1
$\begingroup$

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 $C^\infty(\mathbb R^2;\mathbb R)= C^\infty(\mathbb R;\mathbb R) \otimes C^\infty(\mathbb R;\mathbb R)$?

$\endgroup$
  • 3
    $\begingroup$ 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. $\endgroup$ – Olivier Bégassat May 8 '12 at 14:32
  • $\begingroup$ @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? $\endgroup$ – Hans May 8 '12 at 14:46
  • 2
    $\begingroup$ For smooth functions with compact support, it is discussed here math.stackexchange.com/questions/63416/…. $\endgroup$ – Davide Giraudo May 8 '12 at 14:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.