We all know Rolle’s theorem. It says that if is continuous on , differentiable on and satisfies , then there must be a such that . Few of us, though, know the proof of Rolle’s theorem. Unfortunately, it is given without proof in high school, and it is usually not proven in university courses.

In this post, I intend to give a proof of Rolle’s theorem as well as the mean value theorem, and we will see how this will allow us to prove the famous fact that is increasing if and only if .

**Theorem.** Every real sequence has a monotone subsequence.**Proof.** Let be a sequence of real numbers. Define: .

We have two possibilities: either is finite, or is infinite. If is finite, then it is either empty or non-empty. We will show that in all the three cases, we can construct a monotone subsequence of .

Case 1. . Then, by the definition of , for each , there exists an such that . If , then it’s not true that , so we can say that there exists an such that . For , we have an such that . For , we have an such that . Proceeding inductively, for , we have an such that . We have therefore constructed a decreasing subsequence of .

Case 2. is non-empty but finite, say where . Let . Then, for all , we have , so such that . We then proceed as in case 1 to obtain a decreasing subsequence of .

Case 3. is infinite.

By the well-ordering theorem, has a minimal element, say . Also, by the well-ordering theorem, has a minimal element, say . We have that . Again, by the well-ordering theorem, has a minimal element, say , and we have . Proceeding as such, we may construct a strictly increasing sequence ; in other words, with

For a given , we have , so for all , . In particular, the subsequence of is increasing. QED.

**Proposition.** If is increasing (resp. decreasing) and bounded above (resp. below), then is convergent.**Proof.** We will just show that if is increasing and bounded above then it’s convergent. Suppose that is increasing and that it is bounded above by . Let . Then, . Let . Then, by definition, there is an such that . As is increasing, we get for all . Therefore, is convergent to . QED.

**Corollary. (Bolzano-Weiertrass)** Every bounded real sequence has a convergent subsequence.**Proof.** Let be a bounded sequence. Then, has a monotone subsequence, and this subsequence is bounded and is therefore convergent. QED.

**Proposition.** Let be a function defined on an open interval around a given point , and suppose that is continuous at . Let be a any sequence converging to , and suppose WLOG that . Then, the sequence converges to .**Proof. **Let . As is continuous at , there exists a such that for all , . As , there exists an such that for all , Hence for all , we have . QED.

**Theorem. (Extreme Value Theorem)** Let be a continuous function. Then, attains a maximum and a minimum, i.e. there are points such that for all .**Proof.** We will show that attains a maximum. That attains a minimum follows from having a maximum. Let . We have that is bounded above. Indeed, suppose that it is not bounded above. Then, there is a sequence in such that . Now for each , for some . The sequence is contained in , so it is bounded, and Bolzano-Weierstrass implies that it has a convergent subsequence . We have for some . Hence, as is continuous at , we get . Now for each , . This implies that converges to , contradicting the fact that .

Let and let us show that . As , we can find a sequence converging to . Now for each , for some . The sequence is bounded, and so it has a convergent subsequence converging to some . As is continuous at , we have that . But, , so . As is a subsequence of , we have that . By uniqueness of limit, we have . Therefore, .

Write . Let . We have that , and as , we get . This shows that attains a maximum. QED.

**Proposition.** Let , where is an open interval. Let be an interior point of and suppose that has a local extremum at . If is differentiable at , then .**Proof.** WLOG suppose that is a local maximum. We have:

and

Hence, . QED.

**Theorem. (Rolle’s Theorem)** Let be a continuous function. Suppose that and that is differentiable on . Then, there exists a such that .**Proof.** We know that attains a maximum and a minimum on , say at and , respectively. If , then is constant on and thus for all and there’s nothing to prove. Hence suppose that one of and is in , say . By the above proposition, . QED.

**Theorem. (Mean Value Theorem)** Let be a continuous function. Suppose that is differentiable on . Then, there exists a such that

**Proof.**Let

satisfies the hypothesis of Rolle’s theorem, so there exists a such that . But

so we are done. QED.

**Corollary.** Let , where is an open interval, and suppose that is differentiable. We have that is increasing if and only if .**Proof.** Suppose that is increasing. We have for all :

Conversely, suppose that for all . Let such that . is continuous on and differentiable on , so by the mean value theorem, there exists a such that

Hence, , so . This shows that is increasing. QED.