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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.

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When you write sup_{0\le t\le 1} instead of \sup_{0\le t\le 1}, then you get $\displaystyle sup_{0\le t\le 1}$ instead of $\displaystyle\sup_{0\le t\le 1}$. The proper notation (with a backslash) also results in proper spacing in such things as $a\sup b$. In some other respects your way of using $\TeX$ is very strange; see my edits. – Michael Hardy Oct 6 '12 at 14:36
Thanks for that input. Yeah I muddle up the commands mostly because I am used to LyX which has too many built in comforts :) – Vishesh Oct 7 '12 at 5:08
up vote 2 down vote accepted

The choice to investigate arbitrary $M>M_0$ does not seem too unnatural to me. After all $\sup$ is the least upper bound, hence it is often a good idea to use all other upper bounds. Especially, as it alllows you to define a positive $\epsilon$ just when one needs one.

Closedness of $S$ should be apparent by your explicit argument or because the inverse image of a closed set (like $(-\infty,0]$) under a continuous map (like $t\mapsto ||f(x+t(y-x))-f(x)||-Mt|x-y|$) is always clsed - this is one possible ddefinition of continuous after all.

For the rest you seem to have left out some detail. How is $s$ defined? It cannot be just arbitray. (Otherwise a proof of $s=1$ would be a contradiction). After your Edit: As $s$ is defined as $\sup S$, it seems natural to investigate $t>s$, which by definition of $\sup$ must be outside $S$, and hence a contradiction here shows that there are no $t>s$, i.e. $s=1$. The inequality is of course valid for all $t$ (near $s$), but in the course of the proof, we use it only for $t>s$.

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I have edited the post. Thanks for your answer. I am also a bit unclear on the choice of $ \epsilon $. – Vishesh Oct 6 '12 at 9:00

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