Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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)

share|improve this question

2 Answers 2

up vote 1 down vote accepted

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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.