Help with D&K Proposition 2.3.2: Directional & Partial Derivatives

[Math Amateur]

I am reading "Multidimensional Real Analysis I: Differentiation" by J. J. Duistermaat and J. A. C. Kolk ...

I am focused on Chapter 2: Differentiation ... ...

I need help with another aspect of the proof of Proposition 2.3.2 ... ...

Duistermaat and Kolk's Proposition 2.3.2 and its proof read as follows:
View attachment 7850
https://www.physicsforums.com/attachments/7851
In the above proof we read the following:

" ... ... The partial differentiability of (ii) is a consequence of (i); the formula follows from $$Df(a) \in \text{ Lin} ( \mathbb{R}^n , \mathbb{R}^p )$$ and $$v = \sum_{ 1 \le j \le n } v_j e_j $$ ( see 1.11) ... ... "Can someone please demonstrate explicitly and rigorously how it is that the partial differentiability of (ii) is a consequence of (i) and, further, how exactly it is that the formula follows from $$Df(a) \in \text{ Lin} ( \mathbb{R}^n , \mathbb{R}^p )$$ and $$v = \sum_{ 1 \le j \le n } v_j e_j$$ ... ...
Help will be much appreciated ...

Peter==========================================================================================***NOTE***

It may help readers of the above post to have access to the start of Section "2.3: Directional and Partial Derivatives" ... in order to understand the context and notation of the post ... so I am providing the same ... as follows:
View attachment 7852The above post refers to (1.1) so I am providing text relevant to and including (1.1) ... as follows ...View attachment 7853Hope that the above notes/text help readers of the post understand the context and notation of the post ...

Peter

 

[HOI]

The "partial derivatives" in (ii) are the "directional derivatives" in (i) in the direction of the coordinate axes. The formula follows from the fact that a vector can be written as a sum of its component in each coordinate direction times a unit vector in that direction- that the unit vectors in each coordinate direction form a basis for the vector space.

 

[Math Amateur]

The "partial derivatives" in (ii) are the "directional derivatives" in (i) in the direction of the coordinate axes. The formula follows from the fact that a vector can be written as a sum of its component in each coordinate direction times a unit vector in that direction- that the unit vectors in each coordinate direction form a basis for the vector space.

Thanks Country Boy ...

... appreciate the help ...

Peter

 

[Math Amateur]

I am reading "Multidimensional Real Analysis I: Differentiation" by J. J. Duistermaat and J. A. C. Kolk ...

I am focused on Chapter 2: Differentiation ... ...

I need help with another aspect of the proof of Proposition 2.3.2 ... ...

Duistermaat and Kolk's Proposition 2.3.2 and its proof read as follows:In the above proof we read the following:

" ... ... The partial differentiability of (ii) is a consequence of (i); the formula follows from $$Df(a) \in \text{ Lin} ( \mathbb{R}^n , \mathbb{R}^p )$$ and $$v = \sum_{ 1 \le j \le n } v_j e_j $$ ( see 1.11) ... ... "Can someone please demonstrate explicitly and rigorously how it is that the partial differentiability of (ii) is a consequence of (i) and, further, how exactly it is that the formula follows from $$Df(a) \in \text{ Lin} ( \mathbb{R}^n , \mathbb{R}^p )$$ and $$v = \sum_{ 1 \le j \le n } v_j e_j$$ ... ...
Help will be much appreciated ...

Peter==========================================================================================***NOTE***

It may help readers of the above post to have access to the start of Section "2.3: Directional and Partial Derivatives" ... in order to understand the context and notation of the post ... so I am providing the same ... as follows:
The above post refers to (1.1) so I am providing text relevant to and including (1.1) ... as follows ...Hope that the above notes/text help readers of the post understand the context and notation of the post ...

Peter

Here is an attempt to show $$Df(a) v = \sum_{ 1 \le j \le n } v_j D_j f(a) \ \ \ \ (v \in \mathbb{R}^n ) $$
Now we have ...$$Df(a) v = \begin{pmatrix} D_1 f_1(a) & D_2 f_1(a) & ... & ... & D_n f_1(a) \\ D_1 f_2(a) & D_2 f_2(a) & ... & ... & D_n f_2(a) \\ ... & ... & ... & ... &... \\ ... & ... & ... & ... &... \\ ... & ... & ... & ... &... \\ D_1 f_p(a) & D_2 f_p(a) & ... & ... & D_n f_p(a) \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ ... \\ ... \\ ... \\ v_n \end{pmatrix}$$

