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Is the functional \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 $\lambda>0$ convex? You can assume $\Omega$ convenient such that the expression above make sense. For example, $C^1(\Omega), H^1_0(\Omega)$ or other space such that the functional above be convex. I will appreciate any hint. Thank you.

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    yes, I will add.2012-08-18

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No. Here is a one-dimensional counterexample, but you can adopt the idea to higher dimensions if you want. Let $\Omega=(0,20)$, $A_1=10$ and $A_2=1$. Define functions $u$ and $v$ by $u'=\chi_{[0,1]}-\chi_{[19,20]}\quad \text{ and }\quad v'=\sum_{k=1}^9 (-\chi_{[2k-1,2k]}+\chi_{[2k,2k+1]})$ Both vanish on the boundary of $\Omega$. Since $u\ge 0$ and $v\le 0$, we have $F(u)=20$ and $F(v)=18$. The average $w=(u+v)/2$ is nonnegative because $v\ge -1$ everywhere and $u=1$ on the support of $v$. Since $|w'|\equiv 1/2$, it follows that $F(u)=\int_0^{20}10\cdot\frac14=50$, which is greater than either $F(u)$ or $F(v)$.

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    @timur Indeed, because it becomes $\int \Phi(u)$ with convex $\Phi$.2012-08-18