolution of the volume form
$\partial_{t}dV_{g(t)}=-{\mathcal{R}}_{g(t)}dV_{g(t)}$, the lower scalar bound
${\mathcal{R}}_{g(0)}\geq-\sigma$ (which is preserved along complete bounded
curvature flows), and the condition on the existence time
$\sigma\widehat{T}\leq\log 2$. By (10) and (11), we have the desired bound
$R\leq 2^{n+2}b_{0}c_{0}^{-1}$. ∎

######  Remark 3.3.

Because local lower bounds on the scalar curvature ${\mathcal{R}}_{g(0)}$ are
semi-preserved along all Ricci flows with $\alpha/t$ curvature control [8], it
is clear that Lemma 3.2 can be localized (i.e., for some compact region
contained within a possibly incomplete and/or unbounded curvature flow), but
the stated version will suffice for our purposes.

We end this section by giving a lemma demonstrating how the local curvature
concentration evolves under the Ricci flow. The proof is essentially that of
[2, Lemma 2.1], but we provide a brief sketch to demonstrate explicit
estimates on the coefficients.

######  Lemma 3.4.

Suppose $(M,g(t))$, $n\geq 3$, $t\in[0,T]$ is a complete (possibly unbounded
curvature) Ricci flow and $\phi:M\times[0,T]\to{\mathbb{R}}$ is a smooth
nonnegative function with compact support. If $\square\phi\leq 0$ (here
$\square=\partial_{t}-\Delta^{g(t)}$ is the heat operator), then for any
$\beta>0$ there holds

$\displaystyle\frac{d}{dt}\int\phi^{2}$
$\displaystyle(|\text{Rm}|_{g(t)}^{2}+\beta)^{\frac{n}{4}}\,dV_{g(t)}\leq-\frac{1}{3}\int|\nabla(\phi(|\text{Rm}|_{g(t)}^{2}+\beta)^{\frac{n}{8}})|^{2}\,dV_{g(t)}$
$\displaystyle\;\;\;+10n\int|\nabla^{g(t)}\phi|_{g(t)}^{2}(|\text{Rm}|_{g(t)}^{2}+\beta)^{\frac{n}{4}}\,dV_{g(t)}+5n\int\phi^{2}(|\text{Rm}|_{g(t)}^{2}+\beta)^{\frac{n}{4}+\frac{1}{2}}\,dV_{g(t)}.$

###### Proof.

In this proof, we will denote $\alpha=\frac{n}{4}$ and we will not write the
metric $g(t)$ in any terms except for the volume form $dV_{t}$. We begin by
differentiating under the integral, integrating by parts (since
$\text{supp}\,\phi$ is compact), using the evolution of volume
$\partial_{t}\,dV_{t}=-{\mathcal{R}}\,dV_{t}$, and $|{\mathcal{R}}|\leq
n|\text{Rm}|$ (see for instance [17, Proposition 7.28]) which gives

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

For the $\square$ term, we calculate

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

Clearly $\textbf{II}\leq 0$ and since $\square\phi\leq 0$, we have
$\textbf{I}\leq 0$ as well. Also, by the evolution of the curvature tensor and
Kato’s inequality $|\nabla|T||\leq|\nabla T|$, we have

III
$\displaystyle\leq-2\alpha\int\phi^{2}(|\text{Rm}|^{2}+\beta)^{\alpha-2}|\text{Rm}|^{2}|\nabla|\text{Rm}||^{2}\,dV_{t}+16\alpha\int\phi^{2}(|\text{Rm}|^{2}+\beta)^{\alpha+\frac{1}{2}}\,dV_{t}=\textbf{IIIa}+\textbf{IIIb}.$

Applying Cauchy-Schwarz and Young’s inequality gives

V
$\displaystyle\leq\frac{4\alpha}{\varepsilon}\int|\nabla\phi|^{2}(|\text{Rm}|^{2}+\beta)^{\alpha}\,dV_{t}+4\alpha\varepsilon\int\phi^{2}(|\text{Rm}|^{2}+\beta)^{\alpha-2}|\text{Rm}|^{2}|\nabla|\text{Rm}||^{2}\,dV_{t}=\textbf{Va}+\textbf{Vb}.$

Now gather up the terms IV, IIIa, Vb, and set $\varepsilon=\frac{1}{8}$ so
that the coefficient is $c(\alpha):=\frac{3}{2}\alpha+4\alpha(\alpha-1)$,
which satisfies $-c(\alpha)\in[\frac{2}{3}\alpha^{2},4\alpha^{2}]$. Then using
Cauchy-Schwarz again gives

$\displaystyle c(\alpha)$
$\displaystyle\int\phi^{2}(|\text{Rm}|^{2}+\beta)^{\alpha-2}|\text{Rm}|^{2}|\nabla|\text{Rm}||^{2}\,dV_{t}=c(\alpha)\frac{1}{\alpha^{2}}\int\phi^{2}|\nabla(|\text{Rm}|^{2}+\beta)^{\frac{\alpha}{2}}|^{2}\,dV_{t}$
$\displaystyle\leq
c(\alpha)\frac{1}{\alpha^{2}}\left(\frac{1}{2}\int|\nabla\phi(|\text{Rm}|^{2}+\beta)^{\frac{\alpha}{2}}|^{2}\,dV_{t}-\int|\nabla\phi|^{2}(|\text{Rm}|^{2}+\beta)^{\alpha}\,dV_{t}\right)$
$\displaystyle\leq-\frac{1}{3}\int|\nabla\phi(|\text{Rm}|^{2}+\beta)^{\frac{\alpha}{2}}|^{2}\,dV_{t}+4\int|\nabla\phi|^{2}(|\text{Rm}|^{2}+\beta)^{\alpha}\,dV_{t}.$

The result follows by gathering like terms and simplifying. ∎

##  4\. A bounded curvature Ricci flow existence result

As part of the proof of Theorem 1.4, we will require a short-time Ricci flow
existence result for bounded curvature metrics with the properties that the
flow enjoys $1/t$ curvature control and preservation of the local curvature
concentration (see Theorem 4.2). This result is similar to [3, Theorem 3.1],
which in turn is only a slight generalization of [2, Theorem 1.2]. Because of
these similarities, we may omit some of the details in the proof of Theorem
4.2 when appropriate and refer the reader to those works. Before stating that
however, we first prove an elementary lemma which states that the local
curvature concentration can only grow at most at some uniform rate under
bounded curvature flows.

######  Lemma 4.1.

Let $(M^{n},g(t))$, $t\in[0,T]$, $n\geq 3$ be a complete bounded curvature
Ricci flow such that

$\sup_{x\in
M}\int_{B_{g(0)}(x,r_{1})}|\text{Rm}|_{g(0)}^{\frac{n}{2}}\,dV_{g(0)}\leq\sigma$

for some $r_{1}>0$, $\sigma>0$. Then for any $r_{2}<r_{1}$ and
$\varepsilon>0$, there exists $\eta>0$ such that for each $t\in[0,\eta]$,
there holds

$\sup_{x\in
M}\int_{B_{g(t)}(x,r_{2})}|\text{Rm}|_{g(t)}^{\frac{n}{2}}\,dV_{g(t)}\leq\sigma+\varepsilon.$

###### Proof.

For the sake of brevity, we write $K=\sup|\text{Rm}(x,t)|+1$ where the
supremum is taken over all $(x,t)\in M\times[0,T]$. In what follows, we will
write $C$ to be some constant possibly depending on $n,r_{1},r_{2},\sigma,K$
which may vary line-by-line. Let $\psi:{\mathbb{R}}\to{\mathbb{R}}$ be some
smooth decreasing cutoff function such that $\psi\equiv 1$ on $(-\infty,0]$,
$\psi\equiv 0$ on $[1,\infty)$ and $\psi^{\prime}\geq-2$. Now for any $x\in
M$, let $\phi_{x}:M\to{\mathbb{R}}$ be defined as

$\phi_{x}(y)=\psi\left(\frac{2}{r_{1}-r_{2}}d_{g(0)}(x,y)-\frac{r_{1}+r_{2}}{r_{1}-r_{2}}\right).$

