Analogue of continuous mapping theorem for convergence in $L^2$ Could you help please:
Is there any analogue of continuous mapping theorem for convergence of sequence of random variables in $L^2$?
I mean:
If $g$ is a continuous function $\mathbb{R}\to\mathbb{R}$ (not differentiable in general) and $X_n \stackrel{L^2} \to X$ then $g(X_n) \stackrel{L^2} \to g(X).$
Thanks in advance
 A: Without further assumptions this is false.  If $g$ is unbounded, we might not even have $g(X_n), g(X) \in L^2$.
If $g$ is bounded and continuous, then $g(X_n) \to g(X)$ in measure by the continuous mapping theorem, and also in any $L^p$ by the dominated (bounded) convergence theorem.  So the statement is true in this case.
It also holds when $g$ is unbounded but Lipschitz: if $C$ is the Lipschitz constant, then $g(X_n), g(X) \in L^2$ because $|g(X_n)| \le |g(0)| + C |X_n|$ where the right side is an $L^2$ random variable.  Moreover, we have $$E|g(X_n) - g(X)|^2 \le C^2 E|X_n - X|^2$$ where the latter goes to zero.
A: Suppose $g$ is $\alpha$-Hölder continuous, for $\alpha \geq 0$, and that $X_n \overset{L^m}{\to} X$ for some $m > 0$. If $\alpha > 0$, then $g(X_n) \overset{L^{m/\alpha}}{\to} g(X)$. If $\alpha = 0$ (i.e. bounded), and $g$ is continuous, then $g(X_n) \overset{L^m}{\to} g(X)$. The case $\alpha = 1$ corresponds to Lipschitz continuity. The case $\alpha > 0$ is very easy to prove. I'd be interested in more general results.
A: If there exists constants $c_1$, $c_2$ such that $|g(x)| \leq c_1 + c_2 \cdot |x|$ for all $x \in \mathbf{R}$, then $g(X_n) \stackrel{L^{2}}{\to} g(X)$.
On a closed and bounded interval $[-M, M]$, $g$ is uniformly continuous. Thus for any $\epsilon > 0$, there exists $\delta(M) > 0$ such that
$$
\text{ if } |X_n - X| < \delta(M), \text{ then } |g(X_n) - g(X)| < \sqrt{\epsilon}
$$
Consider sets of the form
$$
B_n(M) = \{x: |X_n| < M \text{ and } |X| < M\}
$$
$$
A_n(M) = \{x \in B_n(M): |X - X_n| < \delta(M)\}
$$
Then
\begin{align*}
E(|g(X_n) - g(X)|^{2}) &= E(|g(X_n) - g(X)|^{2} \cdot \mathbf{1}_{A_n(M)}) + E(|g(X_n) - g(X)|^{2} \cdot \mathbf{1}_{A_n(M)^c}) \\
&\leq \epsilon + E(|g(X_n) - g(X)|^{2} \cdot \mathbf{1}_{A_n(M)^c}) \\
\end{align*}
for all $n$ and $M$.
$\Pr(A_n(M)^c)$ can be made arbitrarily small for big enough $M$ and $n$ using Markov's Inequality and $L^2$-convergence.
The result then follows by showing $\{|g(X_n) - g(X)|^2\}$ is uniformly integrable when the condition on $g$ holds.
