Proving the Mean Value Theorem Without Rolle's Theorem

Can someone point me to a proof of the mean value theorem that doesn't directly utilize the results of Rolle's theorem?

Solution 1:

Curiosity is awesome and definitely welcomed in maths :)

Let $f \in C^1([a,b], \mathbb{R})$. The Mean Value Theorem asserts the equality $f(b)-f(a) = f'(c)(b-a)$ for some $c \in [a, b]$. Equivalently,

$$ f(b)-f(a) - f'(c)(b-a) = 0. $$

Let

$$ h(x) := f(b)-f(a) - f'(x)(b-a). $$

The "standard" argument is to say that, unless $h(a)=0$ or $h(b)=0$, this is equivalent to looking for a (local) extremum $c \in (a, b)$ of the function $x \mapsto (f(b)-f(a))x - f(x)(b-a)$. Indeed, the latter function's derivative is precisely $h$ so then we obtain $h(c) = 0$ as desired. As you pointed out, this is achieved via Rolle's Theorem.

However, one does not need to formulate the problem as an instance of Rolle's Theorem. A more direct proof would only involve $h$. It is easy to show that:

Lemma: If $f \in C([a, b], \mathbb{R})$ and $f$ is differentiable in $(a, b)$ such that $m \le f'(x) \le M$ for all $x \in (a, b)$. Then, $m(b-a) \le f(b) -f(a) \le M(b-a)$.

A fortiori, a function $f \in C^1([a,b], \mathbb{R})$ satisfies the above lemma's hypothesis. Then, for all $x \in [a, b]$

$$ (m - M)(b-a) \le h(x) \le (M-m)(b-a) $$

Since $(m-M)(b-a) \le 0$ and $(M-m)(b-a) \ge 0$, the Intermediate Value Theorem applied to $h(x)$ shows that there exists $c \in [a, b]$ such that $h(c) = 0$ as desired.

EDIT: the assumption $f \in C^1([a, b], \mathbb{R})$ can be relaxed. Indeed, Darboux's Theorem tells us that if $f \in C([a, b], \mathbb{R})$ is differentiable then for all $y \in [f'(a), f'(b)]$, there exists $c \in (a, b)$ such that $f'(c) = y$.

Therefore, since $0 \in [(m-M)(b-a), (M-m)(b-a)]$, there exists $c \in (a, b)$ such that $h(c) = 0$.