On the proof of a $W^{2,p}$ estimate - regularity on eliptic PDE I see this proof studying the paper http://www.math.uiowa.edu/~lwang/cccalderon.pdf and I couldn't understand what he did. If $||f||_{L^p(B_{4})} = \delta$ is small and the measure $|\{ x \in B_1; M(|D^2u|^2)>N_1^2 \}| \leq \epsilon|B_1| $. Then, we have:
$$\sum_{i=1}^{+ \infty} (N_{1})^{ip}|\{M(|f|^2) > \delta^2 (N_{1})^{2i}\}| \leq \frac{pN^{p}}{\delta^{p}(N-1)}||f||_{{L^p}(B_{1})}^{p}$$
Someone could help me? Here M denotes the maximal function of Hardy Littlewood.
 A: That could be written better... Let's begin with layercake formula for $L^1$ norm of a function $g$ on a finite measure space $X$: for $\lambda>1$, the quantity 
$$
\sum_{i=1}^\infty \lambda^i |\{x\in X: |g(x)|>\lambda^i \}|  
$$
is comparable to $\|g\|_{L^1}$. We actually need only one-sided inequality here: 
$$
\sum_{i=1}^\infty \lambda^i |\{x\in X: |g(x)|>\lambda^i \}| 
= \int_X \sum_{\lambda^i<|g(x)|} \lambda^i \,dx \le C \int_X |g(x)|\,dx
$$
because a geometric sum is comparable to its largest term.
Recall that $p>2$ in this section of the paper. Therefore we have the strong type $(p/2,p/2)$ inequality for $M$: 
$$\int M(|\delta^{-1}f|^2)^{p/2} \le C \|(\delta^{-1} f)^2\|_{L^{p/2}}^{p/2} $$
The right hand side simplifies to $C \delta^{-p} \|f\|_{L^p}^p$. Hence 
$$
\sum_{i=1}^\infty \lambda^i |\{ \delta^{-p} M(|f|^2)^{p/2}>\lambda^i \}| \le  C \delta^{-p} \|f\|_{L^p}^p
$$
Put $\lambda = N_1^{p}$ here: 
$$
\sum_{i=1}^\infty N_1^{ip} |\{  M(|f|^2) > \delta^2 N_1^{2i} \}| \le  C \delta^{-p} \|f\|_{L^p}^p
$$
The precise form of constant does not matter for the argument. We  know that $\delta^{-p} \|f\|_{L^p}^p$ is controlled by another $C$, and the rest goes as written. 

One point I haven't discussed is the domain used in the maximal function is taken: what averages are included in $M(|f|^2)$? The exposition in the paper is not clear at this point.
