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I am trying to comprehend the proof of a theorem in the calculus of variations (sketch: the functional $\int\limits_\Omega f(Du_j(x))~dx$ is weak*-sequentially semicontinuous if and only if $f$ is quasiconvex), but I am stuck on a (probably) simple argument which is not explained there.

Let $f:\mathbb{R}^{m} \rightarrow \mathbb{R}$ be continuous and let $u_j \rightarrow u \text{ weak}^*$ in $L^\infty(\Omega,\mathbb{R}^m)$, $\Omega$ a bounded Lipschitz-domain. Does the following hold: $f(u_j) \rightarrow f(u)$ $\text{weak}^*$ $L^\infty(\Omega)$?

Here I identfied $L^{\infty}(\Omega,\mathbb{R}^m)$ with the dual of $L^{1}(\Omega,\mathbb{R}^m)$.

Edit: I added "$\text{weak}^*$" in the last line.

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    Tha$n$ks for your suggestio$n$, Davide. Actually it is a cou$n$terexample, see copper.hat's a$n$swer.2012-12-01

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I think this is a counterexample. It hinges on my understanding of $f(u_k)$.

Suppose $\Omega=[0,1]$, $m=2$. Let $f(x) = \sqrt{x_1^2+x_2^2}$.

Let $v_k(t) = (\cos 2 \pi k t, \sin 2 \pi k t)$, and $u_k(x) = \int_{[0,1]} v_k(t)^T x(t) dt$, with $x \in L^1([0,1], \mathbb{R}^2)$.

I am presuming that $f(u_k)$ is defined as $f(u_k)(\eta) = \int_{[0,1]} f(v_k(t)) \eta(t) dt$, with $\eta \in L^1([0,1], \mathbb{R})$. Then $f(u_k)(\eta) = \int_{[0,1]} \eta(t) dt$, for all $k$, and $f(0) = 0$.

Then we have $u_k(x) \to 0$ for all $x$, but $f(u_k)(\eta) = \int_{[0,1]} \eta(t) \neq f(0)(\eta) = 0$.

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    Thank you. So this is not why the argument in the proo$f$ holds. But I am confident of finding out how it does.2012-11-30