(Theorem) Rolle's and mean value

Question

• Why isn’t differentiability required at endpoints?

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Rolle’s theorem §

Theorem (Tao 10.2.7): Rolle’s theorem

Let $f:[a,b]\to \mathbb{R}$ be a continuous function which is differentiable on $(a,b)$. If $f(a)=f(b)$, then there exists some $x_0\in (a,b)$ such that $g’(x_0)=0$.

^THM-rolles

#wip

Use the fact that differentiable → continuous, extreme value theorem, followed by fact that local extrema are stationary.

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Mean value theorem §

Theorem: Generalized mean value theorem

If $f,g:[a,b]\to \mathbb{R}$ are continuous functions which are differentiable on $(a,b)$, then there exists a point $x_0\in (a,b)$ such that

$$[f(b)-f(a)]g^{\prime }(x_0)=[g(b)-g(a)]f^{\prime }(x_0).$$

Note that by taking $g(x)=x$, we obtain “the” mean value theorem

$$f(b)-f(a)=(b-a)f^{\prime }(x_0)⟹f^{\prime }(x_0)=\frac{f(b)-f(a)}{b-a}.$$

^THM-mean-value

Proof from Modern Analysis I. Uses:

• (Proof pattern) Middle-man trick
• (Proof pattern) Defining a new function

Our general strategy is to define a new function that satisfies the assumptions of Rolle’s theorem. Let $h:[a,b]\to \mathbb{R}$ be given by

$$h(x)=f(x)(g(b)-g(a))-g(x)(f(b)-f(a)).$$

Then $h$ is differentiable, and we have

$$\begin{array}{rl}h(a)& =f(a)(g(b)-g(a))-g(a)(f(b)-f(a))\\ & =f(a)g(b)-f(a)g(a)-f(b)g(a)+f(a)g(a)\\ & =f(b)g(b)-f(b)g(a)-f(b)g(b)+f(a)g(b)\\ & =f(b)(g(b)-g(a))-g(b)(f(b)-f(a))=h(b).\end{array}$$

It follows from Rolle’s theorem that there exists some $x_0\in (a,b)$ such that

$$h^{\prime }(x_0)=f^{\prime }(x_0)(g(b)-g(a))-g^{\prime }(x_0)(f(b)-f(a)),$$

and rearranging gives us the desired conclusion. $\square$