In this [article](https://www.jstor.org/stable/2046117?seq=1) 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](https://en.wikipedia.org/wiki/Big_O_notation#Multiple_variables) 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.



What does $O(N)$ mean in this article and how does it imply this lemma?
The $O(1)$ is indeed an abuse of notations, but I personally find pretty clear what is meant by it, given the context. In fact the statement would have been crystal clear and rigorously meaningful if the author had simply written $\geq CN$ instead of $=O(N)$... This is precisely the definition of $\Omega(N)$. But choosing one or the other is just a matter of aesthetics and personal taste, I guess (for example in France we are never taught the $\Omega(N)$ notation and I only learnt it after 10++ years of professional maths activity). Anyway, as far as the "real" maths are concerned, here is an answer to your question.


Writing it as a Riemann sum, it is straightforward to check that the sum
$$
S\_N:=\sum\_{k=0}^N k^\delta=N^{1+\delta}\sum\limits\_{k=0}^N\left(\frac kN\right)^\delta\frac 1N
\sim N^{1+\delta}\int\_0^1 x^\delta dx =\frac{1}{\delta+1}N^{1+\delta}
$$
as $N\to+\infty$.
It is then easy to check that starting the series fom $k=2\log\_2N$ doesn't change anything to the matter (you would get a small $o(1)$ negligible correction to the integral in the Riemann sum). In other words,
$$
\tilde S\_N:=\sum\_{k=2\log\_2 N}^N k^\delta \sim \frac{1}{\delta+1}N^{1+\delta}
$$
as well.
Putting the pieces together, and recalling the key property that $M\_S v(x)\leq 2$, you get
\begin{multline\*}
[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}
=c\Bigg[\frac{N}{(N+1)^2}\tilde S\_N\Bigg]^{1/\delta}\\
\sim c\_\delta\Bigg[\frac{N}{(N+1)^2}N^{1+\delta}\Bigg]^{1/\delta}\\
 \sim c\_\delta N \geq c\_\delta\frac{N}{2}2
\geq c'\_\delta N M\_Sv(x)
\end{multline\*}


