Proposition: No exact finite-time continuum transfer

In this declared continuum model, $|B(T)|<1$ for finite $T$.

Proof. If $|B(T)|=1$, unitarity forces $\psi (T)=0$. Since $f(\omega )>0$, the packet formula implies

for almost every positive $\omega$. The function $c$ is bounded and integrable. Both sides extend to entire functions. Their equality on a full-measure subset of a real interval gives equality everywhere by continuity and the identity theorem.

At $\omega =\mathrm{i}y$, $y>0$, the left side is $e^{y{t}_0}$, whereas

The right side tends to zero and the left side is at least one, a contradiction.