The proof is by contradiction. So let us assume for some $\epsilon>0$ there is no such $\delta({\rm v},\epsilon)>0$. Thus, we have a sequence of spaces $(M^{4}_{j},g_{j},p_{j})$ with ${\rm Vol}(B_{1}(p_{j}))>{\rm v}>0$, $|{\rm Ric}_{M^{4}_{j}}|\leq\delta_{j}\to 0$ and $|\mathcal{V}_{4}(p_{j})-\mathcal{V}_{1/4}(p_{j})|<\delta_{j}\to 0$, but the conclusions of the theorem fail. After passing to a subsequence we can take a limit

$\displaystyle(M^{4}_{j},d_{j},p_{j})\stackrel{{\scriptstyle d_{GH}}}{{\longrightarrow}}(X,d,p)\,.$

(8.5)

Using the almost volume cone implies almost metric cone theorem of [ChCo1], we then have

$\displaystyle B_{4}(p)=B_{4}\big(y_{0})\,,$

(8.6)

where $y_{0}\in C(Y)$ is the cone vertex and $Y$ some metric space of diameter $\leq\pi$.

Now using Theorem 1.1, we know that away from a set of codimension $4$ in $C(Y)$, the harmonic radius $r_{h}>0$ is bounded uniformly from below. Assume there is some point $y\in Y$ such that $r_{h}(y)=0$ and consider the ray $\gamma_{y}$ in $C(Y)$ through the point $y$. In that case, it would follow that for every point of $\gamma_{y}$, the harmonic radius $r_{h}=0$ vanishes. The ray $\gamma$ has Hausdorff dimension $1$, and therefore its existence would contradict Theorem 1.1. Thus, we conclude that $r_{h}>0$ and that $Y=(Y,g_{Y})$ is a $C^{1,\alpha}\cap W^{2,q}$ manifold for every $\alpha<1$ and $q<\infty$.

Now by writing the formula for the Ricci tensor in harmonic coordinates and using $|{\rm Ric}_{M^{4}_{j}}|\to 0$, it follows that $C(Y)$ is smooth and Ricci flat away from the vertex. In particular, since $C(Y)$ is a metric cone over $Y$, we must ${\rm Ric}_{Y^{3}}=3g^{Y}$. Since in dimension $3$, constant Ricci curvature implies constant sectional curvature, it follows $Y=S^{3}/\Gamma$ has constant sectional curvature $\equiv 1$. Additionally, we know from the volume bound, ${\rm Vol}(B_{1}(p))>{\rm v}>0$, that the order $|\Gamma|<N({\rm v})$ is uniformly bounded. In particular, we have that $C(Y)=\mathds{R}^{4}/\Gamma$ is an orbifold with an isolated singularity.

It now follows that there exists $r_{0}({\rm v})>0$ such that for $y\in\mathds{R}^{4}/\Gamma$ with $|y|=1$, we have

$\displaystyle B_{2r_{0}}(y)=B_{2r_{0}}(0^{4})\,,$

(8.7)

where $0^{4}\in\mathds{R}^{4}$. In particular, for all $j$ sufficiently large, we have from the standard $\epsilon$-regularity theorem, Theorem 2.3, that for all $x\in A_{\epsilon,2}(p_{j})$, the harmonic radius, $r_{h}(x)>r_{0}({\rm v},\epsilon)=r_{0}({\rm v})\epsilon$ is bounded uniformly from below independent of $j$. Thus, if there exists $\epsilon$ as above, for which there is no $\delta({\rm v},\epsilon)$, it must be (2) that fails to hold.

