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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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  • $\begingroup$ If $u_j \in L^\infty(\Omega,\mathbb{R}^m)^*$, then $u_j : L^\infty(\Omega,\mathbb{R}^m) \to \mathbb{R}$. What does $f(u_j)$ mean? $\endgroup$ – copper.hat Nov 26 '12 at 19:16
  • $\begingroup$ $u_j \in L^\infty(\Omega,\mathbb{R}^m)$ - I changed the question to make this clear, too. $\endgroup$ – Ole Burghardt Nov 27 '12 at 12:09
  • $\begingroup$ Have you tried with $\Omega:=(0,1)$, $u_j(x):=\sin(j\pi x)$ and $f(x):=|x|$? $\endgroup$ – Davide Giraudo Nov 27 '12 at 19:52
  • $\begingroup$ Thanks for your suggestion, Davide. Actually it is a counterexample, see copper.hat's answer. $\endgroup$ – Ole Burghardt Dec 1 '12 at 0:03
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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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  • $\begingroup$ Thank you. So this is not why the argument in the proof holds. But I am confident of finding out how it does. $\endgroup$ – Ole Burghardt Nov 30 '12 at 23:58

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