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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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Presumably $\lambda\geq 0$? – copper.hat Aug 18 '12 at 1:21
yes, I will add. – user29999 Aug 18 '12 at 1:29
up vote 3 down vote accepted

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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What is the idea behind the counterexample? – timur Aug 18 '12 at 3:30
@timur One function is positive, with small oscillation. Another is negative, with large oscillation. The average is positive and has at least half of large oscillation. And since positivity is rewarded with $A_1$, the average beats both original functions. – user31373 Aug 18 '12 at 3:34
Thanks! Maybe it is not interesting but if we modify $F$ by replacing $\nabla u$ by $u$, it seems to be convex. – timur Aug 18 '12 at 3:58
@timur Indeed, because it becomes $\int \Phi(u)$ with convex $\Phi$. – user31373 Aug 18 '12 at 4:32

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