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For instance, in my calculus class, all theorems are in the following form:

For example, Rolle's theorem: If $f(x)$ is continuous on $[a,b]$, differentiable on $(a,b)$ ... (etc)

My question is, when presented with a closed interval, why must we talk about its derivatives on an open interval? Why doesn't Rolle's theorem say `differentiable on $[a,b]$' instead of $(a,b)$

Przemysław Scherwentke
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Jason
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  • If it's not defined on $x<a$, then how can you take the limit from the left? – Nishant Oct 28 '14 at 19:49
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    Derivatives are limits, so you need some room around a point in order to talk about the limit. – Adam Hughes Oct 28 '14 at 19:49
  • You can have a closed interval, it just needs to be specified what is the derivative in the endpoints. Either it can be defined using one-sided neighborhoods, or as a derivative of some smooth extension. – Peter Franek Oct 28 '14 at 19:50
  • We can say that $f$ has derivative in close interval $[a,b]$ but it's only right\left derivative on $a$ \ $b$. – brick Oct 28 '14 at 19:51

2 Answers2

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As @PeterFrank says in the comments, one can talk about so-called one-sided differentiability in the end points, e.g.

$$ f'(a)=\lim_{h \downarrow 0} \frac{f(a+h)-f(a)}{h}. $$

But the point is that if you only assume that $f$ has to be differentiable on $(a,b)$, the theorem is better/stronger.

For example, we can apply Rolle to $x\mapsto \sqrt{x}$ on $[0,1]$, although this map is not one-sidedly differentiable in $0$.

If you would assume that $f$ has to be differentiable on $[0,1]$, you could not apply Rolle in this case.

PhoemueX
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  • So, if we have a closed interval $[a,b]$ and we try to find the limit of $a$ we can only talk about the one sided limit (from the right side). But if we have an open interval where $c$ is in $(a,b)$ we can talk about both sides of the limit? – Jason Oct 28 '14 at 20:23
  • Yes, because for such a $c$, the function $f$ is defined "to the left of $c$ and "to the right of $c$", so we can consider the limit from both sides. In contrast, the square root is not defined "left of $0$", so that it makes no sense to consider the left-sided derivative (which intuitively is the slope if you approach the point from the left) at $0$. – PhoemueX Oct 28 '14 at 21:17
  • If we have an open continuous interval $(a, b)$ for a function $f$, is $f(x)$ differentiable at $x\to a^+$? I don't think it is but would like to hear your answer and a brief explanation if possible, please. Thank you. – Lars Smith Jan 11 '21 at 20:40
  • I meant if we have a continuous function $f$ on an open interval $(a, b)$. – Lars Smith Jan 11 '21 at 20:54
  • @LarsSmith - No, the answer already gives a counter-example. – mr_e_man Jun 20 '22 at 18:05
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For this reason: (I talk about Mean Value Theorem here)
If we consider differentiability on the closed interval $\mathbf[a, b\mathbf]$, we are then losing the generality of our theorem; because we're asking for a more strict condition to be satisfied.
Meanwhile, considering differentiability on $\mathbf(a, b\mathbf)$ makes the theorem more general and stronger without affecting the validity of our theorem.

For example, take a look at the graph of the semicircle with radius $1$ $$ f(x) = \sqrt{1 - x^2} $$ where $f(x)$ is continuous on its domain $\mathbf[-1,1\mathbf]$, but it happens to be differentiable only on $\mathbf(-1, 1\mathbf)$

And my point is that, with the stricter closed interval condition; this function is not qualified, even though there really is a point $\mathbf{c}$ in $\mathbf(a, b\mathbf)$ where the slope of the tangent there is equal to that of the secant cutting through the endpoints $\mathbf{a}$ & $\mathbf{b}$.
And as far as I know, and as far as I am convinced, there doesn't exist a function where you'd need such a strict condition for the theorem to remain true.

So, we conclude that there is not any need for considering differentiability on a closed interval instead of an open one; since that'd cause a loss of generality for no acceptable reason.

And as always, Allah S.W.T. knows best. I hope this helps insha'Allah.

potatomandoesphysics
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