Then $\text{supp}\,\phi_{x}\subset B_{g(0)}(x,r_{1})$, $\phi_{x}\equiv 1$ on
$B_{g(0)}(x,\frac{r_{1}+r_{2}}{2})$, and
$|\nabla^{g(0)}\phi_{x}|^{2}_{g(0)}\leq C$. By the fact that the flow has
bounded curvature, we may assume that $\eta>0$ is small enough so that for all
$(x,t)\in M\times[0,\eta]$, we have

$\begin{cases}&B_{g(t)}(x,r_{2})\subset B_{g(0)}(x,\frac{r_{1}+r_{2}}{2}),\\\
&|\nabla^{g(t)}\phi_{x}|^{2}_{g(t)}\leq
2|\nabla^{g(0)}\phi_{x}|^{2}_{g(0)},\text{ and }\\\
&\text{Vol}_{g(t)}B_{g(0)}(x,r_{1})\leq C.\end{cases}$

Now for any $0<\beta\leq K$, we apply [5, Lemma 4.1] to see that

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

Integrating in time then sending $\beta\to 0$, we can apply our properties of
$\phi$ to obtain

$\displaystyle\int_{B_{g(t)}(x_{0},r_{2})}|\text{Rm}|_{g(t)}^{\frac{n}{2}}\,dV_{g(t)}$
$\displaystyle\leq\int\phi^{2}|\text{Rm}|_{g(t)}^{\frac{n}{2}}\,dV_{g(t)}\leq\int\phi^{2}|\text{Rm}|_{g(0)}^{\frac{n}{2}}\,dV_{g(0)}+Ct$
$\displaystyle\leq\int_{B_{g(0)}(x_{0},r_{1})}|\text{Rm}|_{g(0)}^{\frac{n}{2}}\,dV_{g(0)}+Ct.$

The result follows by restricting $\eta\leq\varepsilon/C$. ∎

######  Theorem 4.2.

For all $n\geq 3$, $\lambda\geq 1$, $v_{0}>0$, there exists $C(n)\geq 1$,
$c(n)>0$, $\iota(n,\lambda,v_{0})$, $\delta(n,\lambda)>0$,
$\Lambda(n,\lambda)>1$, and $\widetilde{T}(n,\lambda,v_{0})>0$ such that the
following holds. Suppose that $(M^{n},g(t))$, $t\in[0,T]$ is a complete
bounded curvature Ricci flow and the initial metric $g(0)$ satisfies the
following for all $x\in M$:

  1. (a)

$\text{Rc}_{g(0)}(x)\geq-1$;

  2. (b)

$\text{Vol}_{g(0)}B_{g(0)}(x,2r)\leq
2^{2n+1}\cdot\text{Vol}_{g(0)}B_{g(0)}(x,r)\leq 2^{2n+2}\omega_{n}r^{n}$ for
all $r\leq 4$; and

  3. (c)

$\int_{B_{g(0)}(x,4)}|\text{Rm}|_{g(0)}^{\frac{n}{2}}\,dV_{g(0)}\leq\delta(n,\lambda)v_{0}$.

Moreover for each $x\in M$ and $r\in(0,1]$, assume that

$v_{xr}:=\frac{\text{Vol}_{g(0)}B_{g(0)}(x,r)}{r^{n}}\geq v_{0}$

and that

(12) $\operatorname{Sob}(B_{g(0)}(x,4r),g(0))\leq\left[\lambda
v_{xr}^{-\frac{2}{n}},0\right].$

Then the flow $g(t)$ satisfies each of the following estimates uniformly for
all $(x,t)\in M\times[0,\widetilde{T}\land T]$:

  1. (i)

${\mathcal{R}}_{g(t)}(x)\geq-n$;

  2. (ii)

$|\text{Rm}|_{g(t)}(x)\leq\frac{1}{t}$;

  3. (iii)

$\operatorname{inj}_{g(t)}(x)\geq\iota(n,\lambda,v_{0})\sqrt{t}$;

  4. (iv)

$\text{Vol}_{g(t)}B_{g(t)}(x,r)\geq c(n)\lambda^{-\frac{n}{2}}v_{0}\,r^{n}$
for all $r\in(0,1]$;

  5. (v)

$\int_{B_{g(t)}(x,1)}|\text{Rm}|_{g(t)}^{\frac{n}{2}}\,dV_{g(t)}\leq\Lambda(n,\lambda)\delta(n,\lambda)v_{0}$;
and

  6. (vi)

$\operatorname{Sob}(B_{g(t)}(x,r),g(t))\leq\left[C(n)\lambda
v_{xr}^{-\frac{2}{n}},1\right]$ for all $r\in(0,1]$.

In particular, the flow exists up until time $\widetilde{T}$.

######  Remark 4.3.

The lower Ricci bound assumption (a) can be weakened to a lower bound on the
scalar curvature ${\mathcal{R}}_{g(0)}\geq-1$ in the precense of the volume
doubling assumption (b), at the expense of possibly shrinking
$\delta(n,\lambda)$. We choose to state the theorem like this because it is
more natural in our setting, and because this stronger hypothesis allows us to
substitute a limiting contradiction argument for the Moser Iteration argument
(see Step 2 in the proof).

###### Proof.

Note that since lower scalar curvature bounds are preserved by complete
bounded curvature flows, conclusion (i) is trivially satisfied (we have
included it in the theorem statement for clarity later). We begin the proof of
the other properties by fixing our constants. First, we set

(13)
$\widetilde{T}:=\min\left\\{T_{1},C_{4}^{-1},\frac{1}{16}\beta^{-2},\frac{1}{4}C_{3}^{-1},\frac{\log
2}{10},\frac{r_{0}^{4}}{10^{4}},r_{0}^{2}\widehat{T}\right\\}.$

Here $T_{1}=T_{1}(n,\lambda,v_{0})>0$ is some positive time coming from Step 2
of the proof, $\beta(n)$ is from the Shrinking Balls Lemma 3.1,
$\widehat{T}=\widehat{T}(n,1,1)$ is from the Expanding Balls Lemma 3.2, and
$c_{1}(n)$, $C_{3}(n)$, $C_{4}(n,\lambda)$, $r_{0}(n,\lambda)$ are constants
that will be defined within the proof. Also, we define

(14)
$\delta(n,\lambda)=\Lambda(n,\lambda)^{-1}\left(\frac{1}{10^{7}n^{3}\lambda}\right)^{\frac{n}{2}},$

where $\Lambda(n,\lambda)$ is defined in (28). We will see in Step 1 of the
proof that we may take

$C(n)=1000n\;\;\text{ and }\;\;c(n)=(2^{n+4}\cdot 1000n)^{-\frac{n}{2}}.$

Finally, recall that condition (iii) follows from the conclusions (ii) and
(iv) by [7], so we will assume that the $\iota(n,\lambda,v_{0})>0$ is the
constant given there.

Now we may proceed with the proof. Let $T^{\prime}$ be the maximal time in the
interval $[0,T\land\widetilde{T}]$ such that for all $(x,t)\in
M\times[0,T^{\prime}]$, there holds

(15)
$\int_{B_{g(t)}(x,1)}|\text{Rm}|_{g(t)}^{\frac{n}{2}}\,dV_{g(t)}\leq\Lambda(n,\lambda)\delta(n,\lambda)v_{0}.$

Note that because $(M,g(t))$, $t\in[0,T]$ is a bounded curvature flow which
satisfies (c), Lemma 4.1 guarantees that $T^{\prime}>0$. In advance however,
it appears as though $T^{\prime}$ depends on the curvature bound of the flow
$g(t)$, but we will show that this is not the case.

Our proof is broken down into three steps: (1) Show we have a uniform local
Sobolev inequality (i.e., conclusion (vi)) along the flow on the interval
$[0,T^{\prime}]$ assuming the time interval is possibly shortened so that a
global $\alpha/t$ curvature bound holds, which in particular implies
conclusion (iv); (2) Show that the assumption of the global $\alpha/t$
curvature bound assumption in Step 1 is implied as long as
$T^{\prime}\leq\widetilde{T}$ (this gives conclusions (ii) and (iii)); (3)
Derive a contradiction in the event that $T^{\prime}<T\land\widetilde{T}$.

