I have recently started to study vector valued integration. I wish to prove the following claim:
Assume that $u(t)\in L^2(I;L^2(\Omega))$ where $I\subseteq\mathbb{R}$ is an (open) bounded interval and $\Omega\subseteq\mathbb{R}^n$ is a bounded domain. Let $f:\mathbb{R} \rightarrow\mathbb{R}$ be a lipschitz function. Then $f(u(t)):I\rightarrow L^2(\Omega)$ is strongly measurable.
Much to my surprise, I couldn't come up with a simple proof than this one:
As $u(t):I\rightarrow L^2(\Omega)$ is strongly measurable, the function $u(t,x):I\times\Omega\rightarrow\mathbb{R}$ is measurable (I had already proven this proposition). As $f$ is cts., $f(u(t,x)):I\times\Omega\rightarrow\mathbb{R}$ is measurable. It's easy to show that $f(u(t,x))\in L^2(I\times\Omega)$. Hence, the function $f(u(t)):I\rightarrow L^2(\Omega)$ is strongly measurable (again, I had already proven this).
However, I feel like I'm using a sledgehammer to crack a nut. If anyone could show me an elementary argument (after which I'll probably feel ashamed), I'll be grateful. Thank you
Using the Lipschitz property of $f $, show that the (nonlinear) map $$ T_f : L^2 (\Omega) \to L^2 (\Omega), u \mapsto f \circ u $$ is continuous. For well-definedness, we use that $\Omega $ is bounded (otherwise, assume $f (0)=0$).
Now, if $G : X \to B $ is strongly measurable and if $F : B \to C $ is continuous (with Banach spaces $B,C $ and a measure space $X $), it is not hard to show (do it!) that $F \circ G $ is strongly measurable. Depending on your exact definition of strong measurability, you will also need to assume that $X $ is sigma finite.