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. ∎