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I was looking up the definition of the derivative in several books, and what was making me uneasy was the first sentence, generally along the lines of "let $f$ be defined on...". They don't seem to be able to agree on what $f$ should be defined on. Most books say "interval"; one says "open set"; another just says "let $f$ be a real-valued function"; etc. So I decided to do a little investigation.

At the end of the day, we always define the derivative as the limit of a difference quotient (two versions), so we need to look into limits of functions. It turns out that the limit of $f$ at $a$ is only defined for $a \in (\operatorname{dom} f)'$ (we'll denote the set of cluster points of $A$ by $A'$ for convenience), because otherwise, the $0 < |x - a| < \delta$ part of the definition can be made false by choosing $\delta$ small enough to isolate $a$, thereby making the implication vacuously true, which gives non-unique limits, but we don't want that. So at this point, we know that $a$ must be a cluster point of the domain of the difference quotient.

A slight digression: It's easy to show that if $B$ is a finite set, then $A' = (A \cup B)' = (A \setminus B)'$. (That is, "adding" or "subtracting" a finite number of points doesn't affect the "stickiness" of a set.)

Let $A \subseteq \mathbb R$ and let $f: A \to \mathbb R$. For each $a \in A$, we can define a difference quotient function $q_a: A \setminus \{a\} \to \mathbb R$ by $$q_a(x) = \frac{f(x) - f(a)}{x - a}.$$

Now, (in a rather perverted pseudo-self-referential manner), we can define $$f': \{a \in A \cap A': \lim_{x \to a} q_a(x) \in \mathbb R \} \to \mathbb R \qquad \text{by} \qquad f'(a) = \lim_{x \to a} q_a(x).$$

We need $a \in A$ because $f(a)$ is needed to evaluate $q_a$, and we need $a \in A' = (A \setminus \{a\})' = (\operatorname{dom} q_a)'$ for the limit to be defined. So really, derivatives can be defined on sets that are much more general than intervals: we really only need $\operatorname{dom} f \cap (\operatorname{dom} f)' \neq \varnothing$ in order to talk about the derivative of $f$ at some point (of course, the derivative itself need not exist, but at least we can talk about it not existing).

And then I started getting worried... The function $f: \mathbb R \to \mathbb R$, defined by $$f(x) = \begin{cases} x, & x \in \mathbb Q, \\ 0, & x \notin \mathbb Q, \end{cases}$$ has no derivative anywhere. But if we restrict the domain to $\mathbb Q$, then suddenly every point is differentiable with derivative $1$? Now I'm starting to wonder if I made some stupid mistake in the development above, but I can't seem to find it.

So the question is, is the bolded statement above true?

