# Dense Subspaces of Sobolev Spaces

• January 28th 2013, 06:44 AM
HAL9000
Dense Subspaces of Sobolev Spaces
Hello!

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?