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The Hilbert Space tensor product gives

$L^2(\mathbb R^2,dx\otimes dx;\mathbb R)= L^2(\mathbb R,d x;\mathbb R) \otimes L^2(\mathbb R,dx;\mathbb R)$

My question is: does there exist also a notion of tensor product which gives $C^\infty(\mathbb R^2;\mathbb R)= C^\infty(\mathbb R;\mathbb R) \otimes C^\infty(\mathbb R;\mathbb R)$?

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    The standard tensor product surely fails to accomplish this. I seem to remember something like, as a corollary to the Arzelà-Ascoli theorem, that this holds for continuous functions on a product of compact spaces (that is $C(X_1)\otimes \cdots \otimes C(X_n)$ is dense in $C(X_1\times\cdots\times X_n)$ and thus equality of the completion of the left side with the space on the right.) I wouldn't blelieve this holds for smooth functions on non compact spaces.2012-05-08
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    @Olivier Bégassat. Thanks for your comment. Can you suggest me a reference where these things are discussed? What if I consider the space of smooth functions with compact support?2012-05-08
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    For smooth functions with compact support, it is discussed here http://math.stackexchange.com/questions/63416/tensor-products-of-functions-generate-dense-subspace.2012-05-08

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