Let $\Omega \subset \mathbb{R}^n$ be a bounded domain. If $f:\mathbb R \to \mathbb R$ is continuous and piecewise $C^1$ with $f'$ bounded, and if $u \in L^2(0,T;L^2(\Omega))$ then $f(u) \in L^2(0,T;L^2(\Omega))$.
How to prove this fact? I know that for all $t$ $f(u(t)) \in L^2(\Omega)$ but not sure how the integral over time is finite.
Hint: Since $f$ has bounded derivative, then you know that $$|f(x)|^2 \leq \big(|f(x) - f(0)| + |f(0)|\big)^2 \leq M | x |^2 + C_1|x| + C_2$$ where $|f'|\leq M$ and $C_1$ and $C_2$ are judiciously chosen constants depending on $f(0)$ and $M$.
What you want to show is that $$ \int_0^T \| f(u(t))\|_{L^2(\Omega)}^2 \,dt = \int_0^T \int_{\Omega} \big|f\big(u(t)(x)\big)\big|^2\,dx\,dt < \infty $$ From here, show that the first inequality above implies that $$ \big|f\big(u(t)(x)\big)\big|^2 \leq M \big|u(t)(x)\big|^2 +C_1\big|u(t)(x)\big| + C_2 $$ and remember that since $u(t) \in L^2(\Omega)$, then $u(t) \in L^1(\Omega)$ since $\Omega$ is bounded, and hence has finite measure with respect to Lebesgue measure.