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Yet Further Help ... Differentiabilty & Maxima ... Browder, Proposition 8.14 ... ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,911
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 \(\displaystyle Lv \leq 0\). ... ...


Now ... the above quote implies that

\(\displaystyle \lim_{ t \to 0 } \frac{ r(tv) }{ t } = 0\) ... ...


But why exactly (formally and rigorously) is this the case ... ... ?


I note that we have that \(\displaystyle \lim_{ h \to 0 } \frac{ r(h) }{ |h| } = \lim_{ t \to 0 } \frac{ r(tv) }{ tv } = 0\)


... but this is (apparently anyway) not exactly the same thing ...


we need to be able to demonstrate rigorously that


\(\displaystyle \lim_{ h \to 0 } \frac{ r(h) }{ |h| } = \lim_{ t \to 0 } \frac{ r(tv) }{ t } = 0\) ... ...


... but how do we proceed to do this ...?





Hope someone can help ... ...

Peter




==============================================================================


EDIT:

Just noticed that in the above quote, Browder argues that


\(\displaystyle L(tv) + r(tv) \leq 0\) implies that \(\displaystyle Lv \leq r(tv)/t\) ... ...


... BUT ... i suspect he should have written


\(\displaystyle L(tv) + r(tv) \leq 0\) implies that \(\displaystyle Lv \leq - r(tv)/t\) ... ..


... however ... in either case ... when we let \(\displaystyle t \to 0\) we get the same result ... namely \(\displaystyle Lv \leq 0 \)...


==============================================================================
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,694
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 \(\displaystyle Lv \leq 0\). ... ...


Now ... the above quote implies that

\(\displaystyle \lim_{ t \to 0 } \frac{ r(tv) }{ t } = 0\) ... ...


But why exactly (formally and rigorously) is this the case ... ... ?


I note that we have that \(\displaystyle \lim_{ h \to 0 } \frac{ r(h) }{ |h| } = \lim_{ t \to 0 } \frac{ r(tv) }{ tv } = 0\)


... but this is (apparently anyway) not exactly the same thing ...


we need to be able to demonstrate rigorously that


\(\displaystyle \lim_{ h \to 0 } \frac{ r(h) }{ |h| } = \lim_{ t \to 0 } \frac{ r(tv) }{ t } = 0\) ... ...


... 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 \(\displaystyle \lim_{t\to0}\frac{r(tv)}{t|v|} = 0\), and then you can multiply by the fixed nonzero scalar $|v|$ to get \(\displaystyle \lim_{t\to0}\frac{r(tv)}{t} = 0\).

Just noticed that in the above quote, Browder argues that


\(\displaystyle L(tv) + r(tv) \leq 0\) implies that \(\displaystyle Lv \leq r(tv)/t\) ... ...


... BUT ... i suspect he should have written


\(\displaystyle L(tv) + r(tv) \leq 0\) implies that \(\displaystyle Lv \leq - r(tv)/t\) ... ..


... however ... in either case ... when we let \(\displaystyle t \to 0\) we get the same result ... namely \(\displaystyle Lv \leq 0 \)...
Absolutely correct! Browder has omitted a minus sign. But his conclusion is correct.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,911
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 \(\displaystyle \lim_{t\to0}\frac{r(tv)}{t|v|} = 0\), and then you can multiply by the fixed nonzero scalar $|v|$ to get \(\displaystyle \lim_{t\to0}\frac{r(tv)}{t} = 0\).


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



HI Opalg ...

Your post was most helpful to me ...

Peter