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.

  • 4
    $\begingroup$ That'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). $\endgroup$ Sep 3, 2014 at 16:15
  • $\begingroup$ @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? $\endgroup$
    – Tomás
    Sep 3, 2014 at 16:20
  • $\begingroup$ Maybe we can use the norm $\|f\| = \sum_n \frac{1}{2^n}\min\{\sup_{x\in [a,b]} |f^{(n)}(x)|,1\}$? $\endgroup$ Sep 3, 2014 at 16:24
  • $\begingroup$ Here is one place, where I found this claim @DanielFischer. There are others. $\endgroup$
    – Tomás
    Sep 3, 2014 at 16:29
  • 2
    $\begingroup$ @LiuGang, it does not seem to be a norm. For example, $\|\lambda f\|\neq \lambda \|f\|$. $\endgroup$
    – Tomás
    Sep 3, 2014 at 16:35

2 Answers 2


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.

  • 5
    $\begingroup$ Nice 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 $\alpha_{n}\neq0$. $\endgroup$
    – PhoemueX
    Sep 3, 2014 at 18:29
  • $\begingroup$ 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. $\endgroup$
    – Tomás
    Sep 4, 2014 at 12:53
  • $\begingroup$ 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 $k$. $\endgroup$ Sep 4, 2014 at 13:01
  • $\begingroup$ What is $2^{\aleph_0}$ ? And why the dimension is $2^{\aleph_0}$? $\endgroup$
    – Enhao Lan
    May 11, 2016 at 1:59
  • 1
    $\begingroup$ @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 $\geqslant 2^{\aleph_0}$. $\endgroup$ May 11, 2016 at 8:13

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 $$\rho(f) = \sum_{n=1}^{\infty} \frac{1}{2^n} \cdot \frac{\sup ||f^{(n)}||}{1+\sup ||f^{(n)}||}$$


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