Let $G$ be a finite abelian group of order $n$. Let $\hat{G}$ be the dual group of $G$, i.e., the group of characters on $G$. For $f:G\to \mathbb{C}$, we define its Fourier transform $\hat{f}:\hat{G}\to \mathbb{C}$ by $\hat{f}(\chi):=\sum_{g\in G} f(g)\overline{\chi(g)}$. It's not difficult to prove Plancherel's theorem: $$\sum_{\chi\in \hat{G}}|\hat{f}(\chi)|^2=n\sum_{g\in G}|f(g)|^2.$$ For $E\subseteq G$, let $E(x)$ be its characteristic function. Then Plancherels' theorem implies that $\sum_{\chi\in \hat{G}}|\hat{E}(\chi)|^2=n|E|$.
By $\chi_0$, we denote the trivial character, and we see that $\hat{E}(\chi_0)=|E|$. Let $\Phi(E):=\max\{|\hat{E}(\chi)|:\chi\neq \chi_0\}$. It is easy to see that $\Phi(E)\leq |E|$. What about the lower bound? The lower bound is given by the following lemma, which is easy to derive.
Lemma. Let $E\subseteq G$. If $|E|\leq n/2$, then $\Phi(E)\geq \sqrt{{|E|}/{2}}.$
Question 1: I asked myself what if $|E|>n/2$?
The book says that "The assumption $|E| \leq n/2$ is justified by the following exercise: $\Phi(E)=\Phi(E^c)$, where $E^c:=G\setminus E$". The exercise is pretty easy to prove.
Then $|E^c|<n/2$ and by Lemma $\Phi(E^c)\geq \sqrt{{|E^c|}/{2}}$ and hence $\Phi(E)\geq \sqrt{{|E^c|}/{2}}$. So my question is: what is the motivation behind the assumption that $|E|\leq n/2$? I think I understand all the technicalities, but I did not get the point.
Question 2: In one paper I read, it was stated that the best one can reasonably hope for is $\Phi(E)\lesssim \sqrt{|E|}$. This estimate cannot be improved upon if $|E|=o(n)$. What does it mean and how it follows? Can't understand both statements.
