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I've started reading about distributions in Folland's Real Analysis, and on page 282 the book says the following:

Recall that for $E \subset \mathbb{R}^n$ we have defined $C_c^{\infty}(E)$ to be the set of all $C^{\infty}$ functions whose support is compact and contained in $E$. If $U \subset \mathbb{R}^n$ is open, $C_c^{\infty}(U)$ is the union of the spaces $C_c^{\infty}(K)$ as $K$ ranges over all compact subsets of $U$. Each of the latter is a Frechet space with the topology defined by the norms $$ \phi \mapsto \|\partial^{\alpha} \phi \|_u \qquad (\alpha \in \{0,1,2,\ldots,\}^n), $$

in which a sequence $\{\phi_j \}$ converges iff $\partial^{\alpha} \phi_j \to \partial^{\alpha} \phi$ uniformly for all $\alpha$. (The completeness of $C_c^{\infty}(K)$ is easily proved by the argument in Exercise 9 in $\S 5.1$.)

From what I understand, the notion of completeness requires a norm. If that's the case, then what exactly is the norm with which we are equipping $C_c^{\infty}(K)$? Would it be

$$ \|\phi\| := \sum_{\alpha} \| \partial^{\alpha} \phi \|_u = \sum_{k=0}^{\infty} \sum_{|\alpha| = k} \| \partial^{\alpha} \phi \|_u \quad ? $$

Leonidas
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  • I don't think you can equip this space with a norm that makes it a Banach space, if that's what you are going for, see https://math.stackexchange.com/questions/1737864/can-c-infty-mathbbt-become-a-banach-space – DominikS Oct 23 '23 at 16:11
  • Note that you can define plenty of norms on the space, but it will never be complete w.r.t. those norms. – DominikS Oct 23 '23 at 16:12
  • Completeness does not require a norm. Look up what a Frechet space is. It’s a generalization of a Banach space. – Deane Oct 23 '23 at 16:13
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    Consider the seminorm $|x|k= \sum{|\alpha|=k} |\partial^\alpha x|$. The Fréchet topology is a metric space topology with norm $d(x,y)=\sum_{k\geq0} 2^{-k}\frac{|x-y|_k}{1+|x-y|_k}$. See e.g. https://en.wikipedia.org/wiki/Fr%C3%A9chet_space – Pablo Herrera Oct 23 '23 at 16:14
  • @DominikS: Folland seems to be asserting otherwise when he mentions "the completeness of $C_c^{\infty}(K)$"... – Leonidas Oct 23 '23 at 16:20
  • @Leonidas: As Deane mentioned, completeness does not require a norm. – DominikS Oct 23 '23 at 16:26
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    @DominikS: Ok, I think I sort of get it: completeness requires a metric, which we have, as pointed out by Pablo Herrera. – Leonidas Oct 23 '23 at 16:32
  • @PabloHerrera: Thanks for your comment! So when Folland says that $C_c^{\infty}(K)$ is "complete", he means: For any sequence ${\phi_j }{j=1}^{\infty} \subset C_c^{\infty}(K)$ which is Cauchy in the norm you gave, there exists some $\phi \in C_c^{\infty}(K)$ such that $\lim{j \to \infty} d(\phi_j,\phi) = 0$. Do I have that right? – Leonidas Oct 23 '23 at 16:42
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    Yes that's correct. – Pablo Herrera Oct 23 '23 at 16:52

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