Differentiability of mappings from R^n to R^p .... .... D&K Defn 2.2.2 ....

[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 understanding an aspect of Definition 2.2.2 ... ...

Duistermaat and Kolk's Definition 2.2.2 reads as follows:https://www.physicsforums.com/attachments/7789
https://www.physicsforums.com/attachments/7790Towards the end of the above definition, D&K write the following:

" ... ... In the case where \(\displaystyle n = p = 1\), the mapping \(\displaystyle L \mapsto L(1)\) gives a linear isomorphism \(\displaystyle \text{End} ( \mathbb{R} ) \ \ \tilde{ \rightarrow } \ \ \mathbb{R}\). ... ... "I do not understand the above remark ... never mind why it is true ... can someone please explain what D&K mean and why it is true ...

Peter=========================================================================================I think D&K's preceding notes on Differentiable Mappings may be helpful to MHB members trying the understand the above post ... so I am providing the same as follows ... ...
View attachment 7791
View attachment 7792
I also think D&K's preceding notes on Linear Mappings may be helpful to MHB members trying the understand the above post ... so I am providing the same as follows ... ... https://www.physicsforums.com/attachments/7793
View attachment 7794
View attachment 7795Hope the above text helps ...

Peter

 

[S.G. Janssens]

Towards the end of the above definition, D&K write the following:

" ... ... In the case where \(\displaystyle n = p = 1\), the mapping \(\displaystyle L \mapsto L(1)\) gives a linear isomorphism \(\displaystyle \text{End} ( \mathbb{R} ) \ \ \tilde{ \rightarrow } \ \ \mathbb{R}\). ... ... "I do not understand the above remark ... never mind why it is true ... can someone please explain what D&K mean and why it is true ...

This is an (in my opinion somewhat overly fancy) way of saying that a linear mapping $L : \mathbb{R} \to \mathbb{R}$ may be identified with a real number. Specifically, define
\[
\phi : \text{End}(\mathbb{R}) \to \mathbb{R}, \qquad \phi(L) := L(1),
\]
where the parentheses were added for readability. You can verify that $\phi$ is linear, injective and surjective, so $\phi$ is a linear isomorphism. The remark now says that $f$ is "old-differentiable" at $a \in \mathbb{R}$ if and only if $f$ is "new-differentiable" at $a$, in which case
\[
f'(a) = \phi(Df(a)).
\]
Incidentally, had we defined differentiability of mappings $f : \mathbb{R} \to \mathbb{R}^p$ first in the "old" way as well (see the remark by D&K in the paragraph following (2.8)), as is often done in other textbooks, then the above isomorphism-argument would also have worked to show that the "old" and "new" definitions agree. Instead of $\phi$ as above, we would have considered
\[
\Phi : L(\mathbb{R}, \mathbb{R}^p) \to \mathbb{R}^p, \qquad \Phi(L) := L(1).
\]
In the literature, you will see that this issue is often left implicit, but it is good to be aware of it.

 

[Math Amateur]

This is an (in my opinion somewhat overly fancy) way of saying that a linear mapping $L : \mathbb{R} \to \mathbb{R}$ may be identified with a real number. Specifically, define
\[
\phi : \text{End}(\mathbb{R}) \to \mathbb{R}, \qquad \phi(L) := L(1),
\]
where the parentheses were added for readability. You can verify that $\phi$ is linear, injective and surjective, so $\phi$ is a linear isomorphism. The remark now says that $f$ is "old-differentiable" at $a \in \mathbb{R}$ if and only if $f$ is "new-differentiable" at $a$, in which case
\[
f'(a) = \phi(Df(a)).
\]
Incidentally, had we defined differentiability of mappings $f : \mathbb{R} \to \mathbb{R}^p$ first in the "old" way as well (see the remark by D&K in the paragraph following (2.8)), as is often done in other textbooks, then the above isomorphism-argument would also have worked to show that the "old" and "new" definitions agree. Instead of $\phi$ as above, we would have considered
\[
\Phi : L(\mathbb{R}, \mathbb{R}^p) \to \mathbb{R}^p, \qquad \Phi(L) := L(1).
\]
In the literature, you will see that this issue is often left implicit, but it is good to be aware of it.

Thanks Krylov ... but just a simple check ...

What does a linear mapping from \(\displaystyle \mathbb{R}\) to \(\displaystyle \mathbb{R}\) look like ... must it be of the form \(\displaystyle L(x) = kx\) where \(\displaystyle k\) is some number in \(\displaystyle \mathbb{R}\) ... ?Also ... having a bit of trouble interpreting \(\displaystyle \phi : \text{End}(\mathbb{R}) \to \mathbb{R}, \qquad \phi(L) := L(1) \)particularly \(\displaystyle \phi(L) := L(1)\)Can you help with an example ...?

Peter

 

[S.G. Janssens]

Thanks Krylov ... but just a simple check ...

What does a linear mapping from \(\displaystyle \mathbb{R}\) to \(\displaystyle \mathbb{R}\) look like ... must it be of the form \(\displaystyle L(x) = kx\) where \(\displaystyle k\) is some number in \(\displaystyle \mathbb{R}\) ... ?

Peter

Yes, indeed. Exactly because of the bijectivity of $\phi$ as defined in post #2, every linear mapping from $\mathbb{R}$ to $\mathbb{R}$ is of the form you indicated.

Also ... having a bit of trouble interpreting \(\displaystyle \phi : \text{End}(\mathbb{R}) \to \mathbb{R}, \qquad \phi(L) := L(1) \)particularly \(\displaystyle \phi(L) := L(1)\)Can you help with an example ...?

Peter

Sure. Define $L : \mathbb{R} \to \mathbb{R}$ by $Lh := \pi h$. Then $L$ is linear, so $L \in \text{End}(\mathbb{R})$. Now $\phi(L) = L(1) = \pi \cdot 1 = \pi$. So, $\phi$ serves to "identify" $L$ with its coefficient $\pi$. Generally, $\phi$ allows us to go back and forth between both representations of $L$ with impunity.

As a side note: You will see this kind of thing quite often in different areas of mathematics. Sometimes, the isomorphism in question is not very difficult and intuitive (like here). Other times, proving that an isomorphism exists can be quite challenging. In any case, often the purpose of these kinds of "identifications" is to retain essential information while disregarding less relevant details of a particular representation.

 

[Math Amateur]

Yes, indeed. Exactly because of the bijectivity of $\phi$ as defined in post #2, every linear mapping from $\mathbb{R}$ to $\mathbb{R}$ is of the form you indicated.
Sure. Define $L : \mathbb{R} \to \mathbb{R}$ by $Lh := \pi h$. Then $L$ is linear, so $L \in \text{End}(\mathbb{R})$. Now $\phi(L) = L(1) = \pi \cdot 1 = \pi$. So, $\phi$ serves to "identify" $L$ with its coefficient $\pi$. Generally, $\phi$ allows us to go back and forth between both representations of $L$ with impunity.

As a side note: You will see this kind of thing quite often in different areas of mathematics. Sometimes, the isomorphism in question is not very difficult and intuitive (like here). Other times, proving that an isomorphism exists can be quite challenging. In any case, often the purpose of these kinds of "identifications" is to retain essential information while disregarding less relevant details of a particular representation.

Thanks for all your help on the above matters Krylov ...

It is much appreciated ...

Peter