Easy question on Sobolev spaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T14:06:36Z http://mathoverflow.net/feeds/question/110621 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110621/easy-question-on-sobolev-spaces Easy question on Sobolev spaces Axel 2012-10-25T06:33:05Z 2012-10-25T09:49:28Z <p>I understand that this question would be trivial for experts, sorry for that, I just need to clarify things.</p> <p>So let $S(\mathbb{R}^n)$ denote the Schwartz space on $\mathbb{R}^n$ and $W_p$, $W_q$ are the Sobolev spaces, or in the other words completions of $S(\mathbb{R}^n)$ with respect to $p$ and $q$ - Sobolev norms. </p> <p>Let us assume that $\phi:$ $W_q\rightarrow W_p$ is a map , continuously extending the identity map on $S$ assuming that $p &lt; q$.</p> <p>Question: why is $\phi$ an embedding? It is only clear that $\phi$ is embedding of $S$ ( as it is identity on it).</p> <p>Similar question for Sobolev embedding theorem into $C^k$.</p> <p>P.S. To clarify my question here is what I mean by the Sobolev space. For a real number $p\in\mathbb{R}$ the $p$-Sobolev norm on $S$ is given by ${|f|}^2_p=\int{(1+|\xi|)}^{2p}{|\hat{f}(\xi)|}^2d\xi$, where $\hat{f}(\xi)$ is a Fourier transform of $f(x)\in S$. So $W_p$ is a completion of $S$ with respect to this norm.</p> <p>P.P.S. To make things little more clear I am mostly interested in case of non-integer or negative $p$ when the Sobolev $p$-norm can not be defined through the weak derivatives.</p> http://mathoverflow.net/questions/110621/easy-question-on-sobolev-spaces/110627#110627 Answer by Delio Mugnolo for Easy question on Sobolev spaces Delio Mugnolo 2012-10-25T08:00:54Z 2012-10-25T08:00:54Z <p>Axel,</p> <p>please read my above comment. If by $W^p$ you mean $W^{p,2}$, then the imbedding follows from the very definition of Sobolev space - or rather, if you wish to keep your definition, from the Meyer-Serrin theorem which essentially establishes the equivalence between your definition and mine.</p> <p>If by $W^p$ you mean $W^{1,p}$, then $\phi$ is an embedding only if you have a finite measure space (and in this case the assertion follows from the embedding of $L^q$ ind $L^p$ by Hölder's inequality), otherwise it is in general wrong (think of the Sobolev spaces over $\mathbb R^n$).</p> http://mathoverflow.net/questions/110621/easy-question-on-sobolev-spaces/110631#110631 Answer by András Bátkai for Easy question on Sobolev spaces András Bátkai 2012-10-25T08:34:26Z 2012-10-25T08:34:26Z <p>I would like to expand a bit what Delio said. Your question is a bit confusing as it is, but we may assume that you mean that $p$ and $q$ are the parameters representing the derivatives. Then what you need is $$\|f\|_p \leq C\|f\|_q.$$</p> <p>If you write out the definition of the Sobolev norm on $S$, then you see immediately that this holds. This implies that the identity map defined on $S$ extends uniquely and continuously to the whole $H^q$. This gives you the embedding. (Strictly speaking you have to verify that this extension is the identity map, but this follows easily from continuity considerations.)</p>