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

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

In this community wiki post I want to collect all progress we made so far in answering this question. Please feel free to edit it and to extend it with your ideas (you can also start a new community wiki post if you have a new approaches)

Notes

• To proof the differentiability of the function in a neighborhood of $x$ one might find the following article interesting: How to proof the convergence of functions without knowing the limit

Some approaches

• I know the formula $f''(x) = \lim_{h\rightarrow 0} \frac{f(x+h)-2f(x)+f(x-h)}{h^2}$ but this limit does not provide the existence of the second derivative (see section “limit” of wikipedia article “second derivative”). So we cannot derive an epsilon-delta definition from the limit $\lim_{h\rightarrow 0} \frac{f(x+h)-2f(x)+f(x-h)}{h^2}$.

• There is the idea to apply the above epsilon-delta definition to the first derivative:

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

  But then we already used the existence of the first derivative in our definition. To be a definition we shall conclude from it, that the function is differentiable in a neighborhood of $x$ (This might be hard, I know ;-) ).

• There is the approach by Christian Blatter to define the second derivative from the taylor series $f(x_0+h)=f(x_0)+f'(x)h+\tfrac 12 f''(x) h^2$, so that it should be the limit

  $$\lim_{h\to 0}2\cdot {f(x_0+h)-f(x_0)- a h \over h^2}=b$$

  whereby $a$ shall be the unique number for which the above limit exists. Unfortunately the function $f(x) := \begin{cases} x^3 & ;x\in\mathbb{Q} \\ 0&;x\notin\mathbb{Q} \end{cases}$ is a function which is not differentiable in a neighborhood of $x$ but the above limit exists for $a=0$. (Notice: the above formula gives a definition for the "pointwise second derivative", please see the comment of Leonid Kovalev)

bartgol's approach

bartgol has the following idea (see his answer):

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

• Harald Hanche-Olsen mentioned the double mean value theorem in the comments. For me it seems to have a somehow similar form as bartgol's idea.

• The counterexample $f(x) := \begin{cases} x^3 & ;x\in\mathbb{Q} \\ 0&;x\notin\mathbb{Q} \end{cases}$ does not work for this approach ;-) Let $x=0$, $k+h\in\mathbb{Q}$ with $k,h\notin\mathbb{Q}$. Then we have

  $$\begin{align}\frac{f(x+h+k)-f(x+h)-f(x+k)+f(x)}{hk}&=\frac{(h+k)^3}{hk}\\&=\frac{h^2}{k}+3h+3k+\frac{k^2}{h}\end{align}$$

  If we fix $h<\delta$ and let $k\rightarrow0$ the amount of $\frac{h^2}{k}$ will get arbitrary high, while $3h+3k+\frac{k^2}{h} \rightarrow 3h$ so that $\left|\frac{f(x+h+k)-f(x+h)-f(x+k)+f(x)}{hk}-f''(x)\right|$ cannot be smaller than $\varepsilon$.

• I wanted to find a proof that $f$ has to be differentiable in a neighborhood of $x$. Therefore there has to be an open interval $O$ so that for all $x+h\in D\cap O$ we have:

  $$\forall \epsilon > 0\, \exists \delta > 0\, \forall k,\tilde k\in\mathbb{R} : |k| < \delta \land |\tilde k |<\delta \land x+h+k,x+h+\tilde k\in D \Rightarrow \left|{f(x+h+k)-f(x+h) \over k} - {f(x+h+\tilde k)-f(x+h) \over \tilde k}\right|<\epsilon$$

  With the representation $f(k+h+k)=f(x+h)+f(x+k)-f(x)+f''(x)hk+R(h,k)$ ($|R(h,k)|<|\epsilon h k|$) I got

  $$\begin{align}&\left|{f(x+h+k)-f(x+h) \over k} - {f(x+h+\tilde k)-f(x+h) \over \tilde k}\right|\\=&\left|{f(x+k)-f(x)\over k }-{f(x+\tilde k)-f(x)\over \tilde k}+{R(h,k)\over k}-{R(h,\tilde k)\over \tilde k}\right|\end{align}$$

  If one can prove the differentiability of $f$ at $x$ the term ${f(x+k)-f(x)\over k }-{f(x+\tilde k)-f(x)\over \tilde k}$ can be made arbitrary small for $k,\tilde k\rightarrow 0$. Unfortunately we have $\left|{R(h,k)\over k}\right|<|\epsilon h|$so that we have no control over this term if $k$ goes to zero.