Step 1 (Sobolev inequality)

Fix some $r_{0}(n,\lambda)\in(0,1)$ to be defined later (in hindsight, we
could take $r_{0}=[2^{3n+6}(2^{n+4}1000n\lambda)^{\frac{n}{2}}]^{-1}$). Since
$(M,g(t))$, $t\in[0,T]$ is a bounded curvature flow, we may assume that there
is some $T_{1}>0$ such that

(16) $|\text{Rm}|_{g(t)}\leq\frac{3r_{0}/10}{t}\;\;\text{ for all
}t\in(0,T_{1}].$

In this step of the proof, we will restrict to the interval $[0,T_{1}\land
T^{\prime}]$. We will show in Step 2 that it is necessarily true that $T_{1}$
can be taken to depend only on the variables $n,\lambda,v_{0}$.

Fix some $x\in M$ and $r\in[r_{0},1]$. By our smallness condition on
$\delta(n,\lambda)$ in (14) and our assumptions (12) and (c), as well as the
fact that $v_{x,r}\geq v_{0}$, we may apply Lemma 2.2 (with $a=0$) to obtain

$\nu(B_{g(0)}(x,4r),g(0),2)\geq-\frac{n}{2}\log(C_{1}(n)\lambda
v_{x,r}^{-\frac{2}{n}}).$

Now by our curvature assumption (16), we apply the “almost-monotonicity” of
this local functional along the Ricci flow [10, Lemma 2.5] (with the variables
there being $A=\frac{3r}{10},B=\frac{1}{4}$, $D=0$, $\tau_{1}=1$) to see that
for each $y\in B_{g(0)}(x,r)$ and $t\in[0,T_{1}\land T^{\prime}]$ there holds

$\displaystyle\nu(B_{g(t)}(y,r),g(t),1)$
$\displaystyle\geq\nu(B_{g(0)}(y,3r),g(0),2)-\left(\frac{2}{10(3r/10)^{2}(1/4)^{2}}+e^{-1}\right)(e^{\frac{T^{\prime}}{10(3r/10)^{2}(1/4)^{2}}}-1)$
$\displaystyle\geq\nu(B_{g(0)}(x,4r),g(0),2)-\frac{n}{2}\log
2\geq-\frac{n}{2}\log(2C_{1}(n)\lambda v_{x,r}^{-\frac{2}{n}}).$

Note that in the second inequality, we have used the fact that
$T^{\prime}\leq\widetilde{T}\leq\frac{r_{0}^{4}}{10^{4}}$ so that

$\left(\frac{2}{10(3r/10)^{2}(1/4)^{2}}+e^{-1}\right)(e^{\frac{T^{\prime}}{10(3r/10)^{2}(1/4)^{2}}}-1)\leq\frac{100}{r^{2}}(e^{\frac{50T^{\prime}}{r^{2}}}-1)\leq
10^{4}r_{0}^{-4}\widetilde{T}\leq 1<\frac{n}{2}\log 2.$

Since our smallness condition on $\delta$ in (14) in particular implies that

$\delta(n,\lambda)\leq\Lambda(n,\lambda)^{-1}\left(\frac{1}{250n\lambda}\right)^{\frac{n}{2}},$

then by Lemma 2.3 and Remark 2.4, we obtain (with $C_{2}(n)=1000n$) the
uniform Sobolev inequality

(17) $\operatorname{Sob}(B_{g(t)}(y,r),g(t))\leq[C_{2}(n)\lambda
v_{x,r}^{-\frac{2}{n}},1]$

whenever $y\in B_{g(0)}(x,r)$, which gives conclusion (vi). Note that by Lemma
2.1, we can also say that

(18) $\text{Vol}_{g(t)}B_{g(t)}(y,\rho)\geq
c_{1}(n)\lambda^{-\frac{n}{2}}v_{x,r}\rho^{n}\;\;\text{ whenever }\;y\in
B_{g(0)}(x,r),\;\rho\in(0,r],\;t\in[0,T_{1}\land T^{\prime}].$

This gives conclusion (iv) since $v_{x,r}\geq v_{0}$. Furthermore, note that
$c_{1}(n)\geq(2^{n+4}\cdot C_{2}(n))^{-\frac{n}{2}}$. We also point out that
although the work in this step was valid for any $r\in[r_{0},1]$, we will only
need to apply it with $r=r_{0}$ and $r=1$.

Step 2 (Curvature bound)

In this step of the proof, we will show that the flow $g(t)$ satisfies the
curvature bound (16) as long as (15) holds on the time interval
$[0,T^{\prime}]$, assuming that $T^{\prime}\leq T_{1}$ where $T_{1}$ depends
only on $n,\lambda,v_{0}$. Just like the proof of [2, Proposition 2.1], we
could do this by applying Moser’s iteration, which is well-known to experts
but is a somewhat technical argument, especially in our setting where we must
prove that the smallness of $\int|\text{Rm}|^{\frac{n}{2}}$ need only depend
linearly on $v$. Instead, we will argue by contradiction by employing a
limiting contradiction argument with respect to the existence time. Thus we
may assume that for some $n\geq 3$, $\lambda\geq 1$, $v_{0}>0$, there exists a
sequence of times $T_{i}\to 0$ and associated bounded curvature Ricci flows
$(M_{i}^{n},g_{i}(t))$, $t\in[0,T_{i}]$ each satisfying the hypotheses of the
theorem and such that

$\int_{B_{g_{i}(t)}(x,1)}|\text{Rm}|_{g_{i}(t)}^{\frac{n}{2}}\,dV_{g_{i}(t)}\leq\Lambda\delta
v_{0}$

for all $(x,t)\in M_{i}\times[0,T_{i}]$, but which do not enjoy a $\alpha/t$
curvature bound on the interval $[0,T_{i}]$, where $\alpha=3r_{0}/10$. Since
each $(M_{i},g_{i}(t))$, $t\in[0,T_{i}]$ is assumed to be a bounded curvature
flow, we may assume that there is some _first time_ $t_{i}\in[0,T_{i}]$ which
fails our desired bound. In other words we assume that

$|\text{Rm}|_{g_{i}(t_{i})}(x_{i})=\frac{\alpha}{t_{i}}$

for some $x_{i}\in M_{i}$ and that

$|\text{Rm}|_{g_{i}(t)}\leq\frac{\alpha}{t}\;\text{ on }M_{i}\times(0,t_{i}].$

Moreover by Step 1, we know that the Sobolev inequality (17) and volume ratio
bound (18) hold with $r=1$, $\rho=\sqrt{t}$, $v_{x,1}=v_{0}$. This volume
ratio estimate, together with the assumed $\frac{1}{t}$ curvature bound on
$(0,t_{i}]$ imply (by [7]) the injectivity radius bound

$\operatorname{inj}_{g_{i}(t)}(x)\geq\iota(n,\lambda,v_{0})\sqrt{t}\;\text{
for all }x\in M_{i},\;t\in(0,t_{i}].$

This gives conclusion (iii). Now parabolically rescale the $g_{i}(t)$ so that
$t_{i}\nearrow 1$, which is to say we define the sequence of complete flows
$\tilde{g}_{i}(t)=\frac{1}{t_{i}}g_{i}(t_{i}t)$, each defined on $[0,1]$,
which satisfy

(19) $\begin{cases}&\text{Rc}_{\tilde{g}_{i}(0)}\geq-
t_{i},\;\;{\mathcal{R}}_{\tilde{g}_{i}(t)}\geq-nt_{i},\\\
&|\text{Rm}|_{\tilde{g}_{i}(1)}(x_{i})=\alpha,\\\
&|\text{Rm}|_{\tilde{g}_{i}(t)}\leq\frac{\alpha}{t}\;\text{ for all
}t\in(0,1],\\\
&\int_{\Omega_{i}}|\text{Rm}|_{\tilde{g}_{i}(1)}^{\frac{n}{2}}\,dV_{\tilde{g}_{i}(1)}\leq\Lambda\delta
v_{0},\\\
&\operatorname{inj}_{\tilde{g}_{i}(1)}(x_{i})\geq\iota(n,\lambda,v_{0}),\text{
and }\\\ &\operatorname{Sob}(\Omega_{i},\tilde{g}(1))\leq[C_{2}(n)\lambda
v_{0}^{-\frac{2}{n}},t_{i}].\end{cases}$

