1
$\begingroup$

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.

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

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – maths_student_2000 Jun 23 '14 at 19:59
  • $\begingroup$ A continuous piecewise $C^1$ function is absolutely continuous. $\endgroup$ – Tom Jun 23 '14 at 20:12
  • $\begingroup$ @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. $\endgroup$ – Tom Jun 23 '14 at 20:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.