# Why don't we define the class of $C^{\infty}$ in this way?

$C^{\infty}$ is defined to be the class of functions which have all orders of derivative. But as a convention, as far as the infinity is concerned, we always refer to limit. So why don't consider the function and all orders of it's derivative as a sequence of funtions, define a metric and then let then converge. Thus we all functions on R with such convergence belong to class $C^{\infty}$.

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Don't you need to assume that you have derivatives of all orders of your function in order to be able to speak of the (infinite) sequence of $f$ and its derivatives? So I don't follow how you want to define your metric without defining $C^\infty$ in the usual way. –  t.b. Dec 31 '11 at 1:52
Can those who down-vote questions please specify why they down-voted them. As obvious as it may be to you, it may or may not be apparent to the user, and he/she can make more sense out of these down-votes. –  r.g. Dec 31 '11 at 2:36
What metric? I don't understand the question. –  Jonas Meyer Dec 31 '11 at 3:31
I agree with r.g. I for one didn't downvote, but I would like to see the question clarified before I would upvote. –  Jonas Meyer Dec 31 '11 at 3:43
'I protest against the use of infinite magnitude as something completed, which is never permissible in mathematics. Infinity is merely a way of speaking, the true meaning being a limit which certain ratios approach indefinitely close, while others are permitted to increase without restriction.' -Gauss –  Matt Calhoun Dec 31 '11 at 4:06

$C^{\infty}$ can be seen as a limit of nested sets: $$C^{\infty} = \bigcap_{k=0}^{\infty}C^k=\lim_{n\to\infty} \bigcap _{k=0}^{n}C^k=\lim_{n\to\infty} C^n$$

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but this "limit" is not in the sense of any metric right? –  user12014 Dec 31 '11 at 2:49
@PZZ, no, not any natural matric I can think of at the moment. But it can probably be phrased in the language of nets. The natural place to study convergence is topological spaces, not metric ones. –  lhf Dec 31 '11 at 2:52
If we identify sets of functions with their indicator functions, then the limit here is a limit in the product topology on $\{0,1\}^{(\mathbb R^{\mathbb R})}$. –  Henning Makholm Dec 31 '11 at 3:13
I do believe that $C^\infty(\Omega)$ can also be realized as the inverse limit of the system $$\dotsb \hookrightarrow C^k(\Omega) \hookrightarrow C^{k-1}(\Omega) \hookrightarrow \dotsb \hookrightarrow C^0(\Omega)$$ where the projection morphisms are the inclusions $C^i(\Omega) \hookrightarrow C^j(\Omega)$ for $i\geq j$ and convergence in $C^k(\Omega)$ is uniform convergence of sequences and their first $k$ derivatives on compact subsets of $\Omega \subseteq \mathbb R^n$. –  kahen Dec 31 '11 at 3:36

$\infty$ does not "always refer to limit". This is a use of infinity more as a set.

• $C=C^0$ is the set of functions whose $0^{th}$-derivative is continuous (i.e. continuous functions).

• $C^1$ is the set of functions whose $1^{st}$-derivative is continuous (and thus also its $0^{th}$-derivative is continuous).

• $C^2$ is the set of functions whose $2^{nd}$-derivative is continuous (and thus also its $0^{th}$ and $1^{st}$-derivatives are continuous as well).

• $C^k$ is the set of functions whose $k^{th}$-derivative is continuous. Equivalently this is the set of functions which have continuous derivatives of orders $0,1,\dots,k$.

So naturally if we wish to denote the set of functions whose derivatives of all orders (i.e. $0,1,2,\dots$) exist and are continuous, why not $C^\infty$?

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Consider the smooth function $f(x)=\frac{1}{x^2}$ over $]0,1[$. Denote its $n$-th derivative as $f_n$. Then we have $f_n(x)\rightarrow\infty$ as $n\rightarrow\infty$, with $x$ fixed. So there is no such convergence as you said.

Indeed, class of $C^k$ is a description of how much smooth a function is, rather than the convergence of a series of functions.

But what you said has been broadly and deeply studied in functional analysis. Sometimes, a space of functions having some metric structure is suitable for solving Partial Differential Equations, for example you can refer to Soblev Spaces and $L^2$ theory in second-order PDE.

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