Let's say I have a sequence of random variables $X_n$ such that $X_n \leadsto N(0,1)$ (convergence in distribution). How can I prove that $X_n^2 \leadsto \chi_{(1)}^2$ without using the continuous mapping theorem?
I'm thinking something like, if $X_n \leadsto N(0,1)$ then $P(X_n \leq x) \rightarrow \Phi(x)$ where $\Phi(x)$ is the CDF of $N(0,1)$, so (suppose $X>0$) $P(X_n^2 \leq x) = P(X_n \leq \sqrt{n}) \rightarrow \Phi(\sqrt{x})$ but now I have the problem to prove that $\Phi(\sqrt{x})$ matches the CDF of the $\chi_{(1)}^2$ (where both CDFs are difficult to write it down in the first place).
Now, if I try with Levy's theorem, $X_n \leadsto N(0,1)$ if and only if $\varphi_{X_n}(t) \rightarrow e^{-t^2/2}$ for all $t$. What I need is $$\varphi_{X_n^2}(t) \rightarrow (1-2it)^{-1/2},$$ but again, I don't know how to conclude this using only the fact that $\varphi_{X_n}(t) \rightarrow e^{-t^2/2}$. Thanks
You were almost there with the CDF, for $x\geq 0$ $$ P(X_n^2 \leq x) = P(-\sqrt{x}\leq X_n \leq \sqrt{x}) \to P(-\sqrt{x}\leq X \leq \sqrt{x}) =2\Phi(\sqrt{x})-1. $$ Now it is true that the CDF $2\Phi(\sqrt{x})-1$ is little difficult to see but the PDF is easy, take a derivative and get $\frac{1}{\sqrt{x}}\phi(\sqrt{x})=(2\pi)^{-1/2}x^{-1/2}e^{-x/2}$, when $x>0$, and $0$ otherwise.
If you want to use Levy's theorem, then $$ \lim_{n\rightarrow \infty}\mathbb{E}(e^{itX_n^2}) =\mathbb{E}( e^{itX^2}) = (1-2it)^{-1/2}. $$ The first equality is due to dominated convergence theorem.