# If $f:\mathbb R \to \mathbb R$ is continuous and piecewise $C^1$ with $f'$ bounded and $u \in L^2(0,T;L^2)$ then $f(u) \in L^2(0,T;L^2)$

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.

• Is $\Omega$ a bounded subset of $\Bbb R$? I am unclear what the ordered triple $(0,T; L^2(\Omega))$ means as well. – Alex Schiff Jun 23 '14 at 18:23
• $\Omega$ is bounded domain in $\mathbb{R}^n$. $L^2(0,T;L^2(\Omega))$ is the usual Bochner space. – maths_student_2000 Jun 23 '14 at 18:25
• Thanks, I wasn't familiar with Bochner spaces. I should have read the tags associated with this post, I apologize. – Alex Schiff Jun 23 '14 at 18:27
• @AlexSchiff No problem! – maths_student_2000 Jun 23 '14 at 18:35
• Is it true that $\|f(u)\|_2^2:=\int_0^T\|f(u(t))\|_2^2\,d\lambda(t)=\int_0^T\left(\int_\Omega f(u(t))^2\,d\lambda\right)^{1/2}\,dt$? – Alex Schiff Jun 23 '14 at 18:43

## 1 Answer

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.

• Thanks. Actually $f(0)=0$ so we can forget about that. I see you used Lipschitz constant but I am not sure that it is true because I only found the theorem "if $f$ is absolutely continuous (so it has derivative a.e) then $f$ is Lipschitz with constant $|f'|_{\infty}$." Here $f$ I don't know if it is absolutely continuous. So not sure if I can apply that theorem. – maths_student_2000 Jun 23 '14 at 19:59
• A continuous piecewise $C^1$ function is absolutely continuous. – Tom Jun 23 '14 at 20:12
• @maths_student_2000 ..but absolute continuity is heavier machinery than you need; just use the mean value theorem to convince yourself that if $f$ is $C^1$ with bounded derivative, then it is Lipschitz. – Tom Jun 23 '14 at 20:18