Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\Omega \subset \mathbb{R^{n}}$ be a bounded domain and $u , u_j \in H^{1}(\Omega)$ such that $u_j \rightharpoonup u$ in $H^{1}(\Omega)$ $$ F_{1}(u) = \int_{\{ u > 0\}} \dfrac{1}{2}\langle A_1 \nabla u, \nabla u \rangle +\int_{\{ u \le 0\}}\ \dfrac{1}{2}\langle A_2 \nabla u, \nabla u \rangle $$ and \begin{equation} F_2(u)= \int_{\{ u>0\}} f^{+}u dx + \int_{\{ u \le 0\}} f^{-}u dx \end{equation} where $ f^{+}(x), f^{-}(x) \in L^{2^{*}}(\Omega) $ with $ 2^{*} = 2n / (n-2)$ and the matrices $A_1, A_2$ are $\alpha$-Hölder continuous for some $0 \ < \alpha \ < 1$ satisfying \begin{equation} \lambda \le \langle A_1 (x) \xi, \xi \rangle ,\langle A_2(x)\xi, \xi \rangle \le \Lambda \end{equation} for constants $\lambda, \Lambda$.

I don't know if \begin{equation} F_i(u) \le \liminf_{j} F_i (u_j) \quad i=1,2. \end{equation}

share|cite|improve this question
What is a Hölder continuous matrix? – Davide Giraudo May 29 '12 at 19:33
If the domain $\Omega$ is reasonable (a $W^{1,2}$ extension domain) then weak convergence in $W^{1,2}$ implies strong convergence in $L^2$, which yields $F_2(u_j)\to F_2(u)$. – user31373 May 29 '12 at 20:42
@Davide: $A_1$ and $A_2$ are matrix-valued Hölder continuous functions, I guess. – Hendrik Vogt May 30 '12 at 12:18

Your Answer


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

Browse other questions tagged or ask your own question.