I am currently working through a book on differential topology and Lie groups on my own.This features in the appendix on multi-variable calculus prerequisites. I am trying to go through an outline of the proof given and reason out every step. This question is a bit long, so please bear with me. The statement is as follows:
Given $ f: (a,b) \rightarrow E $ where $ E $ is a Banach space and $ f $ is differentiable, we have ,
$ ||f(y)-f(x)|| \leq |y-x| \sup_{0\leq t \leq 1} ||f'(x+t(y-x))|| $
$ \forall (x,y) \in (a,b) $
Now the proof runs as follows:
The author takes an $ M\gt M_0 =\sup_{0\leq t \leq 1} ||f'(x+t(y-x))|| $ and the set
$ S = \{ t \in [0,1]:||f(x+t(y-x))-f(x)|| \leq Mt|x-y| \} $
This construction didnt seem natural to me.Next the author claims that S is closed. I presume that if we have a limit point $ t' $ of $ S $ and consequently a sequence $ (t_n) $ in $ S $ , then by continuity of $ f $ and the right side of the inequality , we have $ t' $ in $ S $. Is this right?? Then as $ f $ is differentiable on $ (x,y) $, given $ \epsilon \lneq $ $ M-M_0 $, for all $ t $ near $ s $ and $ t \gt s $, we have:
$ ||f(x + t(y-x))-f(x+s(y-x))-f'(x+s(y-x))(t-s)(y-x)|| \leq \epsilon |t-s||y-x| $
I get that this inequality is due the Frechet derivative definition. But why $ t \gt s $? Its then shown that $ t \in S $ and hence $ s=1 $. How's this??
Thanks in advance.
Edit: Sorry, I have left out a glaring detail:that $ s = \sup S $ which exists in $ S $ obviously as $ S $ is closed and bounded.