Does anyone knows a reference, which proves the following:
Let $a,b\in \mathbb{R}$ with $a<b$. There is no norm in the space $C^\infty([a,b])$, which makes it a Banach space.
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4That's not true, there are norms making it a Banach space since there are Banach spaces of the same dimension ($2^{\aleph_0}$). There is no "reasonable" norm making it a Banach space, "reasonable" meaning inducing a topology that is comparable with the standard Fréchet space topology (open mapping theorem). – Daniel Fischer Sep 03 '14 at 16:15
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@DanielFischer, I see. I read in plenty of places, that the claim above were true. So I just lost my time... Do you know any reference for your reply? – Tomás Sep 03 '14 at 16:20
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Maybe we can use the norm $|f| = \sum_n \frac{1}{2^n}\min{\sup_{x\in [a,b]} |f^{(n)}(x)|,1}$? – Petite Etincelle Sep 03 '14 at 16:24
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Here is one place, where I found this claim @DanielFischer. There are others. – Tomás Sep 03 '14 at 16:29
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2@LiuGang, it does not seem to be a norm. For example, $|\lambda f|\neq \lambda |f|$. – Tomás Sep 03 '14 at 16:35
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@Tomás Yeah, you are right. Thanks – Petite Etincelle Sep 03 '14 at 16:44
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@Tomás Strictly, what they wrote is wrong, but what they meant was that in the natural topology the space is not normable, or perhaps even that no norm inducing a finer or coarser topology than the natural topology can make it a Banach space. – Daniel Fischer Sep 03 '14 at 17:22
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@DanielFischer: Do you know an "easy" way to see that the (algebraic) dimension is indeed $2^{\aleph_0}$? – PhoemueX Sep 03 '14 at 17:22
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@PhoemueX See below. – Daniel Fischer Sep 03 '14 at 17:34
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Literally, the claim is wrong, since the space has dimension $2^{\aleph_0}$, and there are Banach spaces with the same dimension (e.g. the $\ell^p(\mathbb{N})$ spaces). Since the rationals are dense in $[a,b]$, there is a linear injection of $C^\infty([a,b])$ into $\mathbb{R}^{\mathbb{Q}\cap [a,b]}$, and the latter space has cardinality
$$\operatorname{card}(\mathbb{R})^{\aleph_0} = \left(2^{\aleph_0}\right)^{\aleph_0} = 2^{\aleph_0},$$
so $\dim C^\infty([a,b]) \leqslant 2^{\aleph_0}$. On the other hand, the functions $t\mapsto e^{ct},\, c\in\mathbb{R}$ are linearly independent, so the dimension is at least $2^{\aleph_0}$.
What is meant, even if it is not said, when that claim is made, is that the space $C^\infty([a,b])$ in its natural topology - the usual Fréchet space topology induced by the seminorms $\lVert f\rVert_k = \sup \{ \lvert f^{(k)}(t)\rvert : t\in [a,b]\}$ for $k\in\mathbb{N}$ - is not normable.
An easy way to see that is to note that $C^\infty([a,b])$ is a Fréchet-Montel space, that is, every closed and bounded subset is compact. That is a repeated application of the Ascoli-Arzelà theorem; the boundedness of $\{f^{(k+1)} : f\in B\}$ implies the equicontinuity of $\{ f^{(k)} : f\in B\}$.
But a normed space has the Montel-Heine-Borel property if and only if it is finite-dimensional.
Daniel Fischer
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5Nice proof, thank you very much. Maybe the following can save others a few minutes: To see that $\left(e^{ct}\right){c\in\mathbb{R}}$ is linearly independent, assume $\sum{i=1}^{n}\alpha_{i}e^{c_{i}t}\equiv0$ on $\left[a,b\right]$ with all $\alpha_{i}\neq0$ and $c_{1}<\dots<c_{n}$. Using the identity theorem (for holomorphic functions, say), this implies $\sum_{i=1}^{n}\alpha_{i}e^{c_{i}t}\equiv0$ on $\mathbb{R}$. Division by $e^{c_{n}t}$ yields $$0\equiv\sum_{i=1}^{n}\alpha_{i}e^{\left(c_{i}-c_{n}\right)t}\xrightarrow[t \to \infty]{}\alpha_{n}$$ in contradiction to $\alpha_{n}\neq0$. – PhoemueX Sep 03 '14 at 18:29
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Am I wrong in think (intuitively), that any norm in $C^\infty([a,b])$, which makes it a Banach space, will have to involve somehow, $f$ and all it's derivatives? I mean something like this $$|f|=H(f,f',f'',...,f^{(n)},...),$$ where $H$ is some function? If the question is to vague, I will delete it. – Tomás Sep 04 '14 at 12:53
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I think that you cannot construct a norm making it into a Banach space explicitly. Finding Hamel bases of $C^\infty([a,b])$ and $\ell^p$ (or some other Banach space of dimension $2^{\aleph_0}$) seems hopeless, and apart from transporting the norm via a linear isomorphism, I don't know how one would make it. But abstractly, or hypothetically, I think you're right, if it involved only finitely many derivatives, I don't see how $C^\infty([a,b])$ could not be dense in $C^k([a,b])$ for a sufficiently large $k$. – Daniel Fischer Sep 04 '14 at 13:01
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1@lanse7pty $\aleph_0$ is the cardinality of $\mathbb{N}$, hence $2^{\aleph_0}$ is the cardinality of the power set of $\mathbb{N}$, which also is the cardinality of $\mathbb{R}$. The dimension is $2^{\aleph_0}$ because a) the whole space has cardinality $2^{\aleph_0}$, so the dimension is $\leqslant 2^{\aleph_0}$, and b) one can find linearly independent families in the space which have cardinality $2^{\aleph_0}$, hence the dimension is $\geqslant 2^{\aleph_0}$. – Daniel Fischer May 11 '16 at 08:13
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Assume there exists a norm generating the topology. Consider the open unit ball $B$ around $0$. Since the seminorms $f \mapsto \sup ||f^n||$ define the same topology there exists
an $\epsilon > 0$ and $n$ so that
$$\{ f \ | \sup ||f^{(i)}|| <\epsilon, i=1,\ldots n\} \subset B$$
The set $\{ f \ | \ \sup ||f^{(n+1)}|| < 1 \} $ is an open neighborhood of $0$ so there exists a $\delta>0 $ so that $\delta B \subset \{ f \ | \ \sup ||f^{(n+1)}|| < 1 \} $. We conclude from the above that
$$\{ f \ | \sup ||f^{(i)}|| <\epsilon\cdot \delta\ ,\ i=1,\ldots n\} \subset \{ f \ | \ \sup ||f^{(n+1)}|| < 1 \} $$ and therefore
$$\sup ||f^{(n+1)}|| \le \frac{1}{\epsilon \cdot \delta} \max ( \sup ||f^{(i)}|| , i=1,\ldots n)$$
for all $f$.
It's not too hard to see that such an inequality is not possible. For instance consider $f$ with $f^{(n)}(x) = \frac{1}{M} \sin (Mx)$ for large $M$.
Note that there exists a metric that defines the topology given by $d(f,g) = \rho(f-g)$ where $$\rho(f) = \sum_{n=1}^{\infty} \frac{1}{2^n} \cdot \frac{\sup ||f^{(n)}||}{1+\sup ||f^{(n)}||}$$
orangeskid
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