Here we have denoted $\Omega_{i}=B_{\tilde{g}(1)}(x,t_{i}^{-\frac{1}{2}})$.
Therefore we may apply Hamilton’s Compactness Theorem [14] and a diagonal
argument to say that $(M_{i},\tilde{g}_{i}(t),x_{i})$ subconverges (in the
pointed Cheeger-Gromov sense) to some complete Ricci flow
$(N,g_{\infty}(t),x_{\infty})$, $t\in(0,1]$, which by virtue of the properties
(19), must satisfy

$\begin{cases}&{\mathcal{R}}_{g_{\infty}(t)}\geq 0\;\text{ for all
}t\in(0,1],\\\ &|\text{Rm}|_{g_{\infty}(1)}(x_{\infty})=\alpha,\\\ &\sup_{x\in
N}|\text{Rm}|_{g_{\infty}(t)}(x)\leq\frac{\alpha}{t}\;\text{ for all
}t\in(0,1],\\\
&\int_{N}|\text{Rm}|_{g_{\infty}(1)}^{\frac{n}{2}}\,dV_{g_{\infty}(1)}\leq\Lambda\delta
v_{0},\;\text{ and }\\\
&\operatorname{Sob}(N,g_{\infty}(1))\leq[C_{2}(n)\lambda
v_{0}^{-\frac{2}{n}},0].\end{cases}$

Because $C_{2}(n)=1000n$, our requirement on $\delta(n,\lambda)$ in (14)
implies that

(20)
$\delta(n,\lambda)\leq\Lambda(n,\lambda)^{-1}\left(\frac{\delta(n)}{C_{2}(n)\lambda}\right)^{\frac{n}{2}}$

where $\delta(n)\geq\frac{1}{10^{4}n^{2}}$ is the dimensional constant from
[5, Theorem 1.2]. Then just as in the introduction (specifically, the proof of
Theorem 1.2), we may conclude that for each $t\in(0,1]$, there holds

(21)
$\lim_{r\to\infty}\frac{\text{Vol}_{g_{\infty}(t)}B_{g_{\infty}(t)}(x_{\infty},r)}{r^{n}}\geq\omega_{n}.$

Now we have the following claim.

######  Claim 4.4.

For any $A\geq 1$, there exists $N_{A}\in{\mathbb{N}}$ such that whenever
$i\geq N_{A}$, there holds

$\inf_{0<t\leq
1}\mu(B_{\tilde{g}_{i}(0)}(x_{i},20A\sqrt{t}),\tilde{g}_{i}(0),t)\geq-A^{-2}.$

This claim will be sufficient derive a contradiction because we may take
$A=A(\alpha/2,n)$ from [21, Theorem 4.5], which, in combination with (19),
would yield the absurd inequality

$\alpha=|\text{Rm}|_{\tilde{g}_{i}(1)}(x_{i})\leq\frac{\alpha}{2}.$

We will now prove the claim.

###### Proof of Claim.

For any given $A\geq 1$ define $\eta=A^{-2}$ and let $\xi(\eta,n,A)>0$ be the
constant from [21, Lemma 4.10]. It will suffice to show that for all $i$
sufficiently large, we in fact have

(22) $\begin{cases}&\text{Rc}_{\tilde{g}_{i}(0)}(x)\geq-\xi\;\text{ for all
}x\in M;\\\
&\text{Vol}_{\tilde{g}_{i}(0)}B_{\tilde{g}_{i}(0)}(x_{i},r)\geq(1-\xi)\omega_{n}r^{n}\;\text{
for all }r\in(0,\xi].\end{cases}$

Indeed, for any $t\in(0,1]$, the metric $\frac{1}{t}\tilde{g}_{i}(0)$ will
then also satisfy these conditions which implies

$-A^{-2}\leq\mu\left(B_{\frac{1}{t}\tilde{g}_{i}(0)}(x_{i},20A),\frac{1}{t}\tilde{g}_{i}(0),1\right)=\mu\left(B_{\tilde{g}_{i}(0)}(x_{i},20A\sqrt{t}),\tilde{g}_{i}(0),t\right)$

where we have applied [21, Lemma 4.10] in the first inequality while the
second follows from the scaling properties of the $\mu$-functional.

First let $i$ be large enough to ensure that $t_{i}\leq\xi$, which shows the
the lower Ricci bound condition. For the volume condition, we begin by
defining $\hat{\xi}>0$ by $(1-\hat{\xi})^{4}=(1-\xi)$. Fix $s\in(0,1]$ small
enough so that

$R^{n}\geq(1-\hat{\xi})(R+\beta\sqrt{\alpha s})^{n}\;\text{ for all
}R\geq\xi^{-1},$

where $\beta=\beta(n)$ from Lemma 3.1. Now by the convergence
$\tilde{g}_{i}(s)\to g_{\infty}(s)$ and the asymptotic volume estimate for
$g_{\infty}(s)$ as in (21), we can fix some $R_{0}\geq\xi^{-1}$ such that

$\text{Vol}_{\tilde{g}_{i}(s)}B_{\tilde{g}_{i}(s)}(x_{i},R_{0})\geq(1-\hat{\xi})\omega_{n}R_{0}^{n}.$

We also assume that $i$ is large enough so that $e^{nst_{i}}\leq 1+\hat{\xi}$.
Finally by Bishop-Gromov Volume Comparison, we can say that for all $i$ large
enough and $r\in(0,R_{0}+\beta\sqrt{\alpha s}]$, there holds

$\frac{\text{Vol}_{\tilde{g}_{i}(0)}B_{\tilde{g}_{i}(0)}(x_{i},r)}{r^{n}}\geq(1-\hat{\xi})\frac{\text{Vol}_{\tilde{g}_{i}(0)}B_{\tilde{g}_{i}(0)}(x_{i},R_{0}+\beta\sqrt{\alpha
s})}{(R_{0}+\beta\sqrt{\alpha s})^{n}}.$

Then for any $r\in(0,\xi^{-1}]$, these conditions on the largeness of $i$
imply

$\displaystyle
r^{-n}\text{Vol}_{\tilde{g}_{i}(0)}B_{\tilde{g}_{i}(0)}(x_{i},r)$
$\displaystyle\geq(1-\hat{\xi})(R_{0}+\beta\sqrt{\alpha
s})^{-n}\text{Vol}_{\tilde{g}_{i}(0)}B_{\tilde{g}_{i}(0)}(x_{0},R_{0}+\beta\sqrt{\alpha
s})$ $\displaystyle\geq(1-\hat{\xi})^{2}(R_{0}+\beta\sqrt{\alpha
s})^{-n}\text{Vol}_{\tilde{g}_{i}(s)}B_{\tilde{g}_{i}(0)}(x_{0},R_{0}+\beta\sqrt{\alpha
s})$
$\displaystyle\geq(1-\hat{\xi})^{3}R_{0}^{-n}\text{Vol}_{\tilde{g}_{i}(s)}B_{\tilde{g}_{i}(s)}(x_{0},R_{0})$
$\displaystyle\geq(1-\hat{\xi})^{4}\omega_{n}=(1-\xi)\omega_{n}.$

This shows that condition (22) holds and complete the proof of the claim, and
thus of Step 2 as well. ∎

Step 3 (Uniform time)

We have already seen in Steps 1 and 2 that conclusions (i), (ii), (iii), (iv),
and (vi) all follow as long as conclusion (v) holds on $[0,T^{\prime}]$ if
$T^{\prime}\leq T\land\widetilde{T}$. We will assume that
$T^{\prime}<T\land\widetilde{T}$ then show that this contradicts the
maximality of $T^{\prime}$ which will complete the proof of the theorem.

