In this article the author proves the following lemma:

LEMMA: $\forall N \in \Bbb N$, there exists $v=v_N$ with compact support so that $$[M_S(M_S v)^\delta(x)]^{1/\delta} \geq c_\delta NM_S v(x), \forall x \in [0,1]^2.$$

Here $M_S$ is the strong maximal operator (averages on rectangles in $\Bbb R^2$), $\delta >0$ and $c_\delta$ is a constant that depends on $\delta$.

In order to prove it, for all $k, j \in \Bbb Z$, the author defines $Q_{k,j}$ as the unit square with lower left corner at the point $(k,j)$ and for $k=1,2,\ldots, N$, $d_k$ is the integral part of $2^k/k$. Then he defines $$v(x)=v_N(x)=\sum_{1 \leq k \leq N} 2^k \chi_{Q_{k,d_k}}(x),$$ and proves that $M_S v(x) \leq 2$ $\forall x \in [0,1]^2$ and that $f=M_S v(x) \geq k/2$ on $Q_{k,j}$ for $j=1,2,\ldots N$ provided that $k \geq 2 \log_2 N$. Finally, he takes the square $R=[0,N+1]^2$ and he obtains that for all $x \in [0,1]^2$, $$[M_S(M_S v)^\delta (x)]^{1/\delta} \geq [M_S(f)^\delta ((0,0))]^{1/\delta} \geq \left( |R|^{-1} \int_R (f)^\delta dy \right)^{1/\delta} \geq c \left[ (N+1)^{-2} \sum_{2 \log_2 N \leq k \leq N} N(k)^\delta \right]^{1/\delta}=O(N).$$

Using this definition of the big O notation, $O(N)$ means that $|O(N)| \leq CN$ for all $x$ such that $x_i$ is sufficiently for some $i$.

I really don't see why the quantity on the left is in $O(N)$ nor why this implies the lemma of the article.

Note: I reposted the question in order to improve it.