However, by using again the diffeomorphism statement of Theorem 8.1, we have that for $j$ sufficiently large, there exists diffeomorphisms

$\displaystyle\Phi_{j}:A_{\epsilon,2}(0)\to M^{4}_{j}\,,$

(8.8)

such that

$\displaystyle\Phi_{j}^{*}g_{j}\stackrel{{\scriptstyle C^{1,\alpha}\cap W^{2,q}}}{{\longrightarrow}}dr^{2}+r^{2}g_{Y}\,.$

(8.9)

For $j$ sufficiently large, this implies that (2) holds; a contradiction. ∎

For $x\in B_{2}(p)$ fixed, we have

$\displaystyle{\rm Vol}(B_{1}(x))\geq C(n)^{-1}{\rm Vol}(B_{3}(x))\geq C^{-1}{\rm Vol}(B_{1}(p))\geq C^{-1}{\rm v}>0\,,$

(8.11)

and so,

$\displaystyle\mathcal{V}^{\delta}_{1}(x)\leq-\ln\Big(C^{-1}{\rm v}\Big)=C(n,{\rm v})\,.$

(8.12)

From the monotonicity of $\mathcal{V}^{\delta}_{r}(x)$, we have

	$\displaystyle C(n,{\rm v})-1\geq\mathcal{V}^{\delta}_{1}(x)-\mathcal{V}^{\delta}_{0}(x)$	$\displaystyle=\sum\Big(\mathcal{V}^{\delta}_{r_{\alpha}}(x)-\mathcal{V}^{\delta}_{r_{\alpha+1}}(x)\Big)$
		$\displaystyle=\sum\Big|\mathcal{V}^{\delta}_{r_{\alpha}}(x)-\mathcal{V}^{\delta}_{r_{\alpha+1}}(x)\Big|\,.$		(8.13)

In particular, there are at most $N=C(n,{\rm v})(\delta^{\prime})^{-1}$ elements $\alpha\in\mathds{N}$ such that

$\displaystyle\Big|\mathcal{V}_{r_{\alpha}}(x)-\mathcal{V}_{r_{\alpha+1}}(x)\Big|>\delta^{\prime}\,,$

(8.14)

as claimed. ∎

Put $\delta^{\prime}=\delta/3$. Then for $x\in B_{2}(p)$, there are at most $N({\rm v},\delta)=C({\rm v})\delta^{-1}$ scales $\alpha$ for which

$\displaystyle\Big|\mathcal{V}_{r_{\alpha}}(x)-\mathcal{V}_{r_{\alpha+1}}(x)\Big|>\frac{\delta}{3}\,.$

(8.16)

Hence, there are at most $3N$ elements $\alpha\in\mathds{N}$ such that

$\displaystyle\Big|\mathcal{V}_{r_{\beta}}(x)-\mathcal{V}_{r_{\beta+1}}(x)\Big|>\frac{\delta}{3}\,,$

(8.17)

for some $\beta\in\{\alpha-1,\alpha,\alpha+1\}$. Therefore, for all other $\alpha$, we must have

$\displaystyle\Big|\mathcal{V}_{4r_{\alpha}}(x)-\mathcal{V}_{r_{\alpha}/4}(x)\Big|<\delta\,,$

(8.18)

which proves the corollary. ∎

Let $M^{n}$ satisfy $|{\rm Ric}_{M^{4}}|\leq 3$ and ${\rm Vol}(B_{1}(p))>{\rm v}>0$. We will prove the estimate for the harmonic radius $r_{h}$. The same argument works in the Einstein case to control the regularity scale.

So let $0<\epsilon<<1$ be fixed with $\delta(n,\epsilon)$ chosen to satisfy Theorem 8.3. Consider the set

$\displaystyle\{x\in B_{1}(p):r_{h}(x)<r\}\,.$

(8.19)

In view of the doubling condition implied by the Bishop-Gromov inequality, we have by a standard construction that there exists a covering $\{B_{r}(x_{j})\}_{1}^{N}$ with

$\displaystyle\{x\in B_{1}(p):r_{h}(x)<r\}\subseteq\bigcup_{j}B_{r}(x_{j})\,,$

(8.20)

but such that $\{B_{r/4}(x_{j})\}$ are disjoint. Such coverings, which we will term “efficient”, will be constructed several times below. Note that

