L'Hospital's Rule .... Bartle and Sherbert Theorem 6.2.3 .... ....

[Math Amateur]

I am reading "Introduction to Real Analysis" (Fourth Edition) by Robert G Bartle and Donald R Sherbert ...

I am focused on Chapter 6: Differentiation ...

I need help in fully understanding the proof of Theorem 6.3.3 (L'Hospital's Rule ... ) ...Theorem 6.3.3 and its proof ... ... read as follows:
View attachment 7302In the above proof we read the following:"... If $$a \lt \alpha \lt \beta \lt b$$, then Rolle's Theorem implies that $$g( \beta ) \neq g( \alpha )$$ ... ... "Can someone please explain EXACTLY how Rolle's Theorem implies that $$g( \beta ) \neq g( \alpha )$$ ... ***Note***

I suspect B&S are using the contrapositive of Rollé's Theorem but i am unsure exactly how to form the contrapositive in this case ...Hope someone can help ...

Peter
The above post refers to Rolle's Theorem ... therefore I am providing B&S's statement of Rolle's Theorem ... as follows ...View attachment 7303

 

[Opalg]

You are right, it is essentially the contrapositive form of Rolle's theorem.

Rolle's theorem says that if $g(\beta) = g(\alpha)$ then there exists $\gamma$ between $\alpha$ and $\beta$ such that $g'(\gamma) = 0.$ The contrapositive says that if there does not exist $\gamma$ between $\alpha$ and $\beta$ such that $g'(\gamma) = 0$ then $g(\beta) \ne g(\alpha)$.

Theorem 6.3.3 contains the condition $g'(x) \ne0$ for all $x$ in $(a,b)$, which (again by using a contrapositive form) implies that there does not exist $\gamma$ between $\alpha$ and $\beta$ such that $g'(\gamma) = 0$.

 

[Math Amateur]

You are right, it is essentially the contrapositive form of Rolle's theorem.

Rolle's theorem says that if $g(\beta) = g(\alpha)$ then there exists $\gamma$ between $\alpha$ and $\beta$ such that $g'(\gamma) = 0.$ The contrapositive says that if there does not exist $\gamma$ between $\alpha$ and $\beta$ such that $g'(\gamma) = 0$ then $g(\beta) \ne g(\alpha)$.

Theorem 6.3.3 contains the condition $g'(x) \ne0$ for all $x$ in $(a,b)$, which (again by using a contrapositive form) implies that there does not exist $\gamma$ between $\alpha$ and $\beta$ such that $g'(\gamma) = 0$.

Thanks for the help, Opalg ...

But ... just a clarification ...

The contrapositive of $$P \Longrightarrow Q$$ is $$\neg Q \Longrightarrow \neg P$$ ... where $$\neg P$$ is more complex than just $$\neg ["f(a) = f(b)" ] $$ ... Indeed it seems to me that $$\neg P$$ is of the form $$\neg [ A \wedge B \wedge C ]$$

where

$$A \equiv \ ,"f \text{ is continuous on } I = [a, b]"$$

$$B \equiv \ "f' \text{ exists at every point of } (a, b)"$$

$$C \equiv \ "f(a) = f(b) = 0"$$Can you comment and clarify?

Peter

 

[Opalg]

The contrapositive of $$P \Longrightarrow Q$$ is $$\neg Q \Longrightarrow \neg P$$ ... where $$\neg P$$ is more complex than just $$\neg ["f(a) = f(b)" ] $$ ... Indeed it seems to me that $$\neg P$$ is of the form $$\neg [ A \wedge B \wedge C ]$$

where

$$A \equiv \ ,"f \text{ is continuous on } I = [a, b]"$$

$$B \equiv \ "f' \text{ exists at every point of } (a, b)"$$

$$C \equiv \ "f(a) = f(b) = 0"$$Can you comment and clarify?

The way I look at it is this. The proof of 6.3.3 takes place in a background environment consisting of functions that are continuous on $[a,b]$ and differentiable on $(a,b)$. (There is also the background condition that $\alpha$ and $\beta$ are points satisfying $a<\alpha <\beta <b$.) Within that environment, Rolle's theorem says that $P\Longrightarrow Q$, where $P$ is the condition "$f(\alpha) = f(\beta) $" (and it is understood that $f$ refers to a function in the given environment) and $Q$ is the condition "there exists $\gamma$ with $\alpha< \gamma < \beta$ such that $f'(\gamma) = 0$". The contrapositive $\neg Q \Longrightarrow \neg P$ then also needs to be interpreted as taking place within the given environment of differentiable functions.