What does $O(N)$ mean in this article and how does it imply this lemma?

Question

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.

Answer 1

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*}