• My interpretation of bartgol's idea for arbitrary functions is (let be $f:D\rightarrow \mathbb{R}$): Forall $\epsilon > 0$, there is a $\delta > 0$, so that for all $x+h+k\in D$ with $|h|+|k|<\delta$ we have

  $$f(x+h+k)=f(x+h)+f(x+k) -f(x)+f''(x)hk+R(h,k)$$

  so that $|R(h,k)|<|\varepsilon h k|$.

  • There is a counterexample for the above interpretation of bartgol's idea (Please notice, that bartgol's idea may still work for functions with open domains): Let $A:=\{q\cdot\pi : q\in\mathbb{Q}\}$ and $B:=\{q\cdot\sqrt{2} : q\in\mathbb{Q}\}$. $A$ and $B$ are abelian groups with $+$ as group operation and $A\cap B=\{0\}$. If we just consider $k,h\in (A \cup B)\setminus\{0\}$ we have $k+h\in A \Rightarrow k\in A \land h \in A$ and $k+h\in B \Rightarrow k\in B \land h \in B$. Now we define $f:A\cup B\rightarrow \mathbb{R}: x \mapsto \begin{cases} x^3 & ;x\in A\setminus\{0\} \\ 0 & ; x\in B\end{cases}$. For $x=0$ we have either:

    $$\begin{align}\left|\frac{f(x+h+k)-f(x+h)-f(x+k)+f(x)}{hk}\right|&=\left|\frac{(h+k)^3-h^3-k^3}{hk}\right|\\&=\left|3h+3k\right|\\&\le 3 (|h|+|k|)\end{align}$$

    or

    $$\left|\frac{f(x+h+k)-f(x+h)-f(x+k)+f(x)}{hk}\right|=\left|\frac{0}{hk}\right|=0$$

    This proves that the above statement is true for $f$ with $f''(0)=0$. Because $A$ and $B$ are dense in $\mathbb{R}$ the concept of derivation is well defined for $f$. But because there are noncontinuous jumps in every neighborhood of a point $x\in A \cup B\setminus \{0\}$ the function $f$ is not differentiable in a neighborhood of $x=0$.

• Actually, this definition does not work for functions defined on open domains either. Consider a $\mathbb{Q}$-linear function $f:\mathbb{R}\to \mathbb{R}$. Then $f(x+h+k)-f(x+h)-f(x+k)+f(x)$ is identically zero, so this definition would give $f''(0)=0$; but such a function need not even be continuous. This was pointed out by Tom Goodwillie here. He did show, however, that this definition does work if we assume that $f$ is differentiable in a neighborhood.

Answer 4

Here is an enhancement of bartgol's idea which also implies that the function is differentiable at $x$. There is a number $a$ such that $\forall \epsilon>0 \;\exists \delta>0$ such that if $h^2 h^k + v^2 <\delta$, then

$$ \Big| f(x+h+k+v) - f(x+h) - f(x+k) + f(x) - c h k - a v \Big| < \epsilon. $$

Letting $v=0$ we recover bartgol's definition. But letting $h=k=0$, we recover the statement that $f$ is differentiable at $x$ with derivative $a$. Therefore, if $f$ satisfies this definition, then it is differentiable at $x$; and furthermore if it is differentiable in some neighborhood of $x$, then it is twice differentiable at $x$ and $f''(x)=c$.