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Let $E=C([0, 1])$ be the space of all real-valued continuous functions on $[0, 1]$, equipped with the uniform norm. $C(E)$ stand for the continuous real-valued functions on $E$.

I am wondering that if $C(E)$, equipped with some proper metric, is a complete and seperable?

I am considering a metric on $C(E)$ which is similar to the Frechet distance: $$|F|_n \triangleq \sup_{|u| \leq n}|F(u)| \in [0, \infty]; $$ $$|F| \triangleq \sum_{n=1}^\infty 2^{-n}\frac{|F|_n}{1+|F|_n}, $$ where we set $\frac{|F|_n}{1+|F|_n}=1$ in case that $|F|_n=\infty$.

Equipped with this metric, I believe that $C(E)$ becomes a complete metric space. But traditional way of taking polynomials seems not working well for separability.

I am appreciate if you tell me the answer. If it does not work well with $C(E)$, answers to similar questions for $C_b(E)$ (bound continuous functions) or $C_L(E)$ (Lipchitz continuous functions) also help me alot.

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  • $\begingroup$ Should the evaluation function $C(E) \times E \to \mathbb{R}$ be continuous? $\endgroup$ Dec 2, 2014 at 7:31
  • $\begingroup$ Also, I do not understand what $u_n$ is in the definition of $|F|_n$, did you mean just $u$? $\endgroup$ Dec 2, 2014 at 7:32
  • $\begingroup$ I am sorry, it should be $u$. And what do you mean by evaluation function $C(E) \times E \rightarrow \mathbb{R}$? $\endgroup$ Dec 2, 2014 at 7:47
  • $\begingroup$ Would it be desirable that the function $C(E) \times E \to \mathbb{R}$ which maps $(F,u) \mapsto F(u)$ is continuous? This is a likely requirement and it constraints the possible topologies. By the way, you're probably out of luck, this sort of thing won't even be metrizable. $\endgroup$ Dec 2, 2014 at 8:05
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    $\begingroup$ To be precise, |F| is not a norm (it's bounded, it's not homogeneous,..). It defines a distance |F-G|, which makes C(E) a complete metric space, though not separable (it contains an uncountable discrete set) and not a topological vector space (it has uncountably many connected components). It's the topology of convergence on bounded sets of E, so the evaluation $(F,u)\mapsto F(u)$ is continuous. The connected component of 0 in this topology is the space of functions which are bounded on bounded sets of E, a Frechét space, topologized by your semi-norms $|\cdot|_n$ $\endgroup$ Dec 2, 2014 at 9:26

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