 Solution

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