 Solution

Parametrize the unit circle by $\omega (\theta )=(\cos \theta ,\sin \theta )$, with $-\pi \le \theta <\pi$ and $d\sigma =d\theta /(2\pi )$. Extract the constant phase at $\omega (0)$:On the support, $|\theta |\le 2\delta /C$. The frequency rectangle and the elementary bounds $|1-\cos \theta |\le {\theta }^2/2$, $|\sin \theta |\le |\theta |$ implyChoose the fixed constant $C$ large enough that this is less than $\pi /3$. Every phase then has real part at least $1/2$. Nonnegativity of $\psi$ prevents cancellation after this phase rotation, while its central plateau gives $\int \psi d\sigma \ge \delta /(\pi C)$. Hence the Fourier transform satisfiesThis is the circle cap Fourier lower bound. No upper bound on the values of $\psi$ is needed here; the support, plateau and nonnegativity suffice. The long radial scale ${\delta }^{-2}$ comes from the quadratic term $\cos \theta -1$, whereas the transverse scale ${\delta }^{-1}$ comes from the linear term $\sin \theta$.