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

and so,

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

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

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

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

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

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

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

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

and thus

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

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

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

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

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

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.

∎