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Is there an inner product on the vector space of continuous real functions $\mathcal{C}(\Bbb R,\Bbb R)$ ? If so what is it and if not why ?

Joachim
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    http://math.stackexchange.com/questions/1466878/inner-product-for-functions?rq=1 – JMoravitz Jan 31 '17 at 20:08
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    Sorry I was not clear enough it is about function from R to R I will edit – Joachim Jan 31 '17 at 20:11
  • thanks for your consideration – Joachim Jan 31 '17 at 20:14
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    It seems clear that the usual "integral" inner product won't work. Of course, no inner product induces the usual topology on $C(\Bbb R,\Bbb R)$, which is given by the sup norm. – Ben Grossmann Jan 31 '17 at 20:50
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    Question is answered at http://math.stackexchange.com/questions/814754/inner-product-on-c-mathbb-r. Yes with the Axiom of Choice, using a Hamel basis. But no "explicit" form possible--it's consistent with ZF that no inner product exists. – Jonas Meyer Jan 31 '17 at 20:57
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    @Omnomnomnom The usual topology on $\mathcal{C}(\mathbb{R},\mathbb{R})$ is not normable. There are of course plenty of non-equivalent inner products on every infinite-dimensional real (or complex) vector space, but for spaces like $\mathcal{C}(\mathbb{R},\mathbb{R})$, one needs quite a lot of choice for that. It is consistent with dependent choice that no norm exists on that space, and a fortiori no inner product. – Daniel Fischer Jan 31 '17 at 20:57
  • @Daniel thanks for that – Ben Grossmann Jan 31 '17 at 21:07

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