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$\Omega\subset\mathbb{R}^n$ is a bounded domain with smooth boundary $\partial\Omega$. Does $$ \liminf_{k\rightarrow\infty} \int_{\Omega} \rho(u_k)\,dx \geq \int_{\Omega} \liminf_{k\rightarrow\infty} \rho(u_k)\,dx $$ hold when $\rho$ is a $C^1$ function, and the sequence of functions $\{ u_k(x) \}_{k=1}^{k=\infty}\in H^1_{0}(\Omega)$ satisfies \begin{equation} u_k \rightarrow u^{\ast} \quad \text{a.e.} \quad \text{in}\quad \Omega \quad \text{as} \quad k\rightarrow \infty? \end{equation}

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  • $\begingroup$ Fatou's lemma is for non-negative functions. $\endgroup$ – J.R. Apr 19 '14 at 10:41
  • $\begingroup$ You are right. Actually I'm asking whether that inequality is true or not. $\endgroup$ – LCH Apr 19 '14 at 10:46
  • $\begingroup$ Probably not even for $\rho(x)=x$: Choose the $u_k$ negative and $u^*\not\in L^1(\Omega)$. $\endgroup$ – J.R. Apr 19 '14 at 10:47
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Making Your Ad Here's example explicit: fix $x_0\in \Omega$ and let $\rho(t)= - t$. Define $u_k(x) = k^{n-1} \max(0, 1-k|x-x_0|)$. The graph of $k$ is the cone with vertex $(x_0,k^{n-1})$ and base $B(x_0,1/k)$. For large $k$ we have $u_k\in H^1_0(\Omega)$.

On one hand, $\int_{\Omega} \rho( u_k)$ is a fixed negative value, independent of $k$. On another hand, $u_k\to 0$ a.e.

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  • $\begingroup$ Thank you for providing this example! I will check it. $\endgroup$ – LCH Apr 23 '14 at 9:40

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