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I'm going through the proof of Corollary 8.11 in Brezis' Functional Analysis, Sobolev Spaces and Partial Differential Equations which states:

Let $G \in C^1(\mathbb{R})$ be such that $G(0) = 0$, and let $u \in W^{1,p}(I)$ with $p \in [1, \infty]$, then $G \circ u \in W^{1,p}(I)$ and $(G \circ u )' = (G' \circ u)u'$.

Following his proof and filling some steps I've shown that $(G \circ u)$ and $(G' \circ u)u'$ are in $L^p(I)$, focusing on the case $p < \infty$ we obtain $(u_n)_n \subseteq C_c^\infty(\mathbb{R})$ such that $u_{n\vert I} \to u$ in $W^{1,p}(I)$ (and hence in $L^\infty(I)$) then he states

Thus $(G\circ u_n)_{\vert I} \to G \circ u$ in $L^\infty(I)$ and $(G' \circ u_n)u'_{n\vert I} \to (G' \circ u)u'$ in $L^p(I)$.

I've managed to prove the first assertion but not the second. So far i've gotten that (to make the notation lighter I'm omitting the restriction to $I$ but it should be there)

\begin{align*} ||(G' \circ u_n)u'_n - (G' \circ u)u'||_{p} &\leq ||(G'\circ u_n)u'_n -(G' \circ u)u'_n||_p + ||(G' \circ u)u'_n - (G' \circ u)u'||_p\\ &\leq ||u'_n||_p ||G' \circ u_n - G' \circ u||_\infty + ||G'\circ u||_{\infty}||u'_n-u'||_{p} \end{align*} since $u'_{n \vert I} \to u'$ in $L^p(I)$ we have that $\sup_{n \in \mathbb{N}}||u'_{n \vert I}||_{p}$ is finite and since $G' \circ u$ is bounded a.e. the second summand vanishes, so it'd be enough to prove $G' \circ u_n \to G' \circ u$ in $L^\infty(I)$, I was thinking on using that $u_n \to u$ uniformly a.e. alongside the continuity of $G'$ but since $I$ could be unbounded that doesn't necessarily work.

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1 Answer 1

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Your idea does work, as $I$ being unbounded does not cause problems.

Since we are in one dimension, by Theorem 8.8 we have $\lVert v \rVert_{L^{\infty}(I)} \leq C \lVert v \rVert_{W^{1,p}(I)}$ for all $v \in W^{1,p}(I)$; note this holds true for any $I$. Since $u_n \to u$ strongly in $W^{1,p}(I)$, we infer that:

  • $u_n \to u$ uniformly on $I$ (applying the inequality to $u_n-u$),

  • there is some $M>0$ such that $\lVert u_n \rVert_{L^{\infty}(I)}, \lVert u \rVert_{L^{\infty}(I)} \leq M$ for all $n$ (since $(u_n)$ is bounded in $W^{1,p}(I)$).

Then since $G'$ is continuous on $\mathbb R$, we have $G'$ is uniformly continuous on $[-M,M]$. Since uniform convergence is preserved by composition with uniformly continuous functions, we infer that $G' \circ u_n \to G' \circ u$ uniformly on $I$.

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