$$ = \begin{pmatrix} D_1 f_1(a) v_1 + D_2 f_1(a) v_2 + \ ... \ ... \ + D_n f_1(a) v_n \\ D_1 f_2(a) v_1 + D_2 f_2(a) v_2 + \ ... \ ... \ + D_n f_2(a) v_n \\ ... \ ... \ ... \ ... \ ... \\ ... \ ... \ ... \ ... \ ... \\ ... \ ... \ ... \ ... \ ... \\ D_1 f_p(a) v_1 + D_2 f_p(a) v_2 + \ ... \ ... \ + D_n f_p(a) v_n \end{pmatrix} $$
Now $$D_j f(a) = D_{e_j} f(a) = D f(a) e_j$$ ...

But ... taking (as an example) $$j = 1$$ ... ... i.e. $$e_j = e_1$$ ... we have

$$ D_1 f(a) = D_{ e_1} f(a) = D f(a) e_1 = \begin{pmatrix} D_1 f_1(a) & D_2 f_1(a) & ... & ... & D_n f_1(a) \\ D_1 f_2(a) & D_2 f_2(a) & ... & ... & D_n f_2(a) \\ ... & ... & ... & ... &... \\ ... & ... & ... & ... &... \\ ... & ... & ... & ... &... \\ D_1 f_p(a) & D_2 f_p(a) & ... & ... & D_n f_p(a) \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 0 \\ ... \\ ... \\ ... \\ 0 \end{pmatrix}
$$
$$= \begin{pmatrix} D_1 f_1 (a) \\ D_1 f_2 (a) \\ D_1 f_3 (a) \\ ... \\ ... \\ ... \\ D_1 f_p (a) \end{pmatrix}
$$
$$= D_{ e_1} f(a) = D_1 f(a) $$
... and similar expressions can be derived for $$D_2 f(a), D_3 f(a) , \ ... \ ...\ , \ D_n f(a)$$ ...

From the above, it is clear that $$\sum_{ 1 \le j \le n} v_j D_j f(a) = Df(a) v$$

Can someone please confirm that the above is basically correct ...?
Peter

 

[Math Amateur]

Here is an attempt to show $$Df(a) v = \sum_{ 1 \le j \le n } v_j D_j f(a) \ \ \ \ (v \in \mathbb{R}^n ) $$
Now we have ...$$Df(a) v = \begin{pmatrix} D_1 f_1(a) & D_2 f_1(a) & ... & ... & D_n f_1(a) \\ D_1 f_2(a) & D_2 f_2(a) & ... & ... & D_n f_2(a) \\ ... & ... & ... & ... &... \\ ... & ... & ... & ... &... \\ ... & ... & ... & ... &... \\ D_1 f_p(a) & D_2 f_p(a) & ... & ... & D_n f_p(a) \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ ... \\ ... \\ ... \\ v_n \end{pmatrix}$$

$$ = \begin{pmatrix} D_1 f_1(a) v_1 + D_2 f_1(a) v_2 + \ ... \ ... \ + D_n f_1(a) v_n \\ D_1 f_2(a) v_1 + D_2 f_2(a) v_2 + \ ... \ ... \ + D_n f_2(a) v_n \\ ... \ ... \ ... \ ... \ ... \\ ... \ ... \ ... \ ... \ ... \\ ... \ ... \ ... \ ... \ ... \\ D_1 f_p(a) v_1 + D_2 f_p(a) v_2 + \ ... \ ... \ + D_n f_p(a) v_n \end{pmatrix} $$
Now $$D_j f(a) = D_{e_j} f(a) = D f(a) e_j$$ ...

But ... taking (as an example) $$j = 1$$ ... ... i.e. $$e_j = e_1$$ ... we have

$$ D_1 f(a) = D_{ e_1} f(a) = D f(a) e_1 = \begin{pmatrix} D_1 f_1(a) & D_2 f_1(a) & ... & ... & D_n f_1(a) \\ D_1 f_2(a) & D_2 f_2(a) & ... & ... & D_n f_2(a) \\ ... & ... & ... & ... &... \\ ... & ... & ... & ... &... \\ ... & ... & ... & ... &... \\ D_1 f_p(a) & D_2 f_p(a) & ... & ... & D_n f_p(a) \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 0 \\ ... \\ ... \\ ... \\ 0 \end{pmatrix}
$$
$$= \begin{pmatrix} D_1 f_1 (a) \\ D_1 f_2 (a) \\ D_1 f_3 (a) \\ ... \\ ... \\ ... \\ D_1 f_p (a) \end{pmatrix}
$$
$$= D_{ e_1} f(a) = D_1 f(a) $$
... and similar expressions can be derived for $$D_2 f(a), D_3 f(a) , \ ... \ ...\ , \ D_n f(a)$$ ...

