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Latest EDIT : This question in its latest form has been solved here

I want to know the existence of a class of functions with the following properties. If they exist, how to construct them ? In case they do not exist, please suggest a proof.

Let $ f : \mathbb{R} \to \mathbb{R} $ be a differentiable function with compact support in [0,1]. Let the notation $ Cf(x) = k$ represent that $ f $ is exactly $ k $ times differentiable at the point $ x \in \mathbb{R} $ where $ k \in \mathbb{N} $. $ f $ should satisfy the following properties.

$ Cf(x) >= q $ $\forall x \in \mathbb{Q}\cap (0,1] $ where $ q \in \mathbb{N} $ is denominator of $ x $ when written in simplest form. $ Cf(0) >= 1 $. $ Cf(x) = \infty \forall x \in (\mathbb{R}\setminus\mathbb{Q})\cap[0,1] $. The last statement means that $ f(x) $ is smooth at every $ x\in (\mathbb{R}\setminus\mathbb{Q})\cap[0,1] $.

I want to study such functions (in case they exist) as they seem to have some very nice properties.Also they might be the most well behaved after smooth functions.

EDIT :

Clarification : $f(x)$ is exactly $k$ times differentiable at $x$ means it is not differentiable more than $k$ times at $x$ .

EDIT 2 :

Clarification : $Cf(x)$ is always finite for $\forall x \in Q\cap (0,1] $

EDIT 3 :

One approach i tried is as given in answer/coments to this question question.

I added such functions starting at every rational point to make up the function and varying the scale of the bump functions $\epsilon$. Making $\epsilon \to 0$ would give a limit function (by making some restrictions) which does not have the desired properties.

EDIT 4 : I would like to know whether you find it interesting due to any reasons or not much due to some obvious reasons. Please comment.

To me it appears very beautiful and interesting concept..if i am allowed to make such statements.

EDIT 5 edited on Feb 12 2011.

I have constructed such a class of functions which happen to be nonempty.More over it has many interesting and important properties.Please contact me if you are interested in knowing more details and the proof.

EDIT : 6

@chandok : thank you for the clarification. I think i should mention something after having a re look at your answer. When i was asking the question, I wanted to rule out a particular condition which seemed not of interest, but later due to my misinterpretation/mistaken view of Darboux Theorem, I thought such a condition would not arise, but strikingly your answer is exactly utilizing such a condition and only after looking at your answer i came to know that such a condition is indeed possible and hence I realize that i made a mistake by not ruling out such a condition in the question itself. Now i would like to mention that condition in the question, but i want you to note that i give full credit to your answer as it brought to notice that such a condition is indeed possible, even though it is not very interesting.

New Condition

when ever $f(x)$ is not differentiable more than $n$ times at a point is is because for the $(n+1)^{th}$ derivative, the left and right limits exist and are not equal. The left and right limits do not diverge.

Keeping the above condition in mind i have constructed such a class of functions and it indeed is not empty and i think it has very important and interesting properties.If interested please contact me for details and the proof. You are also welcome to make any comments.

I once again thank Chandok for taking patience in providing an answer.

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3  
That's a very weird set of conditions... –  Mariano Suárez-Alvarez Jan 22 '11 at 19:47
2  
Not to be too picky but... The class of functions you are considering surely exists: what you want to know is if it is empty or not :) –  Mariano Suárez-Alvarez Jan 22 '11 at 19:58
3  
Can you tell us what nice properties do functions in this class have? –  Mariano Suárez-Alvarez Jan 22 '11 at 20:03
2  
@Rajesh: I would like to know what properties you are referring to, too. –  Jonas Meyer Jan 22 '11 at 20:24
9  
@Rajesh: it is usually best if you add all this information about your question to the question itself, not relegating it to comments. –  Mariano Suárez-Alvarez Jan 22 '11 at 20:44
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1 Answer

up vote 2 down vote accepted

Being $C^n$ is a local property so it's a bit unusual to say $C^n$ "at a point"

Pick your favourite family of periodic functions that are not $C^{n+1}$ on $\frac{1}{n}\mathbb{Z}$, but are $C^{n}$ everywhere. Mine happens to be the family $f_n(x) = (\sin(n \pi x))^{n+1/3}$.

Take note of how big the derivatives of $f_n$ can get for large n. Here, $||f_n^{(k)}||_\infty \leq (n+1)^{2k} $ for $n \ge k$

Now sum these functions with an appropriate weight such that their sum as well as the sum of all their derivative converges uniformly : $$F(x) = \Sigma_{n \ge 1} f_n(x) e^{-n} $$ Now you can prove that for all n,$F$ is $C^n$ exactly on the open set $\mathbb{R} - \{\frac{k}{d}, k \in \mathbb{Z}, 1 \le d \lt n\}$

It is a funny example of a function being $C^n$ for all n around irrational points, but nowhere $C^\infty$. That said, I don't expect it to be anymore interesting than from having this property.

Edit : I should clarify why being $C^k$ should be a local property :

http://en.wikipedia.org/wiki/Smooth_function#Differentiability_classes

When talking about differentiability, we want talk about the behaviour of the function around a point. If $f$ is $C^k$ "at" a point x, we really want to say that $f$ is well-behaved around that point. It is a property of the behaviour of $f$ around that point, and not a property of what the value of $f$ is at that point :

If $f$ is $C^k$ "at" a point, it means that the $k$th derivative of $f$ exists on a neighboorhood of $x$, or else we can't talk about it being continuous at $k$. Then, yes, there is a difference between "being continuous at $x$" and "being continuous on a neighboorhood of $x$". The definition on that wikipedia page uses the latter, while your $Cf(x)$ may use the former.

Here in my construction, the difference doesn't appear (until discussing if $F$ is $C^\infty$ somewhere). Whichever definition you choose, $F$ is $C^n$ "at" the same points. Or rather, $F$ is $C^n$ on the open subset that you wanted it to.

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The differentiability classes make sense on open sets : A function $f$ is $C^k$ on an open set $D$ if $f$ is $k$ times differentiable on $D$ and $f^{(k)}$ is continuous. It is $C^\infty$ on $D$ if it is $C^k$ on $D$ for every $k$. Here, $f_n$ is $C^n$ on $\mathbb{R}$, and $C^\infty$ on $\mathbb{R} - \frac{1}{n}\mathbb{Z}$, which is an open set. But since $\mathbb{Q}$ is dense, there is no open set on which we can argue that the serie should be $C^\infty$ –  mercio Jan 27 '11 at 14:11
    
@Rajesh D : $f_n$ is $\frac{2}{n}$ periodic, and the point is to have $f_n^{(n)}$ not differentiable at the points in $\frac{1}{n}\mathbb{Z}$, while keeping $f_n$ relatively simple. I tried to address your other comment properly in the answer. –  mercio Jan 27 '11 at 18:40
    
@Rajesh D : $(\sin(n\pi x))^{n+1/3}$ is not smooth because its (n+1)th derivative introduces a term of the form $(\sin(n\pi x))^{-2/3}$, which diverges when $\sin(n \pi x) = 0$. (if it is not clear what $x^{-2/3}$ is for negative x, please note that it is consistent with the usual derivation rules to define $x^{a/b}$ as the unique real bth root of $x^a$ for all real x, integer a, and odd positive integer b) –  mercio Feb 12 '11 at 13:14
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