Calculus with Complex Variables I come across the following calculus problem when I am studying probability theory:
Let $\varphi(t):\mathbb R\to \mathbb C$ be a characteristic function of a random variable. (For your reference, this is a bounded and uniformly continuous function which has $\varphi(0)=1$). Assume there is $a\in \mathbb R$ such that 
$$
\varphi(t/n)^n\to e^{iat},\quad \text{for any } t\in\mathbb R. 
$$
Can we conclude that $\varphi'(0)=ia$?
My idea is to take logarithm on both sides. But which branch should I choose? Secondly, how could I use the uniform continuity of $\varphi$ to pass the limit in $n$ to a limit in $h\to 0$?
 A: Since $\phi(0)=1$ and $\phi$ is continuous there is some interval $(-\delta,\delta)$ where $|1-\phi(t)|<1$. There is a continuous logarithm in the disk $|1-z|<1$. So there is a continuous function $l(t)$ (defined at least for $|t|<\delta$) with $l(0)=0$ and $$\phi(t)=e^{l(t)}\quad(|t|<\delta).$$
Now if $t$ is small enough we must have $$\frac{l(t/n)-l(0)}{t/n}=\frac ntl(t/n)\to  ia.$$
If you know that $\phi$ is differentiable then $l$ is differentiable and the above shows that $l'(0)=ia$, hence $\phi'(0)=ia$. Does the above actually imply that $l$ is differentiable? This is not immediately clear to me.
Edit: I believe the answer is yes, it does follow that $l'(0)$ exists. The proof I have is not quite trivial - if it's correct then the answer to your question is yes.
Note that if $\phi$ is the characteristic function of $X$, where $\mathbb E[|X|]<\infty$, then yes $\phi$ is differentiable.
A: I have written a solution on my own, inspired by the posts and credit to:@David C. Ullrich
A classical problem about limit of continuous function at infinity and its connection with Baire Category Theorem
Relationship between the weak law of large numbers and characteristic functions
Taking logarithms on both sides (with the branch $-\pi<\arg z\leq \pi$, and we choose $|t|<\delta$ with $\delta |a|<\pi$), we have
                $n\log \varphi (\frac t n)\to iat$. Since $\log z$ is differentiable at $z=1$, we have
                $$
    n(\varphi(t/n)-1)=\left(\frac{\varphi(t/n)-1}{\log \varphi(t/n)}\right)( n \log \varphi(t/n))\to iat.
    $$
Lastly, we need to bootstrap from the limit in $n$ to a limit in $h\to 0^+$. (By symmetry, the same holds for $h\to 0^-$)
                                Let $\epsilon>0$. The sets
                $$
    E_N:=\left\{t\in [0,\delta]: \sup_{n\geq N}\left|n\left[\varphi\left(\frac t n\right)-1\right]-iat\right|\leq \epsilon\right\}
    $$
                are all closed, by continuity of $\varphi$.                        
Since for any $t$, the above holds for sufficiently large $N$, we have $[0,\delta]=\cup_N E_N$.                
By Baire Category Theorem, there is $N$ such that $E_N$ has nonempty interior. Let $(a,b)\subseteq E_N$, and assume without loss of generality that $0<a<b$. Since $E_N$'s are nested by definition, for all $0<h<\min\{b-a,a/N\}$, there is $n\geq N$ so that $hn\in (a,b)\subseteq E_N\subseteq E_n$, thus
                $$
    \left|\frac{\varphi(h)-1}{h}-ia\right|\leq \epsilon.
    $$
