# Lower semicontinuity of a Bochner integral of a convex function

I'm looking for the following result:

Let $\Omega \subset \mathbb{R}^n$ be a bounded domain. The map $$u \mapsto \int_0^T \int_{\Omega} f(u(t))$$ is lower semicontinuous for $u \in L^2(0,T;L^2)$ where $f:\mathbb R \to \mathbb R$ is convex.

Does anyone know how to prove this, or a reference for this result? Thank you.

I managed to this for a subsequence (so if $u_n \to u$, I showed for $\liminf_{n_j \to \infty}$) but could not show it for the full sequence.

• In the first sentence of your question you started saying something about $f$ but didn't finish it. – k3thomps Jun 11 '14 at 13:58

Use the following definition for lower semi continuity. That is, $f$ is lower semi continuous at $x_0$ is $\liminf_{x\rightarrow x_0} f(x) \geq f(x_0)$. This is equivalent to
For all $\epsilon > 0$ there exists $\delta > 0$ so that $\epsilon \geq f(x_0) - f(x)$ for all $x \in B_\delta(x_0)$
Fix $u \in L^2(0,T;L^2(\Omega))$ and pick $\epsilon > 0$. One has \begin{eqnarray} F(u) - F(v) &=& \int_0^T \int_\Omega \left[ f(u) - f(v) \right]\end{eqnarray} Use the fact that convex functions are Lipshitz to get that $$F(u) - F(v) \leq C \int_0^T \int_\Omega \left| u - v\right|$$ for some constant $C$. From this you can easily conclude what you want by using the fact that you are playing over a bounded domain and using Holder.
• Thanks. How did you get your first displayed equation? I could only get $f(\frac u2) -\frac 12f(\frac v2) \leq \frac 12f(\frac{2u-v}{2})$? Well I guess we don't need that second line anyway. – aca888 Jun 11 '14 at 14:51
• One last thing: is it obvious that the Lipschitz constant is independent of $x \in \Omega$ and $t$? When $f:[a,b] \to \mathbb{R}$ then (I believe) the Lipschitz constant depends on $b$. – aca888 Jun 11 '14 at 15:28
• You are right. $f$ is only locally Lipshitz. That's OK though. We only care that $f$ is Lipshitz around $0$. I am brushing some details aside, but we are thinking of $u-v$ as being close to 0. Of course this is in the $L^2(0,T;L^2)$ sense, but with some fidling this is enough. – k3thomps Jun 11 '14 at 16:07