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Consider $C^{\infty}[0,1]$ space, where $f_n \to f$ iff all its derivatives $f_n^{(k)}(x)$ uniformly converge to $f^{(k)}(x)$. The question is prove that this space is metrizable, bot can not be normalized.

My attempt: it's not hard to see that using $\{\sup_{[0,1]} |f^{(k)}(x) - g^{(k)}(x)|\}_{k\ge0}$ we can construct a metric $\rho$, which replicate convergence in it's natural way. So now we need to find a sequence of function $\{f_n\}$ that converges in $\rho$ sense, but there is no norm for which this sequence also converges? The first idea is to consider something easily differentiable $\{\exp(nx)\}_n$. This sequence looks reasonable, but I don't know why this sequence doesn't converge in any norm.

Actually, it looks curious, that there is no appropriate norm to normalize this space.

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    $\begingroup$ I think that this is a Fréchet space with the Heine Borel property (closed and bounded sets are compact). Indeed, by Ascoli Arzelá, a bounded set in C^k is precompact in C^{k-1}. Now, the only normal spaces with the Heine Borel property are finite dimensional, and the space in this question is clearly not. Just a sketch, see if it convinces you and do not trust my word. $\endgroup$ Sep 14 at 23:44
  • $\begingroup$ There is no complete algebraic norm on the linear space $C^{\infty}[0, 1]$. Note that the operator of the derivative is a linear bounded operator on $C^{\infty}[0, 1]$ . Then there is some $0\lt M$ such that $||f'||\leq Mf$. But for $f(t)=e^{2Mt}$ we have $2M||f||=||f'||\leq M||f||$. That is a contradiction. $\endgroup$
    – ali reza
    Sep 15 at 10:46
  • $\begingroup$ @alireza why should it be bounded. Maybe I miss something. $\endgroup$
    – openspace
    Sep 15 at 12:15
  • $\begingroup$ It is bounded by the closed graph theorem. $\endgroup$
    – ali reza
    Sep 16 at 17:57
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HINT:

The space $C^{\infty}[0,1]$ cannot be normalized because it does not have any bounded neighborhoods of $0$. Indeed, every neighborhood of $0$ contains a set of the form $$\{ f \ | \ p_k(f) \le \delta_k, k = 0, \ldots n\}$$ and on such a set every seminorm $p_{m}$ with $m>n$ is unbounded. Indeed, let $f$ a function in $C^n$ but not in $C^{n+1}$, and $f_m$ a sequence in $C^{\infty}$ converging to $f$ in the $C^n$ topology. If we had an inequality $p_{n+1}< C \max_{0\le k \le n} p_k$, then the sequence $f_m$ being Cauchy in the $C^{n}$ topology, would also be Cauchy in the $C^{n+1}$ topology, so $f_m$ would converge to a function in $C^{n+1}$, contradiction.

$\bf{Added}$ @Jochen's idea: we can write down explicitly a function with first $n$ derivatives small, but $p_{n+1}(f)$ large, for instance $f(x)= \frac{1}{M^{2n+1}} cos M^2 x$, with $M$ large.

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    $\begingroup$ Alternatively, you can write down exlicitly functions whose first $n$ derivatives are small but $f^{(n+1)}(0)$ is bigger than $C$. Try with $f(x)=a_n\exp(b_n x)$. $\endgroup$
    – Jochen
    Sep 15 at 8:47
  • $\begingroup$ @Jochen f.e. $f_n(x) = exp^{nx} / n^{n-1}$. It's derivatives are bounded, but $n + 1$th derivative is unbounded for $n \to \infty$? At would mean that for this functions $\|f^{(n)}_n\|$ doesn't converges? $\endgroup$
    – openspace
    Sep 15 at 13:50
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    $\begingroup$ This shows that you cannot estimate $p_{n+1}$ by the maximum of $p_0,\ldots,p_n$. $\endgroup$
    – Jochen
    Sep 15 at 13:58
  • $\begingroup$ @Jochen: Neat idea, added it! $\endgroup$
    – orangeskid
    Sep 15 at 15:10
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Another reason why $C^\infty$ is not normable is that it satisfies the Heine-Borel property, that is, every closed and bounded set is compact. (Note that $C^\infty$ is a metric space, as already observed in the question, so it makes sense to speak of "bounded sets").

The proof uses the theorem of Ascoli-Arzelá. Let $f_n\in C^\infty$ be a sequence such that $$\sup_{n, k}\lVert f_n\rVert_{C^k}< \infty.$$ Then, since the sequence is $C^1$ bounded, it has a $C^0$-converging subsequence. Such subsequence is $C^2$ bounded, thus it has a $C^1$-converging subsequence. Using the diagonal sequence trick (a.k.a. diagonal argument, Cantor argument, etc...) we can produce a subsequence of $f_n$ that is $C^\infty$ convergent. This proves that $C^\infty$ is Heine-Borel.

To conclude we observe that the only normed spaces that are Heine-Borel are the finite-dimensional ones, which is a standard theorem usually attributed to Riesz. Since $C^\infty$ is infinite-dimensional, it cannot be a normable space.

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    $\begingroup$ This is the same reason for which $\mathcal{H}(\mathbb{D})$ (i.e. the set of holomorphic functions on the open unit disc) with the topology of compact uniform convergence is not normable $\endgroup$
    – Pelota
    Sep 16 at 19:23
  • $\begingroup$ Right. For different reasons, but both spaces are Heine-Borel. $\endgroup$ Sep 16 at 20:07

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