# Surjectivity of Neumann trace in $H^1_0$

Let $\Omega$ be a Lipschitz domain. The Neumann trace operator is surjective as $H^1(\Omega)\rightarrow H^{-\frac 1 2}(\partial\Omega)$ by construction. How about as $H^1_0\rightarrow H^{-\frac 1 2}$?

• $H_0^1$ functions are zero on the boundary. – Christian Remling Sep 9 '16 at 16:20
• I'm thinking of (sorry for not being rigorous enough in the first place) Neumann trace as mapping equivalent to $u \mapsto \boldsymbol n \cdot \nabla u$ for smooth stuff, where $\boldsymbol n$ is the boundary normal. So yes, $H^1_0$ functions are zero on the boundary, but Neumann traces are necessarily not. – jmk Sep 9 '16 at 16:26
• Thanks for clarifying. Still, don't you define this trace as $\lim n\cdot \nabla u_n$ for smooth approximations $u_n\to u$, which can be taken to be zero near the boundary if $u\in H_0^1$, so you're getting zero after all. – Christian Remling Sep 9 '16 at 17:46
• They can, but I'd like to have $n\cdot \nabla u$ to be an arbitrary trace space function. – jmk Sep 9 '16 at 17:55
• OK, so what this shows is that the map $u\mapsto n\cdot\nabla u$ does not extend continuously to $H_0^1$ (from smooth functions). But then how do you define the trace of $u$ for an arbitrary $u\in H_0^1$? – Christian Remling Sep 9 '16 at 20:57