Start by fixing some $x_{0}\in M$ and let $\phi:M\to{\mathbb{R}}$ be defined
as in [2]. That is $\phi(x,t)=e^{-10t}\varphi(\eta(x,t))$ where
$\eta(x,t)=d_{g(t)}(x,x_{0})+C_{3}(n)\sqrt{t}$ and
$\varphi:{\mathbb{R}}\to{\mathbb{R}}$ is a smooth cutoff function with
$\varphi\equiv 1$ on $(-\infty,\frac{1}{2}]$, $\varphi\equiv 0$ on
$[1,\infty)$, $\varphi^{\prime\prime}\geq-10\varphi$,
$0\geq\varphi^{\prime}\geq-10\sqrt{\varphi}$. Moreover $C_{3}(n)$ is chosen
large enough so that $\square\phi\leq 0$. With this choice of $\phi$ in Lemma
3.4, we have

$\begin{split}\frac{d}{dt}\int_{M}&\phi^{2}(|\text{Rm}|_{g(t)}^{2}+\beta)^{\frac{n}{4}}\,dV_{g(t)}\leq-\frac{1}{3}\int_{M}|\nabla^{g(t)}(\phi(|\text{Rm}|_{g(t)}^{2}+\beta)^{\frac{n}{8}})|_{g(t)}^{2}\,dV_{g(t)}\\\
&+5n\int_{M}\phi^{2}(|\text{Rm}|_{g(t)}^{2}+\beta)^{\frac{n}{4}+\frac{1}{2}}\,dV_{g(t)}+1000n\int_{\text{supp}\,\phi}(|\text{Rm}|_{g(t)}^{2}+\beta)^{\frac{n}{4}}\,dV_{g(t)}.\end{split}$

Now to the terms on the RHS, we apply our Sobolev inequality (17) (here $r=1$
and recall that $v_{x,1}\geq v_{0}$) to the first and Hölder’s inequality to
the second to see that

(23)
$\begin{split}\frac{d}{dt}\int_{M}\phi^{2}&(|\text{Rm}|_{g(t)}^{2}+\beta)^{\frac{n}{4}}\,dV_{g(t)}\leq-\frac{1}{3C_{2}(n)\lambda
v_{0}^{-\frac{2}{n}}}\left(\int_{B_{g(t)}(x,1)}|\phi(|\text{Rm}|_{g(t)}^{2}+\beta)^{\frac{n}{8}}|^{\frac{2n}{n-2}}\,dV_{g(t)}\right)^{\frac{n-2}{n}}\\\
&+5n\left(\int_{\text{supp}\phi}(|\text{Rm}|_{g(t)}^{2}+\beta)^{\frac{n}{4}}\,dV_{g(t)}\right)^{\frac{2}{n}}\left(\int_{B_{g(t)}(x,1)}|\phi(|\text{Rm}|_{g(t)}^{2}+\beta)^{\frac{n}{8}}|^{\frac{2n}{n-2}}\,dV_{g(t)}\right)^{\frac{n-2}{n}}\\\
&+1001n\int_{\text{supp}\phi}(|\text{Rm}|_{g(t)}^{2}+\beta)^{\frac{n}{4}}\,dV_{g(t)}.\end{split}$

Then integrate over $[0,t]$ on both sides and send $\beta\to 0$. Because
$\text{supp}\,\phi\subset B_{g(t)}(x_{0},1)$, our restriction on
$\delta(n,\lambda)$ in (14) ensures that the sum of the two terms containing
$\|\phi^{2}|\text{Rm}|^{\frac{n}{2}}\|_{L^{\frac{n}{n-2}}}$ in (23) is
nonpositive. Therefore we have

$\int_{M}\phi|\text{Rm}|_{g(t)}^{\frac{n}{2}}\,dV_{g(t)}\leq\int_{M}\phi|\text{Rm}|_{g(0)}^{\frac{n}{2}}\,dV_{g(0)}+1001n\Lambda\delta
v_{0}t\leq\delta v_{0}+1001n\Lambda\delta v_{0}t=:(1+C_{4}(n,\lambda)t)\delta
v_{0}.$

Now we multiply through by $e^{10t}$ and notice whenever $t\leq
T^{\prime}\leq\widetilde{T}$, the conditions in (13) imply $e^{10t}\leq 2$,
$C_{4}(n,\lambda)t\leq 1$, and $\phi\equiv 1$ on $B_{g(t)}(x_{0},1/4)$. Then
for all such $t\in[0,T^{\prime}]$, there holds

(24) $\int_{B_{g(t)}(x_{0},\frac{1}{4})}|\text{Rm}|^{\frac{n}{2}}\,dV_{t}\leq
4\delta v_{0}.$

Now fix some $(y,t)\in M\times(0,T^{\prime}]$ and let $\\{x_{j}\\}\subset
B_{g(t)}(y,3/2)$ be some subset of points such that
$\\{B_{g(t)}(x_{j},\frac{1}{8})\\}_{j=1}^{N}$ are mutually disjoint, but so
that

(25) $\bigsqcup_{i=1}^{N}B_{g(t)}(x_{i},1/8)\subset B_{g(t)}(y,7/4)\;\;\text{
and }\;\;B_{g(t)}(y,3/2)\subset\bigcup_{i=1}^{N}B_{g(t)}(x_{i},1/4).$

Then by the Shrinking Balls Lemma 3.1, since
$T^{\prime}\leq\widetilde{T}\leq\frac{1}{16\beta^{2}}$, we have

(26) $B_{g(t)}(y,7/4)\subset B_{g(0)}(y,2).$

We would like to be able to say that we a similar reverse containment as well.
For this, we will be enlisting the help of Lemma 3.2. Recall we have a
$\alpha/t$ curvature bound by Step 2, a lower volume bound (18) by Step 1, and
an assumed lower bound on scalar curvature (a). In summary, for all $x\in M$,
there holds

(27) $\begin{cases}&{\mathcal{R}}_{g(0)}(x)\geq-n;\\\
&|\text{Rm}|_{g(t)}(x)\leq\frac{\alpha}{t}\;\text{ for all
}t\in[0,T^{\prime}];\\\
&\text{Vol}_{g(0)}B_{g(0)}(x,r_{0})=:v_{x,r_{0}}r_{0}^{n};\;\text{ and }\\\
&\text{Vol}_{g(t)}B_{g(t)}(y,r)\geq
c_{1}(n)\lambda^{-\frac{n}{2}}v_{x,r_{0}}r^{n}\;\text{ for all }y\in
B_{g(0)}(x,r_{0}),r\in(0,r_{0}],\;t\in[0,T^{\prime}].\end{cases}$

Now we parabolically rescale $g(t)$ by definining

$\tilde{g}(t)=r_{0}^{-2}g\left(r_{0}^{2}t\right)$

where $r_{0}(n,\lambda)=\frac{c_{1}(n)}{2^{3n+6}\lambda^{\frac{n}{2}}}$. Note
that this rescaling has scaled distances $r_{0}\to 1$ and
$\frac{1}{8}\to\frac{2^{3n+3}\lambda^{\frac{n}{2}}}{c_{1}(n)}$. Therefore, if
we write $T_{2}=r_{0}^{-2}T^{\prime}$, we can see from the properties in (27),
and our volume doubling estimate (b), that the flow $\tilde{g}(t)$, in
particular, satisfies

$\begin{cases}&{\mathcal{R}}_{\tilde{g}(0)}(x)\geq-1;\\\
&|\text{Rm}|_{\tilde{g}(t)}(x)\leq t^{-1}\;\text{ for all }t\in[0,T_{2}];\\\
&\text{Vol}_{\tilde{g}(0)}B_{\tilde{g}(0)}(x,2)\leq
2^{2n+1}v_{x,r_{0}};\;\text{ and }\\\
&\text{Vol}_{\tilde{g}(t)}B_{\tilde{g}(t)}(y,r)\geq
c_{1}(n)\lambda^{-\frac{n}{2}}v_{x,r_{0}}r^{n}\;\text{ for all }y\in
B_{\tilde{g}(0)}(x,1),r\in(0,1/2],\;t\in[0,T_{2}].\end{cases}$

Then Lemma 3.2 implies that $B_{\tilde{g}(0)}(x,1)\subset
B_{\tilde{g}(t)}(x,2^{n+2}2^{2n+1}\lambda^{\frac{n}{2}}c_{1}(n)^{-1})$ for all
$x\in M$ and $t\in[0,T_{2}\land\widehat{T}(n,1,1)]$. Scaling back and
recalling the last condition on the smallness of $\widetilde{T}$ in (13), we
can see that for all $i\in\\{1,\dots,N\\}$ there holds

