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.
2
-
$\begingroup$ 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 $\endgroup$– Yemon ChoiCommented Feb 1, 2011 at 7:49
-
$\begingroup$ 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! $\endgroup$– TaQCommented Feb 1, 2011 at 8:57
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
2
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".
-
2$\begingroup$ Could you give a bit more detail on your second point? (My general TVS knowledge/facility is somewhat rusty) $\endgroup$ Commented Jan 31, 2011 at 19:11
-
$\begingroup$ 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. $\endgroup$– TaQCommented Feb 1, 2011 at 10:34
$\begingroup$
$\endgroup$
9
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 $E=\mathbb R^{{\ \mathbb N}_0}$ .
-
$\begingroup$ 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 $p\ge 1$. $\endgroup$– B RCommented Jan 31, 2011 at 22:14
-
$\begingroup$ 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). $\endgroup$– B RCommented Jan 31, 2011 at 22:19
-
$\begingroup$ 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! $\endgroup$– TaQCommented Jan 31, 2011 at 22:27
-
$\begingroup$ 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. $\endgroup$– B RCommented Jan 31, 2011 at 23:49
-
$\begingroup$ 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 ... $\endgroup$– TaQCommented Feb 1, 2011 at 8:50