Estimation the $L^p$ norm of $u$ by using trace and gradient. Given $\Omega\subset \mathbb R^N$ open bounded with nice boundary. Then for $u\in W^{1,p}(\Omega)$, $1\leq p\leq \infty$, we have
$$\|u\|_{L^p(\Omega)}\leq C(\|T[u]\|_{L^p(\partial\Omega)}+\|\nabla u\|_{L^p(\Omega)})  $$
where $T$ denotes the trace operator and constant $C$ depends on $\Omega$, $\partial\Omega$, and $p$, $N$.
I proved this by using density argument, i.e., first show for $(u_n)\subset C^\infty (\bar{\Omega})$ and push to limit. Since $u_n\to u$ in $W^{1,p}(\Omega)$ implies that $T[u_n]\to T[u]$ in $L^p(\partial\Omega)$, so the density argument works. 
I use this argument a lot in weak theorem for Robin's boundary problem, especially in non-linear case.
However, I never see this statement in any text book, at least in H. Brezis, Evans, Evans & Gariepy, Adams, and Leoni's, nor some exercises... 
So I was wondering is my statement true? It intuitively makes sense, and if you want, I can write some details. Also, if you know this statement was stated in some books, please kindly direct me there. Thank you!
 A: This is a known result. For example,  Theorem 4.2.1 in   Ziemer's Weakly Differentiable Functions, specialized to first-order Sobolev spaces, says that for any linear functional $\varphi:W^{1,p}\to\mathbb R$ that takes  value $1$ on the constant function $1$, we have 
$$
\|u-\varphi(u)\|_{p;\Omega} \le C\|\nabla u\|_{p;\Omega}
$$
Take $\varphi( u)=\frac{1}{|\partial \Omega|}\int_{\partial \Omega}T[u]$ to obtain your result. 
A stronger statement (a form of the Poincaré-Sobolev inequality) can be found later in the same book: 

4.4.7. Corollary If $\Omega$ is a bounded Lipschitz domain  and $u\in W^{1,p}(\Omega)$, $p>1$, then $$\|u\|_{p^*;\Omega}\le C\left[\|\nabla  u\|_{p;\Omega}+\left|\int_{\partial\Omega} u\right|\right]$$

(The statement in the book has a typo: the absolute value around the boundary integral is omitted.) Here subscripts mean Lebesgue norms, and $p^*=np/(n-p)$ is the Sobolev embedding  exponent for $p$. The author then remarks that the inequality extends to $BV$, in particular to $W^{1,1}$. 
