Weak convergence using characteristic functions

This problem is from Billingsley's "Probability and measure" book.

Let $a_n \to a$, $b_n \to b$ and $\{X_n\}$,$X$ be a sequence of random variables such that $X_n \to^w X$ (weak convergence). Prove that $$a_nX_n + b_n \to^w aX+b$$ using characteristic functions.

I was able to reduce the problem into proving the following:

If $a_n \to 0$ and $X_n \to^w X$ then $a_nX_n \to^w 0$.

To begin with, we have $$|E[e^{ita_nX_n}] - 1| \leq E[|e^{ita_nX_n} - 1|] = 2E|\sin\left(\frac{ta_nX_n}{2}\right)|$$

The trouble I am having is that the sequence $X_n$ is in the way. If $X_n$ are integrable, then the proof is complete using $|\sin(x)| \leq |x|$. However since that is not given, I don't know how to proceed. Any help would be appreciated.

Edit: I just saw an answer here. However that used Skorokhod's theorem. Isn't there some simplification that could be done to RHS?

If $\phi_n$ is the characteristic function for $X_n$ and $\phi$ the characteristic function for $X$ then you know $\phi_n(t) \to \phi(t)$ for all $t$. Let $\epsilon > 0$. Choose $N_1$ large enough so that $n \geq N_1$ implies $$|\phi_n(at) - \phi(at)| < \frac{\epsilon}{2}.$$ Since the $\phi_n$ are uniformly continuous, for each $n$ there is a $\delta_n > 0$ such that $|x-y| < \delta_n$ implies $|\phi_n(x) - \phi_n(y)| < \epsilon$. Without loss of generality, assume that $\delta_n$ is a decreasing sequence (if $\delta_n$ is not smaller than the $\delta_i$'s with $i\leq n$, taking it to be less than or equal to the preceding values does not make the condition false).
Now, since $a_n \to a$ you can choose $N_2 \in \mathbb{N}$ such that $n \geq \max\{ N_2,m\}$ implies $|a_n - a| < \delta_m$ whence $$|\phi_m(a_n t) - \phi_m(at)| < \frac{\epsilon}{2}.$$ Since $\delta_m$ is decreasing this means that $|\phi_n(a_n t) - \phi_n(at)| < \frac{\epsilon}{2}$. Therefore, $n \geq \max\{N_1,N_2\}$ implies $$|\phi_n(a_nt) - \phi(at) \leq |\phi_n(a_n t) - \phi_n(at)| + |\phi_n(at) - \phi(at)| < \epsilon.$$ This says that the characteristic functions for $a_nX_n$ converge to the characteristic function for $aX$.
• I see. So what you did is essentially decouple $X_n$ and $a_n$ using uniform continuity. Then bring them together. Thanks. Additionally, do you think in my approach the RHS could be manipulated to give the result? – Gautam Shenoy Apr 24 '14 at 4:16
• Something seems off... In math.stackexchange.com/a/570519/35983, it was shown that uniform continuity of a sequence of functions, whose ptwise limit is also continuous does not imply $f_n(a_n) \to f(a)$. Has something additional been used here? – Gautam Shenoy Apr 24 '14 at 6:22
• Hey Gautam. So I think the difference here is that each $\phi_n$, and $phi$, is uniformly continuous. That doesn't seem to be the case in the link! – user139388 Apr 24 '14 at 15:57
• Continuous functions on compact sets are UC. Moreover I think the problem is in This step: Since $\delta_m$ is decreasing, we have... – Gautam Shenoy Apr 24 '14 at 16:02