$Z_n\stackrel{L^2}{\to} Z$ imply $Z_n^2\stackrel{L^1}{\to} Z^2$

Question

Let $(\Omega,\mathcal A,P)$ be a probability space, $Z_n,Z:\Omega\to \Bbb R$ $\in L^2(\Omega,\mathcal A,P)$ such that $Z_n\to Z$ in $L^2$.

How can I prove that $Z_n^2\to Z^2$ in $L^1$?

I tried in many ways but I wasn't able to conlude.

Answer 1

Using the Cauchy-Schwarz inequality we get

$$\begin{align}0 \leq||Z_n^2-Z^2||_1&=\int|Z_n^2-Z^2| \\&=\int|Z_n+Z| |Z_n-Z| \\&\leq ||Z_n-Z||_2 ||Z_n+Z||_2\leq \underbrace{||Z_n-Z||_2}_{\to 0}\underbrace{\left(||Z_n||_2+||Z||_2\right)}_{\text{bounded}} \end{align}$$

and by the squeeze theorem we conclude that $Z_n^2$ converges in $L^1$ to $Z^2$.