= 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 $\omega(0)$:
$$
e^{i\xi_1}\widehat{\psi\,d\sigma}(\xi)
=\frac1{2\pi}\int\psi(\omega(\theta))
e^{-i[\xi_1(\cos\theta-1)+\xi_2\sin\theta]}\,d\theta.
$$
On the support, $|\theta|\le2\delta/C$. The frequency rectangle and the elementary bounds $|1-\cos\theta|\le\theta^2/2$, $|\sin\theta|\le|\theta|$ imply
$$
|\xi_1(\cos\theta-1)+\xi_2\sin\theta|
\le\frac2{C^2}+\frac2C.
$$
Choose 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> satisfies
$$
\boxed{|\widehat{\psi\,d\sigma}(\xi)|
\ge\operatorname{Re}\!\left(e^{i\xi_1}\widehat{\psi\,d\sigma}(\xi)\right)
\ge\frac12\int\psi\,d\sigma
\ge\frac{\delta}{2\pi C}.}
$$
This 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 $\sin\theta$.
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