(of Theorem 1.3) Let $(M^{n},g,p)$ satisfy $|{\rm Ric}_{M^{n}}|\leq n-1$ and ${\rm Vol}(B_{1}(p))>{\rm v}>0$. We will first show that for every $q<2$ there exists $C=C(n,{\rm v},q)>0$ such that

$\displaystyle\fint_{B_{1}(p)}r_{h}^{-2q}\leq C\,.$

(7.3)

Simultaneously, we will show that if $M^{n}$ is Einstein, then this can be improved to

$\displaystyle\fint_{B_{1}(p)}r_{x}^{-2q}\leq C\,,$

(7.4)

where $r_{x}$ denotes the regularity scale at $x$.

Let $q<2$ and set $\eta=4-2q$. Consider Theorem 7.3 with $\epsilon=\epsilon(n)>0$ chosen from Theorem 6.1 and $\eta$ as above. Thus, there exists $C(n,{\rm v},q)$ such that

$\displaystyle{\rm Vol}(T_{r}(\{x\in\mathcal{S}^{n-4}_{\epsilon,2r}\cap B_{1}(p)\}))<Cr^{4-\eta}\,.$

(7.5)

Note that by rescaling, we may regard the $\epsilon$-regularity theorem (Theorem 6.1) as stating that if $x$ is $(n-3,\epsilon,2r)$-symmetric then $r_{h}>r$, and if $M^{n}$ is Einstein then $r_{x}>r$. In fact, we have that if $x$ is $(n-3,\epsilon,s)$-symmetric for any $s\geq 2r$, then $r_{h}>r$. This is to say that if $x\not\in\mathcal{S}^{n-4}_{\epsilon,2r}$, then $r_{h}>\frac{s}{2}>r$. The contrapositive gives the inclusion

$\displaystyle\{x\in B_{1}(p):r_{h}\leq r\}\subseteq\mathcal{S}^{n-4}_{\epsilon,2r}\cap B_{1}(p)\,.$

(7.6)

which by (7.5) gives us the desired estimate

$\displaystyle{\rm Vol}(T_{r}(\{x\in B_{1}(p):r_{h}\leq r\}))<Cr^{4-\eta}\leq Cr^{2p}\,.$

(7.7)

If $M^{n}$ is Einstein, then Theorem 6.1 allows us to replace $r_{h}$ with $r_{x}$, as claimed.

Now, for $q<2$, let us prove the $L^{q}$ bound on the curvature from Theorem 1.3. For this note that if $r_{h}(x)>r$ then by definition there exists harmonic coordinates $\Phi:B_{r}(0^{n})\to M$ with $\phi(0)=x$ and such that

$\displaystyle||g_{ij}-\eta_{ij}||_{C^{0}(B_{r}(0))}+r||\partial_{k}g_{ij}||_{C^{0}(B_{r}(0))}<10^{-3}\,,$

(7.8)

where $g_{ij}=\Phi^{*}g$ is the pullback metric. Since the Ricci curvature satisfies the bound $|{\rm Ric}_{M^{n}}|\leq n-1$, this implies that

$\displaystyle|\Delta_{x}g_{ij}|<C(n)r^{-2}\,,$

(7.9)

where $\Delta_{x}$ denotes the Laplacian written in coordinates. In particular, for every $\alpha<1$ and $s<\infty$, we have the scale invariant estimates

		$\displaystyle r^{1+\alpha}||\partial_{k}g_{ij}||_{C^{\alpha}(B_{\frac{3r}{4}}(0))}\leq C(n,\alpha)\,,$
		$\displaystyle r^{2}||g_{ij}||_{W^{2,s}(B_{\frac{3r}{4}}(0))}\leq C(n,s)\,.$		(7.10)

In particular, applying this to $s=q$ we get

$\displaystyle r^{2q}\fint_{B_{r/2}(x)}|{\rm Rm}|^{q}\leq C(n)r^{2q}\fint_{B_{3r/4}(0)}|\Phi^{*}{\rm Rm}|^{q}<C(n,q)\,.$

(7.11)

Let $\eta=2-q$ be chosen so that $q+\frac{\eta}{2}<2$. Then we have already shown that

$\displaystyle{\rm Vol}(T_{r}(\{x\in B_{1}(p):r_{h}\leq r\}))<Cr^{2q+\eta}\,,$

(7.12)

for $C(n,{\rm v},q)>0$. Consider the covering $\{B_{r_{h}(x)}(x)\}$ of $B_{1}(p)$, and a subcovering $\{B_{r_{j}}(x_{j})\}$ by mutually disjoint balls, such that

1. (1)

   $B_{1}(p)\subseteq\bigcup B_{r_{j}}(x_{j})$ with $r_{j}=\frac{1}{2}r_{h}(x)$.