$\displaystyle T_{r}\big(\{x\in B_{1}(p):r_{h}(x)<r\}\big)\subseteq\bigcup_{j}B_{2r}(x_{j})\,,$

(8.21)

and thus

$\displaystyle{\rm Vol}\Big(T_{r}\big(\{x\in B_{1}(p):r_{h}(x)<r\}\big)\Big)\leq\sum_{1}^{N}{\rm Vol}(B_{2r}(x_{j}))\leq C(n)N\cdot r^{4}\,.$

(8.22)

Hence, our goal is to control the number of balls $N$ in the covering. Denote by $\mathcal{C}\equiv\{x_{j}\}_{1}^{N}$ this collection of points.

Now note the following: if $x_{j}$ is one of our ball centers and $T_{\alpha}(x_{j})=0$, then by Theorem 8.3 we have for every $x\in A_{r_{\alpha}/2,2r_{\alpha}}(x_{j})$ that $r_{h}(x)>\bar{r}({\rm v})\cdot r_{\alpha}$. In particular, if $r_{\alpha}>\bar{r}^{-1}r$, this implies that

$\displaystyle x_{k}\not\in A_{r_{\alpha}/2,2r_{\alpha}}(x)\,.$

(8.23)

Now let us inductively build a sequence of decreasing subsets $\mathcal{C}^{k+1}\subseteq\mathcal{C}^{k}\subseteq\cdots\subseteq\mathcal{C}$ and associated radii $s_{k}=r_{\alpha_{k}}>0$ with ${\rm diam}(\mathcal{C}^{k})<4s_{k}$. There are three key inductive properties that will be proved about these sets:

1. (1)

   There exists $C(n)>0$ such that the cardinality of $\mathcal{C}^{k}$ satisfies

   $\displaystyle\big|\#\mathcal{C}^{k}\big|\geq C^{-k}\big|\#\mathcal{C}\big|=C^{-k}N\,.$

   (8.24)

2. (2)

   For every $x^{k}_{j}\in\mathcal{C}^{k}$ we have

   $\displaystyle\sum_{0\leq\alpha\leq\alpha_{k}}T^{\delta}_{\alpha}(x^{k}_{j})\geq k\,.$

   (8.25)

3. (3)

   If $\big|\#\mathcal{C}^{k}\big|>1$ and $s_{k}>\bar{r}^{-1}r$ then $\mathcal{C}^{k+1}\neq\emptyset$.
   

Before constructing the sequence of sets, let us see that once the construction is complete, we will have proved our desired estimate on $N$. Indeed, let $k$ be the largest index such that $\mathcal{C}^{k}\neq\emptyset$. By the third property we must have either $|\#\mathcal{C}^{k}|=1$ or $s_{k}\leq\bar{r}^{-1}r$, at which point we get by a covering argument that $|\#\mathcal{C}^{k}|<C(n)$. By Lemma 8.5 and the second property we have that $k\leq k(n,{\rm v},\delta)=k(n,{\rm v})$, and thus by the first property we have

$\displaystyle N\leq C(n)^{k(n,{\rm v})}\cdot|\#\mathcal{C}^{k}|\leq C(n,{\rm v})\,,$

(8.26)

which proves the result.