From the above, it is clear that $$\sum_{ 1 \le j \le n} v_j D_j f(a) = Df(a) v$$

Can someone please confirm that the above is basically correct ...?
Peter

l will now attempt to, explicitly, complete the demonstration that $$Df(a) v = \sum_{ 1 \le j \le n } v_j D_j f(a) \ \ \ \ (v \in \mathbb{R}^n ) $$In the previous post we have demonstrated that

$$D_1 f(a) = D_{ e_1} f(a) = D f(a) e_1 = \begin{pmatrix} D_1 f_1 (a) \\ D_1 f_2 (a) \\ D_1 f_3 (a) \\ ... \\ ... \\ ... \\ D_1 f_p (a) \end{pmatrix}$$

... ... and in general

$$D_j f(a) = D_{ e_j} f(a) = D f(a) e_j =\begin{pmatrix} D_j f_1 (a) \\ D_j f_2 (a) \\ D_j f_3 (a) \\ ... \\ ... \\ ... \\ D_j f_p (a) \end{pmatrix}$$
So ...$$v_1 D_1 f(a) = v_1 \begin{pmatrix} D_1 f_1 (a) \\ D_1 f_2 (a) \\ D_1 f_3 (a) \\ ... \\ ... \\ ... \\ D_1 f_p (a) \end{pmatrix} = \begin{pmatrix} v_1 D_1 f_1 (a) \\ v_1 D_1 f_2 (a) \\ v_1 D_1 f_3 (a) \\ ... \\ ... \\ ... \\ v_1 D_1 f_p (a) \end{pmatrix} = \begin{pmatrix} D_1 f_1 (a) v_1 \\ D_1 f_2 (a) v_1 \\ D_1 f_3 (a) v_1 \\ ... \\ ... \\ ... \\ D_1 f_p (a) v_1 \end{pmatrix}
$$
and in general ...$$v_j D_j f(a) = \begin{pmatrix} D_j f_1 (a) v_j \\ D_j f_2 (a) v_j \\ D_j f_3 (a) v_j \\ ... \\ ... \\ ... \\ D_1 f_p (a) v_j \end{pmatrix}$$
So ... ...$$ \sum_{ 1 \le j \le n } v_j D_j f(a)$$ $$= \begin{pmatrix} D_1 f_1 (a) v_1 \\ D_1 f_2 (a) v_1 \\ D_1 f_3 (a) v_1 \\ ... \\ ... \\ ... \\ D_1 f_p (a) v_1 \end{pmatrix} + \begin{pmatrix} D_2 f_1 (a) v_2 \\ D_2 f_2 (a) v_2 \\ D_2 f_3 (a) v_2 \\ ... \\ ... \\ ... \\ D_2 f_p (a) v_2 \end{pmatrix} + \ ... \ ... \ \begin{pmatrix} D_j f_1 (a) v_j \\ D_j f_2 (a) v_j \\ D_j f_3 (a) v_j \\ ... \\ ... \\ ... \\ D_j f_p (a) v_j \end{pmatrix} + \ ... \ ... \ + \begin{pmatrix} D_n f_1 (a) v_n \\ D_n f_2 (a) v_n \\ D_n f_3 (a) v_n \\ ... \\ ... \\ ... \\ D_n f_p (a) v_n \end{pmatrix}
$$

= $$ = \begin{pmatrix} D_1 f_1(a) v_1 + D_2 f_1(a) v_2 + \ ... \ ... \ + D_n f_1(a) v_n \\ D_1 f_2(a) v_1 + D_2 f_2(a) v_2 + \ ... \ ... \ + D_n f_2(a) v_n \\ ... \ ... \ ... \ ... \ ... \\ ... \ ... \ ... \ ... \ ... \\ ... \ ... \ ... \ ... \ ... \\ D_1 f_p(a) v_1 + D_2 f_p(a) v_2 + \ ... \ ... \ + D_n f_p(a) v_n \end{pmatrix} $$$$= Df(a) v$$
Can someone please either confirm the above demonstration is correct or point out the errors and shortcomings ...

Peter