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I am dealing with the proof of proposition $9.5$ given in Haim Brezis' Functional analysis, Sobolev Spaces and Partial differential equations. I quote it here:

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How does one conclude

$G \circ u_n \to G \circ u \in L^p(\Omega)$ and that $(G'\circ u_n) \frac{\partial u_n}{\partial x_i} \to (G'\circ u) \frac{\partial u}{\partial x_i}$ in $L^p(\omega)$?

We are supposed to use Dominated convergence theorem. Note that the sequence $u_n$ is defined as $u_n= \xi_n (\rho_n * u)$, where

  1. $ \rho_n$ is the standard mollifier sequence

  2. $\xi_n$ is defined as $\xi_n=\xi(\frac{x}{n})$ with $\xi \in C_c^{\infty}(\mathbb{R}^N)$, $0 \leq \xi \leq 1$ and $$\xi(x)= \left\{\begin{matrix} 1 & |x| \leq 1 \\ 0 & \ |x| \geq 2. \end{matrix}\right. $$

  3. $\overline{u}$ is the extension to be $0$ outside $\Omega$.

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For the first question, since $|G'(s)|\le M$ for all $s\in\mathbb{R}$, it follows that $|G(u)-G(u_n)|\le M|u-u_n|$, and hence $\|G\circ u - G\circ u_n\|_{L^p(\Omega)}\le M\|u-u_n\|_{L^p(\Omega)}\rightarrow 0$.

For the second question, notice that $$ (G'\circ u)\frac{\partial u}{\partial x_i} - (G'\circ u_n)\frac{\partial u_n}{\partial x_i} = (G'\circ u - G'\circ u_n)\frac{\partial u}{\partial x_i}+(G'\circ u_n)\left(\frac{\partial u}{\partial x_i} - \frac{\partial u_n}{\partial x_i}\right). $$ Since $|G'(u)-G'(u_n)|\le 2M$, it follows that $$ \left|(G'(u) - G'(u_n))\frac{\partial u}{\partial x_i}\right|^p\le (2M)^p\left|\frac{\partial u}{\partial x_i}\right|^p $$ pointwise, and the right-hand side is integrable on $\omega$. Since $u_n\rightarrow u$ a.e. pointwise, it follows that $(G'\circ u - G'\circ u_n)\frac{\partial u}{\partial x_i}\rightarrow 0$ a.e. pointwise (since $G$ is $C^1$), and hence by the Dominated Convergence Theorem $$\left\|(G'\circ u - G'\circ u_n)\frac{\partial u}{\partial x_i}\right\|_{L^p(\omega)}^p = \int\limits_{\omega}{\left|(G'\circ u - G'\circ u_n)\frac{\partial u}{\partial x_i}\right|^p}\rightarrow 0. $$ Now, since $|G'(u_n)|\le M$, it follows that $\left|G'(u_n)\left(\frac{\partial u}{\partial x_i} - \frac{\partial u_n}{\partial x_i}\right)\right|\le M\left|\frac{\partial u}{\partial x_i} - \frac{\partial u_n}{\partial x_i}\right|$, and hence $$\left\|(G'\circ u_n)\left(\frac{\partial u}{\partial x_i} - \frac{\partial u_n}{\partial x_i}\right)\right\|_{L^p(\omega)}\le M\left\|\frac{\partial u}{\partial x_i} - \frac{\partial u_n}{\partial x_i}\right\|_{L^p(\omega)}\rightarrow 0.$$

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  • $\begingroup$ How do you conclude $(G'\circ u - G'\circ u_n)\frac{\partial u}{\partial x_i}\rightarrow 0$ almost everywhere? $\endgroup$ – D1X Jul 12 '16 at 13:04
  • $\begingroup$ @D1X if $x\in\Omega$ satisfies $u_n(x)\rightarrow u(x)$, then $G'(u_n(x))\rightarrow G'(u(x))$ by continuity of $G'$. By assumption, $u_n(x)\rightarrow u(x)$ for almost every $x\in\Omega$. $\endgroup$ – Joey Zou Jul 13 '16 at 15:36

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