Rolle's theorem proof in Apostol: meaningfulness of interior

Question

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 ′() is strictly positive on (,), then the function is strictly increasing on [,]. 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 and as valid places for the derivative zero! It feels unproven that the function () is increasing on [,], when it can be for example decreasing at a point , and further increasing on the interior.

Answer 1

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

Answer 2

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.

Answer 3

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 $f$.