Now let $\mathcal{C}^{0}\equiv\mathcal{C}$ with $s_{0}=1$. Clearly, the inductive properties hold for $\mathcal{C}^{0}$. Assume we have built $\mathcal{C}^{k}\subseteq\mathcal{C}$ with $s_{k}>0$ satisfying the inductive properties, and let us build $\mathcal{C}^{k+1}$. First note that if $|\#\mathcal{C}^{k}|=1$ or $s_{k}\leq\bar{r}^{-1}r$, then we let $\mathcal{C}^{k+1}=\emptyset$. Our construction will otherwise give us a nonempty $\mathcal{C}^{k+1}$, so that the third inductive property will automatically be satisfied. So let us denote $s^{\prime}_{k}={\rm diam}(\mathcal{C}^{k})\cdot 2^{-10}$. Choose an efficient covering $\{B_{s^{\prime}_{k}}(x^{k}_{j})\}$, where $x^{k}_{j}\in S^{k}$, so that the balls in $\{B_{s^{\prime}_{k}/4}(x^{k}_{j})\}$ are disjoint. Note that because ${\rm diam}(\mathcal{C}^{k})<4s_{k}$, the usual doubling estimates imply that there are at most $C(n)$ balls in this covering. We choose the ball $B_{s^{\prime}_{k}}(y)$ such that $\mathcal{C}^{k}\cap B_{s^{\prime}_{k}}(y)$ has the largest cardinality of any ball from the covering. Then we define $\mathcal{C}^{k+1}=\mathcal{C}^{k}\cap B_{s^{\prime}_{k}}(y)$.

By that by our choice of ball, $B_{s^{\prime}_{k}}(y)$, we have

$\displaystyle\big|\#\mathcal{C}^{k+1}\big|$

$\displaystyle=\big|\#\mathcal{C}^{k}\cap B_{s^{\prime}_{k}}(y)\big|\geq C(n)^{-1}\big|\#\mathcal{C}^{k}\big|\geq C^{-(k+1)}N\,,$

(8.27)

so that $\mathcal{C}^{k+1}$ satisfies the first inductive property. To find $s_{k+1}$ and prove the second inductive property, let us define the following. For each $x^{k+1}_{j}\in\mathcal{C}^{k+1}$ if

$\displaystyle T^{\delta}_{\alpha_{k}+7}(x^{k+1}_{j})=1\,,$

(8.28)

then let us set $\beta_{j}=\alpha_{k}+7$, and otherwise let $\beta_{j}\geq\alpha_{k}+8$ be the largest integer such that $T^{\delta}_{\beta_{j}-1}(x^{k+1}_{j})=0$ but $T^{\delta}_{\beta_{j}}(x^{k+1}_{j})=1$. Note that $B_{s^{\prime}_{k}}(y)\subseteq B_{r_{\alpha_{k}+7}}(x^{k}_{j})$. Let $\alpha_{k+1}\equiv\max\{\beta_{j},\lceil-\log_{2}\big(\bar{r}r^{-1}\big)\rceil\}$ with $x^{k+1}\in\mathcal{C}^{k+1}$ the associated element which attains the maximum, and note by (8.23) that

$\displaystyle\mathcal{C}^{k+1}=\mathcal{C}^{k}\cap B_{s^{\prime}_{k}}(y)=\mathcal{C}^{k}\cap B_{2^{-\alpha_{k+1}+1}}(x^{k+1})\,.$

(8.29)

In particular, with $s_{k+1}=r_{\alpha_{k+1}}$ then ${\rm diam}(\mathcal{C}^{k+1})<4s_{k+1}$, and the second inductive property holds, which completes the induction step of the construction, and hence, the proof.

∎

We will fix $\epsilon({\rm v})>0$ later. For the moment let any $\epsilon_{1}>0$ be arbitrary with $\delta_{1}({\rm v},\epsilon_{1})>0$ the corresponding number from Theorem 8.3. If $T^{\delta_{1}}_{\alpha_{1}}(x)=0$ then there exists a diffeomorphism

$\displaystyle\Phi_{\alpha_{1}}:A_{r_{\alpha_{1}}/2,2r_{\alpha_{1}}}(0)\to U_{\alpha_{1}}\,,$

(8.31)

where $0\in\mathds{R}^{4}/\Gamma_{\alpha_{1}}$ and $A_{r_{\alpha_{1}}/2,2r_{\alpha_{1}}}(x)\subseteq U_{\alpha}\subseteq A_{(1-\epsilon)r_{\alpha_{1}}/2,(1+\epsilon)r_{\alpha_{1}}}(x)$, such that

