I have a problem on the following question. I can easily prove that $max_j|\hat f(j)|\geq\sqrt{n}$ by using the formula $\sum_j |f(j)|^2=\frac{1}{n}\sum_j |\hat f(j)|^2$ but I don't know how to exclude the equality.
Problem: $f\in l^2(Z_n)$, and suppose that $|f| = X_E, E\subset Z_n$. Suppose that the support of $\hat f$ is contained in a set of size at most $|E|$. Show that $max_j|\hat f(j)|>\sqrt{n}.$
Here, $\hat f(k)=\sum_{j}f(j)e^{-2\pi ijk/n}, X_E=1_E.$
Unfinished proof: If $\max_{j}|\hat f(j)|\leq\sqrt{n}$ then we have $$ \frac{1}{n}\sum_{j\in\mathbb{Z_n}}|\hat f(j)|^2=\frac{1}{n}\sum_{j\in supp{\hat f}}|\hat f(j)|^2\leq\sum_{j\in supp{\hat f}}1 =|supp \hat f| \leq|E|.$$ On the other hand, $$|E|=\sum_{j\in\mathbb{Z_n}}|f(j)|^2=\frac{1}{n}\sum_{j\in\mathbb{Z_n}}|\hat f(j)|^2,$$ so the equality holds, which means $$\frac{|\hat f(j)|}{\sqrt{n}}=1_{supp \hat f}.$$ $$\cdots$$