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.

Continuity is an intuitive concept. I will not dwell on the precise definitions of continuity and the rest here. Note that differentiability is a more restrictive condition than continuity, while analyticity for complex-valued functions is even more restrictive than differentiability.

To some extent, I understand the motivation behind defining these terms as they are defined right now. My question is: what is the next condition in the sequence

continuity, differentiability(scalar/vector/left-right/partial:all), analyticity $\cdots$?

Does there exist a next term in this sequence? If yes, in what context? if no, what is the reason? are all possible restrictions on functions' behavior covered in some sense? As one goes on in higher dimensions, is there some behavior that prompts any further restriction, so to speak?

PS: I am talking in very general terms, with their usual connotations.

share|improve this question
    
Complex differentiability implies analyticity (for functions $\mathbb C \to \mathbb C$. –  lhf Jan 6 '12 at 20:32
6  
In contexts where they make sense, the sequence may continue: polynomial, affine/linear, constant. –  Jonas Meyer Jan 6 '12 at 20:32
    
@lhf hmm I see. Then I would like to know whether a condition more restrictive than complex differentiability-this and nothing else :) JonasMeyer Sorry I didn't get you. –  Nikhil Bellarykar Jan 6 '12 at 20:36
2  
Nikhil: Polynomials are a very special subclass of analytic functions, affine or linear functions (e.g. $z\mapsto az+b$ on $\mathbb C$) are a very special subclass of polynomial functions, constant functions are a very special subclass of these. –  Jonas Meyer Jan 6 '12 at 20:38
3  
@Nikhil: It isn't clear to me precisely what the difference is. A function satisfies the condition of being analytic if and only if it is an analytic function. Is the property of being analytic separate from talking about the set of analytic functions? In what way is this different from talking about the condition of being a polynomial function? There are contexts in which the class of polynomial maps is of great importance, e.g. in algebraic geometry. –  Jonas Meyer Jan 6 '12 at 20:46

4 Answers 4

up vote 5 down vote accepted

No, at least not ones I consider analogous, but you've skipped (infinitely) many steps. All these properties of functions are examples of what is called the differentiability class of a function. For example, a continuous function is of class $C^0$, and a differentiable function is of class $C^1$. More generally, a function is of class $C^k$ if it has continuous $k^{th}$ derivatives. This extends to $C^\infty$, the class of functions with continuous $k^{th}$ derivatives for all $k\geq 0$, i.e. the smooth functions, and analytic functions are said to be of class $C^\omega$.

share|improve this answer
    
Thanks for pointing out the intermediaries. But the question still remains:- does there exist anything beyond $C^\omega$ in the same vein or asking this is superflous? –  Nikhil Bellarykar Jan 6 '12 at 20:39
    
Like I said, there are no more that I consider analogous, but the question is somewhat subjective. In fact, the class $C^\omega$ is of somewhat different character than the rest (note that it is strictly contained in $C^\infty$). However, you certainly can't go any further without significant modification because after the ordinal $\omega$ you run out of derivatives to take (as there are only $\omega$ derivatives). –  Alex Becker Jan 6 '12 at 20:42
    
hmm I see. Thanks for the clarification. –  Nikhil Bellarykar Jan 6 '12 at 20:45

I think Jonas Meyer's comment deserves to be an answer. The progression in the question can be looked at many ways depending on context, but one is as increasing levels of rigidity:

Suppose $f:\mathbb{C}\rightarrow\mathbb{C}$ is a function.

If $f$ is continuous, then $f$'s values in an open set are determined by its values on a dense subset of the set. (But outside of the open set, a lot of different things could be going on.)

If $f$ is differentiable as a function $\mathbb{R}^2\rightarrow\mathbb{R}^2$, then we can deduce more about it from less information about its values. The more times continuously differentiable, the easier it is to deduce things about $f$. (But it is still true that we have nearly no knowledge about what it's doing outside of whatever open set we know something about.)

If $f$ is analytic, then its values in the entire plane are determined by its values on any set that has an accumulation point. For example, its values on any dense subset of any nonempty open set (no matter how small) determine its values everywhere.

In this light, polynomial - linear - constant is a natural continuation of the sequence. Each step represents a class of functions such that the amount of information needed about $f$'s values in order for $f$ to be determined completely, is less than the previous.

If $f$ is polynomial, then its values in the entire plane are determined by its values at any infinite set of points, whether the set has an accumulation point or no. In fact, by a finite set of points, but without further information we don't know how many; but if we have a bound on the degree of the polynomial, for example if we know a polynomial bound on its growth rate, then we can say $f$ is determined by its value at a specific (finite) number of points.

If $f$ is linear, then its values everywhere are determined by its values at 2 points.

If $f$ is constant, evaluation at one point suffices to know everything about $f$.

share|improve this answer
    
so you say that the 'rigidity' enables one to know more about the function based on lesser information. Interesting. But don't you think that this hierarchy is 'somewhat too specific'? what I want to say is that we can have complicated functions e.g. exponentials and whatnot satisfying ever so restrictive criteria, so why restrict the 'determining' set? Although what you say would also be a valid progression, no doubt about that. –  Nikhil Bellarykar Jan 6 '12 at 21:34
    
Can you rephrase your question? I'm having trouble making sense of it. –  Ben Blum-Smith Jan 6 '12 at 22:12
    
hmm the comment doesnt really have a question. what I wanted to say is that you are restricting the set of points, which determine the value of function whereas I am talking about restricting the behaviour of a function-both are different-thats all. I agree that the progression according to your definition is also valid. –  Nikhil Bellarykar Jan 6 '12 at 22:19
    
Ah, thanks for the clarification. Like the conv. with Jonas Meyer in the comments above, I think there's a connection here you may be missing. The restriction on the function's behavior is exactly the thing that allows the function to be determined from a smaller data set. –  Ben Blum-Smith Jan 6 '12 at 22:38
    
ok, but aren't there functions whose values cant be determined from a smaller data set but are analytic, exponential for example? –  Nikhil Bellarykar Jan 7 '12 at 6:28

For complex functions, there is a finer hierarchy for entire functions based on order of growth.

share|improve this answer
    
by 'finer hierarchy' do you mean the classes $C^0,C^1,\cdots C^\infty$ given by Alex becker? –  Nikhil Bellarykar Jan 6 '12 at 20:43
    
No, I mean nested subclasses of entire functions based on the order of growth. For instance, entire functions that have polynomial growth are actually polynomials. But this is not something beyond analyticity... –  lhf Jan 6 '12 at 20:45
    
I see. Thanks for clarification. –  Nikhil Bellarykar Jan 6 '12 at 20:46

In a conversation today my advisor mentioned Gevrey class, and I immediately remembered this question. It turns out that functions of this class are always $C^\infty$, but may be non-analytic; conversely, there are $C^\infty$ functions which are not Gevrey. See here: http://en.wikipedia.org/wiki/Gevrey_class

share|improve this answer
    
this is interesting. –  Nikhil Bellarykar Jan 25 '12 at 10:01

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.