User:Dionyziz/Rolle

This is an alternative proof of Rolle's theorem, based on Intermediate Value Theorem.

Theorem 1
Let h : Dh &rarr; $$ \mathbb{R} $$ for $$ Dh \subseteq \mathbb{R} $$ be a continuous function in closed interval $$ [a, b] \subseteq Dh $$, a < b and let h(a) < h(b). Then there exists some $$ C = [c, d] \subseteq [a, b] $$ for which h is monotonically increasing.

(In the same manner, if h : Dh &rarr; $$ \mathbb{R} $$ for $$ Dh \subseteq \mathbb{R} $$ is a continuous function in closed interval $$ [a, b] \subseteq Dh $$, a < b with h(a) > h(b), then there exists some $$ C = [c, d] \subseteq [a, b] $$ for which h is monotonically decreasing.)

Proof
Assume that h is monotonically non-increasing in [a, b]. Then, because it is continuous, h(a) ≥ h(b), which is a contradiction. Therefore, there exists some $$ C = [c, d] \subseteq [a, b] $$ for which h is monotonically increasing.

Theorem 2
Let h : Dh &rarr; $$ \mathbb{R} $$ for $$ Dh \subseteq \mathbb{R} $$ be a differentiable function in open interval $$ (a, b) \subseteq Dh $$ for a < b, then h' is continuous in (a, b).

(This theorem is incorrect)

Proof
Assume that h' is not continuous in (a, b). Then there must either exist some r $$\in$$ (a, b) for which:

$$ \lim_{x \to r}{h'(x)} \not= h'(r) $$

or the limit $$ \lim_{x \to r}{h'(x)} $$ must be undefined.

We have: $$ \lim_{x \to r}{h'(x)} \not= h'(r) $$

$$ \Rightarrow \lim_{x \to r}{[\lim_{y \to x}{\frac{h(y) - h(x)}{y - x}}]} \not= h'(r) $$ (first mistake: we need to prove that the inner limit is well-defined)

$$ \Rightarrow \lim_{x \to r}{[\lim_{y \to x}{\frac{h(y) - h(x)}{y - x}}]} \not= \lim_{x \to r}{\frac{h(x) - h(r)}{x - r}} $$

by limits definition this becomes:

$$ \Rightarrow \lim_{y \to r}{\frac{h(y) - h(r)}{y - r}} \not= \lim_{x \to r}{\frac{h(x) - h(r)}{x - r}} $$

which is a contradiction.

If $$ \lim_{x \to r}{h'(x)} $$ is not well-defined, then:

$$ \lim_{x \to r}{h'(x)} $$ is not well-defined

$$ \Rightarrow \lim_{x \to r}{[\lim_{y \to x}{\frac{h(y) - h(x)}{y - x}}]} $$ is not well-defined

$$ \Rightarrow \lim_{x \to r}{\frac{h(y) - h(r)}{y - r}} $$ is not well-defined

which is a contradiction because of our assumption that h is differentiable in (a, b).

Therefore h' must be continuous in (a, b).

Rolle's Theorem
Let h : Dh &rarr; $$ \mathbb{R} $$ for $$ Dh \subseteq \mathbb{R} $$ be a continuous function in $$ [a, b] \subseteq Dh $$ for a < b and differentiable in open interval (a, b) with h(a) = h(b). Then there exists some $$ r \in (a, b) $$ for which $$ h'(r) = 0 $$.

Proof
(this is an alternative proof of Rolle's theorem; I do not know if it is valid, so please use with care)

If $$ h(x) = h(a) = h(b) \forall x \in [a, b] $$, then h is constant in (a, b) and therefore $$ f'(x) = 0 \forall x \in (a, b) $$.

If not, then there must exist some $$ q \in (a, b) $$ for which $$ h(q) \not= h(a) $$.

If h(a) < h(q) we have due to Theorem 1: $$ \exists C = [c1, c2] \subseteq [a, q] $$ in which h is monotonically increasing and

$$ \exists D = [d1, d2] \subseteq [q, b] $$ in which h is monotonically decreasing.

Therefore $$ \exists c \in C, d \in D $$ for which h'(c) > 0 and h'(d) < 0.

$$ h'(c)h'(d) < 0 $$

h' is continuous in (a, b) because of Theorem 2, so we can apply Intermediate Value Theorem (aka Bolzano's Theorem) in [c, d] and we have:

$$ \exists r $$ for which h'(r) = 0.

In the same manner, if h(a) > h(q) then h is monotonically decreasing in some

$$C = [c1, c2] \subseteq [a, q] $$

and monotonically increasing in some

$$D = [d1, d2] \subseteq [q, b] $$

taking $$c \in C, d \in D$$ leads to

$$ h'(c)h'(d) < 0 $$

which again leads to the equation

$$ h'(r) = 0 $$