$\displaystyle||\Phi_{\alpha_{1}}^{*}g_{ij}-\delta_{ij}||_{C^{0}(A_{r_{\alpha_{1}}/2},2r_{\alpha_{1}})}+r_{\alpha_{1}}||\partial_{k}\Phi_{\alpha_{1}}^{*}g_{ij}||_{C^{0}(A_{r_{\alpha_{1}}/2},2r_{\alpha_{1}})}<\epsilon_{1}\,.$

(8.32)

In particular, if $\epsilon>0$ is fixed and $2\delta(n,\epsilon)$ is the corresponding number from Theorem 8.3, then we can choose $\epsilon_{1}=\epsilon_{1}(\epsilon,{\rm v})$ sufficiently small so that

$\displaystyle\mathcal{V}^{\delta}_{r_{\alpha_{1}}}(x)<\ln|\Gamma_{\alpha_{1}}|+\delta\,.$

(8.33)

Thus, if $\alpha_{2}$ is such that

$\displaystyle\mathcal{V}^{\delta}_{r_{\alpha_{2}}/2}(x)\geq\ln|\Gamma_{\alpha_{1}}|-\delta\,,$

(8.34)

then for all $\alpha_{1}\leq\alpha\leq\alpha_{2}$ we have $T^{2\delta}_{\alpha}(x)=0$.

By Theorem 8.3, there exists for each $\alpha_{1}\leq\alpha\leq\alpha_{2}$, a diffeomorphism

$\displaystyle\Phi_{\alpha}:A_{r_{\alpha}/2,2r_{\alpha}}(0)\to U_{\alpha}\,,$

(8.35)

where $0\in\mathds{R}^{4}/\Gamma_{\alpha}$ and $A_{r_{\alpha}/2,2r_{\alpha}}(x)\subseteq U_{\alpha}\subseteq A_{(1-\epsilon)r_{\alpha}/2,2(1+\epsilon)r_{\alpha}}(x)$, such that

$\displaystyle||\Phi_{\alpha}^{*}g_{ij}-\delta_{ij}||_{C^{0}(A_{r_{\alpha}/2},2r_{\alpha})}+r_{\alpha}||\partial_{k}\Phi_{\alpha}^{*}g_{ij}||_{C^{0}(A_{r_{\alpha}/2},2r_{\alpha})}<\epsilon\,.$

(8.36)

In particular this implies that $\Gamma_{\alpha}=\Gamma$ is independent of $\alpha$.

Next we focus on the inverse maps

$\displaystyle\Phi_{\alpha}^{-1}:U_{\alpha}\to A_{r_{\alpha}/2,2r_{\alpha}}(0)\,.$

(8.37)

Observe that by (8.36), after possibly composing $\Phi_{\alpha}$ with a rotation of $\mathds{R}^{4}/\Gamma$ we can assume for $x\in U_{\alpha}\cap U_{\beta}$ that

$\displaystyle|\Phi_{\alpha}^{-1}(x)-\Phi_{\beta}^{-1}(x)|<\epsilon r_{\alpha}\,.$

(8.38)

Now let $\epsilon<\epsilon({\rm v})$ be sufficiently small, so that if $x\in\mathds{R}^{4}/\Gamma$, then $B_{\epsilon|x|}(x)\subseteq\mathds{R}^{4}/\Gamma$ is isometric to the standard Euclidean ball $B_{\epsilon|x|}(0^{4})\subseteq\mathds{R}^{4}$. Note in particular that if $\{x_{i}\}\in B_{\epsilon|x|}(x)$ is a collection of points, then any convex combination is well defined.

