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Another Question ... Differentiabilty & Maxima ... Browder, Proposition 8.14 ... ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

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:




Browder - Proposition 8.14 ... .....png




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