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The question I'd like to ask is this:

If $f''(0)$ exists, does $f'$ exist in a neighborhood of $0$?

Of course, under the standard definition of $f''(0)$, we have already assumed that $f'$ exists in a neighborhood of $0$. So instead:

Question: Is there a standard way to define $f''(0)$ as a limit expression that does not include $f'$ in it, and if so, can we deduce from the fact that $f''(0)$ exists that $f'$ exists in a neighborhood of $0$?

Details

If I know what $f'(0)$ is, I can make $f''(0)$ be the constant (if it exists) such that $$ \lim\limits_{h \to 0} \frac{f(h) - \left[ f(0) + f'(0) h + \frac{1}{2} f''(0) h^2\right]}{h^2} = 0. $$ i.e., the Taylor polynomial approximates $f$ to second order. Then, I could just plug in for $f'(0)$ the expression $\frac{f(h) - f(0)}{h}$. But this doesn't work; everything cancels out. Is there a different standard way to define $f''$ without using $f'$?

I should probably put some more work into answering this myself, but first I wanted to see if this is a standard or well-known question.

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    $\begingroup$ Check out this answer : math.stackexchange.com/a/776271/97045 where he gets to "Comment on Bos's work", there is some consideration made towards the definition of a second derivative. $\endgroup$
    – DanielV
    May 4, 2014 at 1:26
  • $\begingroup$ @DanielV: Thank you! You are quite right in some sense. If a curve is already known to be smooth, then the limit I gave in the answer there does connect to the second derivative, though only directly for $h_1=h_2$. But I have to agree with Paramanand Singh, commenting on the answer by Mauricio G Tec, that these fractions suggest a method for calculating the second derivative rather than defining it. Perhaps such a definition is possible, perhaps not. I am still contemplating :) $\endgroup$
    – String
    May 6, 2014 at 9:38
  • $\begingroup$ Just to clarify, Leibniz and his contemporaries did not consider the issue of differentiality as a whole. They were well aware of issues regarding singularities, but a fully-fledged concept of differentiability seems to me to have been irrelevant to the concepts of the 17th century mathematics. They only studied fairly well behaved relations, after all. $\endgroup$
    – String
    May 6, 2014 at 9:44

2 Answers 2

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To the question in the title the answer is Yes. One answer is often used in the theory of numerical differentiation, specifically in finite difference methods. For example:

$$ f''(x) = \lim_{h\to 0} \frac{f(x+h)-2f(x)+f(x-h)}{h^2} $$

In response to the very interesting comments: Indeed, the problem with this centered definition is that it has the same problem with defining $f'(x)$ using the expression $$f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x-h)}{2h}$$ as for any odd function this limit will be zero even if the function is not continuous. Here is another try:

$$f''(x) = \lim_{h\to 0} \frac{f(x)-2f(x+h)+f(x+2h)}{h^2}$$

I wonder if the following formula can solve this issue. In the literature of finite differences, these are known as forward difference and the former formula as a centered difference. A similar "backward" expression can be give, and of course, one expects them to be the same.

About the other question: it might be possible to prove that $f'(0)$ exists when the set of points of discontinuity of $f''$ is of measure zero in an interval containing $0$, since this characterizes Riemann-integrable functions.

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    $\begingroup$ However, from the fact that this limit exists at $x=0$ we can not conclude that $f'$ exists anywhere, let alone in a neighbourhood of $0$. $\endgroup$
    – Robert Israel
    Mar 24, 2014 at 6:11
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    $\begingroup$ In fact, $f$ might not even be continuous at $0$. Define $f$ arbitrarily on $(0,\infty)$, take $f(0) = 0$ and $f(-x) = -f(x)$. Then $\dfrac{f(0+h) - 2 f(0) + f(0-h)}{h^2} = 0$. $\endgroup$
    – Robert Israel
    Mar 24, 2014 at 6:16
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    $\begingroup$ Hmm @RobertIsrael. Perhaps if we we make two variables, $h_1$ and $h_2$, and take the limit (in $\mathbb{R}^2$) as both go to zero. $\endgroup$
    – 6005
    Mar 24, 2014 at 6:21
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    $\begingroup$ The formulas which you have mentioned can't be used to define the derivative (first and second order). Instead they can be used to calculate derivatives on the assumption that the function is differentiable. Hence we may say that there is no way to define second derivative directly in terms of original function, but there is a way to calculate it (provided it exists) via a direct formula involving the original function. $\endgroup$
    – Paramanand Singh ♦
    Mar 24, 2014 at 8:37
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    $\begingroup$ Indeed some additional requirements like suggested by the last comment in the answer seem to be necessary. What if we define an equivalence relation $$x\sim y\iff \exists n\in\mathbb Z:x=2^n y$$ then we may define the function $$f(x)=\max\{k\in[0,1)\ \mid\ k\sim x\}\cdot x^3$$ This function has $$\lim_{h\rightarrow 0}\frac{f(0)-2 f(h)+f(2h)}{h^2}=0$$ but has a very strange behaviour with infinitely many discontinuity points accumulating around $0$, I think ... $\endgroup$
    – String
    May 3, 2014 at 23:25
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This is considered in section 20 of Spivak's Calculus book. The definition the OP proposes for $f''(0)$ fails because, as they already suggest in the question, we won't be able to show that $f'$ exists in a neighborhood of $0$.

The example Spivak gives is $$f(x)=\begin{cases} 0 & x\in\mathbb{Q}\\ x^3 & x\in\mathbb{R}\setminus\mathbb{Q}\end{cases},$$ and the polynomal $p(x)=0$. Then, since $$\lim_{x\rightarrow 0}\frac{f(x)-p(x)}{x^2}=0,$$the OP's definition would imply that $f''(0)=0$. This is however not true since $f$ is not differentiable anywhere but at $0$.

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