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Differentiabilty & Maxima of Vector-Valued Functions ... 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 some 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:


"... ... Let \(\displaystyle L = \text{df}_p\); then \(\displaystyle f(p + h) - f(p) = Lh + r(h)\) where \(\displaystyle r(h)/|h| \to 0\) as \(\displaystyle h \to 0\) ... .. "



My question is as follows:


Can someone please formally and rigorously demonstrate how Browder's definition of differentiability (Definition 8.9) ...

... ... leads to the equation \(\displaystyle f(p + h) - f(p) = Lh + r(h)\) where \(\displaystyle r(h)/|h| \to 0\) as \(\displaystyle h \to 0\) ... ..




Help will be much appreciated ...

Peter



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NOTE:

The above post mentions Browder's Definition 8.9 ... Definition 8.9 reads as follows:


Browder - Definition 8.9  ... Differentials ....png




Hope that helps ...

Peter
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,681
In the above proof by Browder we read the following:


"... ... Let \(\displaystyle L = \text{df}_p\); then \(\displaystyle f(p + h) - f(p) = Lh + r(h)\) where \(\displaystyle r(h)/|h| \to 0\) as \(\displaystyle h \to 0\) ... .. "



My question is as follows:


Can someone please formally and rigorously demonstrate how Browder's definition of differentiability (Definition 8.9) ...

... ... leads to the equation \(\displaystyle f(p + h) - f(p) = Lh + r(h)\) where \(\displaystyle r(h)/|h| \to 0\) as \(\displaystyle h \to 0\) ... ..
If you define $r(h)$ by $r(h) = f(p + h) - f(p) - Lh$ then Definition 8.9 says that \(\displaystyle \lim_{h\to0}\frac{r(h)}h = 0\), or equivalently \(\displaystyle \lim_{h\to0}\frac{r(h)}{|h|} = 0\)
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
If you define $r(h)$ by $r(h) = f(p + h) - f(p) - Lh$ then Definition 8.9 says that \(\displaystyle \lim_{h\to0}\frac{r(h)}h = 0\), or equivalently \(\displaystyle \lim_{h\to0}\frac{r(h)}{|h|} = 0\)



Thanks Opalg ..

I appreciate the help ...

Peter