Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
Presumably $\lambda\geq 0$? –  copper.hat Aug 18 '12 at 1:21
    
yes, I will add. –  user29999 Aug 18 '12 at 1:29
add comment

1 Answer

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 $F(v)$.

share|improve this answer
    
What is the idea behind the counterexample? –  timur Aug 18 '12 at 3:30
1  
@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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.