$B_{g(0)}(x_{i},r_{0})\subset B_{g(t)}(x_{i},1/8).$

Now starting from $r=r_{0}$, we apply our volume doubling estimate (b) $k$
times, where $k(n,\lambda)$ is the unique integer so that so that

$4\leq 2^{k}r_{0}<8.$

Recalling that $x_{i}\in B_{g(t)}(y,3/2)$ and the containment (26), the
doubline estimate yields

$2^{(2n+1)k}\text{Vol}_{g(0)}B_{g(0)}(x_{i},r_{0})\geq\text{Vol}_{g(0)}B_{g(0)}(x_{i},2^{k}r_{0})\geq\text{Vol}_{g(0)}B_{g(0)}(y,2)\geq\text{Vol}_{g(0)}B_{g(t)}(y,7/4).$

On the other hand, the $\\{B_{g(0)}(x_{i},r_{0})\\}_{i=1}^{N}$ are disjoint
and each is contained within $B_{g(t)}(y,7/4)$ by (25). Therefore

$N2^{-(2n+1)k}\text{Vol}_{g(0)}B_{g(t)}(y,7/4)\leq\sum_{i=1}^{N}\text{Vol}_{g(0)}B_{g(0)}(x_{i},r_{0})\leq\text{Vol}_{g(0)}B_{g(t)}(y,7/4)$

from which we can readily conclude that $N\leq 2^{(2n+1)k}$. So if we define
$\Lambda$ by

(28) $\Lambda(n,\lambda):=8N\leq
8\left(\frac{2^{3n+9}\lambda^{\frac{n}{2}}}{c_{1}(n)}\right)^{2n+1}\leq
8\left(2^{3n+9}\lambda^{\frac{n}{2}}(2^{n+4}1000n)^{\frac{n}{2}}\right)^{2n+1},$

then we can combine (24) and (25) to see that for all $(y,t)\in
M\times[0,T^{\prime}]$, there holds

(29)
$\int_{B_{g(t)}(y,3/2)}|\text{Rm}|^{\frac{n}{2}}\,dV_{t}\leq\sum_{i=1}^{N}\int_{B_{g(t)}(x_{i},\frac{1}{4})}|\text{Rm}|^{\frac{n}{2}}\,dV_{t}\leq
4N\delta v_{0}=\frac{1}{2}\Lambda\delta v_{0}.$

But since $(M,g(t))$ is a bounded curvature flow, condition (29) (via an
application of Lemma 4.1) is in contradiction to the maximality of
$T^{\prime}$ when $T^{\prime}<T\land\widetilde{T}$. This completes the proof.
∎

##  5\. Proof of Theorem 1.4

Before beginning the proof of Theorem 1.4, we need to recall some of the
details from [3]. The following is taken out of the proof of their main
theorem.

######  Theorem 5.1 ([3, Proof of Theorem 1.1]).

Let $(M^{n},g_{0},x_{0})$ be some complete pointed noncompact Riemannian
manifold of dimension $n\geq 4$ (not assumed bounded curvature), which
exhibits a smooth function $\rho:M\to{\mathbb{R}}$ such that for some $C>0$
and all $x\in M$, there holds:

(30) $\begin{cases}&C^{-1}d_{g_{0}}(x_{0},x)-C\leq\rho(x)\leq
d_{g_{0}}(x_{0},x);\\\ &|\nabla\rho(x)|^{2}+|\Delta\rho(x)|\leq C;\text{ and
}\\\ &\int_{B_{g_{0}}(x,1)}|\nabla^{2}\rho|^{n}\,dV_{g_{0}}\leq C.\end{cases}$

Then there exists a sequence sets of $U_{j}\subset M$ exhausting $M$, and
metrics $g_{j}$ on $U_{j}$ such that: each $(U_{j},g_{j})$ is compete with
bounded curvature; for any compact $\Omega\subset M$,
$g_{j}|_{\Omega}=g_{0}|_{\Omega}$ for all $j$ large enough; and for each $x\in
U_{j}$, the metric $g_{j}$ satisfy the locally equivalent condition

(31) $c_{x,j}^{2}g_{0}\leq g_{j}\leq C_{x,j}^{2}g_{0}\text{ on
}B_{g_{j}}(x,1)\text{ for some }C_{x,j}\geq c_{x,j}\geq 1\text{ satisfying
}\left(\frac{C_{x,j}}{c_{x,j}}\right)^{n}\leq 2.$

Moreover, if $\inf_{z\in M}\text{Rc}_{g_{0}}(z)>-\infty$ then for any
$\lambda,L,v_{0},\alpha$, we may assume that each of the following
implications holds (by possibly deleting finitely many of the $g_{j}$):

  1. (a)

If $\text{Rc}_{g_{0}}(x)\geq-\lambda$ for all $x\in M$, then
$\text{Rc}_{g_{j}}(x)\geq-2\lambda$ for all $x\in U_{j}$;

  2. (b)

If $\text{Vol}_{g_{0}}B_{g_{0}}(x,2r)\leq
L\cdot\text{Vol}_{g_{0}}B_{g_{0}}(x,r)\leq Lv_{0}r^{n}$ for all $x\in M$ and
$r\leq 1$, then $\text{Vol}_{g_{j}}B_{g_{j}}(x,2r)\leq
2L^{2}\cdot\text{Vol}_{g_{j}}B_{g_{j}}(x,r)\leq 4L^{2}v_{0}r^{n}$ for all
$x\in U_{j}$, $r\leq 1/2$; and

  3. (c)

If
$\int_{B_{g_{0}}(x,1)}|\text{Rm}|_{g_{0}}^{\frac{n}{2}}\,dV_{g_{0}}\leq\alpha$
for all $x\in M$, then
$\int_{B_{g_{j}}(x,1)}|\text{Rm}|_{g_{j}}^{\frac{n}{2}}\,dV_{g_{j}}\leq
4\alpha$ for all $x\in U_{j}$.

######  Remark 5.2.

See [3, Claims 4.2, 4.4] for the proofs of properties (b) and (c) respectively
(note that we have assumed the $\kappa$ there is small enough so that
$(1+c_{2}\kappa)^{n}\leq 2$ which is valid because $c_{2}$ was an absolute
constant). Note that in [3], implication (a) is replaced with the implication
on scalar curvature
${\mathcal{R}}_{g_{0}}\geq-\lambda\Rightarrow{\mathcal{R}}_{g_{j}}\geq-2\lambda$,
but the implication on Ricci curvature is more natural in our setting and
follows in identical fashion. In particular, using the notation from [3] of
$g_{j}=g_{R,0}=e^{2F_{R}}g_{0}$ for $R$ sufficiently large, and the formula
for conformal change of Ricci curvature, we have

$\displaystyle\text{Rc}_{g_{R,0}}(v,v)$
$\displaystyle=\text{Rc}_{g_{0}}(v,v)-(n-2)\left(\nabla^{g_{0}}_{v}\nabla^{g_{0}}_{v}F_{R}-\nabla^{g_{0}}_{v}F_{R}\nabla^{g_{0}}_{v}F_{R}\right)-\left(\Delta^{g_{0}}F_{R}+(n-2)|dF_{R}|_{g_{0}}^{2}\right)g_{0}(v,v)$
$\displaystyle\geq-\lambda
g_{0}(v,v)-\frac{C_{n}}{R^{2}}e^{2F_{R}}g_{0}(v,v)=\left(-e^{-2F_{R}}\lambda-\frac{C_{n}}{R^{2}}\right)e^{2F_{R}}g_{0}(v,v)\geq-2\lambda
g_{R,0}(v,v).$

We will be applying Theorem 5.1 with the underlying manifold $(M,g_{0})$
having nonnegative Ricci curvature and positive asymptotic volume ratio. In
order to even get started, we clearly we must first demonstrate the existence
of a distance-like function $\rho$ with the desired properties.

######  Lemma 5.3.

Let $(M^{n},g,x_{0})$, $n\geq 4$ be a complete pointed Riemannian manifold
with $\text{Rc}_{g}\geq 0$, $\operatorname{AVR}(g)>0$, and $|\text{Rm}|_{g}\in
L^{\frac{n}{2}}(M,g)$. Then there exists a smooth $\rho:M\to{\mathbb{R}}$ and
a constant $C>0$ such that condition (30) holds uniformly for all $x\in M$.