  • I think that your function, $$f(x) = \begin{cases} x, & x \in \mathbb Q, \ 0, & x \notin \mathbb Q, \end{cases}$$ has two derivatives defined on two different sets. $f|{\mathbb{Q}}$ has a derivative $1$ on $\mathbb{Q}$. On the other hand $f|{\mathbb{R}-\mathbb{Q}}$ has a derivative of $0$ on $\mathbb{R}-\mathbb{Q}$. Since these values agree nowhere, the function is not differentiable on , $\mathbb{R}$. – Baby Dragon Jun 23 '12 at 01:20
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    @BabyDragon I disagree. Derivative is unique when it exists. The following statements are true: $f$ has no derivative anywhere; $f_{|\mathbb Q}$ is differentiable according to chester's definition; $f_{|\mathbb R\setminus \mathbb Q}$ is also differentiable according to chester's definition. These are three different functions. (Recall that by definition, a function is just a set of ordered pairs $(x,y)$ that satisfies certain conditions. Restricting the values of argument removes some elements from this set, and thus makes it a different set.) –  Jun 23 '12 at 01:24
  • @BabyDragon I don't think I understand what you're trying to say. Consider $g(x) = x^2$. The derivative of every point in $g|\mathbb Q$ is rational, and the derivative of every point in $g|{\mathbb R \setminus \mathbb Q}$ is irrational, so they agree nowhere, but the function is differentiable everywhere...? –  Jun 23 '12 at 01:27
  • @LeonidKovalev Yes, that's what I'm picturing right now. And I guess that's what scares me: gluing up two everywhere-differentiable functions gives a nowhere-differentiable one? Should this bother me/anyone? –  Jun 23 '12 at 01:29
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    It does not bother me. After all, $|x|$ is obtained by gluing $x$ and $-x$; you just have a more extreme form of this phenomenon, due to a very permissive definition of derivative. By the way: you can talk about derivatives on $\mathbb Q$ and such sets all you want. But if you want to use derivatives, you need Mean Value Theorem. And the proof of MVT requires an interval. No interval, no MVT, no calculus. –  Jun 23 '12 at 01:35
  • @Leonid You are correct, these are three different functions. I was being a bit sloppy with my language when I said "I think your function... has two different derivatives on two different sets". – Baby Dragon Jun 23 '12 at 01:48
  • @Chester Consider the function $$g(x) = \begin{cases} x^2, & x \in \mathbb Q, \ 0, & x \notin \mathbb Q, \end{cases}$$. The derivative of $g$ on $\mathbb{Q}$ is $2x$, and on $\mathbb{R}-\mathbb{Q}$ the deriviative of of $g$ is zero. Thus the values "match" at zero in the sense that the function which we shall call $g'$ (the one cobbled together by the derivatives on the two different set) is continuous at zero. So, one might conjecture that the function $g$ is differentiable at zero when $g$ is defined on $\mathbb{R}$. – Baby Dragon Jun 23 '12 at 01:57
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    @Baby Dragon: the function $g$ is differentiable at zero with $g'(0)=0$, and this is true for any function in the universe satisfying $\mid g(x)\mid \leq x^2$ . – Georges Elencwajg Jun 23 '12 at 06:55
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    The point of a derivative (to me) is to characterize behavior of $f$ locally. Local (again, to me) seems to imply some significant portion of a neighborhood. Significant would seem to at least include intervals so some value can be derived from the derivative, so to speak... – copper.hat Jun 23 '12 at 07:12
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    @chester The definition of derivative you gave is just the definition I learned at the university ;-) – Stephan Kulla Jun 23 '12 at 19:49
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    From S. Lang, Undergraduate Analysis, 2nd ed., p. 66: "Let $f$ be a function defined on an interval having more than one point, say, $I$. We shall say that $f$ is differentiable at $x$ if the limit of the Newton quotient $$\lim_{h\to 0} \frac{f(x+h)-f(x)}h$$ exists. It is understood that the limit is taken for $x+h \in I$. Thus if $x$ is, say, a left end point of the interval, we consider only values of $h > 0$. We see no reason to limit ourselves to open intervals." [last emphasis mine] – cardinal Jun 24 '12 at 18:00

6 Answers6

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I personally think it is a bad idea to define the derivative of a function at a point if no neighborhood of that point is contained in the domain. The whole point is that we want a linear approximation $$f(a + h) = f(a) + f'(a) h + o(|h|)$$

to hold in a neighborhood of $a$ (hence for $h$ in some open ball) and if there are no such neighborhoods you aren't really talking about the derivative in a conventional sense, but in some weirder sense.

In other words, my inclination would be to restrict to open sets.

