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

• MHB
• Math Amateur
In summary, Duistermaat and Kolk define Differentiability in their book "Multidimensional Real Analysis I: Differentiation" as a mapping from $\mathbb{R}$ to $\mathbb{R}^p$, which can be identified with a real number through a linear isomorphism. This allows for the "old" and "new" definitions of differentiability to be equivalent. The purpose of this identification is to retain essential information while disregarding less relevant details of a particular representation.
Math Amateur
Gold Member
MHB
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

Last edited:
Peter said:
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.

Krylov said:
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

Peter said:
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.

Peter said:
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.

Last edited:
Krylov said:
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

## 1. What is the definition of differentiability of mappings from R^n to R^p?

The definition of differentiability of mappings from R^n to R^p, as stated in D&K Defn 2.2.2, is that a function f:R^n --> R^p is differentiable at a point x in R^n if there exists a linear transformation T:R^n --> R^p such that the following limit exists:
lim_{h->0} {f(x+h)-f(x)-T(h)}/ |h| = 0. This linear transformation T is called the derivative of f at x and is denoted by Df(x).

## 2. What does it mean for a mapping to be differentiable?

A mapping is differentiable if it satisfies the definition of differentiability as stated in D&K Defn 2.2.2. This means that at every point x in the domain of the mapping, there exists a linear transformation (the derivative) that approximates the change in the function as the input changes. In other words, the function is locally linear at every point in its domain.

## 3. How is differentiability related to continuity?

Differentiability implies continuity, but continuity does not necessarily imply differentiability. If a function is differentiable at a point, it must also be continuous at that point. However, a function can be continuous at a point without being differentiable at that point. For example, a function with a sharp corner or cusp is continuous but not differentiable at that point.

## 4. Can a function be differentiable at some points but not others?

Yes, a function can be differentiable at some points but not others. Differentiability is a local property, meaning that it is determined by the behavior of the function in a small neighborhood around a point. So, a function can be differentiable at one point but not at a neighboring point.

## 5. How is differentiability of a mapping from R^n to R^p tested?

The differentiability of a mapping from R^n to R^p can be tested using the definition of differentiability (D&K Defn 2.2.2). This involves checking if the limit lim_{h->0} {f(x+h)-f(x)-T(h)}/ |h| exists for each point in the domain of the mapping. Alternatively, the differentiability of a mapping can also be tested using partial derivatives and the Jacobian matrix, as outlined in D&K Thm 2.2.3.

Replies
6
Views
2K
Replies
3
Views
1K
Replies
2
Views
1K
Replies
2
Views
1K
Replies
11
Views
2K
Replies
2
Views
1K
Replies
4
Views
1K
Replies
2
Views
1K
Replies
5
Views
2K
Replies
2
Views
1K