 Solution

Let $B={\mathbb{F}}_2^n\setminus (A+A)$ have density $\beta$. Since $1_A\ast 1_A$ vanishes on $B$,The zero-frequency contribution is ${\alpha }^2\beta$, soPut $\Gamma ={\mathrm{Spec}}_{\alpha /2}(1_B)$. Outside $\Gamma$, Parseval identity bounds the contribution byThereforebecause $|\widehat{1_B}(t)|\le \beta$. Chang's theorem places $\Gamma$ in a subspace $W$ of dimensionSet $V=W^{\perp }$. Then $V$ has this codimension, $V^{\perp }=W$, and adding the zero-frequency term $|\widehat{1_A}(0){|}^2={\alpha }^2$ gives