Qiaochu Yuan
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    I think it is worth adding that this generalized notion of "derivative" may be interesting for sufficiently nice subsets of $\mathbb{R}$ (or $\mathbb{R}^n$) that are not necessarily open. For example, it is useful to define a notion of "derivative" for a function defined on a rectangle $\prod_{i=1}^{n} [a_i,b_i]\subseteq \mathbb{R}^n$ at a boundary point of the rectangle especially in the theory of smooth manifolds with boundary. I am sure that there are other similar situations as well. – Amitesh Datta Jun 23 '12 at 05:37
  • @Amitesh: sure, but in the theory of smooth manifolds with boundary you at least have half-neighborhoods. The OP wants to define derivatives on $\mathbb{Q}$! – Qiaochu Yuan Jun 23 '12 at 06:13
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    @QiaochuYuan Keep in mind that if you consider a function with domain $\mathbb{Q}$ you also have to consider the subspace topology on $\mathbb{Q}$ derived from the topology on $\mathbb{R}$. So (for example) every set $]a-\epsilon;a+\epsilon[\cap \mathbb{Q}$ is an open neighborhood for every $a\in\mathbb{Q}$ (although $]a-\epsilon;a+\epsilon[\cap \mathbb{Q}$ is not open in $\mathbb{R}$ - but our underlying set is $\mathbb{Q}$). Reminder that concepts like "open and closed sets" and "neighborhood" depend on the underlying topology space. – Stephan Kulla Jun 23 '12 at 20:16
  • @tampis: I am well aware of this. But those neighborhoods are not homeomorphic to open intervals. – Qiaochu Yuan Jun 23 '12 at 21:14
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    @QiaochuYuan Why need they to be homeomorphic? But I might get your point: If the domain $D$ of $f$ is open in $\mathbb{R}$ you cannot change the differentiability of $f$ at points in $D$ by extending $f$ in $\mathbb{R}$. But you can change the differentiability of $f$ at those points by extending it to a complex function... – Stephan Kulla Jun 23 '12 at 22:19
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    @tampis: I just don't see it as capturing my intuitive notion of differentiability. – Qiaochu Yuan Jun 24 '12 at 01:03
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    @QiaochuYuan: I realise that this is a very old post, but shouldn't the correct remainder in the linear approximation be $o(|h|)$, not $O(|h|^2)$? – Joe Aug 09 '21 at 10:47
  • @Joe: oops. Yes, thank you. – Qiaochu Yuan Aug 17 '21 at 01:09
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The definition you gave is the definition of derivative I learned at the university ;-)

Here are some links to papers which also use your definition of derivative:

As you already said different authors give different definitions of the derivation and you will also find references with other definitions. The definition you gave is the most general one.

As you mentioned: The derivative of a function $f:A\rightarrow\mathbb{R}$ can be defined for every point of $a\in A$ for which exists at least one sequence $(a_n)$ in $A\setminus\{a\}$ with $\lim_{n\rightarrow\infty} a_n = a$.

As Leonid Kovalev already said in a comment, there are important theorems in calculus (like the mean value theorem) where the differentiable function need to be defined on an interval. That may be the reason why many authors for textbooks in calculus just define the differentiability of functions with an interval as domain.

To have a concept of differentiability for functions like $\mathbb{R}\setminus\{0\}\rightarrow \frac 1x$ too, you have to extend your definition to open subsets of $\mathbb{R}$ (because $\mathbb{R}\setminus\{0\}$ is no interval and formaly two functions differs if their domain is different). Another step of generalization leads you to the definition with limit points.

To your given example: Let $g : \mathbb{Q} \rightarrow \mathbb{R} : x \rightarrow x$. For every $x,h\in\mathbb{Q}$ we have:

$$g(x+h)=g(x)+1\cdot h$$

So every $x,h\in\mathbb{Q}$ fulfills

$$g(x+h)\approx g(x)+1\cdot h+ O(|h|^2)$$

and therefore for every $x\in\mathbb{Q}$ the derivative of $g$ at $x$ is $1$ (as you can see from the above approximiation). There is no reason why we should worry that $g$ is not defined on all $\mathbb{R}$. The approximation $g(x+h)\approx g(x)+1\cdot h+ O(|h|^2)$ is well defined (and you can think of cases where it is useful too).

Now we extend $g$ to your function $f(x) = \begin{cases} x, & x \in \mathbb Q, \\ 0, & x \notin \mathbb Q, \end{cases}$. Because there are new arguments we have to worry about, the approximation fails for every $x\in\mathbb{R}$ and therefore the extended function is not differentiable.

You have this kind of “problem” also if you just define differentiability for functions on open sets in $\mathbb{R}$. $f(x) = \begin{cases} \frac 1x & ;x\neq 0 \\ 0 & ; x=0\end{cases}$ is not differentiable everywhere but $g:\mathbb{R}\setminus\{0\}:x\mapsto \frac 1x$ is differentiable everywhere.

Or a better example from complex analysis: $f:\mathbb{R}\rightarrow\mathbb{R}:x\mapsto 1$ is differentiable but $g: \mathbb{C}\rightarrow\mathbb{R}:a+b\cdot i\mapsto\begin{cases} 1&; b=0 \\0&;b\neq 0 \end{cases}$ is not differentiable for all real numbers.

