\begin{align*} \int_L|u|^p\,\mathrm{d}x' &< \int_{\{x_n=0\}} T|u|^p\,\mathrm{d}x=-\int_{B^+}(T|u|^p)_{x_n}\,\mathrm{d}x\\ &=-\int_{B^+}(|u|^pT_{x^n}+p|u|^{p-1}(sgn(u))u_{x_n}T)\,\mathrm{d}x\\ &< C\int_{B^+}(|u|^p+|Du|^p)\,\mathrm{d}x, \end{align*} where $x'= (x_1,x_2,\cdots,x_{n-1})\in\mathbb{R}^{n-1}$.
How does it make sense?
The above is from the book Partial Differential Equation by Evens on page 274. Can someone help me understand this? I can understand how to get from the first line to the second, but I don't know how to obtain the last inequality(the third line). The book says the last inequalit is obtained through Young's inequality, but I don't see how Young's inequality is applied here.
The second line is obtained by the product rule: $$ (T|u|^p)_{x_n} = |u|^pT_{x_n} + T(|u|^p)_{x_n} $$ followed by the chain rule (twice): $$ (|u|^p)_{x_n} = p |u|^{p-1}(|u|)_{x_n} = p |u|^{p-1}\operatorname{sgn}(u) u_{x_n} $$ The reason for $\operatorname{sgn}$ appearing is that it's the derivative of the absolute value function, $|x|' = \operatorname{sgn}x$.
Also, your assumptions on $T$ (called $\zeta$ in the book) are not correct, it should be $1$ on most of the ball (which includes the set $L$, which is not the whole boundary of $B^+$). This is why we have the first inequality in the chain. The proof would make more sense to you if you got the assumptions correctly.