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.
$Z_n\stackrel{L^2}{\to} Z$ imply $Z_n^2\stackrel{L^1}{\to} Z^2$
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probability-theory
lp-spaces
1 Answers
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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 $Z^2$.
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0I agree with your proof. Only thing, in the last term, you need to stress that is bounded rather than finite. This will allow you to conclude that the RHS goes to 0. – 2017-01-15
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0@Kolmo Thank you for your remark, of course you are correct, I'll add this. – 2017-01-15