# Why $C^{\infty}[0,1]$ with natural topology can not be normalized.

Consider $$C^{\infty}[0,1]$$ space, where $$f_n \to f$$ iff all its derivatives $$f_n^{(k)}(x)$$ uniformly converge to $$f^{(k)}(x)$$. The question is prove that this space is metrizable, bot can not be normalized.

My attempt: it's not hard to see that using $$\{\sup_{[0,1]} |f^{(k)}(x) - g^{(k)}(x)|\}_{k\ge0}$$ we can construct a metric $$\rho$$, which replicate convergence in it's natural way. So now we need to find a sequence of function $$\{f_n\}$$ that converges in $$\rho$$ sense, but there is no norm for which this sequence also converges? The first idea is to consider something easily differentiable $$\{\exp(nx)\}_n$$. This sequence looks reasonable, but I don't know why this sequence doesn't converge in any norm.

Actually, it looks curious, that there is no appropriate norm to normalize this space.

• I think that this is a Fréchet space with the Heine Borel property (closed and bounded sets are compact). Indeed, by Ascoli Arzelá, a bounded set in C^k is precompact in C^{k-1}. Now, the only normal spaces with the Heine Borel property are finite dimensional, and the space in this question is clearly not. Just a sketch, see if it convinces you and do not trust my word. Sep 14 at 23:44
• There is no complete algebraic norm on the linear space $C^{\infty}[0, 1]$. Note that the operator of the derivative is a linear bounded operator on $C^{\infty}[0, 1]$ . Then there is some $0\lt M$ such that $||f'||\leq Mf$. But for $f(t)=e^{2Mt}$ we have $2M||f||=||f'||\leq M||f||$. That is a contradiction. Sep 15 at 10:46
• @alireza why should it be bounded. Maybe I miss something. Sep 15 at 12:15
• It is bounded by the closed graph theorem. Sep 16 at 17:57

HINT:

The space $$C^{\infty}[0,1]$$ cannot be normalized because it does not have any bounded neighborhoods of $$0$$. Indeed, every neighborhood of $$0$$ contains a set of the form $$\{ f \ | \ p_k(f) \le \delta_k, k = 0, \ldots n\}$$ and on such a set every seminorm $$p_{m}$$ with $$m>n$$ is unbounded. Indeed, let $$f$$ a function in $$C^n$$ but not in $$C^{n+1}$$, and $$f_m$$ a sequence in $$C^{\infty}$$ converging to $$f$$ in the $$C^n$$ topology. If we had an inequality $$p_{n+1}< C \max_{0\le k \le n} p_k$$, then the sequence $$f_m$$ being Cauchy in the $$C^{n}$$ topology, would also be Cauchy in the $$C^{n+1}$$ topology, so $$f_m$$ would converge to a function in $$C^{n+1}$$, contradiction.

$$\bf{Added}$$ @Jochen's idea: we can write down explicitly a function with first $$n$$ derivatives small, but $$p_{n+1}(f)$$ large, for instance $$f(x)= \frac{1}{M^{2n+1}} cos M^2 x$$, with $$M$$ large.

• Alternatively, you can write down exlicitly functions whose first $n$ derivatives are small but $f^{(n+1)}(0)$ is bigger than $C$. Try with $f(x)=a_n\exp(b_n x)$. Sep 15 at 8:47
• @Jochen f.e. $f_n(x) = exp^{nx} / n^{n-1}$. It's derivatives are bounded, but $n + 1$th derivative is unbounded for $n \to \infty$? At would mean that for this functions $\|f^{(n)}_n\|$ doesn't converges? Sep 15 at 13:50
• This shows that you cannot estimate $p_{n+1}$ by the maximum of $p_0,\ldots,p_n$. Sep 15 at 13:58
• @Jochen: Neat idea, added it! Sep 15 at 15:10

Another reason why $$C^\infty$$ is not normable is that it satisfies the Heine-Borel property, that is, every closed and bounded set is compact. (Note that $$C^\infty$$ is a metric space, as already observed in the question, so it makes sense to speak of "bounded sets").

The proof uses the theorem of Ascoli-Arzelá. Let $$f_n\in C^\infty$$ be a sequence such that $$\sup_{n, k}\lVert f_n\rVert_{C^k}< \infty.$$ Then, since the sequence is $$C^1$$ bounded, it has a $$C^0$$-converging subsequence. Such subsequence is $$C^2$$ bounded, thus it has a $$C^1$$-converging subsequence. Using the diagonal sequence trick (a.k.a. diagonal argument, Cantor argument, etc...) we can produce a subsequence of $$f_n$$ that is $$C^\infty$$ convergent. This proves that $$C^\infty$$ is Heine-Borel.

To conclude we observe that the only normed spaces that are Heine-Borel are the finite-dimensional ones, which is a standard theorem usually attributed to Riesz. Since $$C^\infty$$ is infinite-dimensional, it cannot be a normable space.

• This is the same reason for which $\mathcal{H}(\mathbb{D})$ (i.e. the set of holomorphic functions on the open unit disc) with the topology of compact uniform convergence is not normable Sep 16 at 19:23
• Right. For different reasons, but both spaces are Heine-Borel. Sep 16 at 20:07