2
$\begingroup$

Is $H^{s+1} (\Bbb R^n)$ dense in $H^s(\Bbb R^n)$ for $s = 0,1,2, \cdots$ ? ($H^s$ : general sobolev space)

2 Answers 2

1

Note that $C_0^{s+1}(\mathbb{R}^n)$ is a subset of $H^{s+1}(\mathbb{R}^n)$, where $C_0^{s+1}(\mathbb{R}^n)$ is the space of $s+1$ differentiable functions with compac support. On the other hand you have that $C_0^{s+1}(\mathbb{R}^n)$ is dense in $H^{s}(\mathbb{R}^n)$. Hence you can conclude.

4

There is this chain of continuous embedding:\begin{equation} \mathcal{S}(\mathbb{R}^d)\hookrightarrow H_s\hookrightarrow H_t\hookrightarrow\mathcal{S}'(\mathbb{R}^d), (s>t) \end{equation} where $\mathcal{S}(\mathbb{R}^d)$ is the Shwartz class and $\mathcal{S}'(\mathbb{R}^d)$ is the space of tempered distributions.

$\mathcal{S}(\mathcal{R}^d)$ is dense in $H_s$ so the answer to your question follows.