Is the functional $F:H^{1}_{0}(\Omega) \longrightarrow \mathbb{R}$ given by \begin{equation} F(u) = \int_{\Omega} \langle (A_1(x)\chi_{\{u>0\}}+A_2(x)\chi_{\{u\le0\}}) \nabla u, \nabla u \rangle \, dx \end{equation} where $A_i,i=1,2$ is a matrix satisfying \begin{equation} \lambda |\xi|^2 \le \langle A_i(x) \xi,\xi \rangle \le \Lambda |\xi|^2, i=1,2. \end{equation} with $\lambda>0$ weak lower semicontinuous? I will appreciate any hint. Thank you.
Is the functional $F:H^{1}_{0}(\Omega) \longrightarrow \mathbb{R}$ given below weak lower semicontinuous?
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functional-analysis
sobolev-spaces
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0At this rate of setting bounties you are going to spend all your reputation pretty fast. More to the point: what is the dependence of $A_i$ on $x$: smooth, continuous, or just measurable? – 2012-08-21
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0Yf you get with just measurable, great. If not, you can assume $A_i$ Holder or what you need. In fact, this is a good question. What regularity on $A_i$ is possible to obtain a satisfactory answer? – 2012-08-21
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0Let's start with the simplest case: is $\int |\nabla u^+|^2$ weakly lower semicontinuous? – 2012-08-21
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0I think the answer is no in general. Let us say $\Omega$ is 2-dimensional, $A_1=1$, and $A_2=2$. Take $u$ to be nonpositive (or $u=0$ for concreteness), and $u_k=u+\phi_k$, where $\phi_k$ is an appropriately chosen positive function of height $1$, supported in a disk of radius $\varepsilon=1/k$. One can check that $u_k$ converges to $u$ weakly in $H^1$, but $F(u_k)$ stays strictly below $F(u)$. – 2012-08-23
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0On a more positive note, there maybe some hope for weak lower semicontinuity along minimizing sequences. The fact that $A_i$ are matrix functions complicates matters. Would it be worth studying what happens when $A_i$ are just constants? – 2012-08-23
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0@timur I don't understand your counterexample. In particular, if $u=0$ (for concreteness), then $F(u)=0$ and there is no way for $F(u_k)$ to be below $0$. – 2012-08-23
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0@LVK: That is my mistake. So $u$ cannot be constant. Still, take $u$ to be a sawtooth function in one direction, make it flat near the support of $\phi_k$ (maybe the flattening is unnecessary). The idea is to make $u$ positive on a very small region, in such a way that the energy it takes away stays finite as you shrink the region. – 2012-08-23
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0@timur I still don't see this. If $u$ is flat on the support of $\phi_k$, then $E(u+\phi_k)\ge E(u)$. Maybe you could write your example as an answer, since it would answer the question posed here? – 2012-08-23
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0On OP's update, it is possible that $\int_{E_n}|\nabla u_m|$ is not small, because $\nabla u_m$ could be concentrating on a small region. This was my motivation to try to construct a counterexample. – 2012-08-23
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0This is not very important, but why the homogeneous Dirichlet condition? The zero function will be a minimizer. – 2012-08-24
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0Is true.A priori, I put $H^{1}_{0}$ for simplicity, but in fact is $H^{1}_{\phi}$ where $\phi$ changes sing on boundary of $\Omega$. – 2012-08-24
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0@LVK: The fate of my "counterexample" is getting darker now. – 2012-08-24