# Rolle's theorem proof in Apostol: meaningfulness of interior

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 $$\exists c\in[a,b]$$.

• 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. – John Jul 3 '19 at 22:03
• I'm glad I could help. – 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.

• 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? – John Jul 3 '19 at 22:01
• @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 $f'(c)=0$" – 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 $$f$$.

• 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. – John Jul 3 '19 at 22:16
• @John Yes you did remove it. You really shouldn't do that after it's been answered. – David C. Ullrich Jul 3 '19 at 22:17
• 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! :) – John Jul 3 '19 at 22:22
• @John Oh. Sorry... – David C. Ullrich Jul 4 '19 at 0:31