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

#### Peter

##### Well-known member
MHB Site Helper
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 ...

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 $$\displaystyle Lv \leq 0$$

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... 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