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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.

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    Complex differentiability implies analyticity (for functions $\mathbb C \to \mathbb C$.2012-01-06
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    In contexts where they make sense, the sequence may continue: polynomial, affine/linear, constant.2012-01-06
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    @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.2012-01-06
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    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.2012-01-06
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    oh I see. But I am asking about conditions to be satisfied by the functions, not functions themselves.2012-01-06
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    @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.2012-01-06
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    @JonasMeyer and Srivatsan I agree. Its just that you did not talk in terms of 'analyticity' etc. and thus I was confused. The mapping function of a type<-->property satisfied by it didn't occur to me then, hence the confusion, which is cleared now :)2012-01-06
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    I like this question a lot, and I intuitively (I think!) understand the question, but formally - not so well. For example, we can talk about conditions which are somehow stronger than analyticity, like harmonicity, for example. But is it in some sense "smoother" than an analytic function? No. In fact, harmonicity is not a "smoothness" criterion (in the sense that analyticity is built in, and additional properties are algebraic). So, my question is: in what sense do you wish to continue the sequence?2012-01-07
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    Harmonicity you say...hmm yes, that could be a direction to continue, kind of close to what I vaguely have in mind.2012-01-07

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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

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    this is interesting.2012-01-25
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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 $f$.

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    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.2012-01-06
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    Can you rephrase your question? I'm having trouble making sense of it.2012-01-06
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    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.2012-01-06
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    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.2012-01-06
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    ok, but aren't there functions whose values cant be determined from a smaller data set but are analytic, exponential for example?2012-01-07
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    @NikhilBellarykar: No finite set of points determines a function among the class of analytic functions, or even among polynomials unless you restrict the degree. If we are talking about functions in one (real or complex) variable for example, then if $n$ values are given, there is a unique polynomial of degree less than $n$ having those values (en.wikipedia.org/wiki/Lagrange_polynomial), but there are infinitely many analytic functions (even polynomials) having those values. E.g., there exists a degree 99 polynomial function $f$ such that $f(n)=e^n$ for $n\in\{1,2,3,\ldots,100\}$.2012-01-07
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    @Nikhil: (sorry that last comment didn't answer your comment) Every complex analytic function on (an open subset of) $\mathbb C$ is determined by its values on any set having a limit point in the domain. E.g., if $f$ is an analytic function on $\mathbb C$ such that $f\left(\frac{1}{n}\right)=\sqrt[n]{e}$ for each positive integer $n$, then $f(z)=e^z$ for all $z\in\mathbb C$.2012-01-07
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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 $C^\omega$.

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    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?2012-01-06
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    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).2012-01-06
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    hmm I see. Thanks for the clarification.2012-01-06
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For complex functions, there is a finer hierarchy for entire functions based on order of growth.

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    by 'finer hierarchy' do you mean the classes $C^0,C^1,\cdots C^\infty$ given by Alex becker?2012-01-06
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    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...2012-01-06
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    I see. Thanks for clarification.2012-01-06