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The title says it all. I find the apparent neglect of this idea rather unfortunate because (1) the notion of a function being differentiable on a closed interval is intuitively reasonable (and ought to be discussed), and (2) several theorems from calculus can be strengthened if they're stated in terms of closed interval differentiability.

Before proceeding further, it's worth taking time to make sure we agree on what we mean we say that a function is "differentiable over a closed interval". Here's the definition I had in mind:

A function $f:[a,b]\to\mathbb{R}$ is differentiable over $[a,b]$ if it is differentiable (in the ordinary sense) over $(a,b)$, right-differentiable at $a$, and left-differentiable at $b$.

As you can probably guess, a function $f$ is said to be left-differentiable at a number $x_0$ if the limit

$$\lim_{h\to 0^-}\frac{f(x_0+h)-f(x_0)}{h}$$

exists, with a similar definition applying to right-differentiability. The limits are then defined to be the left-hand and right-hand derivatives of $f$, respectively.

With that said, here's an example of a theorem (FTC1) that, under its usual hypotheses, can be strengthened (albeit slightly) if the notion of closed-interval differentiability is used:

  • If $f:[a,b]\to\mathbb{R}$ is continuous, then the function $F:[a,b]\to\mathbb{R}$ defined by $F(x)=\int_{a}^{x}f(t)\text{ }dt$ is differentiable over $[a,b]$. Moreover, for all $x\in[a,b]$, $F'(x)=f(x)$ (at the endpoints, $F'(x)$ is understood to be a left or right-hand derivative)

The return you get by applying this notion to this example is obviously minimal. That said, I truly believe that making these extra definitions is worthwhile anyway, namely because the resulting concepts strongly align with our intuitions.

To bring everything back together, I'll restate my question one last time: why don't introductory calculus textbooks usually introduce the notion of differentiability on closed intervals? Any and all responses are greatly appreciated.

Note: the "theorems" I mentioned earlier do not include existence theorems like Rolle's theorem or the MVT. I understand why these only require differentiability in the interior and why we shouldn't strengthen that requirement.

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    $\begingroup$ It seems like this sort of notion is more useful for integration, especially when you want to deal with things like intgeration by parts on piecewise functions. But pedagogically these are usually introduced much later than more introductory differentiability theorems. $\endgroup$ Feb 17 at 7:57
  • $\begingroup$ See this question: math.stackexchange.com/questions/1977588/… $\endgroup$ Feb 17 at 11:12
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    $\begingroup$ I think the short answer is that while there's nothing stopping anybody from doing it, it involves a corner case as opposed to a key idea and it's not needed as a hypothesis in many theorems. Note that even just limits can be defined more generally than for functions defined on open subsets of $\mathbb{R}$, or functions whose domains always contain at least a half interval at least little to the left or a little to the right of each point, but most calculus textbooks don't do that either. Almost nothing in calculus books (e.g. the FTC) are stated at maximum levels of generality. $\endgroup$ Feb 18 at 23:43
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Your question is not new in this forum. See for example Why does the concept of differentiation have the meaning only on open sets? and Most general $A \subseteq \mathbb R$ to define derivative of $f: A \to \mathbb R$?

For each $f : J \to \mathbb R$ living on an arbitrary interval $J \subset \mathbb R$ you can of course define that $f$ is differentiable in $x_0 \in J$ if the limit $$\lim_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0} $$ exists and denote this limit by $f'(x_0)$. This means, as expected, that $$\forall \epsilon > 0 \phantom {x} \exists \delta > 0 \phantom {x} \forall x \in J : 0 < \lvert x - x_0 \rvert < \delta \implies \left\lvert \frac{f(x)-f(x_0)}{x-x_0} - f'(x_0) \right\rvert < \epsilon .$$ If $x_0$ is a right / left boundary point of such an interval $J$, then you can say that $f$ is left /right differentiable at $x_0$ with left /right derivative $f'(x_0)$ because you can approach $x_0$ only from the left / right side. But this is implicit in the definition and in my opinion there is no need to emphasize it.

Doing so, the concepts of left and right derivative of $f$ nevertheless make sense for an interior point $x_0$ of $J$ because you can approach $x_0$ from both sides.

So why do most calculus textbooks introduce the notion of differentiability on open intervals only? In my opinion it is just tradition. See my answer to Why does the concept of differentiation have the meaning only on open sets? But note that there are also textbooks with a more general approach.

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Of course one could do this, it's convention. One issue I can see with your definition is as follows:

Consider $f: \mathbb R \rightarrow \mathbb R, f(x):=\lvert x \rvert $. With your definition, we would now say to the students "$f$ is differentiable on $(− \infty,0]$ and on $[0,\infty)$", but "$f$ is not differentiable on $(-\infty, \infty)$".

Whereas, as long as we insist the endpoints are special and say "$f$ is differentiable on both $(−\infty,0)$ and $(0,\infty)$, and also has a left and a right derivative at $0$", then anyone would see a red flag before concluding it were differentiable on all of $\mathbb R$.

Actually, this "gluing-compatibility" in my eyes is one of the strongest points in favour of definining differentiability on open sets. If $f$ is differentiable on $U$ and differentiable on $V$, then it should be differentiable on $U \cup V$. That is a beautiful property that breaks down too often when one extends the notion of differentiability to all kinds of non-open sets.

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