Is there a 'certainty' principle?

By request, I add a comment as an answer with some additional details; but what I meant is really straightforward. The simplest realization is as follows: take any $\psi\in L^2$. Split its support into finitely many parts to obtain a representation $\psi=\sum_{k= 0}^N\psi_k$ where $\psi_0$ is small in $L^2$ (the infinite tail) and $\psi_k$ for $k>0$ are small (less than $\varepsilon$) in $L^1$ (short intervals). Now multiply each $\psi_k$ with $k>0$ by $e^{2\pi i Mkx}$ with $M$ chosen so that $\sup_{\lvert y\rvert>M,1\le k\le N}\lvert\widehat\psi_k(y)\rvert\le \frac{\varepsilon}N$ (it exists by Riemann–Lebesgue). Then the Fourier transform of the resulting function at any point $y$ will be bounded by $\lvert\widehat\psi_0(y)\rvert+3\varepsilon$. The first part doesn't influence anything because its $L^2$-norm is small and the rest is uniformly small and, therefore, spread wide.

If $\psi\in L^1\cap L^2$, then no special treatment of $\psi_0$ is needed. Also, you can get the true uniform smallness by splitting into countably many parts and choosing the phase shifts inductively instead of just using an arithmetic progression. And so on, and so forth.

Edit: Now about convexity. Take $f$ to be the characteristic function on $[0,1]$ and consider $g(k)$ where $k\in\mathbb Z$ (in this case the point values are continuous functionals). Clearly, every sequence with all zeroes and one $1$ is admissible ($\psi(x)=e^{2\pi ik_0x}$ on $[0,1]$). Thus, if the convexity had held, we would be able to construct a function on $[0,1]$ that is identically $1$ (or, at least, as close to that as we would like) such that $g(0)=g(1)=\frac 12$ and all other $g(k)=0$. However, that would be just a two-term polynomial with equal coefficients, so it would vary quite a bit in absolute value on $[0,1]$. This proves at least that sometimes convexity does not hold. I suspect that this trick can be generalized quite a bit but the details are elusive yet.