# Is there an infinite−dimensional Banach subspace in C^∞([0,1]) ?

More specifically, with $I=[0,1]$ let $E=(X,\mathcal T\ )=C^\infty(I)$, where $X$ is the underlying (say real) vector space and $\mathcal T\$ is the (standard projective limit) topology of uniform convergence of each fixed derivative. One asks whether there is a $\mathcal T\$−closed infinite−dimensional vector subspace $S$ in $X$ such that $S$ with the induced vector operations and the induced topology is Banach(able). This requires existence of a zero neighbourhood $V$ in $E$ such that $S\cap V$ be bounded in $E$. Still more explicitly, we should have some natural number $k$ and a sequence $i\mapsto M_i$ of positive reals such that whenever $x\in S$ is such that $|D^ix(t)|\le 1$ for all natural $i$ up to $k$ and all $t\in I$, then we also have $|D^ix(t)|\le M_i$ for all $i$ up to $\infty$ and $t\in I$. My guess after trying some examples is that there is not such an $S$, but I have no proof for this.

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Since the author appears to have answered his or her own question (as has Silver) I think he or she should accept one of these two answers –  Yemon Choi Feb 1 '11 at 7:49
I would like to accept my own answer because it uses more elementary facts but I must wait for 32 hours before I can do so! –  TaQ Feb 1 '11 at 8:57

## 2 Answers

I guess that the answer is NO.

Since

• $C^\infty(I)$ is nuclear,
• any closed subspace of a nuclear space is nuclear,
• any nuclear Banach space is finite-dimensional.

Edit : for another proof, replace "nuclear" by "Fréchet-Montel".

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Could you give a bit more detail on your second point? (My general TVS knowledge/facility is somewhat rusty) –  Yemon Choi Jan 31 '11 at 19:11
In Topological uniform structures by Warren Page, Wiley 1988, on page 237, Definition 20.8 says that a Hausdorff locally convex space is semi-Montel iff bounded subsets are relatively compact. By requiring also infrabarrelledness, we get Montel spaces. On page 238, Theorem 20.10 says that closed (vector)subspaces of semi-Montel spaces are semi-Montel. Also (arbitrary) products of semi-Montel spaces are semi-Montel. In particular, we get the result that no semi-Montel space can contain an infinite-dimensional normable subspace. –  TaQ Feb 1 '11 at 10:34

This was a many years lasting problem to me, but now that I began to think of it anew, I found the solution: Since $E$ has the Heine−Borel property, taking $V$ to be $\mathcal T\$−closed, we get $S\cap V$ also such, and hence $\mathcal T\$−compact. Having a compact zero neighbourhood, so $S$ must be finite−dimensional. No such complicated things as nuclearity are needed. The same argument also gives the same result for the space $E=\mathbb R^{{\ \mathbb N}_0}$ .

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If by $E={\bf R}^{\bf N}$ you mean the Banach space of bounded sequences, this is incorrect, as $E$ does not have the Heine-Borel property (as no infinite-dimensional Banach space has it) AND $E$ contains many infinite-dimensional Banach subspaces (even if you don't include itself), such as $\ell^p$, for $p\ge 1$. –  B R Jan 31 '11 at 22:14
Concerning the main thrust of your argument, it seems you are assuming that you can choose a bounded neighborhood of $E$ (so that its closure is bounded and the intersection of that with $S$ will be closed and bounded, so compact by Heine-Borel). However, $E$ does not have any bounded neighborhoods (a tvs with the Heine-Borel property and a bounded neighborhood is automatically finite-dimensional). –  B R Jan 31 '11 at 22:19
To BR: By ${\mathbb R}^{{\ \mathbb N}_0}$ is meant, as is usual, the Fréchet space of all real valued sequences equipped with the topology of pointwise convergence. I do not assume $E$ to have any bounded zero neighbourhood. Only $S\cap V$ will be bounded, not $V$ . You should think the matter more accurately! –  TaQ Jan 31 '11 at 22:27
You could help by explaining why $S\cap V$ must be bounded in $E$. Not every neighborhood of a Banach space is bounded, so without restrictions on $V$, you can't say $S\cap V$ is bounded in $S$. And I just don't know what needs to be said to go from "bounded in $S$" to "bounded in $E$". If you are using specifics about your $E$, you might want to mention them. Of course, you are answering your own question, so it may not matter to you how clear it is to others. –  B R Jan 31 '11 at 23:49
Again to BR: I am new here (but not in mathematics) and I thought MO should be a place (mainly) for research level mathematicians. So not every detail of beginning undergraduate level should be explained. Nevertheless, ... . To be explicit, write $F$ for the Banachable space obtained by equipping the (abstract) set $S$ with the induced vector space operations and topology from $E$. To be continued ... –  TaQ Feb 1 '11 at 8:50