= Solution
Since $a$ is an <indicator function> of <density of a finite subset> $\alpha$, the <triangle inequality> gives $|\widehat a(r)|\leq\mathbb E_xa(x)=\alpha$ for every frequency. Also, <character orthogonality> and the <Parseval identity on a finite group> give
$$
\sum_r|\widehat a(r)|^2=\mathbb E_x|a(x)|^2=\alpha.
$$
For completeness, the <character orthogonality> used here is
$$
\sum_{r\in\mathbb Z_n}\omega^{r(y-x)}=
\begin{cases}n,&x=y,\\0,&x\ne y,\end{cases}
$$
which follows by summing a finite <geometric series>. Expanding the squared <Fourier coefficients on a finite abelian group> and using this identity proves the displayed <Parseval identity on a finite group> directly.
Combining the uniform bound with that identity gives the \b[fourth-moment bound]
$$
\boxed{\|\widehat A\|_4^4=\sum_r|\widehat a(r)|^4\leq\alpha^2\sum_r|\widehat a(r)|^2=\alpha^3.}
$$
In particular, the <Lp norm> on the frequency side here is a sum, not a normalized average. This is the <fourth Fourier moment bound for an indicator function>.
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