I am trying to follow the reasoning in the proof of Mann's Theorem: $$d(C) \ge \min(d(A)+d(B),1)$$ I am clear that we can assume that:
- $d(A) + d(B) \le 1$
- We only need to prove that for every $n \ge 1$, $\exists m \in [1,n]$ such that $C(n) - C(n-m) \ge (d(A) + d(B))m$
- For any $C(n)$, we can assume $n \notin C$
Here's where I am not clear.
Using the proof from Gelfond & Linnik Elementary Methods in the Analytic Theory of Numbers, they assume that $C$ is what they term normal:
We shall describe the system of numbers $H\subset[0,n]$ as normal if, for any numbers $f\in[1,n];f'\in[1,n];f\notin(H);f'\notin{H}$, we have $f + f' - n \notin H$. If the sequence $C$ possesses the property that segments of it in the segment $[0,n]$ form a normal system, the lemma is easily proved for $C(n)$.
Then, they state the following conclusion:
In fact, let $m$ be the least positive integer not belonging to $C$; then $m < n$, since $n \notin C$ by hypothesis.
What about the situation where $C(n) = n-1$. In that case, wouldn't $m=n$. So, is the assumption that $C(n) < n-1$? How can we be sure that $m < n$?
Here's my understanding of Schnirelmann Density.
$A,B$ are infinite sequences of integers starting with $0$ with in sequential order such as $0, a_1, a_2, \cdots$ where $0 < a_1 < a_2 < \cdots$
Schnirelmann density is defined as: $$d(A) = \inf\limits_{n}\frac{A(n)}{n}$$
where: $$A(n) = \sum\limits_{0<a_i\le{n}}{1}$$
So, it is clear that: $$0 \le \frac{A(n)}{n} \le 1$$
$C = A+B$ where $+$ is the sumset.
Edit: Added detail on normal since this may be part of the answer.
It looks like it is a typo and it should read $m \le n$.
Linnik and Gelfond say in their footnote that the proof is taken from A. Ya. Khinchin, Three Pearls of Number Theory (1952). This book is now published as a Dover edition which I tracked down. In A. Ya. Khinchin's version: $m \le n$.
If $m=n$ with $n \notin C$, their argument applies since: $$C(n) - C(n-m) = C(n) - C(0) = n - 1$$ Let $a_1, a_2, \cdots a_r$ be the numbers in $A(n)$.
Let $b_1, b_2, \cdots b_s$ be the numbers in $B(n)$.
From the assumption that $n \notin C$, we can see that $a_1, a_2, \cdots, a_r, n-b_1, n-b_2, \cdots, n-b_s, n$ are all distinct in $[1,n]$ which gives us: $$A(n) + B(n) \le n-1$$ So, it follows that: $C(n) - C(n-m) = C(n) - C(0) \ge A(m) + B(m) \ge (d(A) + d(B))m$