For each $\alpha$ let $\varphi^{\prime}_{\alpha}:U_{\alpha}\to\mathds{R}$ be a smooth cutoff function such that

$\displaystyle\varphi^{\prime}_{\alpha}(x)=\begin{cases}&1\text{ if }x\in A_{3r_{\alpha}/8,15r_{\alpha}/8}(x)\,,\\ &0\text{ if }x\not\in A_{r_{\alpha}/2,2r_{\alpha}}(x)\,,\end{cases}$

and such that $|\nabla\varphi^{\prime}_{\alpha}|\leq 10r_{\alpha}^{-1}$. If we set $\varphi^{\prime}(x)=\sum_{\alpha}\varphi^{\prime}_{\alpha}(x)$ then $1\leq\varphi^{\prime}(x)\leq 4$. In particular,

$\displaystyle\varphi_{\alpha}=\frac{\varphi^{\prime}_{\alpha}(x)}{\varphi^{\prime}(x)}:U_{\alpha}\to\mathds{R}\,,$

(8.39)

sarisfies $\sum\varphi_{\alpha}(x)=1$, and so, is a partition of unity, with $|\nabla\varphi_{\alpha}|\leq 40r_{\alpha}^{-1}$.

Define the map

$\displaystyle\Phi^{-1}:U=\bigcup_{\alpha}U_{\alpha}\to A_{r_{\alpha_{2}}/2,2r_{\alpha_{1}}}(0)\,,$

(8.40)

given by

$\displaystyle\Phi^{-1}(x)=\sum_{\alpha}\varphi_{\alpha}(x)\Phi^{-1}_{\alpha}(x)\,.$

(8.41)

(As previously noted, the convex combination is well defined since the $\Phi^{-1}_{\alpha}(x)$ all live in a ball which is isometric to a Euclidean ball.) On each domain, $U_{\alpha}$, we have by (8.32) and (8.38) that $\Phi^{-1}$ and $\Phi^{-1}_{\alpha}$ are $C^{1}$-close. Hence, $\Phi^{-1}$ is a diffeomorphism, and a quick computation using (8.32) and (8.38) verifies the desired estimates:

$\displaystyle||\Phi^{*}g_{ij}-\delta_{ij}||_{C^{0}(A_{r_{\alpha}/2},2r_{\alpha})}+r_{\alpha}||\partial_{k}\Phi_{\alpha}^{*}g_{ij}||_{C^{0}(A_{r_{\alpha}/2},2r_{\alpha})}<C\epsilon\,.$

(8.42)

By choosing $\epsilon$ appropriately small, we complete the proof. ∎

First note by Theorem 8.3 that if $\delta$ is fixed, then there exists $\delta^{\prime}({\rm v},\delta)$ such that if $|{\rm Ric}|\leq 3\delta^{\prime}$ and if $T^{\delta^{\prime}}_{0}=0$, then

$\displaystyle\big|\mathcal{V}^{\delta^{\prime}}_{1}(x)-\ln|\Gamma_{0}|\big|<\delta\,.$

(8.43)

By rescaling this inequality, we see that in the context of this lemma, the following holds. If $x\in B_{1}(p)$, $\alpha>\bar{\alpha}(v,\delta)$ and

$\displaystyle\big|\mathcal{V}^{\delta}_{4r_{\alpha}}(x)-\mathcal{V}^{\delta}_{r_{\alpha}/4}(x)\big|<\delta^{\prime}\,,$

(8.44)

then we have

$\displaystyle\big|\mathcal{V}^{\delta}_{r_{\alpha}}(x)-\ln|\Gamma_{\alpha}|\big|<\delta\,.$

(8.45)

In particular, for $x\in B_{1}(p)$, we can apply Lemma 8.5 to see that there exists a scale $\alpha\leq\bar{\alpha}(v,\delta)$ such that

$\displaystyle\big|V^{\delta}_{r_{\alpha}}(x)-\mathcal{V}^{\delta}_{r_{\alpha+1}}(x)\big|<\delta^{\prime}\,,$

(8.46)

and hence

$\displaystyle\big|\mathcal{V}^{\delta}_{r_{\alpha}}(x)-\ln|\Gamma_{\alpha}|\big|<\delta\,.$

(8.47)

However, if

$\displaystyle\mathcal{V}^{\delta}_{r_{\alpha}}(x)<\ln N-\delta\,,$

(8.48)

this implies $|\Gamma_{\alpha}|<N$, which completes the proof. ∎