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Differentials/Total Derivatives in R^n ... Browder, Proposition 8.12 ...

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 help in formulating a proof of Proposition 8.12 ...

Proposition 8.12 reads as follows:




Browder - Proposition 8.12 ... .png




Can someone please help me to demonstrate a formal and rigorous proof of Proposition 8.12 using on the definitions and propositions preceding the above proposition ...


I am most interested in how/why we know that


\(\displaystyle \text{df} (h) = \text{df}_1 (h), \ ... \ ... \ ... \ \text{df}_m (h) )\)


... and also that ...


\(\displaystyle f' (p) = \begin{bmatrix} f'_1 (p) \\ f'_2 (p) \\ . \\ . \\ . \\ f'_n (p) \end{bmatrix} \)


... ... ...


The definitions and propositions pertaining to the differential preceding the above proposition read as follows:




Browder - 1 - Prelim to 8.12 ... PART 1 ... .png
Browder - 2 - Prelim to 8.12 ... PART 2 ... .png



Hope that someone can help ...

Help will be much appreciated ...

Peter
 
Last edited:

GJA

Well-known member
MHB Math Scholar
Jan 16, 2013
260
Hi Peter ,

Here is a hint to hopefully help move things along.

Assuming $\bf{f}$ is differentiable at $\bf{p}$, use the inequality $0\leq \max_{1\leq j\leq m}|a_{j}|\leq |\bf{a}|$ and the definition of differentiability to establish that the $f_{j}$ are also differentiable at $\bf{p}.$

When assuming the $f_{j}$ are differentiable at $\bf{p}$, use the other inequality provided by the author to conclude that $\bf{f}$ is as well.

Let me know if anything is unclear.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
Hi Peter ,

Here is a hint to hopefully help move things along.

Assuming $\bf{f}$ is differentiable at $\bf{p}$, use the inequality $0\leq \max_{1\leq j\leq m}|a_{j}|\leq |\bf{a}|$ and the definition of differentiability to establish that the $f_{j}$ are also differentiable at $\bf{p}.$

When assuming the $f_{j}$ are differentiable at $\bf{p}$, use the other inequality provided by the author to conclude that $\bf{f}$ is as well.

Let me know if anything is unclear.


Thanks GJA ... appreciate your help ...

Think I have gotten the idea ... so ... will try to proceed ...



Proof of ...

... \(\displaystyle f\) is differentiable at \(\displaystyle p \Longleftrightarrow\) each \(\displaystyle f_j\) is differentiable at \(\displaystyle p\) ...


Assume f is differentiable at \(\displaystyle p\) ...


... now ...


... \(\displaystyle f\) is differentiable at \(\displaystyle p\)


\(\displaystyle \Longrightarrow \lim_{ h \to 0 } \frac{1}{ |h| } ( f(p + h) - f(p) - Lh ) = 0\)


\(\displaystyle \Longrightarrow \lim_{ h \to 0 } \frac{ | f(p + h) - f(p) - Lh | }{ |h| } = 0\)


\(\displaystyle \Longrightarrow\) for every \(\displaystyle \epsilon \gt 0 \ \exists \ \delta \gt 0\) such that


\(\displaystyle |h| \gt 0 \Longrightarrow \frac{ | f(p + h) - f(p) - Lh | }{ |h| } \lt \epsilon\)


But ...


\(\displaystyle Lh = \text{df} (h) = ( \text{df}_1 (h), \ldots , \text{df}_m (h) )\)

... and hence ...

\(\displaystyle | f(p + h) - f(p) - Lh | = | f(p + h) - f(p) - \text{df} (h) |\)


... so ... we also have ... for \(\displaystyle |h| \lt \delta\) ... since \(\displaystyle |a_j | \leq |a|\) ... ...


\(\displaystyle \frac{ | f_j(p + h) - f_j(p) - \text{df}_j (h) | }{ |h| } \leq \frac{ | f(p + h) - f(p) - Lh | }{ |h| } \leq \epsilon\)


... therefore ... each \(\displaystyle f_j\) is differentiable at \(\displaystyle p\) ...




Now prove ...

... each \(\displaystyle f_j\) is differentiable at \(\displaystyle p\) ... \Longrightarrow ... f is differentiable at \(\displaystyle p\) ...


Assume ... each \(\displaystyle f_j\) is differentiable at \(\displaystyle p\) ...

Then ... we have ...


\(\displaystyle \frac{ | f_j(p + h) - f_j(p) - \text{df}_j (h) | }{ |h| } \lt \epsilon/m\) for \(\displaystyle |h| \lt \delta\) for \(\displaystyle j = 1, \ldots, m\)


Therefore ...


\(\displaystyle \frac{ | f_1(p + h) - f_1(p) - \text{df}_1 (h) | }{ |h| } + \ldots + \frac{ | f_m(p + h) - f_m(p) - \text{df}_m (h) | }{ |h| } \lt \epsilon\)


But since ... \(\displaystyle |a|\leq \sum_{ j = 1 }^m |a_j|\) ... we have


\(\displaystyle \frac{ | f(p + h) - f(p) - \text{df} (h) | }{ |h| } \leq \frac{ | f_1(p + h) - f_1(p) - \text{df}_1 (h) | }{ |h| } + \ldots + \frac{ | f_m(p + h) - f_m(p) - \text{df}_m (h) | }{ |h| } \lt \epsilon\)


... so ... \(\displaystyle f\) is differentiable at \(\displaystyle p\) ...



Can someone please confirm that the above is correct and/or point out the errors and shortcomings ...


Peter
 
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