Directional and Partial Derivatives .... Another Question ....

[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

 

[fresh_42]

Can someone please demonstrate explicitly and rigorously how it is that the partial differentiability of (ii) is a consequence of (i) ...

If we have directional derivatives in all directions, we automatically have those in coordinate directions, which are exactly the partial derivatives.
$$
\left. \frac{\partial}{\partial x_i}\right|_{a}f = (D_af) (0,\ldots , 1\text{ (i-th position) } , \ldots , 0) = (D_af)(e_i) )= Df(a)(e_i)$$

... and, further, how exactly it is that the formula follows from $Df(a) \in \text{ Lin} ( \mathbb{R}^n , \mathbb{R}^p )$ and $v = \sum_{ 1 \le j \le n } v_j e_j$# ... ...

We want to compute $Df(a)(v)$. What do you get, if you put in $v=\sum_{j=1}^nv_je_j$ and write $Df(a)$ as matrix $((Df(a))_{i,j})_{1 \leq i,j \leq n}\,?$ (Note: $D_jf(a) = (\,Df(a)_{1,j}\, , \,Df(a)_{2,j}\, , \,\ldots \, , \,Df(a)_{n,j} \,)\,$)

 

[Math Amateur]

Hi fresh_42 ...

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$D f(a) e_1 = Df(a) v = \begin{pmatrix} D_1 f_1(a) & D_2 f_1(a) & ... & ... & D_n f_1(a) \\ D_1 f_2(a) & D_2 f_2(a) & ... & ... & D_n f_2(a) \\ ... & ... & ... & ... &... \\ ... & ... & ... & ... &... \\ ... & ... & ... & ... &... \\ D_1 f_p(a) & D_2 f_p(a) & ... & ... & D_n f_p(a) \end{pmatrix}$ $\begin{pmatrix} 1 \\ 0 \\ 0 \\ ... \\ ... \\ ... \\ 0 \end{pmatrix}$

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

$= D_{ e_1} f(a) = D_1 f(a)$... 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

 

[fresh_42]

Basically, this is correct. You should have written it in a compact form, but o.k. ...

Hi fresh_42 ...

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}$

Here are plus signs missing. We have $D_1 f_1(a) v_1 + D_2 f_1(a) v_2 + ... + ... + D_n f_1(a) v_n$ etc.

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$D f(a) e_1 = Df(a) v = \begin{pmatrix} D_1 f_1(a) & D_2 f_1(a) & ... & ... & D_n f_1(a) \\ D_1 f_2(a) & D_2 f_2(a) & ... & ... & D_n f_2(a) \\ ... & ... & ... & ... &... \\ ... & ... & ... & ... &... \\ ... & ... & ... & ... &... \\ D_1 f_p(a) & D_2 f_p(a) & ... & ... & D_n f_p(a) \end{pmatrix}$ $\begin{pmatrix} 1 \\ 0 \\ 0 \\ ... \\ ... \\ ... \\ 0 \end{pmatrix}$

Here you've lost the coefficients: $v=v_1\cdot e_1 +\ldots + v_n\cdot e_n$, so in the first row as in your example, it is not $1$ but $v_1$. I see why and that you inserted it below again, which is a bit confusing, but o.k. However, in this case you shouldn't have mentioned $v$ here at all, or second best, set $v=e_1\,.$Second best, because this is also a bit confusing.

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

$= D_{ e_1} f(a) = D_1 f(a)$... 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]

Hi fresh_42 ...

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$D f(a) e_1 = Df(a) v = \begin{pmatrix} D_1 f_1(a) & D_2 f_1(a) & ... & ... & D_n f_1(a) \\ D_1 f_2(a) & D_2 f_2(a) & ... & ... & D_n f_2(a) \\ ... & ... & ... & ... &... \\ ... & ... & ... & ... &... \\ ... & ... & ... & ... &... \\ D_1 f_p(a) & D_2 f_p(a) & ... & ... & D_n f_p(a) \end{pmatrix}$ $\begin{pmatrix} 1 \\ 0 \\ 0 \\ ... \\ ... \\ ... \\ 0 \end{pmatrix}$

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

$= D_{ e_1} f(a) = D_1 f(a)$... 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

Basically, this is correct. You should have written it in a compact form, but o.k. ...

Here are plus signs missing. We have $D_1 f_1(a) v_1 + D_2 f_1(a) v_2 + ... + ... + D_n f_1(a) v_n$ etc.

Here you've lost the coefficients: $v=v_1\cdot e_1 +\ldots + v_n\cdot e_n$, so in the first row as in your example, it is not $1$ but $v_1$. I see why and that you inserted it below again, which is a bit confusing, but o.k. However, in this case you shouldn't have mentioned $v$ here at all, or second best, set $v=e_1\,.$Second best, because this is also a bit confusing.

Hi fresh_42 ...

I think the following is correct ... having noted the errors that you pointed out ...

Another 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 $\sum_{ 1 \le j \le n} v_j D_j f(a) = Df(a) v$ ( a few explicit steps to go though!) Hope that corrects the errors ...

Peter

 

[Math Amateur]

Hi fresh_42 ...

I think the following is correct ... having noted the errors that you pointed out ...

Another 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 $\sum_{ 1 \le j \le n} v_j D_j f(a) = Df(a) v$ ( a few explicit steps to go though!)Hope that corrects the errors ...

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 ...$v_j D_j f(a) = \begin{pmatrix} D_j f_1 (a) v_j \\ D_j f_2 (a) v_j \\ D_j f_3 (a) v_j \\ ... \\ ... \\ ... \\ D_j f_p (a) v_j \end{pmatrix}$
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

 

[fresh_42]

Yes, these are correct.

As a final note, I like to mention (and hope that I haven't confused indexes here), that the abbreviation $D_jf(a)$ can be done as it is clear from the context what is meant. However, it adds another unnecessary convention to all of those which are already there anyway. Better would have been to write it as $Df(a)(e_j)$ or $(Df(a)_{i,j})_{1\leq i \leq p}$ or at least mention that the matrix entry is
$$(Df(a))_{i,j} = D_jf_i(a)$$ Of course this would have shrunk the main part of the second part of the proposition to what it actually is, namely a convention for the matrix which represents $Df(a)$ rather than an assertion:
\begin{align*}
Df(a)\, : \,\mathbb{R}^n \rightarrow \mathbb{R}^p \text{ linear } &\Rightarrow Df(a) \text{ matrix, if given a basis } \\
&\Rightarrow (Df(a))_{i,j} \text{ matrix entries } \\
&\Rightarrow (Df(a))_{1 \leq i \leq p, j} = D_jf(a) \text{ convention for the matrix columns } \\
&\Rightarrow (Df(a))_{i,j}=(D_jf(a))_i=D_jf_i(a) \\
&\text{ given by the definition of the function as } f=(f_1,\ldots,f_p)
\end{align*}