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Is there an epsilon-delta definition for the second derivative?

I know that there is such a definition for the first derivate $f'(x)$ which can be derived from the limit $f'(x) = \lim_{y\rightarrow x} \frac{f(y)-f(x)}{y-x}$ for a function $f:D\rightarrow \mathbb{R}$:

$$\forall \epsilon > 0\, \exists \delta > 0\, \forall y \in D\setminus \{x\}:|y-x|<\delta \Rightarrow \left|\frac{f(y)-f(x)}{y-x}-f'(x)\right|<\epsilon$$

So $f'(x)$ can be described as the number which fulfills the above statement. Is there a similar statement for the second derivative?

Update: This MSE thread shows that there are different definitions for the derivative (and thus for the second derivative). So I want to make my question more concrete:

My definition of derivation: Let be $f:D\rightarrow\mathbb{R}$ with $D\subseteq\mathbb{R}$ arbitrary. Let $D^*$ be the set off all points $x\in D$ for which there is at least one sequence $(x_n)$ in $D\setminus\{x\}$ with $\lim_{n\rightarrow\infty} x_n=x$. I define the limit $\lim_{y\rightarrow x\ ,y\in D\setminus\{x\}} {f(y)-f(x) \over y-x}$ as the first derivation for a given $x\in D^*$ (if the limit exists).

My definition of the second derivative: Let be $f:D\rightarrow\mathbb{R}$ with $D\subseteq\mathbb{R}$ arbitrary. We call $f''(x)$ the second derivative if there exists an open interval $x\in O\subseteq \mathbb{R}$ so that $f$ is differentiable on $O\cap D$ and $f''(x)$ is the first derivative of the function $f': (O\cap D)\rightarrow\mathbb{R}:x\mapsto f'(x)$ at the point $x$ (which also means that $x\in(O\cap D)^*$).

My question: Is there a statement $\forall \epsilon > 0: \exists \delta > 0: A(\epsilon, \delta, f, x, c)$ for $f:D\rightarrow \mathbb{R}$ ($D\subseteq \mathbb{R}$) and $c,x\in\mathbb{R}$ which is equivalent to the statement that $f$ is differentiable on a set $x\in O\cap D$ where $O$ is an open interval and that $c$ is the second derivative of $f$ at $x$?

I also will accept answers where you need more restrictions to the question. For example you might want to use the value of the first derivative $f'(x)$ (at the same point where you want to define the second derivative) in your statement or you want to restrict $f$ on functions with open domains or domains which are intervals. In this case I will accept your answer and open a new thread asking for a more general solution.

Please notice that there is a community wiki post where I want to collect all the progress we made so far.

