MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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

I guess that the answer is NO.


  • $C^\infty(I)$ 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".

share|cite|improve this answer
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
up vote 0 down vote accepted

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 $E=\mathbb R^{{\ \mathbb N}_0}$ .

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.