# Density of smooth functions in Sobolev space, respecting nonnegative traces

I am looking for a reference for the following result (if it is true, which I would expect):

Let $\Omega\subset\mathbb{R}^n$ be a bounded Lipschitz domain. Let $\Gamma_0\subset\partial\Omega$ be sufficiently regular.

Let $V_1:=\left\{\phi\in C^\infty\left(\overline{\Omega}\right):\phi\geq 0 ~\text{on}~ \Gamma_0\right\}$.

Let $V_2$ be the set of functions $u\in W^{1,p}(\Omega)$ with non-negative trace on $\Gamma_0$.

Then $V_1$ is dense in $V_2$.

If we consider the case of "$\phi = 0$" instead of "$\phi \geq 0$", of course an analogous result for $\Gamma_0=\partial\Omega$ is well known, and there is a result that it stays true if $\Gamma_0\subset\partial\Omega$ is relatively open and has relative Lipschitz boundary.

It would already be very helpful to have a reference for the result with $\phi \geq 0$ and $\Gamma_0=\partial\Omega$.

I don't know a reference, but perhaps you could prove it as follows. For $u \in W^{1,p}(\Omega)$ with nonnegative trace, write $u = u^+ - u^-$ which are both in $W^{1,p}(\Omega)$. Since $u^+$ is nonnegative everywhere, you should be able to approximate it by nonnegative smooth functions $\phi_n$. And since $u^-$ has zero trace, by the result you quote, you can approximate it by smooth functions $\psi_n$ vanishing on the boundary (maybe even compactly supported). Then $\phi_n - \psi_n$ are nonnegative on the boundary and converge to $u$.
• Actually, a Theorem due to Stampacchia says that if $\phi$ is a Lipschit functin, then $u\mapsto\phi\circ u$ is a Lipschitz function from $W^{1,p}$ into itself. Apply this to $\phi(s)=s^+$. – Denis Serre Mar 10 '15 at 15:29
• @DenisSerre: That will certainly work. You can also do it by hand, by approximating $\phi(s) = s^+$ pointwise by $C^\infty$ functions with controlled derivatives. – Nate Eldredge Mar 10 '15 at 15:31