Question

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.

Answer 1

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

Answer 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".