MHB Help Understanding Andrew Browder's Proof of Proposition 8.14 from Math Analysis

[Math Amateur]

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:
View attachment 9409

In the above proof by Browder, we read the following:" ... ... For any $$v \in \mathbb{R}^n$$, and $$t \gt 0$$ sufficiently small, we find (taking $$h = tv$$ above) that $$L(tv) + r(tv) \leq 0$$, or $$Lv \leq r(tv)/t$$, so letting $$t \to 0$$ we have $$Lv \leq 0$$. ... ... Now ... the above quote implies that

$$\lim_{ t \to 0 } \frac{ r(tv) }{ t } = 0$$ ... ... But why exactly (formally and rigorously) is this the case ... ... ?I note that we have that $$\lim_{ h \to 0 } \frac{ r(h) }{ |h| } = \lim_{ t \to 0 } \frac{ r(tv) }{ tv } = 0$$... but this is (apparently anyway) not exactly the same thing ... we need to be able to demonstrate rigorously that$$\lim_{ h \to 0 } \frac{ r(h) }{ |h| } = \lim_{ t \to 0 } \frac{ r(tv) }{ t } = 0$$ ... ... ... but how do we proceed to do this ...?Hope someone can help ... ...

Peter

==============================================================================EDIT:

Just noticed that in the above quote, Browder argues that $$L(tv) + r(tv) \leq 0$$ implies that $$Lv \leq r(tv)/t$$ ... ... ... BUT ... i suspect he should have written $$L(tv) + r(tv) \leq 0$$ implies that $$Lv \leq - r(tv)/t$$ ... ..... however ... in either case ... when we let $$t \to 0$$ we get the same result ... namely $$Lv \leq 0 $$... ==============================================================================

 

[Opalg]

In the above proof by Browder, we read the following:" ... ... For any $$v \in \mathbb{R}^n$$, and $$t \gt 0$$ sufficiently small, we find (taking $$h = tv$$ above) that $$L(tv) + r(tv) \leq 0$$, or $$Lv \leq r(tv)/t$$, so letting $$t \to 0$$ we have $$Lv \leq 0$$. ... ... Now ... the above quote implies that

$$\lim_{ t \to 0 } \frac{ r(tv) }{ t } = 0$$ ... ... But why exactly (formally and rigorously) is this the case ... ... ?I note that we have that $$\lim_{ h \to 0 } \frac{ r(h) }{ |h| } = \lim_{ t \to 0 } \frac{ r(tv) }{ tv } = 0$$... but this is (apparently anyway) not exactly the same thing ... we need to be able to demonstrate rigorously that$$\lim_{ h \to 0 } \frac{ r(h) }{ |h| } = \lim_{ t \to 0 } \frac{ r(tv) }{ t } = 0$$ ... ... ... but how do we proceed to do this ...?

This is another case where you have to distinguish between fixed and variable vectors. Here, $v$ is fixed, but $tv$ varies, and goes to $0$ as $t\to0$. So $$\lim_{t\to0}\frac{r(tv)}{t|v|} = 0$$, and then you can multiply by the fixed nonzero scalar $|v|$ to get $$\lim_{t\to0}\frac{r(tv)}{t} = 0$$.

Just noticed that in the above quote, Browder argues that $$L(tv) + r(tv) \leq 0$$ implies that $$Lv \leq r(tv)/t$$ ... ... ... BUT ... i suspect he should have written $$L(tv) + r(tv) \leq 0$$ implies that $$Lv \leq - r(tv)/t$$ ... ..... however ... in either case ... when we let $$t \to 0$$ we get the same result ... namely $$Lv \leq 0 $$...

Absolutely correct! Browder has omitted a minus sign. But his conclusion is correct.

 

[Math Amateur]

This is another case where you have to distinguish between fixed and variable vectors. Here, $v$ is fixed, but $tv$ varies, and goes to $0$ as $t\to0$. So $$\lim_{t\to0}\frac{r(tv)}{t|v|} = 0$$, and then you can multiply by the fixed nonzero scalar $|v|$ to get $$\lim_{t\to0}\frac{r(tv)}{t} = 0$$.Absolutely correct! Browder has omitted a minus sign. But his conclusion is correct.

HI Opalg ...

Your post was most helpful to me ...

Peter