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I have read that the answer is no, but I am unable to prove it.

Give $C^{\infty}([0,1])$ the metric $$d(f,g) = \sum_{j=0}^{\infty} 2^{-j} \frac{||(f-g)^{(j)}||}{1 + ||(f-g)^{(j)}||}$$ associated to the collection of seminorms $||f^{(j)}||$, the sup-norm of the $j$-th derivative. This makes $C^{\infty}([0,1])$ into a complete metric space.

How do you show that the topological vector space $C^{\infty}([0,1])$ is not a Banach space? That is, there does not exist a norm which induces the same topology.

It does not appear to be enough to show merely that the metric does not arise from a norm, since it is very non-canonical.

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Hints: 1)There is no norm on on $C^{\infty}[0,1]$ which makes the derivative continuous. This is because, $||\frac{d}{dx}(e^{nx})|| \leq c||e^{nx}||$ for all values of $n$ isn't possible.

2) Show that in the above metric the derivative map is continuous.

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  • $\begingroup$ Do you know how to tackle the general case, that is, that there is no norm in $C^\infty([a,b])$, which make it a Banach space? $\endgroup$
    – Tomás
    Commented Sep 2, 2014 at 20:50
  • $\begingroup$ @Tomás: No I do not, unfortunately. If you know the proof- please paste a link (or give us the argument)- it will be very helpful for me. $\endgroup$
    – voldemort
    Commented Sep 2, 2014 at 20:54
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    $\begingroup$ No, I do not know too, however, I think that should be some contradiction, with respect to the derivative operator, if we assumed it to be true. $\endgroup$
    – Tomás
    Commented Sep 2, 2014 at 20:57
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    $\begingroup$ I, too, am also of that opinion, although I, as well, do not have anything with which to back it up. $\endgroup$ Commented Sep 2, 2014 at 21:01
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    $\begingroup$ I am sorry, but why doesn't the same argument work for $C^{\infty}[a,b]$ ? $\endgroup$
    – Optional
    Commented Apr 10, 2016 at 16:06

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