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

[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 some help in formulating a proof of Proposition 8.12 ...

Proposition 8.12 reads as follows:

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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 $$\text{df} (h) = \text{df}_1 (h), \ ... \ ... \ ... \ \text{df}_m (h) )$$... and also that ...$$f' (p) = \begin{bmatrix} f'_1 (p) \\ f'_2 (p) \\ . \\ . \\ . \\ f'_n (p) \end{bmatrix} $$... ... ... The definitions and propositions pertaining to the differential preceding the above proposition read as follows:

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Hope that someone can help ...

Help will be much appreciated ...

Peter

 

[GJA]

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.

 

[Math Amateur]

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 ...

... $$f$$ is differentiable at $$p \Longleftrightarrow$$ each $$f_j$$ is differentiable at $$p$$ ...Assume f is differentiable at $$p$$ ...... now ... ... $$f$$ is differentiable at $$p$$$$\Longrightarrow \lim_{ h \to 0 } \frac{1}{ |h| } ( f(p + h) - f(p) - Lh ) = 0$$


$$\Longrightarrow \lim_{ h \to 0 } \frac{ | f(p + h) - f(p) - Lh | }{ |h| } = 0$$ $$\Longrightarrow$$ for every $$\epsilon \gt 0 \ \exists \ \delta \gt 0$$ such that$$|h| \gt 0 \Longrightarrow \frac{ | f(p + h) - f(p) - Lh | }{ |h| } \lt \epsilon$$But ...$$Lh = \text{df} (h) = ( \text{df}_1 (h), \ldots , \text{df}_m (h) )$$

... and hence ...

$$| f(p + h) - f(p) - Lh | = | f(p + h) - f(p) - \text{df} (h) |$$... so ... we also have ... for $$|h| \lt \delta$$ ... since $$|a_j | \leq |a|$$ ... ...$$\frac{ | f_j(p + h) - f_j(p) - \text{df}_j (h) | }{ |h| } \leq \frac{ | f(p + h) - f(p) - Lh | }{ |h| } \leq \epsilon$$ ... therefore ... each $$f_j$$ is differentiable at $$p$$ ...

Now prove ...

... each $$f_j$$ is differentiable at $$p$$ ... \Longrightarrow ... f is differentiable at $$p$$ ...Assume ... each $$f_j$$ is differentiable at $$p$$ ...

Then ... we have ...$$\frac{ | f_j(p + h) - f_j(p) - \text{df}_j (h) | }{ |h| } \lt \epsilon/m$$ for $$|h| \lt \delta$$ for $$j = 1, \ldots, m$$Therefore ...$$\frac{ | f_1(p + h) - f_1(p) - \text{df}_1 (h) | }{ |h| } + \ldots + \frac{ | f_m(p + h) - f_m(p) - \text{df}_m (h) | }{ |h| } \lt \epsilon$$ But since ... $$|a|\leq \sum_{ j = 1 }^m |a_j|$$ ... we have $$\frac{ | f(p + h) - f(p) - \text{df} (h) | }{ |h| } \leq \frac{ | f_1(p + h) - f_1(p) - \text{df}_1 (h) | }{ |h| } + \ldots + \frac{ | f_m(p + h) - f_m(p) - \text{df}_m (h) | }{ |h| } \lt \epsilon$$ ... so ... $$f$$ is differentiable at $$p$$ ...
Can someone please confirm that the above is correct and/or point out the errors and shortcomings ...Peter