I am doing this exercise:
Let $S_{n}:=\sum_{k=1}^{n}X_{k}$ where the i.i.d random variables $\{X_{k}\}$ have each the characteristic function $\Phi_{X}(\cdot).$
Prove that $S_{n}/n\longrightarrow_{p}a$ implies, for all $t>0$, that
$(1)$ $\Phi_{X_{1}}(x_{k})^{\left \lfloor{t/x_{k}}\right \rfloor}\longrightarrow\exp(iat)\ \text{as}\ x_{k}\searrow 0,$
$(2)$ $\Phi_{X_{1}}(x_{k})^{{\left \lfloor{t/x_{k}}\right \rfloor}}\longrightarrow\exp(-iat)$ as $x_{k}\nearrow 0$.
I know that if $S_{n}/n\longrightarrow_{p} a$ then $S_{n}/n\Rightarrow a$ in distribution, but so that $$\Phi_{S_{n}/n}(t)=\Phi_{X_{1}}(t/n)^{n}\longrightarrow e^{iat},$$ but then I don't know what to do next to connect to this exercise..
Any idea? Thank you!
Edit 1:
Okay, I figured it out. Please read my answer below.
Okay, I figured it out.
If $x_{k}\searrow 0$ and $t>0$, then $$\Phi(x_{k})^{[t/x_{k}]}=\Phi_{S_{[t/x_{k}]}}(x_{k})=\mathbb{E}e^{ix_{k}S_{[t/x_{k}]}}\longrightarrow e^{itb}$$ by bounded convergence theorem since $x_{k}[t/x_{k}]\longrightarrow t$.
If $t<0$, then just take inverse and the $x_{k}\nearrow 0$c case follows by taking complex conjugates.
If we choose $t>0$ such that $|tb|<\dfrac{\pi}{4}$, we make take (principal) logarithms and deduce that $$\dfrac{1}{x}\log\Phi(x)=\dfrac{1}{t}\Big((\dfrac{t}{x}-\Big[\dfrac{t}{x}\Big]\Big)\log\Phi(x)+\log\Phi(x)^{[t/x]}\Big)\longrightarrow ib\ \text{as}\ x\searrow 0,$$ and thus $$\dfrac{1}{x}(\Phi(x)-1)=\dfrac{1}{x}\Big(\log\Phi(x)+O(|\Phi(x)-1|^{2})\Big)\longrightarrow ib.$$
Repeat the argument for $x\searrow 0$ to deduce that $\Phi'(0)=ib$.