5

Let $Y$ denote the space of smooth functions $[0,1]\to\mathbb{R}$, equipped with the sup norm (i.e., topologized as a subspace of the usual space $C([0,1])$ of continuous functions on $[0,1]$). Is $Y$ $\sigma$-compact?

(Note that it follows from the Arzela-Ascoli theorem that for any $M$, the subset of $Y$ consisting of functions $f$ such that $|f(x)|\leq M$ and $|f'(x)|\leq M$ everywhere is precompact as a subset of $C([0,1])$, so $Y$ is $\sigma$-precompact as a subset of $C([0,1])$. However, these subsets are not closed, so they are not compact, and I don't see any way to construct "large" subsets of $Y$ that are compact. A negative answer to this question would complete my answer to another question.)

Community
  • 1
Eric Wofsey
  • 342,377

1 Answers1

3

Consider the topology $\tau$ of uniform convergence of all derivatives on $C^\infty=C^\infty([0,1])$ which is its natural Frechet space topology. If $C^\infty=\bigcup K_n$ with compact (hence closed) subsets of $C([0,1])$ Baire's theorem implies that there is one $K_n$ containing $\tau$-interior points which can be assumed to be $0$. Hence there are $m\in\mathbb N$ and $r>0$ such that $$B=\lbrace f\in C^\infty: |f^{(k)}(x)|\le r \text{ for all $x\in [0,1]$ and $k\le m$}\rbrace \subseteq K_n.$$

In particular, $\overline B^{C([0,1])} \subseteq K_n \subseteq C^\infty$. It should be quite elementary to see that this is wrong, e.g. by taking a function $f\in C^m\setminus C^{m+1}$ so that its $m$-th derivative is small, approximating this $m$-th derivative by a polynomial and integrating $m$ times to approximate $f$ by a polynomial whose first $m$ derivatives are small.

Jochen
  • 13,235
  • By a similar argument, every $\sigma$-compact Baire space is locally compact. But a Hausdorff topological vector space is locally compact if and only if it is finite-dimensional. But this argument crucially relies on the Baire property and does not apply to Eric's question. – MaoWao Nov 17 '15 at 13:15