Does proximity of moment generating functions implies proximity of characteristic functions? Let's assume that $U$ and $V$ are non-negative random variables.     Suppose that
\begin{align}
 \sup_{t \ge 0 } \frac{| M_U(-t) - M_V(-t)|}{t} \le \epsilon 
\end{align}
where $M_U(t)$ and $M_V(t)$ are moment generating functions.
A few facts:

*

*Technically $M(-t)$ is known as Laplace transform.

*$M(t)$ unique on an open interval. Therefore, this question is well defined.

*$ t \to M(-t)$ is decreasing.

Question: Does this imply that
\begin{align}
 \sup_{t \in \mathbb{R} } \frac{| \phi_U(t) - \phi_V(t)| }{|t|}\le f(\epsilon) 
\end{align}
where $\phi_U(t)$ and  $\phi_V(t)$ are characteristic functions, and $f$ is some function that goes to zero as $\epsilon \to 0$.
I was thinking of using that $\phi(t)=M(it)$, but this doesn't work out.
 A: No. Consider $U_n=\frac{1}{2}\delta_{n-1}+\frac{1}{2}\delta_{n+1}$, $V_n=\delta_{n}$. Then
\begin{align*}
\frac{\vert M_{U_n}(-t)-M_{V_n}(-t)\vert}{t}=\frac{\vert \frac{e^{-(n+1)t}}{2}+\frac{e^{-(n-1)t}}{2}-e^{-nt}\vert}{t}=\frac{e^{-(n-1)t}}{2}\frac{(1-e^{-t})^2}{t}. 
\end{align*}
Say there are an infinite sequence of $\{t_{n_k}\}_k$ and an $\epsilon>0$ such that \begin{align*}
\frac{e^{-(n_k-1)t_{n_k}}}{2}\frac{(1-e^{-t_{n_k}})^2}{t_{n_k}}>\epsilon.
\end{align*}
Then we must have $t_{n_k}\rightarrow 0$ as $k\rightarrow\infty$. But
\begin{align*}
\lim_{k\rightarrow\infty}\frac{e^{-(n_k-1)t_{n_k}}}{2}\frac{(1-e^{-t_{n_k}})^2}{t_{n_k}}\leq\frac{1}{2}\lim_{k\rightarrow\infty}\frac{(1-e^{-t_{n_k}})^2}{t_{n_k}}=0
\end{align*}
by a simple application of L'Hopital's Rule.
However,
\begin{align*}
\frac{\vert \phi_{U_n}(t)-\phi_{V_n}(t)\vert}{\vert t \vert}&=\frac{\sqrt{\left(\frac{\cos((n-1)t)}{2}+\frac{\cos((n+1)t)}{2}-\cos(nt)\right)^2+\left(\frac{\sin((n-1)t)}{2}+\frac{\sin((n+1)t)}{2}-\sin(nt)\right)^2}}{\vert t\vert} \\
&=\frac{2}{\pi}
\end{align*}
when $t=\pi$. So for any $\epsilon,f$ I have a pair $U_n,V_n$ that satisfy your MGF constraint but not your CF constraint.
