## Dense Subspaces of Sobolev Spaces

Hello!

Let $\displaystyle u,v \in L^p(\Omega)$. Then $\displaystyle v$ is called the $\displaystyle \alpha$-th weak derivative of $\displaystyle u$ iff

$\displaystyle \int_{\Omega}v(x)\varphi(x)\,dx = (-1)^{|\alpha|} \int_{\Omega}u(x)D^{\alpha}\varphi(x)\,dx$

for all $\displaystyle \varphi \in C^{\infty}(\mathbb{R}^n)$ with compact support in $\displaystyle \Omega$.

Now for $\displaystyle 1 \le p \le \infty, m \in \mathbb{N}_0$ consider the Sobolev space $\displaystyle H^p_m(\Omega)$ of functions $\displaystyle f \in L^p(\Omega)$ which have weak derivatives $\displaystyle D^{\alpha}f \in L^p(\Omega)$ up to order $\displaystyle |\alpha| \le m$,

endowed with the norm

$\displaystyle ||f||_{p,m,\Omega}=\sum_{|\alpha| \le m}||D^{\alpha}f||_{p,\Omega}$.

Consider also for $\displaystyle 1 \le p \le \infty, m \in \mathbb{N}_0 \cup \{\infty\}$ the subspace $\displaystyle C^p_m(\Omega)$ in $\displaystyle L^p(\Omega)$, which consists of all real-valued m-times continuously differentiable functions on $\displaystyle \Omega$ with weak derivatives $\displaystyle D^{\alpha}f \in L^p(\Omega)$
of all orders up to $\displaystyle m$, i.e. $\displaystyle C^p_m(\Omega)=C_m(\Omega) \cap H^p_m(\Omega)$.

There is a theorem which says that the closure $\displaystyle \overline{C^p_{\infty}(\Omega)}$ with respect to the norm above is exactly $\displaystyle H^p_m(\Omega)$ for all $\displaystyle 1 \le p < \infty$, so that $\displaystyle p=\infty$ is the exception.

Before it is shown by example that for $\displaystyle p=\infty$ one inclusion is not fulfilled it is said that $\displaystyle \overline{C^{\infty}_m(\Omega)}=C^{\infty}_m( \Omega )$

with respect to the above norm with $\displaystyle p=\infty$.

This is clear because $\displaystyle L^{\infty}$ is complete.

The author then unexpectedly concludes that $\displaystyle \overline{C^{\infty}_{\infty}(\Omega)}=C^{\infty}_ 1(\Omega)$ with respect to $\displaystyle ||.||_{1,\infty}$

While switching to m=1 he apparently simultaneously shrinks the class of functions on the left, but the norm with respect to which the closure is taken seems to have no influence. I cannot see at a glance why he has the right to do it! So the question is simply: WHY?