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

  • Thread starter Thread starter Math Amateur
  • Start date Start date
  • Tags Tags
    Theorem
Math Amateur
Gold Member
MHB
Messages
3,920
Reaction score
48
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
 
Physics news on Phys.org
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 $g'(\gamma) = 0$.
 
Opalg said:
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
 
Peter said:
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.
 
Back
Top