- Thread starter
- #1

- Jun 22, 2012

- 2,891

I am currently reading Chapter 8: Differentiable Maps and am specifically focused on Section 8.2 Differentials ... ...

I need yet further help in fully understanding the proof of Proposition 8.14 ...

Proposition 8.14 reads as follows:

In the above proof by Browder, we read the following:

" ... ... For any \(\displaystyle v \in \mathbb{R}^n\), and \(\displaystyle t \gt 0\) sufficiently small, we find (taking \(\displaystyle h = tv\) above) that \(\displaystyle L(tv) + r(tv) \leq 0\), or \(\displaystyle Lv \leq r(tv)/t\), so letting \(\displaystyle t \to 0\) we have \(\displaystyle Lv \leq 0\); replacing \(\displaystyle v\) by \(\displaystyle -v\), we also find that \(\displaystyle Lv \geq 0\), so \(\displaystyle Lv = 0\). ... ... "

Now the argument for \(\displaystyle Lv \leq 0\) is as follows:

\(\displaystyle L(tv) + r(tv) \leq 0\)

\(\displaystyle \Longrightarrow tLv \leq - r(tv)\)

\(\displaystyle \Longrightarrow Lv \leq - r(tv)/ t\)

... so taking the limit as \(\displaystyle t \to 0\) we have \(\displaystyle \lim_{ t \to 0 } -r(tv)/t = 0\) ...

Thus \(\displaystyle Lv \leq 0\)

------------------------------------------------------------------

... and ... now put \(\displaystyle v = -v\) ...

then

\(\displaystyle L(t(-v)) + r(t (-v)) \leq 0\)

\(\displaystyle \Longrightarrow -t Lv + r(t (-v)) \leq 0\)

\(\displaystyle \Longrightarrow -t Lv \leq -r(t (-v))\)

\(\displaystyle \Longrightarrow Lv \geq r(- tv)/t\)

... then

... ... so taking the limit as \(\displaystyle t \to 0\) we have \(\displaystyle \lim_{ t \to 0 } r(-tv)/t = 0\) ...

Thus \(\displaystyle Lv \geq 0\)

... BUT ...

why exactly is \(\displaystyle \lim_{ t \to 0 } r(-tv)/t = 0\) ... how do we formally and rigorously demonstrate this is the case ...i.e. true ...

Help will be much appreciated ...

Peter