Let $\Omega$ be a bounded Lipschitz domain.
Let $u \in H^1(\Omega) \cap L^\infty(\Omega)$, and suppose that $\lVert u \rVert_{L^\infty(\Omega)} \leq A$.
Let $T:H^1(\Omega) \to L^2(\partial\Omega)$ be the trace mapping.
Is it true that $Tu \in L^\infty(\partial\Omega)$ with $\lVert Tu \rVert_{L^\infty(\partial\Omega)} \leq A'$ for some constant $A'$?
I think so, since we can find functions $u_n \in C^0(\bar \Omega)$ bounded by $A$ such that $u_n \to u$ in $H^1$ and $Tu = \lim Tu_n$ in $L^2$.
Consider the function $u + A$. It belongs to $H^1(\Omega)$ and is non-negative. A standard procedure yields a sequence $\{v_n\} \in C(\bar\Omega) \cap H^1(\Omega)$ with $v_n \ge 0$ and $v_n \to u + A$ in $H^1(\Omega)$. Now, $T v_n \ge 0$, since it corresponds with the usual trace of $v_n$. Since $T$ is continuous, you have $T v_n \to T(u + A)$ and $T(u+A) \ge 0$. Now, you can easily show $T(u + A) = T u + A$ and this yields $T u \ge -A$. Similarly, $T u \le A$ follows.