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Mar 20, 2013 at 15:52 comment added Claudio Gorodski I mean, wrt $C^k$-norm (uniform convergence of functions and their first $k$ derivatives).
Mar 19, 2013 at 19:45 comment added Claudio Gorodski @Peter: Piotr's comment would be fine for functions of class $C^k$, $k finite$, right?
Mar 19, 2013 at 19:27 vote accept Nevermind
Mar 19, 2013 at 18:58 comment added Peter Michor @Piotr: That is true for continuous functions. Spaces of smooth functions are not normable or Banach spaces.
Mar 19, 2013 at 18:37 answer added Peter Michor timeline score: 4
Mar 19, 2013 at 18:35 comment added Piotr Achinger Assume $M$, $N$ compact. Isn't then $C^\infty(M\times N, \mathbb{R})$ the completion of $C^\infty(M, \mathbb{R}) \otimes C^\infty(N, \mathbb{R})$ with respect to the supremum norm?
Mar 19, 2013 at 18:29 history asked Nevermind CC BY-SA 3.0