Which convergence can be deduced for $\sqrt{X_n}$?

Question

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space (but the question is interesting for a general measure space).

Let $X_n$ be a sequence of random variables converging in $L^1(\Omega)$ to a random variable $X_{\infty}.$

My question is: which convergence can be deduced for the random variables $\sqrt{X_n}$ to the variable $\sqrt{X_\infty}$?

I can't believe there is none, and I tried to verify it with the $L^2$ and $L^1$ norms, but I couldn't handle terms like $\mathbb{E}[\sqrt{X_nX_\infty}].$

Thanks for any help!

Answer 1

There is indeed convergence in $\mathbb L^2$. We use the following result: if a sequence $\left(Y_n\right)_{n\geqslant 1}$ of random variables is uniformly integrable and converges to $0$ in probability then $\mathbb E\left\lvert Y_n\right\rvert\to 0$.

Let $Y_n:=\left(\sqrt{X_n}-\sqrt{X_\infty}\right)^2$. Then $0\leqslant Y_n\leqslant 2X_n+2X_\infty$ and due to the convergence in $\mathbb L^1$of $\left(X_n\right)$, we deduce that $\left(Y_n\right)_{n\geqslant 1}$ is uniformly integrable.

Using the fact that convergence in probability is preserved by continuous maps, we derive that $\left(Y_n\right)_{n\geqslant 1}$ converges to $0$ in probability.