###### Proof.

First note by Corollary 2.11 and Lemma 2.2, there holds

$\bar{\nu}(M,g)\geq\log(C_{1}(n)^{-1}\operatorname{AVR}(g))=:-A>-\infty.$

Now since $|\text{Rm}|_{g}\in L^{\frac{n}{2}}$, [3, Lemma 4.2] implies the
existence of some $r\in(0,1]$ such that

$\sup_{x\in
M}\int_{B_{g}(x,r)}|\text{Rm}|_{g}^{\frac{n}{2}}\,dV_{g}\leq\delta(n,A,1)^{\frac{n}{2}}$

where $\delta(n,A,1)$ is from [3, Theorem 2.1]. Then that result applied to
the scaled up metric $\tilde{g}=r^{-2}g$ implies the existence of a smooth
$\tilde{g}$-distance-like function $\tilde{\rho}$ which satisfies the
properties in (30) relative to $\tilde{g}$ for some $C>0$. Then by taking
$\rho=r\tilde{\rho}$ we can see that $\rho$ is a $g$-distance-like function
satisfying (30) relative to $g$ for some (possibly different) $C>0$, albeit
with the integral condition replaced by

$\int_{B_{g}(x,r)}|\nabla^{2}\rho|^{n}\,dV_{g}\leq C.$

Now for any $x\in M$, let $\\{x_{i}\\}_{i=1}^{N}\subset B_{g}(x,1)$ be some
set of points such that $\\{B_{g}(x_{i},r/2)\\}_{i=1}^{N}$ are mutually
disjoint but so that

$\bigsqcup_{i=1}^{N}B_{g}(x_{i},r/2)\subset B_{g}(x,2)\;\;\text{ and
}\;\;B_{g}(x,1)\subset\bigcup_{i=1}^{N}B_{g}(x_{i},r).$

Then by the assumption of $\text{Rc}_{g}\geq 0$ and volume comparison, we have

$\omega_{n}2^{n}\geq\text{Vol}_{g}B_{g}(x,2)\geq\sum_{i=1}^{N}\text{Vol}_{g}B_{g}(x_{i},r/2)\geq
N\omega_{n}\operatorname{AVR}(g)(r/2)^{n}.$

Therefore by the nature of the covering, there holds

$\int_{B(x,1)}|\nabla^{2}\rho|^{n}\,dV_{g}\leq\sum_{i=1}^{N}\int_{B_{g}(x,r)}|\nabla^{2}\rho|^{n}\,dV_{g}\leq
NC\leq 4^{n}r^{-n}\operatorname{AVR}(g)^{-1}C.$

This completes the proof of the lemma. ∎

We therefore have the following approximation result with _bounded curvature_
metrics, from which we will readily apply our short-time existence result
Theorem 4.2.

######  Proposition 5.4.

For any $n\geq 4$ and $v>0$, there exists a dimensional constant $C(n)\geq 1$
such that the following holds. Let $(M^{n},g)$ be a complete noncompact
Riemannian manifold (not assumed bounded curvature) with $\text{Rc}_{g}\geq
0$, $\operatorname{AVR}(g)\geq v$, and

$\int_{M}|\text{Rm}|^{\frac{n}{2}}_{g}\,dV_{g}\leq\alpha$

for some $\alpha>0$. Then there exists a sequence sets of $U_{j}\subset M$
exhausting $M$, and metrics $g_{j}$ on $U_{j}$ such that each $(U_{j},g_{j})$
is compete with bounded curvature and for any compact $\Omega\subset M$,
$g_{j}|_{\Omega}=g|_{\Omega}$ for all $j$ large enough. Moreover each of the
following conditions holds uniformly for every $x\in U_{j}$:

  1. (i)

$\text{Rc}_{g_{j}}(x)\geq-1$;

  2. (ii)

$\text{Vol}_{g_{j}}B_{g_{j}}(x,2r)\leq
2^{2n+1}\text{Vol}_{g_{j}}B_{g_{j}}(x,r)\leq 2^{2n+2}\omega_{n}r^{n}$ for each
$r\leq 4$;

  3. (iii)

$\int_{B_{g_{j}}(x,4)}|\text{Rm}|_{g_{j}}^{\frac{n}{2}}\,dV_{g_{j}}\leq
4\alpha$;

  4. (iv)

$\frac{\text{Vol}_{g_{j}}B_{g_{j}}(x,r)}{r^{n}}\geq\frac{\omega_{n}}{2}v$ for
each $r\leq 4$; and

  5. (v)

$\operatorname{Sob}(B_{g_{j}}(x,r),g_{j})\leq\left[C(n)\left(\frac{\text{Vol}_{g_{j}}B_{g_{j}}(x,r)}{r^{n}}\right)^{-\frac{2}{n}},0\right]$
for each $r\leq 4$.

###### Proof.

By the scale invariance of the hypotheses and each of the conclusions save
(i), we may prove the theorem with $g_{0}=\frac{1}{32}g$ which allows us to
only require that conditions (ii), (iii), (iv), (v) only hold for scales
$r\leq 1/2$ (or larger). Clearly Lemma 5.3 allows us to apply Theorem 5.1,
which gives the existence of the $(U_{j},g_{j})$. Moreover, properties (i),
(ii), (iii) follow immediately from Theorem 5.1 as well because
$\text{Rc}_{g_{0}}\geq 0$ and volume comparison implies

$\frac{\text{Vol}_{g_{0}}B_{g_{0}}(x,2r)}{(2r)^{n}}\leq\frac{\text{Vol}_{g_{0}}B_{g_{0}}(x,r)}{r^{n}}\leq\omega_{n}$

for all $r>0$. Now fix some $j\in{\mathbb{N}}$, $x\in U_{j}$, $r\in(0,1]$ and
write $\frac{\text{Vol}_{g_{j}}B_{g_{j}}(x,r)}{r^{n}}=v_{xjr}$. By the locally
equivalent condition (31) and the fact that $\operatorname{AVR}(g_{0})\geq v$,
we obtain

$v_{xjr}\geq c_{x,j}^{n}\frac{\text{Vol}_{g_{0}}B_{g_{j}}(x,r)}{r^{n}}\geq
c_{x,j}^{n}\frac{\text{Vol}_{g_{0}}B_{g_{0}}(x,C_{x,j}^{-1}r)}{r^{n}}\geq\frac{1}{2}\frac{\text{Vol}_{g_{0}}B_{g_{0}}(x,C_{x,j}^{-1}r)}{(C_{x,j}^{-1}r)^{n}}\geq\frac{\omega_{n}}{2}v,$

which proves (iv). Now it remains to show (v). Just as above, we may obtain an
upper volume ratio bound:

$v_{xjr}\leq C_{x,j}^{n}\frac{\text{Vol}_{g_{0}}B_{g_{j}}(x,r)}{r^{n}}\leq
C_{x,j}^{n}\frac{\text{Vol}_{g_{0}}B_{g_{0}}(x,c_{x,j}^{-1}r)}{r^{n}}\leq
2\frac{\text{Vol}_{g_{0}}B_{g_{0}}(x,c_{x,j}^{-1}r)}{(c_{x,j}^{-1}r)^{n}}.$

Because $\text{Rc}_{g_{0}}\geq 0$ and $C_{x,j}^{-1}r\leq c_{x,j}^{-1}r$, we
may once again use volume comparison to say that

$\frac{1}{2}\frac{\text{Vol}_{g_{0}}B_{g_{0}}(x,c_{x,j}^{-1}r)}{(c_{x,j}^{-1}r)^{n}}\leq
v_{xjr}\leq
2\frac{\text{Vol}_{g_{0}}B_{g_{0}}(x,c_{x,j}^{-1}r)}{(c_{x,j}^{-1}r)^{n}}\;\iff\;\frac{\text{Vol}_{g_{0}}B_{g_{0}}(x,c_{x,j}^{-1}r)}{(c_{x,j}^{-1}r)^{n}}\in\left[\frac{1}{2}v_{xjr},2v_{xjr}\right].$

