Given $\Omega\subset \mathbb R^N$ is open bounded, nice boundary, and $u\in H^1(\Omega)$. We say that $u\leq \alpha$ on $\partial \Omega$ for a constant $\alpha$ if $(u-\alpha)^+\in H_0^1(\Omega)$, and we define $$ \text{ess sup}_{\partial \Omega} u:=\inf\{\alpha\in R, \,u\leq \alpha\text{ on }\partial \Omega\} \tag 1)$$
Also, as $\Omega$ contain a nice boundary, we have $T[u]$, the trace operator, is well defined. Since $T[u]$ is defined on $\partial \Omega$ and hence we could talk about $$\text{ess sup}_{\partial\Omega}T[u]\tag 2$$ as well. Be aware that here $\text{ess sup}_{\partial\Omega}T[u]$ is defined in the usual way, with respect to $H^{N-1}$ measure, as we define $\text{ess sup}_{\Omega}u$ with respect to $L^{N}$, i.e., the $N$ dimension Lebesgue measure.
Now, have $(1)$ and $(2)$ is well defined. Could we conclude that
$$ \text{ess sup}_{\partial \Omega} u=\text{ess sup}_{\partial\Omega}T[u]$$
If yes, how?
Thanks for helping here!
Yes, this is true. The main point is that the kernel of the trace operator is precisely $H_0^1(\Omega)$. (Theorem 3.44 in Hunter's notes, or in other places). Thus, your formula (1) can be written as $$ \operatorname{ess\,sup}_{\partial \Omega} u =\inf\{\alpha\in \mathbb R : T[(u-\alpha)^+]=0 \} \tag{1*} $$ It remains to show that $T[(u-\alpha)^+]=(T[u]-\alpha)^+$. This is a special case of the following: $$T[\phi\circ u]=\phi\circ T[u]\tag{C}$$ for every Lipschitz function $\phi$. Indeed, both sides of (C) are equal for functions that are smooth in the closure of the domain (since for them the trace operator is just the restriction to the boundary), and such functions are dense, both in $L^p(\partial \Omega)$ and in $H^1(\Omega)$. Composition with a Lipschitz function is a continuous nonlinear operator on both of those spaces too. Thus, the argument "by density" works.