Let $G$ be a finite abelian group, $f:G\to \mathbb C$ not identically zero. Donoho-Stark uncertainty principle say that $|\textrm{supp} f||\textrm{supp} \hat{f}|\geq |G|$. The prove I have goes as follows:
$$\sup_{\gamma\in \widehat{G}}|\hat{f}(\gamma)|=|\hat{f}(\gamma_0)|=\left|\frac{1}{|G|}\sum_{b\in G}f(b)\overline{\gamma_0(b)}\right|\leq\frac{1}{|G|}\sum_{b\in G}f(b)\chi_{\textrm{supp}f}(b)\leq\left(\frac{1}{|G|}\sum_{b\in G}|f(b)|^2\right)^{1/2}\left(\frac{1}{|G|}|\textrm{supp}f|\right)^{1/2}=\left(\sum_{\gamma\in \widehat{G}}|\hat{f}(\gamma)|^2\right)^{1/2}\left(\frac{1}{|G|}|\textrm{supp}f|\right)^{1/2}\leq\left(\frac{1}{|G|}|\textrm{supp}f|\right)^{1/2}\left(|\textrm{supp}\hat{f}||\hat{f}(\gamma_0)|^2\right)^{1/2}.$$
The three inequalities in the proof follow, in this order, from triangle inequality, from Cauchy-Schwarz and, finally, simply from the fact that we take $\gamma_0\in\widehat{G}$ to be a function for which $\hat{f}$ reaches its maximum. Now, I want to show that if equality holds (in particular, if equality holds in these three inequalities), then $f=c\chi_{gH}$ for some $c\in\mathbb C$, $g\in G$ and $H\leq G$. From equality on Cauchy-Schwarz, we may already deduce that $f$ is some multiple of a characteristic function, i.e., there is $c\in\mathbb C$ such that $f=c\chi_{\textrm{supp}f}$. Furthermore, if equality holds in triangle inequality, all elements of the sum must be in the same complex ray. That is to say, for every $b, b'\in G$ there are $a_1,a_2\geq0$ such that $$a_1f(b)\overline{\gamma_0(b)}=a_2f(b')\overline{\gamma_0(b')}.$$ If $b$ or $b'$ are not in the support of $f$, this inequality is trivial. However, for $b,b'\in\textrm{supp}f$, using also the fact that $f$ is a multiple of a characteristic function, we get that $a_1=a_2$ and, hence $\gamma_0(b)=\gamma_0(b')$. Let us call this element $\gamma_0(b)=z_{\gamma_0}$. Then, $\textrm{supp}f\subset \gamma_0^{-1}(z_{\gamma_0})$. Finally, mind that if equality holds in the third inequality, $\hat f(\gamma)=\hat f(\gamma_0)$ for all $\gamma\in\textrm{supp}\hat f$, so the argument above is valid for any such $\gamma$. I feel like this is really close to the solution. In fact if we could prove $\textrm{supp f}=\gamma^{-1}(z_\gamma)$ for some $\gamma\in\textrm{supp}\hat f$, the adequate translate of $\textrm{supp} f$ would be the kernel of such $\gamma$, and hence a subgroup. However, I do not know how prove this. How could I choose the proper $\gamma$? Perhaps there is some standard group theory argument that I am missing...