If the domain $D$ of $f$ is open in $\mathbb{R}$ you cannot change the differentiability of $f$ at points in $D$ by extending $f$ in $\mathbb{R}$. But as the above example shows you can change the differentiability of $f$ at those points by extending it to a complex function...

Community
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Stephan Kulla
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    Well, these sources indeed introduce derivative in greater generality, but do not use this extra generality to any purpose. If anything, it only leads to confusion. For example, the functions $u=0$ (defined only on $\mathbb Q$) and $v=0$ (defined only on $\mathbb Q\cdot \sqrt{2}$) are "differentiable" at every point of their domain. The sum $u+v$ is defined only at $0$ and is not differentiable there. So much for the differentiation rules stated on page 11 of your 3rd source. –  Jun 23 '12 at 22:30
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    @LeonidKovalev I would say that $u+v$ is just defined if the domain of $u$ and $v$ is the same (see this article on planetmath.org) – Stephan Kulla Jun 23 '12 at 22:38
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    @LeonidKovalev But you are also right: You can define $u+v$ on $\mathrm{domain}(u)\cap\mathrm{domain}(v)$ and then you will have problems in defining the right rules for differentiability. – Stephan Kulla Jun 23 '12 at 22:43
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    @LeonidKovalev But isn't it the task of mathematics to worry about the right assumptions for a theorem and to provide the most general definitions for a concept? – Stephan Kulla Jun 23 '12 at 22:51
  • You are welcome to take on this task and write articles and books about differentiable functions on sets such as $\mathbb Q$. See if your readers will thank you. –  Jun 23 '12 at 22:59
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    Very nice answer! – Jonas Dahlbæk Oct 17 '14 at 10:38
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It is fairly common in real analysis literature (and even in some textbooks) to define the derivative of a function $f:A \rightarrow {\mathbb R},$ where $A \subseteq {\mathbb R},$ at all points in $A$ that are also limit points of $A$ (but at no other points). The earliest reasonably general treatment of this notion I know of is

Earle Raymond Hedrick, On derivatives over assemblages, Transactions of the American Mathematical Society 8 #3 (July 1907), 345-353. JFM 38.0419.02

Dave L. Renfro
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This more general notion of a derivative does appear in certain branches of real analysis. For instance, an important technical notion is that of an approximately differentiable function. This notion is defined as the approximate limit of the usual differential quotient and as such is defined at any density point of the domain.

Approximately differentiable functions arise especially in the theory of nonabsolute integration (i.e., integrals of Denjoy-Khinchin-Perron-Kurzweil-Henstock). Russell Gordon's book is a very nice introduction to this circle of ideas.

hardmath
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Pete L. Clark
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Consider the statement:

Let $A, B$ be subsets of the domain of $f$. If $f$ is differentiable on $A$ and on $B$, then $f$ is differentiable on $A \cup B$.

Sounds reasonable. It is true as long as we define differentiability only on open sets, but becomes wrong as soon as we allow even just endpoints (which are of course accumulation points) of intervals: Consider $f(x) =\lvert x \rvert$ and $A=(-\infty, 0]$, $B=[0, \infty)$.

Of course one can have reason to disregard this issue. But one should be aware of it.

Torsten Schoeneberg
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One someone studies Analytic Geometry in a broader way and has studied before the concept of derivative of a function $f$, a natural question is: Why is the concept of derivative that? Derivatives are originally used to "measure" the rate of change... Why do not defining the concept of deriative as $\lim_{x\to a} \frac{d(f(x),f(a))}{d(x,a)}$? It would be a interesting generalization, we could derivate easily on metric spaces... But they are not as interesting as the usual definition... The usual definition is one of the definitions which works when working with integrals, is the one which fits in the Fundamental Theorem of Calculus... There are plenties of defintions,and each one has related a particular version of the Fundamental Theorem of Calculus. Consider the Radom-Nikodym which is one of the derivatives which work with the Lebesgue integral. Derivatives and Integral are two sides of the same coin.

mamurillogarcia
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