Definition: A subset A of the natural numbers is said to have positive upper density if $\ \displaystyle\limsup _{n\to \infty }\frac{\lvert A\cap \{1,2,3,\dotsc ,n\}\rvert}{n}>0.$
Let $\ A\subset \mathbb{N}\ $ have positive upper density, $\ B\ $ be an infinite subset of $\ \mathbb{N},\ $ and $\ k\in\mathbb{N}.\ $ Then does there exist $\ c\in\mathbb{Z}\ $ such that $\ \lvert \{ c+a: a\in A\} \cap B \rvert > k ? $
Motivation:
- It is true if $\ A = b\mathbb{N}\ $ for any $\ b\in\mathbb{N}:\ $ since $\ B\ $ is infinite, due to the pigeonhole principle, there exists $\ d\in\{0,1,\ldots, b-1\}\ $ such that $\ \lvert\{ d + bn: n\in\mathbb{N} \} \cap B\rvert = + \infty.\ $ But I don't think we can use this argument if $\ A \neq b\mathbb{N}.\ $ Or perhaps we can use an analagous argument?
- The role of $\ c\ $ is to avoid obvious counter-examples such as $\ A = 2\mathbb{N},\ B = \{ 2n+1: n\in\mathbb{N}\}.$
Perhaps a back-and-forth counterexample between $\ A\ $ and $\ B\ $ exists, but this seems complicated and I'm not sure of it.
Fix each $n$, write $A_n=A\cap \{1,\dots,n\}$, and define $B_n$ similarly. For each integer $c$, let $d_n(c)$ be the number of ways to write $c=b-a$ for $b\in B_n$ and $a\in A_n$. We have (i) that $d_n$ is supported on $[-n+1,n-1]$ and (ii) that $$\sum_{c\in\mathbb Z}d_n(c)=|A_n||B_n|.$$ So, there exists some $c$ for which $$d_n(c)\geq \frac{|A_n||B_n|}{2n}.$$ If we can show that this quantity grows arbitrarily large as $n$ varies, we'll be done.
Now, since $A$ has positive upper density, there exists some $\epsilon>0$ and a sequence $n_1,n_2,\dots$ for which $$|A_{n_i}|=|A\cap \{1,\dots,n_i\}|\geq \epsilon n_i$$ for each $i$. Furthermore, such $n_i$ can be chosen such that $|B_{n_i}|=|B\cap \{1,\dots,n_i\}|\geq i$, since $B$ is infinite. This gives $$\frac{|A_{n_i}||B_{n_i}|}{2n_i}\geq \frac{\epsilon n_i\cdot i}{2n_i}\geq \frac{\epsilon i}2.$$ As $i$ grows, this grows arbitrarily large, as desired.