I am trying to prove the rule of differentiation of a composition for weak derivatives in Sobolev spaces following the proof given in Corolary 8.11 in Functional Analysis, Sobolev Spaces and Partial Differential Equations, by Haim Brezis.
Let $G \in C^1(\mathbb{R})$ with $G(0)=0$ and $u \in W^{1,p}(I)$ (i.e $u \in L^p(I)$ and has a weak derivative which is in $L^p$). Then:
$G \circ U \in W^{1,p}$ and $(G \circ u)'= (G' \circ u)u' $.
Where I is an open interval in $\mathbb{R}$ (not necessarily bounded) and $1\leq p \leq \infty$.
The first thing I need to know is whether $(G' \circ u)u' $ belongs to $L^p$. I have already shown that $G \circ u \in L^p(I)$, but I cannot see how can I conclude that $(G' \circ u)u' \in L^p(I)$.
I have tried using Hölder's inequality so that we have $\int_{I}|(G' \circ u)^p(u')^p| \leq ||(G' \circ u)^p||_{\infty}||(u')^p||_1$
$||(u')^p||_1$ is finite because $u \in W^{1,p}$ but I need $(G' \circ u)^p$ to be bounded.
Let $M=\|u\|_{\infty}$. Since $G'$ is continuous, there exists a constant $K$ such that $|G'(s)|\leq K$ for all $s\in [-M,M]$. Thus $|G'\circ u|\leq K$; it follows that $|G'\circ u|^p$ is bounded by $K^p$.
The desired conclusion follows from Hölder's inequality as you explained with a little correction (you have to change parenthesis by norms):
$$\int_{I}|(G' \circ u)u'|^p=\int_{I}|G' \circ u|^p|u'|^p \leq \big\||G'\circ u|^p\big\|_\infty\big\||u'|^p\big\|_1<\infty.$$