= Solution
Let $B=\mathbb F_2^n\setminus(A+A)$ have density $\beta$. Since $1_A*1_A$ vanishes on $B$,
$$
0=\langle1_A*1_A,1_B\rangle
=\sum_t\widehat{1_A}(t)^2\widehat{1_B}(t).
$$
The zero-frequency contribution is $\alpha^2\beta$, so
$$
\alpha^2\beta
\le\sum_{t\ne0}|\widehat{1_A}(t)|^2|\widehat{1_B}(t)|.
$$
Put $\Gamma=\operatorname{Spec}_{\alpha/2}(1_B)$. Outside $\Gamma$, <Parseval identity> bounds the contribution by
$$
\frac{\alpha\beta}{2}\sum_t|\widehat{1_A}(t)|^2
=\frac{\alpha^2\beta}{2}.
$$
Therefore
$$
\sum_{t\in\Gamma\setminus\{0\}}|\widehat{1_A}(t)|^2\ge\frac{\alpha^2}{2},
$$
because $|\widehat{1_B}(t)|\le\beta$. Chang's theorem places $\Gamma$ in a subspace $W$ of dimension
$$
O(\alpha^{-2}\log(\beta^{-1})).
$$
Set $V=W^\perp$. Then $V$ has this codimension, $V^\perp=W$, and adding the zero-frequency term $|\widehat{1_A}(0)|^2=\alpha^2$ gives
$$
\boxed{\sum_{t\in V^\perp}|\widehat{1_A}(t)|^2\ge\frac{3\alpha^2}{2}.}
$$
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