# The constant in the Sobolev trace theorem inequality

The trace theorem for nice enough domains states that there is a operator $T:H^1(\Omega) \to L^2(\partial \Omega)$ such that $$|Tu|_{L^2(\partial \Omega)} \leq C|u|_{H^1}.$$

My question, is there an expression for the constant $C$? I want to see exactly how it depends on the domain $\Omega.$ This is because I want to see how the constant varies (eg. continuously) if I vary the domain.

It basically says $$\delta \|u\|_{L^2(\partial\Omega)}^2 \le \|\mu\|_{C^1(\bar\Omega)} \left(\epsilon^{1/2}\|\nabla u\|_{L^2(\Omega)}^2 + (1+\epsilon^{-1/2}) \|u\|_{L^2(\Omega)}^2 \right)$$ for all $\epsilon\in(0,1)$, $u\in H^1(\Omega)$. The vector field $\mu$ has to be chosen to be $C^1(\bar\Omega,\mathbb R^n)$ such that $$\mu \cdot \nu \ge \delta$$ on $\partial \Omega$ with $\nu$ the outer normal vector.
(1) The trace theorem is actually sharper; you can bound even the $H^{1/2}$ norm (not just the $L^2$ norm) of the trace on $\partial \Omega$ in terms of the $H^1$ norm of the function on $\Omega$.