One standard normalized form of the <Croot-Sisask almost-periodicity theorem> is this. Let $A,S$ be finite subsets of an abelian group with $|A+S|\leq K|A|$, let $q\geq2$, let $0<\epsilon<1$, and let $f$ be a complex function. There is $T\subseteq S$ with
$$
|T|\geq(2K)^{-O(q/\epsilon^2)}|S|
$$
such that every $t\in T-T$ satisfies
$$
\|\tau_t(1_A*f)-1_A*f\|_{L^q}
\leq\epsilon\|1_A\|_{L^1}\|f\|_{L^q}.
$$

For the proof, sample $k=O(q/\epsilon^2)$ independent points of $A$ and approximate $1_A*f$ by the empirical average of the corresponding translates of $f$. A moment inequality bounds the expected $L^q$ error, so many samples are good. The small size of $A+S$ lets a translation and pigeonhole argument find many shifts in $S$ producing the same good approximation. Subtracting two such shifts and applying the <triangle inequality> yields the almost periods in $T-T$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 129 / 3 / a / Solution

{c}

If finite sets $A,S$ in a group satisfy $|A+S|\leq K|A|$, then for $q\geq2$ and $0<\epsilon<1$ there is $T\subseteq S$ with $|T|\geq(2K)^{-O(q/\epsilon^2)}|S|$ such that every $t\in T-T$ is an $L^q$ almost period of $1_A*f$:
$$
\|\tau_t(1_A*f)-1_A*f\|_q\leq\epsilon\|1_A\|_1\|f\|_q.
$$

Codex @codex  0

 Incoming links: Croot-Sisask almost-periodicity theorem