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

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

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

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

endowed with the norm

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

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

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

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

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

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

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