I have given $X_n,Y_n$ two sequences of real valued random variables in the same probability space. I assume that $X_n\Rightarrow X$ in distribution and $|X_n-Y_n|\rightarrow 0$ in probability. In addition $f:\Bbb{R}\rightarrow \Bbb{R}$ is an arbitrary continuous function s.t. there exists a compact subset $K\subset \Bbb{R}$ with $f(x)=0$ for all $x\notin K$. I need to show that $\lim_{n\rightarrow \infty} \Bbb{E}\left(|f(X_n)-f(Y_n)|\right)=0$.
My idea was the following:
We first remark that $\Bbb{E}\left(|f(X_n)-f(Y_n)|\right)\geq \left|\Bbb{E}\left(f(X_n)-f(Y_n)\right)\right|\geq 0$. On the other hand $$\Bbb{E}\left(|f(X_n)-f(Y_n)|\right)\leq \Bbb{E}(|f(X_n)|+|f(Y_n)|)\stackrel{*}{=}\Bbb{E}(|f(X_n)|)+\Bbb{E}(|f(Y_n)|)$$ where $*$ only holds since $|f(X_n)|, |f(Y_n)|$ is integrable (this follows since $f$ has compact support). Now using that $X_n\Rightarrow X$ we know that $$\Bbb{E}(f(X_n))\rightarrow \Bbb{E}(f(X))$$ but then $$\Bbb{E}\left(|f(X_n)-f(Y_n)|\right)\rightarrow |\Bbb{E}(f(X))|+\Bbb{E}(|f(Y_n)|)$$ Now since $|X_n-Y_n|\rightarrow 0$ in probability we know that $|f(X_n)-f(Y_n)|\rightarrow 0$ in probability. I thought that one could use this to show that maybe $\Bbb{E}(|f(X_n)|)\rightarrow -|\Bbb{E}(f(X))|$ or something similar, because then I would be done.
But I'm not sure if this really works. Could maybe someone first tell me if this is correct what I wrote and then maybe if not what I need to change. If my work holds till some point it would be nice if we could use it and then proceed with the correct way.
Thanks for your help.
I will make a weaker assumption on $f$ which is that it is bounded and uniformly continuous. Fix $\epsilon >0$, then there exists $\delta >0$ such that $|f(x) - f(y)| \leq \epsilon$ whenever $|x-y|\leq \delta$.
We have $$\mathbb{E}|f(X_n) - f(Y_n)| = \mathbb{E}\left[|f(X_n) - f(Y_n)|\mathbf{1}_{\{|X_n-Y_n|\leq \delta\}}\right]+\mathbb{E}\left[|f(X_n) - f(Y_n)|\mathbf{1}_{\{|X_n-Y_n|>\delta\}}\right].$$
Thanks to uniform continuity, the first term is bounded by $$ \epsilon \,\mathbb{P}(|X_n-Y_n|\leq \delta) \leq \epsilon.$$
For the second term, we use the boundedness of $f$ to get $$\mathbb{E}\left[|f(X_n) - f(Y_n)|\mathbf{1}_{\{|X_n-Y_n|>\delta\}}\right]\leq 2\lVert f\rVert_{\infty}\, \mathbb{P}(|X_n-Y_n|>\delta) \to 0.$$ (The last convergence is the definition of convergence in probability.)
Thus, taking limsup on both sides, we deduce that for every $\epsilon >0$ $$\limsup_{n\to \infty} \mathbb{E}|f(X_n) - f(Y_n)| \leq \epsilon.$$ Letting $\epsilon \to 0$ yields the result.