A completion equality in product of Sobolev spaces

Question

Let $\Omega$ be a nice bounded domain of $\mathbb{R}^n$. For $s\ge 0$ we define $$H_{0}^{s}(\Omega):=\overline{C_{c}^{\infty}(\Omega)}^{H^{s}(\Omega)}.$$ My question: is the following equality true? $$\overline{\{ (f,f|_{\partial \Omega}), f\in C^{\infty}(\overline{\Omega}) \}}^{H^{s}(\Omega)\times {H^{s}(\partial\Omega)}}=H^{s}_0(\Omega)\times {H^{s}(\partial\Omega)}.$$ I expect that the result is true, but I am still stuck in giving a rigorous proof. Any hint would be useful.

Answer 1

This is certainly not true, at least for two reasons:

1. since you are considering all smooth functions $f\in\mathcal C^\infty(\overline\Omega)$ without restrictions on the behaviour at the boundary, there is no resaon why in the first factor you should expect to retrieve the "zero-boundary space" $H^s_0$ instead of $H^s$

2. since the approximating pairs $(f,f|_{\partial\Omega})$ are really special structurally -- namely the second factor $f|_{\partial\Omega}$ must be the trace of the first factor in your product space before completeion -- you cannot expect to retrieve the whole product space $H^s_0(\Omega)\times H^s(\partial\Omega)$ upon completion. For example, how on earth would you approximate the constant (pair of) functions $(0,1)\in H^s_0(\Omega)\times H^s(\partial\Omega)$ by pairs of the form $(f_n,f_n|_{\partial\Omega})$? By continuity of the trace operator $\gamma:H^s(\Omega)\rightarrow H^{s-1/2}(\partial\Omega)$ (see @Liviu Nicolaescu's comment) for any such limit pair $(f,g)=\lim (f_n,f_n|_{\partial\Omega})$ you should get that $g=\lim f_{n}|_{\partial\Omega}=\lim \gamma( f_n) = \gamma(\lim f_n)=\gamma(f)=0$ (the limits hold in $H^{s-1/2}(\partial\Omega)$) so clearly the second factor $g$ should vanish.