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why the set of continuous complex-valued functions on an open set of $R^n$ is not normable?

Gregory Grant
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user244115
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    With what topology? –  May 28 '15 at 12:35
  • Well, I think that we can safely guess that he means the topology of pointwise convergence. – Alex M. May 28 '15 at 13:05
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    @AlexM. Unlikely, the usual topology considered is the topology of locally uniform convergence. – Daniel Fischer May 28 '15 at 13:05
  • The question by @G.Sassatelli is crucial, which topology? Similar to the reasoning here, we can endow the space with norms making it a Banach space, but for a non-empty open set, these norms don't play well with the usual notions of convergence. – Daniel Fischer May 28 '15 at 13:10

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