According to the statement of the Rolle's theorem in Apostol calculus 1, we need to have a continuous function on $S = [a, b]$, and this function should have a derivative on the interior of $S$. I do not get this condition.

a) Why is the derivative restricted to the interior? Doesn't the right derivative ensure the right continuity?

b) As a result, the proof ensures that there is some $c$, s.t. $a < c < b$, where $f'(c) = 0$. However, why not try to prove $a \leq c \leq b$?

Historical part of the question: In Apostol Rolle's theorem is used to prove the mean-value theorem, which is in turn used to prove convexity properties of derivatives, and there is a big problem with endpoints: suppose the derivative 𝑓′(𝑥) is strictly positive on (𝑎,𝑏), then the function is strictly increasing on [𝑎,𝑏]. This is the conclusion from the Rolle's -> mean-value theorems above in Apostol next section. But the Rolle's theorem does not specify the endpoints 𝑎 and 𝑏 as valid places for the derivative zero! It feels unproven that the function 𝑓(𝑥) is increasing on [𝑎,𝑏], when it can be for example decreasing at a point 𝑎, and further increasing on the interior.


a) Since assuming only that the restriction of $f$ to $(a,b)$ is differentiable is enough to prove Rolle's theorem, why would someone add the extra hypothesis that $f$ is also differentiable at $a$ and at $b$?

b) Note that $\exists c\in(a,b)$ is stronger than $\exists c\in[a,b]$.

  • $\begingroup$ Yep, I thought for a few minutes about it. I actually did not think about it this way. If it indeed works for (a, b), it should work for [a, b] too. Somehow my brain did not make this connection. Thx. $\endgroup$ – John Jul 3 '19 at 22:03
  • $\begingroup$ I'm glad I could help. $\endgroup$ – José Carlos Santos Jul 3 '19 at 22:10

The requirement that $f'(a)$ and $f'(b)$ also exist makes the premise unnecessarily strong. Allowing $c=a$ or $c=b$ makes the conclusion unnecessarily weak. Hence both your changes weaken the theorem.

  • $\begingroup$ Could you elaborate a little bit about weakening of the conclusion? Did not get the second part of the answer: why is it weak to set c = a? $\endgroup$ – John Jul 3 '19 at 22:01
  • 1
    $\begingroup$ @John Rolle's theorem is of the form $P\to Q$. Replacing $P$ with $P\land R$ and/or $Q$ with $Q\lor S$ is a weakening. Here, $P$ is "$f$ is continous on $[a,b]$ and differentiable on $(a,b) and $f(a9=f(b)$", $Q$ is $there exists $c\in (a,b)$ with $f'(c)=0$, $R$ is "$f$ is (also) differentoiable at $a$ and $b$", and $S$ is "there exists $c\in\{a,b\}$ with $f'(c)=0$" $\endgroup$ – Hagen von Eitzen Jul 4 '19 at 5:39

Regarding the historical part, in particular "It feels unproven that the function $f(x)$ is increasing on $[a,b]$, when it can be for example decreasing at a point $a$, and further increasing on the interior."

How it "feels" really doesn't matter. What's the actual problem with the proof?

Assuming the proof in Apostol is the standard one:

Theorem Suppose $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f'>0$ on $(a,b)$. Then $f$ is increasing on $[a,b]$.

Proof: We need to show that if $a\le\alpha<\beta\le b$ then $f(\alpha)< f(\beta)$. Since $f$ is continuous on $[\alpha,\beta]$ and differentiable on $(\alpha,\beta)$, MVT shows there exists $c\in(\alpha,\beta)\subset(a,b)$ with $$f(\beta)-f(\alpha)=(\beta-\alpha)f'(c).$$Since $\beta-\alpha>0$ and $f'(c)>0$ this shows that $f(\beta)-f(\alpha)>0$. QED.

The fact that $f$ is perhaps not differentiable at $a$ and $b$ simply doesn't matter.

Yes, if we had $f$ satisfying all those conditions and also $f'(a)<0$ that would be a problem. But there is no such $f$.

  • $\begingroup$ Yep, I removed that part of the question. It is exactly this proof that is given in Apostol. I just thought sth along "what if c falls on a or b". But it is exactly as you write: if MVT is proven for the interior, it is automatically true for the closed interval too. Then everything falls in place. $\endgroup$ – John Jul 3 '19 at 22:16
  • $\begingroup$ @John Yes you did remove it. You really shouldn't do that after it's been answered. $\endgroup$ – David C. Ullrich Jul 3 '19 at 22:17
  • $\begingroup$ Unfortunately, your answer wasn't yet posted when I was removing that part. I returned it so that your answer fully corresponds to the question, and thx for helping! :) $\endgroup$ – John Jul 3 '19 at 22:22
  • $\begingroup$ @John Oh. Sorry... $\endgroup$ – David C. Ullrich Jul 4 '19 at 0:31

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