= Solution
Choose $\epsilon=1/6$ and a sufficiently large absolute constant $k$. Part (iii)'s size bound gives
$$
|Y|/|G|\ge\alpha^{O(1)}.
$$
Let
$$
\Gamma=\{\gamma\in\widehat G:|\widehat{\mu_Y}(\gamma)|\ge1/2\}
$$
and choose a maximal <dissociated set> $\Lambda\subseteq\Gamma$. The entropy form of the <Chang theorem> gives
$$
|\Lambda|=O(\log(|G|/|Y|))=O(\log(\alpha^{-1})),
$$
and maximality gives $\Gamma\subseteq\operatorname{Span}(\Lambda)$.
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 gives
$$
|\gamma(y)-1|\le1/6.
$$
For $f=1_{-A}*\mu_A$, <Parseval identity> gives
$$
\widehat f(\gamma)=\frac{|\widehat{1_A}(\gamma)|^2}{\alpha},
\qquad
\sum_\gamma|\widehat f(\gamma)|=1.
$$
Also $\widehat\mu(\gamma)=|\widehat{\mu_Y}(\gamma)|^{2k}$. <Fourier inversion theorem> therefore yields
$$
|f*\mu(x+y)-f*\mu(x)|
\le\frac16\sum_{\gamma\in\Gamma}|\widehat f(\gamma)|
+2^{1-2k}\sum_{\gamma\notin\Gamma}|\widehat f(\gamma)|
\le\frac13
$$
once $k$ 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 $\|f*\mu-f\|_\infty\le1/6$. Applying this at $x$ and $x+y$ and using the last estimate gives
$$
\boxed{|f(x+y)-f(x)|\le2/3.}
$$
Finally $f(0)=1$. Hence $f(y)\ge1/3>0$ for every $y\in B$, while the support of $f$ is $A-A$. Consequently
$$
\boxed{B\subseteq A-A.}
$$
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