6

Let us define a norm $\|f\|=\sup_{k\geq 0}\sup_{t\in [0,1]}|f^{(k)}(t)|$. It is easy to see that in the polynomial space $\mathbb{P}[0,1]$ space, the norm $\|\|$ is well-defined.

What is the dimension of the completion of $(\mathbb{P}[0,1],\|\|)$ under this norm ?

It seems to be finite dimensional, because one can show that the unit ball in this new Banach space $X$ is compact as follows:

We only need to show that every bounded sequence $\{f_n\}$ in $X$ has a convergent subsequence. That is a repeated application of the Ascoli-Arzelà theorem and Cantor's diagonal argument; the boundedness of $\{f^{(k+1)}_n\}$ implies the equicontinuity of $\{ f^{(k)}_n\}$. For details of this argument, we can refer to

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

Since it is a normed space with Montel-Heine-Borel property, it has to be finite dimensional.

On the other hand, $\mathbb{P}[0,1]$ is already infinite dimensional. Then $X$ has to be at least infinite dimensional.

Something must be wrong.

references:

Every normed space has a completion?

Yuhang
  • 1,575
  • By Ascoli-Arzelà theorem, one can show that any bounded sequence in such space has a convergent subsequence. – Yuhang Nov 28 '17 at 13:26
  • Why must a space with the "Montel-Heine-Borel property" be finite-dimensional ? – Gabriel Romon Nov 28 '17 at 13:41
  • Because in such normed space, the unit ball is compact. – Yuhang Nov 28 '17 at 13:42
  • 2
    Convergence in this norm does not only mean that all derivatives converge uniformly, but also that they must all converge "uniformly fast". The map $p \mapsto (p, p', p'', \dotsc)$ is an isometric embedding of $(\mathbb{P}[0,1], \lVert\cdot\rVert)$ into $\ell^{\infty}\bigl(\mathbb{N}, C([0,1])\bigr)$. So the completion is the closure of its image in $\ell^{\infty}\bigl(\mathbb{N}, C([0,1])\bigr)$, which is $$\bigl{ F \in c_0\bigl(\mathbb{N}, C([0,1])\bigr) : (\forall n)(F_{n+1} = F_n')\bigr}.$$ – Daniel Fischer Nov 28 '17 at 14:28

1 Answers1

3

I think the problem is that the unit ball in $X$ is not compact. The Ascoli-Arzelà theorem and Cantor's diagonal argument are not strong enough to yields the convergent subsequence in $X$. Indeed, let us consider $f_n=\frac{x^n}{n!}$. On can show that $\|f_n\|=1$, and $\|f_n-f_m\|=1$ if $n\neq m$. Therefore, it does not have any convergent subsequence. However, one can find that for any $k$, it holds that $\|f_n\|_{C^k[0,1]}\to 0$ as $n\to\infty$.

Ice sea
  • 1,222