From PDE Evans, page 272. My question is towards the bootom of this post.
THEOREM 1 (Trace Theorem). Assume $U$ is bounded and $\partial U$ is $C^1$. Then there exists a bounded linear operator $$T : W^{1,p}(U) \rightarrow L^p(\partial U)$$ such that
$\quad$(i) $Tu=u|_{\partial U}$ if $u \in W^{1,p}(U) \cap C(\bar{U})$
and
$\quad$(ii) $$\|Tu\|_{L^p(\partial U)} \le C \| U \|_{W^{1,p} (U)},$$ for each $u \in W^{1,p}(U)$, with the constant $C$ depending only on $p$ and $U$.
DEFINITION. We call $Tu$ the trace of $u$ on $\partial U$.
Proof. 1. Assume first $u \in C^1(\bar{U})$. As in the first part of the proof of Theorem 1 in §5.4 let us also intially suppose $x^0 \in \partial U$ and $\partial U$ is flat near $x^0$, lying in the plane $\{x_n=0\}$. Choose an open ball $B$ as in the previous proof and let $\hat{B}$ denote the concentric ball with radius $r/2$.
$\quad$Select $\zeta \in C_c^\infty(B)$, with $\zeta \ge 0$ in $B$, $\zeta \equiv 1$ on $\hat{B}$. Denote by $\Gamma$ that portion of $\partial U$ within $\hat{B}$. Set $x'=(x_1,\ldots,x_{n-1}) \in \mathbb{R}^{n-1} = \{x_n=0\}$. Then \begin{align} \int_\Gamma |u|^p \, dx' &\le \int_{\{x_n=0\}} \zeta |u|^p \, dx' \\ &= -\int_{B^+} (\zeta |u|^p)_{x_n} \, dx \\ &= -\int_{B^+} |u|^p \zeta_{x_n} + p|u|^{p-1} (\text{sgn} u)u_{x_n} \zeta \, dx \\ &\le C \int_{B^+} |u|^p + |Du|^p \, dx, \end{align} where we employed Young's inequality, from §B.2.
The proof continues on in the textbook but I cut it short here. My question was how was Young's inequality used exactly?
For reference, the textbook's appendix (§B.2) says:
Young's inequality. Let $1 < p, q < \infty, \frac 1p + \frac 1q = 1$. Then $$ab \le \frac{a^p}{p}+\frac{b^q}{q} \quad (a,b > 0).$$
I have a strong suspicion that Young's inequality must be used in the last step, because it is one of the only two steps with the $\le$ sign (the other step with $\le$ is the very first one on top). I do know in the third line, the product rule was used to expand $(\zeta |u|^p)_{x_n}$.
