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Let $T(R)$ denote the space of tempered functions on the line,
i.e. the smooth functions that give Schwartz function after a
multiplication by any Schwartz function, equipped with the natural
nuclear topology. e.g. the topology induced from the strong
(convergence on bounded sets) topology on the endomorphism space of
the space of Schwartz functions.

Is it true that tempered functions on the plane is the completed tensor square
of tempered functions on the line, i.e. $T(R) \hat{\otimes} T(R) =T(R^2)$?

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1 Answer 1

up vote 4 down vote accepted

This is proved in 4.1 of: Michel Dubois-Violette, Andreas Kriegl, Yoshiaki Maeda, Peter W. Michor: Smooth *-algebras. Progress of Theoretical Physics Supplement 144 (2001), 54-78. arXiv:math.QA/0106150. pdf

According to L. Schwartz, there are two kind of tempered spaces, $\mathcal O_M$, and $\mathcal O_C$. See the paper, where your question is proved for both of them.

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Thank you very much –  Rami Oct 18 '12 at 22:13
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