Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 129 / 3

One normalized form of the Croot-Sisask almost-periodicity theorem is as follows. If finite sets $A,S$ in a group satisfy $|A+S|\le K|A|$, $p\ge 2$, and $0<ϵ<1$, then for every function $f$ there is $T\subseteq S$ withsuch that

Apply the Croot-Sisask almost-periodicity theorem with the sampling set $A$, $S=G$, $f=1_{-A}$, and error $ϵ/k$. Since $A+G=G$, the doubling parameter is $K={\alpha }^{-1}$, and $\Vert 1_{-A}{\Vert }_p\le 1$. We obtain $X\subseteq G$ withsuch thatfor every $t\in X-X$. Every $x\in kX-kX$ is a sum of $k$ elements of $X-X$. Telescoping these $k$ shifts and using translation invariance and the triangle inequality for the Lp norm gives

Put $f=1_{-A}\ast {\mu }_A$. The probability measureis supported on $kY-kY$. Convolution by $\mu$ averages translates by points in this support, so convexity of the supremum norm givesThus

Choose $ϵ=1/6$ and a sufficiently large absolute constant $k$. Part (iii)'s size bound givesLetand choose a maximal dissociated set $\Lambda \subseteq \Gamma$. The entropy form of the Chang theorem givesand maximality gives $\Gamma \subseteq \mathrm{Span}(\Lambda )$.

Put $B=B(\Lambda ,1/(6|\Lambda |))$. If $y\in B$ and $\gamma \in \Gamma$, expressing $\gamma$ as a product of characters in $\Lambda$ and their inverses givesFor $f=1_{-A}\ast {\mu }_A$, Parseval identity givesAlso $\hat{\mu }(\gamma )=|\widehat{{\mu }_Y}(\gamma ){|}^{2k}$. Fourier inversion theorem therefore yieldsonce $k$ is large enough.

If $\Lambda$ is empty, interpret $B$ as $G$; the same Fourier estimate, using only the second term, is even stronger.

Part (iii) gives $\Vert f\ast \mu -f{\Vert }_{\infty }\le 1/6$. Applying this at $x$ and $x+y$ and using the last estimate givesFinally $f(0)=1$. Hence $f(y)\ge 1/3>0$ for every $y\in B$, while the support of $f$ is $A-A$. Consequently