Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$C^{\infty}$ 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 $C^{\infty}$.

share|improve this question
2  
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
4  
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
1  
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

3 Answers 3

up vote 3 down vote accepted

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

share|improve this answer
1  
but this "limit" is not in the sense of any metric right? –  user12014 Dec 31 '11 at 2:49
1  
@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
1  
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
1  
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 $C^\infty$?

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.