I understood almost everything. But I got stuck in two steps that seems to be very simple.
The first doubt is in the first inequality, in the beginning of the proof. In the first equality, for example, I understood that using the fundamental theorem of calculus, for continuous functions and for bounded domain, in this case $\mathscr{supp} (u)$ is compact and $u$ is continuous hence I can write: $$ \begin{align*} u(x)&= \int_{\mathscr{supp}(u)} ux_i( x_1, ..., x_{i-1}, y_{i}, x_{i+1}, ..., x_n) dy_{i}\\ &= \int_{-\infty}^{x_i} ux_i( x_1, ..., x_{i-1}, y_{i}, x_{i+1}, ..., x_n) dy_{i}\\ \end{align*} $$ where in the last equality is because once $supp(u)$ is compact then is there exists an open subset, like $(-\infty, x_i)$ that $supp(u) \subset (-\infty, x_i)$. Is that right? However I coudn't understood why in the first inequality the limits of integration is considered from $(-\infty, \infty)$, that is $$|u(x)|\leq\int\limits_{-\infty}^{\infty} |Du(x_1, ..., y_i, ...., x_n)|dy_i \quad i\in \{1, 2, ..., n\}$$
Second doubt is: In the first equality that appears in (14). How to proof that $|\nabla (|u|^{\gamma})| = \gamma |u|^{\gamma -2}u\nabla u$?
Sorry for this very simple question. I know that is a very simple question and think is only need to use chain rule, however, I couldn't get this equality. For example, I tryed to calculate, considering $u(x)= u(x_1, x_2, ..., x_n) \in \mathbb{R}$ then
$$ \begin{align*} \left(\frac{\partial |u|^{\gamma}}{\partial x_i}\right)&=\gamma |u|^{\gamma -1}.\frac{\partial |u|}{\partial x_i}.\frac{\partial u}{\partial x_i}\\ &\implies\gamma |u|^{\gamma -1} \nabla |u| \nabla u \end{align*} $$
Thanks in advance.
Since $\gamma > 1$, the function $f(x) = |x|^{\gamma}$ is differentiable everywhere, even at $0$, with $f'(x) = \gamma|x|^{\gamma - 1}\text{sgn}(x) = \gamma|x|^{\gamma - 2}x$ since $|x|\text{ sgn}(x) = x$. Hence $$\partial_{x_i}|u|^{\gamma} = \gamma|u|^{\gamma - 1}\text{sgn}(u)u_{x_i} = \gamma|u|^{\gamma - 2}u u_{x_i}.$$
For the first question, let $[-R, R]^n$ be a box such that $(-R/2, R/2)^n$ contains $\text{supp}(u)$. Then by the fundamental theorem of calculus, \begin{align} u(x) &= u(x) - u(x_1, \dots, -R, \dots, x_n) \\ &= \int_{-R}^{x_i}u_{x_i}(x_1, \dots, y_i, \dots, x_n)\,dy_i \\ &= \int_{-\infty}^{x_i}u_{x_i}(x_1, \dots, y_i, \dots, x_n)\,dy_i. \end{align} Also \begin{align} |u(x)| &= |\int_{-\infty}^{x_i}u_{x_i}(x_1, \dots, y_i, \dots, x_n)\,dy_i| \\ &\leq \int_{-\infty}^{x_i}|u_{x_i}(x_1, \dots, y_i, \dots, x_n)|\,dy_i \\ &\leq \int_{-\infty}^{\infty}|u_{x_i}(x_1, \dots, y_i, \dots, x_n)|\,dy_i. \end{align}