In what follows, $g:\lbrack 0,\infty)\to \mathbb{R}$ will be a continuous function obeying the following constraints: $g(x)=1$ for $x\leq 1-\delta$, $g(x)=0$ for $x\geq 1+\delta$, and $g(1-t)=1-g(1+t)$ for $0\leq t\leq \delta$. (Otherwise put: $g'(t)$ is a symmetric function around $1$ of support $\lbrack 1-\delta,1+\delta\rbrack$.) Other than that, we are free to choose $g$, and for that matter $0<\delta<1$ as well.
Write $G(s)$ for the Mellin transform of $g$.
Let $T>0$, $0<\sigma<1$ be given. Let $\epsilon = \max_{t\geq T} |\sigma+ i t| \left|G(\sigma+it)\right|$. How small can $$\frac{\int_{1}^{1+\delta} |g(t)|^2 dt}{(1-\epsilon)^2}$$ be? Is the minimum reached by a nice function $g$?
(I take there has to be a non-zero infimum, by the uncertainty principle. It is also easy to get a crude upper bound, by choosing any particular $g$ (such as, say, $g(1+t) = \frac{1}{2} - \frac{t}{\delta}$ for $-\delta\leq t\leq \delta$).)
