How to show that $E(\lvert e^{itY_n}-e^{itc}\rvert)$ goes to $0$ where $Y_n\xrightarrow{p}c$

Question

I'm trying to prove this statement using Levy's theorem.

If $X_n\xrightarrow{d}X$ and $Y_n\xrightarrow{p} c$ then $X_n+Y_n\xrightarrow{d} X+c$

I have for the distance of the characteristic functions, which needs to go to $0$: $$\begin{aligned} \left\lvert E\left(e^{it(X_n+Y_n)}\right) - E\left(e^{it(X+c)}\right) \right\rvert &\le \left\lvert E\left(e^{it(X_n+Y_n)}\right) - E\left(e^{it(X_n+c)}\right)\right\rvert + \left\lvert E\left(e^{it(X_n+c)}\right) - E\left(e^{it(X+c)}\right)\right\rvert \\ &=D_1 + D_2 \end{aligned}$$ Here $D_2$ vanishes somewhat trivially because $X_n\xrightarrow d X$. However I'm struggling with $D_1$. Using the triangle inequality we have $$\begin{aligned} \left\lvert E\left(e^{it(X_n+Y_n)}\right) - E\left(e^{it(X_n+c)}\right)\right\rvert &=\left\lvert E\left(e^{it(X_n+Y_n)}-e^{it(X_n+c)}\right)\right\rvert\\ &\le E\left(\left\lvert e^{it(X_n+Y_n)}-e^{it(X_n+c)}\right\rvert\right)\\ &= E\left(\left\lvert e^{it(X_n)}\right\rvert\left\lvert e^{itY_n} - e^{itc}\right\rvert\right)\\ &= E\left(\left\lvert e^{itY_n} - e^{itc}\right\rvert\right) \end{aligned}$$

How can we show that this goes to $0$?

Answer 1

wlog $c=0$ (just multiply by $e^{-itc}$)

Now, fix $\epsilon>0$. For $n$ sufficiently big $\mathbb{P}(|Y_n|>\epsilon) < \epsilon$.

Thus, $\mathbb{E}[|e^{itY_n}-1|] \leq 2\epsilon + |e^{it\epsilon}-1|$ for any such $n$ (why?).

This means that $\limsup\limits_{n \to \infty} \mathbb{E}[e^{itY_n}-1|] = 0$ for any $t$ (even though the convergence "depends on $t$"), but I guess that's fine.