# $L^2$-lower semicontinuity of an integral operator on $G(x,\nabla w(x))$

In the paper "A posteriori error estimates for variable time-step discretizations of nonlinear evolution equations" by Nochetto, Savaré, Verdi we find the following claim in Example 2.4:

Let $\mathcal H:=L^2(\Omega),\ p>1$ $$\phi(w):=\int_\Omega G(x,\nabla w(x))dx,\quad D(\phi):=L^2(\Omega)\cap W_0^{1,p}(\Omega)$$ If $G(x,\xi):\Omega\times\mathbb R^d\rightarrow\mathbb R$ is a Carathéodory function convex and continuously differentiable in $\xi$ for a.e. $x\in\Omega$ such that $$G(x,\xi)\geqslant\alpha_0|\xi|^p-\alpha_1,\quad|\nabla_\xi G(x,\xi)|\leqslant\alpha_2 (1+|\xi|^{p-1}), \quad\forall\xi\in\mathbb R^m,a.e. x\in\Omega$$ with some positive constants $\alpha_i$, then $\phi$ is convex and lower semicontinuous in $\mathcal H=L^2(\Omega)$.

Note that l.s.c. is stated in $L^2$. (At least for $p=2$) I am interested in

1. either a reference to such a result with proof or a hint of how to prove this. I have dificulties to relate this to classical results like the theorem of Serrin and generalizations as they yield $${\lim \inf}_{n\rightarrow\infty} \phi(u_n)\geqslant\phi(u)$$ only for sequences $u_n\rightarrow u$ wrt the $L^1$ norm where the $u_n$ and $u$ are in $W^{1,1}(\Omega)$ while assumptions are usually weaker.
2. whether or not such a result carries over to the vector-valued case, i.e. $\mathcal H=L^2(\Omega,\mathbb R^N)$.

NB: I'm a numerics guy with little practice in analysis...

Ad 1: I will give a possible idea of a proof. Take a sequence $w_k \to w$ in $L^2(\Omega)$. If $\liminf\phi(w_k) = +\infty$, we are fine. Otherwise take a bounded subsequence (still denoted by $w_k$). Now you have $$C \ge \phi(w_k) \ge \alpha_0 \int_\Omega \lvert \nabla w_k \rvert dx - \alpha_1 \, \int_\Omega dx.$$ Hence, $w_k$ is bounded in $W^{1,p}(\Omega)$ and you can select a weakly convergent subsequence (converging still towards $w$). Finally, you obtain the lower semicontinuity, since $G$ is convex in $\xi$ (use: convex + continuous gives weakly lower semicontinuous).
• I think there's the exponent $p$ missing in the estimate (which i may not edit because of too little changed characters...). Otherwise I think I will be able to use this sketch successfully - thanks a lot! (sadly I can neither vote up your answer for lack of rep) – juisoo Jun 11 '13 at 10:18