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

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.

• Just get the first derivative, then use the definition on that one. =)
– Pedro
Jun 17, 2012 at 14:45
• @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... Jun 17, 2012 at 14:54
• @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? Jun 17, 2012 at 15:41
• 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? Jun 19, 2012 at 15:55
• @RagibZaman And I thought that MSE was the ONE place I could go to get away from political BS. I guess I was wrong. Jun 19, 2012 at 15:58

I'm not sure, but I think there are two problems with the formula you used to approximate $f''(x_0)$:

• it uses the same discretization step for the approximation of the first and second derivative (it's like computing one directional derivative for a function of two variables: it might exist, but that does not imply that the differential exists).

• it's a centered finite difference formula, which therefore vanishes for a function that is odd around $x_0$ (or even gives infinity if $f$ is odd but $f(x_0)\neq0$, in which case $f$ is for sure discontinuous).

But I think the idea would work if the increments used in the approximation of the first and second derivative were different and the discretization formulas were not centered. Namely

$$f''(x)\simeq \frac{f'(x+h)-f'(x)}{h}$$

and then

$$f'(x+h)\simeq \frac{f(x+h+k)-f(x+h)}{k}$$ $$f'(x)\simeq \frac{f(x+k)-f(x)}{k}$$ which gives

$$f''(x)\simeq \frac{\dfrac{f(x+h+k)-f(x+h)}{k}-\dfrac{f(x+k)-f(x)}{k}}{h}=$$

$$f''(x)\simeq \frac{f(x+h+k)-f(x+h)-f(x+k)+f(x)}{hk}$$

Now, for the example reported in the link you gave, this formula does not give a finite result as $h,k$ go to $0$ independently.

To summarize, I would say that $f''$ exists and it's equal to $f''(x)$ if

$$\forall \varepsilon>0, \exists \delta>0 : \forall \underline{h}\in\mathbb{R}^2\cap \mathcal{B}(\underline{0},\delta)\setminus\underline{0}$$ $$\left|\frac{f(x+h_1+h_2)-f(x+h_1)-f(x+h_2)+f(x)}{h_1h_2}-f''(x)\right|<\varepsilon.$$

I'm not $100\%$ sure of this statement (in particular of the fact that the two increments have to independent), but it looks right to me. For sure you need not centered schemes though.

• The centered difference formula is not a problem: for the first derivative to exist it can't be odd around our point $x_0$ unless it is zero. But you are right that $h$ and $k$ need to be able to vary independently. The counterexample OP cites will fail in this case. Jun 19, 2012 at 18:08
• I think that if you use centered schemes, even if $h$ and $k$ are independent, the counterexample that tampis posted still holds (if I did my calculations correctly). But I will double check it. Jun 19, 2012 at 21:26
• I'm starting to believe that my definition of $f''$ in terms of $f$ is too restrictive. Indeed, if the condition above is satisfied $\forall\underline{h}\in\mathcal{B}(\underline{0},\delta)\setminus\underline{0}$, taking the limit as $h_1$ goes to zero, we have the definition of $f''$ as the limit of the increment quotient of $f'$. So if that definition is satisfied, then we are fine. However, it might be too restrictive. Indeed, $f''$ is defined as the limit of the increment quotient of $f'$, that is when '$h_1$ has been already sent to zero'...in some sense. Jun 19, 2012 at 21:36
• The increments $h_1$ and $h_2$ do have to be independent; otherwise you just reduce to $\lim_{h\to 0} (f(x+2h)-2f(x+h)+f(x)/h^2$, which is not sufficient for the second derivative to exist; consider e.g. $f(x) = x^3\sin(1/x)$ (with $f(0)=0$). May 13, 2014 at 16:57
• However, even with independent increments, this definition does not even imply that $f$ is continuous at $x$, as pointed out by Tom Goodwillie in his answer to mathoverflow.net/questions/165704/… --- if $f$ is a $\mathbb{Q}$-linear map $\mathbb{R}\to \mathbb{R}$, then the "second-order difference quotient" is identically zero. May 13, 2014 at 16:59

If we are not allowed to talk about $f'(x)$ for $x\ne x_0$ it is not possible to talk about $f''(x_0)$ in the proper sense. One could, however, approach the idea of $f''(x_0)$ via the Taylor expansion of $f$ at $x_0$:

The function $f$, defined in a neighborhood of $x_0$ has second derivative $b$ at $x_0$ if there is an $a\in{\mathbb R}$ such that $$\lim_{h\to 0}{f(x_0+h)-f(x_0)- a h \over h^2}={b\over2}\ .$$ This $\lim$-condition can obviously be expanded into $\epsilon$-$\delta$-language.

Note, however, that the function $f(x):=x^3$ $(x\in{\mathbb Q})$ and $:=0$ $(x\notin{\mathbb Q})$ would have $f''(0)=0$ according to this definition.

• Indeed, this defines what is known as "pointwise second derivative" , which exists more often than $f''(x)$. However, I question why you say "not possible". @bartgol suggested a plausible definition in terms of a limit in $\mathbb R^2$. So far we have no proof that it works, but no counterexample either.
– user31373
Jun 19, 2012 at 16:48