# $f_n \rightarrow 0$ in $L^1$ $\implies \sqrt{f_n} \rightarrow 0$ also?

Let $(X,\Sigma,\mu)$ be a finite measure space, and let $\{f_n : n \in \mathbb{N} \}$ be a sequence of non-negative measurable functions converging in the $L^1$ sense to the zero function. Show that the sequence $\{\sqrt{f_n}:n \in \mathbb{N} \}$ also converges in the $L^1$ sense to the zero function.

So I have to somehow show that

$$\lim_{n \to \infty}\int_X\lvert\sqrt{f_n(x)}\rvert\;\mathbb{d}\mu(x) = 0$$

If I'm honest I don't really know where to start. I think it's an easy question, but I'm new to this stuff. Any help appreciated!

• Have you covered Hölder's inequality? If so, the form and proof of that result will give you a very useful hint. – Simon S Dec 5 '14 at 16:33
• So I tried using $\|u.v\|_1 \leq \|u\|_p \|v\|_q$ with $u=v=\sqrt{f_n}$ and $p=q=\frac{1}{2}$, but that didn't seem to get me anywhere. What am I missing? – Tom Offer Dec 5 '14 at 16:55

You have the right choice of $p = q = 2$. However, choose $u = 1$, $v = \sqrt{f_n}$, then
$$\int_X \left|\sqrt{f_n}\right| \ d\mu = \left\| 1.\sqrt{f_n} \right\|_1 \ \leq \ \left\| 1 \right\|_2 \ \ \left\| \sqrt{f_n}\right\|_2 = \ ...$$
• $= \sqrt{\mu(X)} . \sqrt{\|f_n\|_1} \rightarrow 0$ by assumption (and since $\mu(X) < \infty$). Thanks very much! – Tom Offer Dec 5 '14 at 17:19
Apply Schwarz's inequality with $u=\sqrt {f_n}$ and $v=1$ and remember that $\mu(X)<\infty$.