Is this proof that $u^+\in W^{1,p}(U)$ correct?

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Evans - Edition 1 - Chapter 5 - Problem 17 - p292

I want to show that $u^+$ is in $W^{1,p}(U)$.

Outline. 1) I define $u^+$ as the limit of some collection of functions. 2) I show that it's derivative is bounded. 3) I make use of three facts, that composition of $L^p$ functions is $L^p$, that the product of two $L^p$ functions is $L^p$ and that $u_{x_i}$ is $L^p$.

So we have $u^+=\lim_{\epsilon\to 0} F_\epsilon(u)$ for: $$F_\epsilon(z)=\begin{cases}(z^2+\epsilon^2)^{1/2}-\epsilon&z\geq 0\\0&z<0\end{cases},\qquad F'_\epsilon(z)=\begin{cases}\frac{z}{\sqrt{z^2+\epsilon^2}}&z\geq 0\\0&z<0\end{cases}$$ and we see that $F_\epsilon'\leq 1$, so it is bounded. We see that $F_\epsilon(0)=0$ and so composition of this with any $L^P(U)$ function will be in $L^P(U)$.

So now since we have $u^+=\lim_{\epsilon\to 0} (F_\epsilon \circ u)$, we consider $\phi\in C^\infty_c(U)$: $$\int_U u^+ \phi_{x_i}dx=\lim_{\epsilon\to 0}\int_U (F'_\epsilon\circ u)u_{x_i}\phi dx$$ So the weak derivative is $(F_\epsilon'\circ u)u_{x_i}$. Since $L^p$ space is an algebra, and $u\in W^{1,p}(U)\implies u_{x_i}\in L^p(U)$, then we need only check that $(F'_\epsilon\circ u)$ is in $L^p(U)$, but also, since composition of $L^p(U)$ functions is in $L^p(U)$, we need only check that $F'_\epsilon\in L^p(U)$ right?

So then since we know this is bounded, we just take: $$\left(\int_U |(F_\epsilon')|^p dx\right)^{1/p}\leq \int_U |1|dx<\infty$$ so $F'_\epsilon\in L^p(U)$ and we are done?

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You need to use Lebesgue dominated convergence theorem to pass to the limit. You have $0\le F_\varepsilon(z)\le \sqrt{z^2+1}\le |z|+1$ and so $|F_\varepsilon(u)\phi_{x_i}|\le\Vert \phi_{x_i}\Vert_\infty (|u|+1)$. Assuming that $U$ is bounded, the right-hand side is integrable, and so by LDCT $$\int_U F_\varepsilon(u)\phi_{x_i}\,dx\to \int_U u^+\phi_{x_i}\,dx.$$ Similarly, you have to apply LDCT for $F'_\varepsilon(u) u_{x_i}$. You have
$|F'_\varepsilon(u) u_{x_i}\phi_{x_i}|\le 1\Vert \phi_{x_i}\Vert_\infty|u_{x_i}|$, since $0\le F'_\varepsilon(z) \le 1$, and so by LDCT $$\int_U F'_\varepsilon(u) u_{x_i}\phi\,dx\to \int_{\{u>0\}} u_{x_i}\phi\,dx.$$ You also need to prove that $F_\varepsilon(u)\in W^{1,p}(U)$. You need to use density of smooth functions. If $u$ is smooth, since $F_\varepsilon$ is $C^\infty$, the composition is smooth. For the general case you need to use density of smooth functions in $W^{1,p}$.