Now by Lemma 2.5, we obtain the Sobolev inequality

$\operatorname{Sob}(B_{g_{0}}(x,rc_{x,j}^{-1}),g_{0})\leq\left[C_{1}(n)\left(\frac{v_{xjr}}{2}\right)^{-\frac{2}{n}},0\right]$

for some dimensional constant $C_{1}(n)\geq 1$. Since $B_{g_{j}}(x,r)\subset
B_{g_{0}}(x,rc_{x,j}^{-1})$, we may again apply the locally equivalent
condition (31) to obtain Sobolev inequality for $g_{j}$:

$\displaystyle\left(\int_{B_{g_{j}}(x,r)}|u|^{\frac{2n}{n-2}}\,dV_{g_{j}}\right)^{\frac{n-2}{n}}$
$\displaystyle\leq\left(\frac{C_{x,j}}{c_{x,j}}\right)^{n-2}C_{1}(n)\left(\frac{v_{xjr}}{2}\right)^{-\frac{2}{n}}\int_{B_{g_{j}}(x,r)}|\nabla^{g_{j}}u|^{2}_{g_{j}}\,dV_{g_{j}}$
$\displaystyle\leq
2C_{1}(n)v_{xjr}^{-\frac{2}{n}}\int_{B_{g_{j}}(x,r)}|\nabla^{g_{j}}u|^{2}_{g_{j}}\,dV_{g_{j}}$

for all $u\in W_{0}^{1,2}(B_{g_{j}}(x,r))$. This gives (v) and completes the
proof. ∎

###### Proof of Theorem 1.4.

Given $n\geq 4$ and $v\in(0,1]$, let $(M^{n},g)$ satisfy the hypotheses of
Theorem 1.2 with this choice of $n,v$ and fix some $x_{0}\in M$. For any
$R\geq 1$, denote $g_{R}=R^{-2}g$, which still satisfies
$\operatorname{AVR}(g_{R})\geq v$ and

$\int_{M}|\text{Rm}|_{g_{R}}^{\frac{n}{2}}\,dV_{g_{R}}\leq\varepsilon(n)v.$

Then by Proposition 5.4, we obtain a sequence of sets $U_{R,j}\subset M$
exhausting $M$ and complete bounded curvature metrics $g_{R,j}$ on $U_{R,j}$
which satisfy $g_{R,j}=g_{R}$ on $B_{g_{R}}(x_{0},1)$. Moreover each of the
following conditions holds uniformly for every $x\in U_{R,j}$:

  1. (i)

$\text{Rc}_{g_{R,j}}(x)\geq-1$;

  2. (ii)

$\text{Vol}_{g_{R,j}}B_{g_{R,j}}(x,2r)\leq
2^{2n+1}\text{Vol}_{g_{R,j}}B_{g_{R,j}}(x,r)\leq 2^{2n+2}\omega_{n}r^{n}$ for
each $r\leq 4$;

  3. (iii)

$\int_{B_{g_{R,j}}(x,4)}|\text{Rm}|_{g_{R,j}}^{\frac{n}{2}}\,dV_{g_{R,j}}\leq
4\varepsilon(n)v$;

  4. (iv)

$\frac{\text{Vol}_{g_{R,j}}B_{g_{R,j}}(x,r)}{r^{n}}\geq\frac{\omega_{n}}{2}v$
for each $r\in(0,1]$; and

  5. (v)

$\operatorname{Sob}(B_{g_{j}}(x,r),g_{j})\leq\left[C_{1}(n)\left(\frac{\text{Vol}_{g_{j}}B_{g_{j}}(x,r)}{r^{n}}\right)^{-\frac{2}{n}},0\right]$
for each $r\in(0,4]$.

Writing $v_{0}=\frac{\omega_{n}}{2}v$, we replace condition (iv) with

$v_{xRjr}:=\frac{\text{Vol}_{g_{R,j}}B_{g_{R,j}}(x,r)}{r^{n}}\geq
v_{0}\;\text{ for each }r\in(0,1],$

and replace condition (v) with

$\operatorname{Sob}(B_{g_{j}}(x,4r),g_{j})\leq\left[4^{2}C_{1}(n)v_{xRjr}^{-\frac{2}{n}},0\right]\;\text{
for each }r\in(0,1]$

If we write $\lambda=4^{2}C_{1}(n)$ and define

(32) $\varepsilon(n)=\frac{\delta(n,\lambda)\omega_{n}}{8},$

then we may apply Theorem 4.2 to each metric $g_{R,j}$. Taking the diagonal
scale normalized sequence $g_{j}=j^{2}g_{j,j}$, the result is a sequence of
flows $g_{j}(t)$, $t\in[0,j^{2}\widetilde{T}]$ such that $g_{j}(0)=g$ on
$B_{g}(x_{0},j)$ and for each $j\in{\mathbb{N}}$, all of the following are
satisfied for every $(x,t)\in M\times[0,j^{2}\widetilde{T}]$:

  1. (i)

${\mathcal{R}}_{g_{j}(t)}(x)\geq-nj^{-2}$;

  2. (ii)

$|\text{Rm}|_{g_{j}(t)}(x)\leq\frac{1}{t}$;

  3. (iii)

$\operatorname{inj}_{g_{j}(t)}(x)\geq\iota(n,v)\sqrt{t}$;

  4. (iv)

$\text{Vol}_{g_{j}(t)}B_{g_{j}(t)}(x,r)\geq c_{1}(n)v\,r^{n}$ for all
$r\in(0,j]$;

  5. (v)

$\int_{B_{g_{j}(t)}(x,j)}|\text{Rm}|_{g_{j}(t)}^{\frac{n}{2}}\,dV_{g_{j}(t)}\leq\Lambda(n,\lambda)\delta(n,\lambda)v_{0}$;
and

  6. (vi)

$\operatorname{Sob}(B_{g_{j}(t)}(x,j),g_{j}(t))\leq[C_{2}(n)\lambda
v_{0}^{-\frac{2}{n}},\frac{1}{j^{2}}]$.

Then just as in the proof of [3, Theorem 1.1], we may apply [8, Corollary
3.2], Shi’s modified interior estimates [11, Theorem 14.16], the Arzelà-Ascoli
Theorem, and a diagonal argument, we may conclude that some subsequence of
$g_{j}(t)$ converges in $C^{\infty}_{loc}(M\times[0,\infty))$ to a long-time
Ricci flow $g(t)$ with $g(0)=g$ satisfying the following for all
$t\in[0,\infty)$:

  1. (i)

${\mathcal{R}}_{g(t)}\geq 0$;

  2. (ii)

$|\text{Rm}|_{g(t)}\leq\frac{1}{t}$;

  3. (iii)

$\operatorname{inj}_{g(t)}\geq\iota(n,v)\sqrt{t}$;

  4. (iv)

$\text{Vol}_{g(t)}B_{g(t)}(x_{0},r)\geq c_{1}(n)v\,r^{n}$ for all
$r\in(0,\infty)$;

  5. (v)

$\int_{M}|\text{Rm}|_{g(t)}^{\frac{n}{2}}\,dV_{g(t)}\leq\Lambda(n,\lambda)\delta(n,\lambda)v_{0}$;
and

  6. (vi)

$\operatorname{Sob}(M,g(t))\leq[C_{2}(n)\lambda v_{0}^{-\frac{2}{n}},0]$.

Finally, obtaining the faster than $1/t$ curvature decay follows by applying
[5, Theorem 1.2] to the bounded curvature metric $g(1)$, which is validated by
our smallness assumption on $\delta(n,\lambda)$ in (20). This completes the
proof of Theorem 1.4. ∎

######  Remark 5.5.

By recalling our definitions of $\varepsilon(n)$ in (32), of
$\delta(n,\lambda)$ in (14), and of $\Lambda(n,\lambda)$ in (28), as well as
the fact that $\lambda\leq 32(10n)^{3n}$ by Remark 2.10, we obtain the
explicit bound

$\varepsilon(n)\geq\frac{\omega_{n}}{8}\left(\frac{1}{8\left(2^{3n+9}[32(10n)^{3n}]^{\frac{n}{2}}(2^{n+4}1000n)^{\frac{n}{2}}\right)^{2n+1}}\right)\left(\frac{1}{10^{7}n^{3}[32(10n)^{3n}]}\right)^{\frac{n}{2}}\geq(10n)^{-10n^{3}}.$

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