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Differentiabilty and Continuity of Vector-Valued Functions ... Browder, Proposition 8.13 ... ...

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 8.13 ...

Proposition 8.13 reads as follows:



Browder - 1 - Proposition 8.13... PART 1 ........png
Browder - 2 - Proposition 8.13... PART 2 ... ....png



I think that a fully detailed proof of Proposition 8.13 reads somewhat as follows:


Browder's Definition 8.9 essentially means that \(\displaystyle \text{df}_p\) exists if


\(\displaystyle \lim_{ h \to 0 } \frac{1}{| h | } (f(p + h) - f(p) - \text{df}_p h ) = 0\)



Thus ... if we take \(\displaystyle \epsilon = C - \| df_p \|\) then we can find \(\displaystyle \delta\) such that ...


\(\displaystyle | | h | \lt \delta \Longrightarrow \frac{1}{ | h | } | (f(p + h) - f(p) - \text{df}_p h ) - 0 | \leq \epsilon\) ... ... ... (1)


so that


\(\displaystyle | h | \lt \delta \Longrightarrow | (f(p + h) - f(p) - \text{df}_p h ) | \leq \epsilon | h | \) ... ... ... (2)


Now the reverse triangle inequality (Duistermaat & Kolk Lemma 1.1.7 (iv) ) states that


\(\displaystyle \| x - y \| \geq | \ \| x \| - \| y \| \ |\)


Using the reverse triangle inequality we have


\(\displaystyle | (f(p + h) - f(p) ) - ( \text{df}_p h ) | \geq | \ | f(p + h) - f(p) | - | \text{df}_p h ) | \ | \) ... ... ... (3)


Now (2) and (3) \(\displaystyle \Longrightarrow \)


\(\displaystyle | \ | f(p + h) - f(p) | - | \text{df}_p h ) | \ | \leq \epsilon |h|\)


\(\displaystyle \Longrightarrow | f(p + h) - f(p) | \leq | \text{df}_p h | + \epsilon |h|\)



Now \(\displaystyle | \text{df}_p h ) | \leq | \text{df}_p | | h ) |\) (Is that correct? )


... so that ...


\(\displaystyle | f(p + h) - f(p) | \leq | \text{df}_p | | h ) | + \epsilon |h|\)



\(\displaystyle \Longrightarrow | f(p + h) - f(p) | \leq ( \| \text{df}_p \| + \epsilon ) |h| + C |h|\)



Is that correct?





Now ... my specific problem is how to rigorously and validly make the move


\(\displaystyle | \ | f(p + h) - f(p) | - | \text{df}_p h ) | \ | \leq \epsilon |h|\)


\(\displaystyle \Longrightarrow | f(p + h) - f(p) | \leq | \text{df}_p h | + \epsilon |h|\)


... ... since I have effectively ignored the modulus signs around \(\displaystyle \ | f(p + h) - f(p) | - | \text{df}_p h ) |\) ...



... that is I have assumed that \(\displaystyle | f(p + h) - f(p) | \geq | \text{df}_p h ) |\) ....



Can someone please explain how i deal with this issue ...




Help will be much appreciated ...

Peter
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,678
Now ... my specific problem is how to rigorously and validly make the move


\(\displaystyle | \ | f(p + h) - f(p) | - | \text{df}_p h ) | \ | \leq \epsilon |h|\)


\(\displaystyle \Longrightarrow | f(p + h) - f(p) | \leq | \text{df}_p h | + \epsilon |h|\)


... ... since I have effectively ignored the modulus signs around \(\displaystyle \ | f(p + h) - f(p) | - | \text{df}_p h ) |\) ...



... that is I have assumed that \(\displaystyle | f(p + h) - f(p) | \geq | \text{df}_p h ) |\) ....



Can someone please explain how i deal with this issue ...
For real numbers $X$ and $Y$, $|X|\leqslant Y$ means $-Y\leqslant X\leqslant Y$. In particular, $|X|\leqslant Y \Longrightarrow X\leqslant Y$.

In this case, $| f(p + h) - f(p) | - | \text{df}_p h | \leqslant\bigl| \ | f(p + h) - f(p) | - | \text{df}_p h | \bigr| \leqslant | \text{df}_p h ) |$.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
For real numbers $X$ and $Y$, $|X|\leqslant Y$ means $-Y\leqslant X\leqslant Y$. In particular, $|X|\leqslant Y \Longrightarrow X\leqslant Y$.

In this case, $| f(p + h) - f(p) | - | \text{df}_p h | \leqslant\bigl| \ | f(p + h) - f(p) | - | \text{df}_p h | \bigr| \leqslant | \text{df}_p h ) |$.



Thanks for a most helpful Post, Opalg

Peter