Good morning, I am looking for a reference about the following fact that seems to be folklore. Define the Sobolev (Beppo Levi?) space $$ D^{1,p}(\mathbb{R}^N) = \left\{ u \in L^{p^*}(\mathbb{R}^N) \mid \nabla u \in L^p(\mathbb{R}^N) \right\}. $$ Where can I find a (complete) proof that $D^{1,p}(\mathbb{R}^N)$ coincides with the closure of $C_0^\infty(\mathbb{R}^N)$ under the norm $$ \|u\|^p = \int_{\mathbb{R}^N} |\nabla u|^p ? $$
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$\begingroup$ What is $p^*$, a typo? $\endgroup$– DanielCommented Nov 29, 2014 at 13:39
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$\begingroup$ A standard reference is the book "Sobolev Spaces" by Adams (and Fournier). You can get this by proving the Sobolev inequality for a compactly supported smooth function, where the constant in the inequality depends only on $n$ and $p$ (and not the support of the function). $\endgroup$– Deane YangCommented Nov 29, 2014 at 16:46
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$\begingroup$ $p^* = np/(n-p)$. $\endgroup$– Deane YangCommented Nov 29, 2014 at 16:46
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$\begingroup$ I guess the matter here is exactly to avoid arguments like "this follows easily by adapting..." The OP probably needs a book (or a survey paper) where this characterization is proved. Actually it seems hard to find such a reference. $\endgroup$– SiminoreCommented Nov 29, 2014 at 16:47
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1$\begingroup$ @DeaneYang It seems to me that the OP wants a piece of bibliography. For example "See this book, Theorem X, page Y". $\endgroup$– SiminoreCommented Nov 30, 2014 at 9:33
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1 Answer
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I guess the theorem you are searching for is the theorem by Meyers-Serrin, see
http://www.pnas.org/cgi/reprintframed/51/6/1055
The norm you are using is a bit different, but you can easily prove the equivalence of the topologies using the Poincare-Inequality.
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$\begingroup$ I thought the OP's question was more to do with the density part of the statement: the first definition is an "all objects with finite norm" and the second is "closure of some nice space of well-behaved objects wrt an equivalent norm" $\endgroup$ Commented Nov 29, 2014 at 14:13
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$\begingroup$ The question appears to be about the decay of the function at infinity. $\endgroup$ Commented Nov 29, 2014 at 16:17
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$\begingroup$ No, I don't think this answers my question. $\endgroup$– PaperinoCommented Nov 29, 2014 at 17:50
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3$\begingroup$ Could you clarify what your question is? $\endgroup$ Commented Nov 29, 2014 at 20:26
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$\begingroup$ If I am allowed, I'd say the @Paperino does not want any sentence like "but you can easily prove". $\endgroup$– SiminoreCommented Dec 1, 2014 at 9:46