Is there no norm in $C^\infty ([a,b])$?

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.

Solution 1:

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.

Solution 2:

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