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My question is related to this one: Take the vector space of infinitely differentiable functions on $[0,1]$. The standard norm of $C^k([0,1])$ is just the $\ell^1$-norm of the vector $(\|f\|_\infty, \|f'\|_\infty,\ldots,\|f^{(k)}\|_\infty)$, but of course this idea cannot be further pursued to define a norm on $C^\infty([0,1])$.

However, what if one would consider the space $$ \{f\in C^\infty([0,1]):(\|f^{(n)}\|_\infty)_{n\in\mathbb N}\in \ell^p \} $$ for $p\in [1,\infty]$? This space is certainly small - in particular, it contains neither the exponential function, nor $\sin$ and $\cos$ - but at least it does contain the polynomials and it seems to be a Banach space - in fact even a Banach lattice algebra. Does this space appear in applications (PDEs?)? Has anybody ever studied its functional analytical properties and if this is not the case, what are this space's obvious drawbacks?

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    $\begingroup$ It seems that there are no partitions of unity in this space -- certainly a serious drawback for analytical applications. $\endgroup$
    – Jochen
    Apr 28, 2016 at 8:55

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If your goal is to ultimately show that this is a norm on $ C^\infty([0,1])$ there are many obstacles which you won't overcome:

  1. Take a Cauchy sequence in your subspace. Even if you prove that it converges in you $ \ell^p $-norm, how do you prove that this limit is actually $ (\|f^{(n)}\|_\infty)_{n\in\mathbb N} $ for some $ f \in {C}^{\infty} $

  2. Your subspace does include polynomials, but its closure in your $ \ell^p $-norm amy be strictly smaller than $ C^\infty([0,1])$

So, if you fail any of these two, you won't be able to define it on whole of $ C^\infty([0,1])$ as you wanted.

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  • $\begingroup$ Uh? I don't understand. As "my" spaces are Banach spaces, the closure of any of them is exactly the same space. $x\mapsto e^{2x}$ is an example of a function which lies in $C^\infty([0,1])$ without being in any of my spaces, not even for $p=\infty$. My question is precisely whether any of my spaces is nice enough to have been used in applications. $\endgroup$
    – DeM
    May 5, 2016 at 13:40

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