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    Just get the first derivative, then use the definition on that one. =)2012-06-17
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    @PeterTamaroff Yeah, this would be fine. Is there a epsilon-delta definition without using the first derivative?2012-06-17
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    @tampis It may be hard/unnatural considering the second derivative is *defined* as the derivative of the derivative. Perhaps one could nest one epsilon-delta definition into another but that is the kind of thing they only do to prisoners at Guantanamo Bay...2012-06-17
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    I don't have the time to try it, but I wonder if the [double mean value theorem](http://www.math.ntnu.no/~hanche/blog/doubly-mean.pdf) (pdf) could be used to make such a definition?2012-06-17
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    @tampis: You must use at least the _value_ of $f'(x)$. Otherwise how do you distinguish between functions with the same second derivative but not the same first derivative?2012-06-17
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    In order that the second derivative exist at $x$, it is necessary that the first derivative exist in an entire neighborhood of $x$. How can an epsilon-delta statement ensure that?2012-06-19
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    @RagibZaman And I thought that MSE was the ONE place I could go to get away from political BS. I guess I was wrong.2012-06-19
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    @tampis I don't understand the last example. How is the function defined on $\mathbb R\setminus (A\cup B)$? Why you do not consider all pairs $h,k\in \mathbb R$? For example, take a fixed $k\in A$ and a very small $h\notin A$. Then the difference quotient is $k^3/(kh)\to\infty$ as $h\to 0$.2012-06-22
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    @LeonidKovalev That's the main point. If I would define $f$ for all $h,k\in\mathbb{R}$ the counterexample wouldn't work as you mentioned. Because the function is just defined on $A \cup B$ there is no possibility to fix an $k\in A$ and take a very small $h\notin A$ so that $k+h\in A \cup B$. If $k+h\in A$ then we have automatically $k\in A$ and $h\in A$. Because every point $x$ of $A\cup B$ is a limit point of $A\cup B\setminus \{x\}$ the concept if derivation is well defined for $f$.2012-06-23
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    @tampis. Interesting, but I think your $f$ is differentiable at $x=0$. Indeed, both the expression for the 'double increment quotient' are approaching $0$ as $|(h,k)|\to0$. Yes, $f$ has lots of discontinuity jumps around $0$, but that doesn't matter. Also the function $f(x)=x^3\chi_{\mathbb{Q}}$ is discontinuous everywhere outside the origin, but it's two times differentiable at $x=0$...2012-06-23
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    @bartgol You are right: $f$ is differentiable at $x=0$, but just in $x=0$ and therefore in no neighborhood of $x=0$. So the first derivative just exists in $x=0$ and is thus the function $f':\{0\}\rightarrow \mathbb{R}:0\mapsto 0$. Because there is no derivative defined for a function which is just defined in one point, the second derivative does not exists (which is defined as the derivative of the first derivative). What you mean is the concept of "pointwise second derivative" which Leonid Kovalev already mentioned in a comment to Christian's answer.2012-06-23
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    If a function is not defined on an open interval containing $0$, it automatically fails the definition given by @bartgol. The definition says, in part ... $\forall \underline{h}\in\mathbb{R}^2\cap \mathcal{B}(\underline{0},\delta)\setminus\underline{0}$ (some inequality holds). You can't claim that an inequality holds when its left-hand side is not even defined.2012-06-23
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    @LeonidKovalev I read the definition $\forall \underline{h}\in\mathbb{R}^2\cap \mathcal{B}(\underline{0},\delta)\setminus\{\underline{0}\} : A(\underline{h}, \delta, \epsilon)$ as $\forall \underline{h}\in(D^2\cap \mathcal{B}(\underline{0},\delta))\setminus\{\underline{0}\} : x+h_1+h_2 \in D \Rightarrow A(\underline{h}, \delta, \epsilon)$ where $D$ is the domain of the function. So somehow your are right. IMHO we have a new question: What is the expression of bartgol's approach for functions which are not defined on an open interval containing the considered point $x$?2012-06-23
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    @tampis I don't think there is an established concept of 2nd derivative at such a point. In fact, I'd say that the definition of 1st derivative that you stated in the question is not standard. IMHO it's overly permissive; there is a related [MSE thread](http://math.stackexchange.com/questions/161910/most-general-a-subseteq-mathbb-r-to-define-derivative-of-f-a-to-mathbb-r)2012-06-23
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    @LeonidKovalev I have concreted my question. Now it's also okay to give an answer in the case where the function has to have an open domain. I also cleared thinks up in the community wiki post. I made clear that bartgol's idea still might work for functions defined on open subsets of $\mathbb{R}$.2012-06-24
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    @LeonidKovalev: of course, if the function is not well definied in a neighborhood of the point where you compute the limit, then it does not make sense to compute the limit. However, I don't see how this applies to the counterexample you showed. Can you provide a function $f$ and a point $x_0$ such that my definition would give a wrong answer for the second derivative? I really can't see where it fails.2012-07-01
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    @bartgol I did not say that your definition fails. I wrote that "a function fails the definition", meaning it does not satisfy the definition. I'm pretty sure that your definition is a correct characterization of the second derivative. Someone should write a proof of that; someone with more time than I have at the moment.2012-07-01
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    I think you need to be more explicit about what sort of statement you want $A(\epsilon,\delta,f,x,c)$ to be. Obviously you don't want it to be something like "$f$ is twice differentiable at $x$ and $f''(x)=c$." But it has to be allowed to use quantifiers, so I'm not sure how to exclude that.2014-05-13

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