TITLE: If $\left(\sqrt{\gamma_n}\right)_{n\in\mathbb N}$ is $L^2(\mu)$-Cauchy, does $\left(\gamma_n\right)_{n\in\mathbb N}$ converge in $L^1(\mu)$?
QUESTION [5 upvotes]: Let $(\Omega,\mathcal A,\mu)$ be a measurable space

Suppose $(g_n)_{n\in\mathbb N}$ is a sequence of bounded $\mathcal A$-measurable functions $g_n:\mathbb R\to\mathbb R$ such that $\left(\sqrt{\gamma_n}\right)_{n\in\mathbb N}$ is $L^2(\mu)$-Cauchy. Are we able to conclude that there is a unique $\gamma\in L^1(\mu)$ with $\left\|\gamma_n-\gamma\right\|_{L^1(\mu)}\xrightarrow{n\to\infty}0$?

REPLY [4 votes]: If $\sqrt{\gamma_n}$ is $L^2(\mu)$-Cauchy, there exists $\delta\in L^2(\mu)$ such that
$$
\lim_{n\to\infty}\|\sqrt{\gamma_n}-\delta\|_{L^2}=0.
$$ Observe that
$$\begin{align*}
\int_\Omega |\gamma_n-\delta^2|d\mu &= \int_\Omega |\sqrt{\gamma_n}-\delta| |\sqrt{\gamma_n}+\delta|d\mu\\&\le\left(\int_\Omega |\sqrt{\gamma_n}-\delta|^2 d\mu\right)^{1/2}\left(\int_\Omega |\sqrt{\gamma_n}+\delta|^2d\mu\right)^{1/2}\\
&\le \|\sqrt{\gamma_n}-\delta\|_{L^2}\|\sqrt{\gamma_n}+\delta\|_{L^2}\\&\le
\|\sqrt{\gamma_n}-\delta\|_{L^2}\left(\|\sqrt{\gamma_n}\|_{L^2}+\|\delta\|_{L^2}\right).
\end{align*}$$ Since $\|\sqrt{\gamma_n}\|_{L^2}\le M$, $\|\delta\|_{L^2}\le M$ for some $M>0$, we have
$$\begin{align*}
\|\gamma_n -\delta^2\|_{L^1}=\int_\Omega |\gamma_n-\delta^2|d\mu \ \le\ 2M \|\sqrt{\gamma_n}-\delta\|_{L^2}\xrightarrow{n\to\infty} 0
\end{align*}$$ as wanted. This $L^1$-limit is of course unique.