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This question is about the proof of Theorem 1 in Chapter 8 of Evans' PDE book (p. 468 in the 2nd edition). Let $u,u_k\in\mathrm{W}^{1,q}(U)$ for all $k\in\mathbb{N}$, $U\subset\mathbb{R}^n$ be open, bounded, $L$ smooth, $1<q<\infty$ and $$G_\epsilon=E_\epsilon\cap F_\epsilon$$ where $E_\epsilon$ measurable s.t. $|U-E_\epsilon|\le \epsilon$ and $$F_\epsilon=\left\{x\in U\,|\,|u(x)|+|Du(x)|\le\frac{1}{\epsilon}\right\}.$$ If $$u_k\rightarrow u\;\;\text{uniformly in}\;\;E_\epsilon$$ why does the limit $$ \lim_{k\rightarrow\infty}\int_{G_\epsilon} L(Du,u_k,x) dx = \int_{G_\epsilon}L(Du,u,x)dx, $$ hold? I suppose this is the reason why $F_\epsilon$ is defined as it is, but I don't see the exact connection. I would appreciate if somebody could look it up and help me out :-).

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What are $u_k$? – Davide Giraudo Sep 15 '12 at 16:50
Ooops, forgot the most important part :D. – MrPDE Sep 15 '12 at 19:00
up vote 1 down vote accepted

It's a consequence of mean value inequality, as $u_k(x)$ and $u(x)$ is uniformly bounded (says by $C$). Then \begin{align} \small\left|\int_{G_\varepsilon}L(Du,u_k,x)dx-\int_{G_\varepsilon}L(Du,u,x)dx\right|&\leqslant \int_{G_\varepsilon}\left|\int_{u_k(x)}^{u(x)}\partial_2L(Du,t,x)dt\right|dx\\ &\leqslant |G_{\varepsilon}|\sup_{-R\leqslant t_1,t_2,t_3\leqslant R}|\partial_2L(t_1,t_2,t_3)|\cdot \sup_{x\in G_\varepsilon}|u_k(x)-u(x)|. \end{align} As $G_\varepsilon$ is bounded it has a finite measure and the supremum is finite (as $L$ is smooth hence $\partial_2L$ is continuous on the compact set $[-R,R]^3$).

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Is there a term $|u-u_k|$ missing in the last step? – MrPDE Sep 15 '12 at 20:06
Yes, I will add it readily. – Davide Giraudo Sep 15 '12 at 20:08
Thanks for solving, should have thought of that myself :). – MrPDE Sep 15 '12 at 20:11

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