Estimation for modulus of characteristic function It is well-known that if $\xi$ is an absolutely continuous random variable with characteristic function $\phi(t)$, then for each $\epsilon>0$ one has $\sup\limits_{|t|>\epsilon}|\phi(t)|<1$ (sometimes it is called Cramer's theorem). However, if we can say something about existence of moments of the r.v. then this result can be improved. For instance, if $\mathsf E \xi=0,\, \mathsf E \xi^2<\infty$, then one can conclude(if I do not mistake) that exists such $K>0$ that for all $s \in (0,1)$ $\sup_{|t|>s}|\phi(t)|<e^{-Ks^2}$.
The questions are the following: is the statement above true? How one can prove it(I tried to do something with Taylor's expansion and the estimation like $|e^{is}-1-is|\leq |s|^2/2$, but I did not manage to prove it)? Can there be generalization for r.v. that possesses higher moments?
Thanks
 A: It is the "smoothness" of the random variable $\xi$, not its moments, 
that is related to the decay at infinity of its characteristic function  $\phi$.
For instance, if $\phi$ is absolutely integrable over the line, then $\xi$ has a
continuous density function given by the inversion formula:
$$
f(x)={1\over 2\pi} \int_{-\infty}^\infty \exp(-itx)\, \phi(t)\,dt.
$$
In addition, if
$$\int_{-\infty}^\infty |t^n \phi(t)|\,dt<\infty,$$
then the density $f$ is $n$ times continuously differentiable.
On the other hand, if $\xi=1$ with probability one, then 
$\phi(t)=\exp(it)$ and so $\phi(t)=1$ whenever $t$ is an integer
multiple of $2\pi$. Although $\xi$ has finite moments of all order,
 $|\phi(t)|$ does not decay for large $t$.  
A: The idea that your inequality should hold for k based on moments being bounded, which seems to be the implicit claim, cannot be right because $\epsilon X$ always converges to $0$, and so its ch. fctn converges to 1, yet all moments can be taken to be bounded etc., for , eg. normal r.v.s.  This clearly happens for any class of r.v.s
 in which the variance can go to 0.