2. (2)

   $\{B_{r_{j}/4}(x_{j})\}$ are disjoint.

By using (7.12), we see for each $\alpha\in\mathds{N}$ that

$\displaystyle\sum_{2^{-\alpha-1}<r_{j}\leq 2^{-\alpha}}{\rm Vol}(B_{r_{j}}(x_{j}))\leq Cr_{j}^{2q+\eta}=C\,r_{j}^{2q}\,2^{-\eta\alpha}\,.$

(7.13)

Summing over $\alpha$ this gives

$\displaystyle\sum r_{j}^{-2q}{\rm Vol}(B_{r_{j}}(x_{j}))\leq C\sum 2^{-\eta\alpha}\leq C(n,{\rm v},q)\,.$

(7.14)

Finally, combining this with (7.11) we get

	$\displaystyle\fint_{B_{1}(p)}|{\rm Rm}|^{q}$	$\displaystyle\leq C(n,{\rm v})\sum\int_{B_{r_{j}}(x_{j})}|{\rm Rm}|^{q}$
		$\displaystyle\leq C(n,{\rm v},q)\sum r_{j}^{-2q}{\rm Vol}(B_{r_{j}}(x_{j}))\leq C(n,{\rm v},q)\,,$		(7.15)

which finishes the proof of Theorem 1.3. ∎

Note that by the Cheng-Yau gradient estimate, we have

$\displaystyle\sup_{B_{3/2}(p)}|\nabla u|\leq C(n)\,.$

(7.17)

Now using Theorem 1.3 we know for each $\epsilon>0$ that

$\displaystyle{\rm Vol}(T_{r}(\{x\in B_{1}(p):r_{h}(x)\leq r\}))\leq C_{\epsilon}(n,{\rm v},\epsilon)r^{4-\epsilon}\,.$

(7.18)

In particular, let us consider the sets

$\displaystyle\mathcal{C}_{\alpha}\equiv\{x\in B_{1}(p):r_{\alpha}\leq r_{h}(x)\leq r_{\alpha-1}\}\,,$

(7.19)

where $r_{\alpha}\equiv 2^{-\alpha}$. For the set $\mathcal{C}_{\alpha}$, we have the cover $\{B_{r_{\alpha}}(x)\}_{x\in\mathcal{C}_{\alpha}}$. We can choose a finite subcovering $\{B_{r_{\alpha}/2}(x_{i})\}_{1}^{N_{\alpha}}$ such that the balls $B_{r_{\alpha}/8}(x_{i})$ are mutually disjoint. Using (7.18) we have

$\displaystyle N_{\alpha}\leq C_{\epsilon}r_{\alpha}^{4-n-\epsilon}\,.$

(7.20)

On each ball $B_{r_{\alpha}/2}(x_{j})$ we can use standard elliptic estimates along with the gradient bound $|\nabla u|\leq C(n)$ to get the scale-invariant estimate

$\displaystyle r_{\alpha}^{q}\fint_{B_{r_{\alpha}/2}(x_{j})}|\nabla^{2}u|^{q}\leq C(n,{\rm v},q)\,,$

(7.21)

for any $q<\infty$. In particular, if we choose $q<4$ and pick $\epsilon=\frac{4-q}{2}$, then we have

$\displaystyle\int_{B_{r_{\alpha}/2}(x_{j})}|\nabla^{2}u|^{q}\leq C(n,{\rm v})\,r_{\alpha}^{n-4+2\epsilon}\,.$

(7.22)

Combining this with (7.18) gives us

$\displaystyle\int_{\mathcal{C}_{\alpha}}|\nabla^{2}u|^{q}\leq C(n,{\rm v})\,r_{\alpha}^{n-4+2\epsilon}\cdot N_{\alpha}\leq C(n,{\rm v},q)\,r_{\alpha}^{\epsilon}\,.$

(7.23)

Finally, by summing over $\mathcal{C}_{\alpha}$ we get the estimate

$\displaystyle\int_{B_{1}(p)}|\nabla^{2}u|^{q}\leq C(n,{\rm v},p)\lambda^{q}\,\sum_{\alpha}r_{\alpha}^{\epsilon}=C\,\sum 2^{-\epsilon\alpha}=C(n,{\rm v},q)\,,$

(7.24)

as claimed. ∎