If $|a-b| \geq \sqrt{N}$ for every $a \neq b \in \text{supp}(\hat{f})$, what can we say about $f : \mathbb{Z}/N\mathbb{Z} \to \mathbb{C}?$ Let $f : \mathbb{Z}/N\mathbb{Z} \to \mathbb{C}$ with non-empty support and let $\hat{f} : \mathbb{Z}/N\mathbb{Z} \to \mathbb{C}$ be its Fourier transform given by $\hat{f}(m) = \sum_{j \in \mathbb{Z}/N\mathbb{Z}} f(j) e_N(jm)$, where $e_N(x) := e^{2\pi i x/N}$. 
Question: If $|a - b| \geq \sqrt{N}$ for every $a,b \in \text{supp}(\hat{f})$ with $a \neq b$, is there anything interesting we can say about the support of $f$?
Remark 1: This is a variant of the question in Discrete fourier transform on $\mathbb{Z}/N\mathbb{Z}$ vanishing on an interval of size at least $\sqrt{N}$ where in our question here we make the assumption there stronger.
Remark 2: It is well-known that the sizes of the supports of $f,\hat{f}$ are related by the inequality $\#\text{supp}(f) \#\text{supp}(\hat{f}) \geq N$. Thus we could obtain a lower bound for $\#\text{supp}(f)$ using the information given. However I wonder what else we could say about the support of $f$ given that for every $a \in \text{supp}(\hat{f})$, $\hat{f}$ vanishes on intervals $(a, a + K]$ with $K \geq \sqrt{N}$? For instance, can local densities of $\text{supp}(\hat{f})$ in intervals of size $\sqrt{N}$ affect local densities of $\text{supp}(f)$ also in intervals of the same size? Thanks.
 A: Here is something that can be said (and is probably well-known), but I suspect there should be other things as well that could be said. 
Proposition: Suppose for every $a,b \in \text{supp}(\hat{f})$, with $a \neq b$, that $|a - b| \geq \delta N$ for some fixed $\delta > 0$. Let $I$ be an interval in $\mathbb{Z}/N\mathbb{Z}$. Then
$$
\left| \sum_{m \in I}|f(m)|^2  - \dfrac{|I|}{N}\sum_{m \in \mathbb{Z}/N\mathbb{Z}} |f(m)|^2\right| \leq \dfrac{\delta^{-1} - 1}{N} \sum_{m \in \mathbb{Z}/N\mathbb{Z}} |f(m)|^2.
$$
Proof: We have
\begin{align*}
\sum_{m \in I}|f(m)|^2 &= \sum_{m \in I} \left| \dfrac{1}{N} \sum_{n \in \mathbb{Z}/N\mathbb{Z}}\hat{f}(n) e_N(-mn)\right|^2 \\
&= \dfrac{1}{N^2}\sum_{n,n'\in \mathbb{Z}/N\mathbb{Z}}\hat{f}(n)\overline{\hat{f}(n')}\sum_{m \in I}e_N(m(n' - n))\\
&= \dfrac{|I|}{N^2}\sum_{n \in \mathbb{Z}/N\mathbb{Z}}\left| \hat{f}(n)\right|^2 + \dfrac{1}{N^2}\sum_{\substack{n,n'\in \mathbb{Z}/N\mathbb{Z} \\ n \neq n'}}\hat{f}(n)\overline{\hat{f}(n')}\sum_{m \in I}e_N(m(n' - n))\\
&= \dfrac{|I|}{N}\sum_{n \in \mathbb{Z}/N\mathbb{Z}}\left| f(n)\right|^2 + \dfrac{1}{N^2}\sum_{\substack{n,n'\in \mathbb{Z}/N\mathbb{Z} \\ n \neq n'}}\hat{f}(n)\overline{\hat{f}(n')}\sum_{m \in I}e_N(m(n' - n)).
\end{align*}
The last equality holds by Parseval's identity. Thus
$$
\left| \sum_{m \in I}|f(m)|^2  - \dfrac{|I|}{N}\sum_{m \in \mathbb{Z}/N\mathbb{Z}} |f(m)|^2\right| = \dfrac{1}{N^2}\left|\sum_{\substack{n,n'\in \mathbb{Z}/N\mathbb{Z} \\ n \neq n'}}\hat{f}(n)\overline{\hat{f}(n')}\sum_{m \in I}e_N(m(n' - n))\right|.
$$
Since for every $n,n' \in \text{supp}(\hat{f})$, $n \neq n'$, we have $|n - n'| \geq \delta N$ by assumption, the large additive sieve inequality (see for instance Lemma 8 in here) gives
\begin{align*}
\dfrac{1}{N^2}\left|\sum_{\substack{n,n'\in \mathbb{Z}/N\mathbb{Z} \\ n \neq n'}}\hat{f}(n)\overline{\hat{f}(n')}\sum_{m \in I}e_N(m(n' - n))\right| &\leq \dfrac{\delta^{-1} - 1}{N^2}\sum_{n \in \mathbb{Z}/N\mathbb{Z}}|\hat{f}(n)|^2 \\
&= \dfrac{\delta^{-1} - 1}{N}\sum_{m \in \mathbb{Z}/N\mathbb{Z}}|f(m)|^2,
\end{align*}
as required.
Corollary: Let $f : \mathbb{Z}/N\mathbb{Z} \to \mathbb{C}$ with non-empty support satisfy the assumptions of the proposition above. Then for every interval $I$ of $\mathbb{Z}/N\mathbb{Z}$ with $|I| \geq \delta^{-1}$, we have that $\text{supp}(f) \cap I \neq \emptyset$. 
Proof: From the proposition above we have that 
$$
\dfrac{|I| - \delta^{-1} + 1}{N}\sum_{m \in \mathbb{Z}/N\mathbb{Z}}|f(m)|^2 \leq \sum_{m \in I}|f(m)|^2.
$$
The result now follows. 
Remark: The above argument probably works if we replace the interval $I$ with an arithmetic progression. 
