For a discrete distribution the characteristic function $|\psi(u)|=1$ for other values of $u$ than $0$. We also know that $|\psi(u)|\leq 1$.
How does this imply that for a continuous distribution we need to have $|\psi(u)|\leq e^{-c}$ for some $c>0$? Why $e^{-c}$?
Edit I forgot to mention that the above condition is satisfied for $|u|>T$, for some large "T". That is the condition is of the same type as $|\psi(u)|=o(|u|^{-n})$ for $|u|\to \infty$ showed in Lemma 4, Ch XV.5 of Feller-"An Introduction to Probability Theory and Its Applications" Vol 2.
The result is used in proves related with the remainder for expansions of density functions(See Gnedenko- "Limit distributions for sums of independent random variables" p.229 for example)
When the density $f$ is of class $C^1$, this follows from an integration by parts.
For the general case, we may approximate $f$ in $\mathbb L^1(\mathbf R,\lambda)$ by a $C^1$ function.