Total Derivatives and Linear Mappings .... D&K Example 2.2.5 ....

[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 an aspect of Example 2.2.5 ... ...

Duistermaat and Kolk's Example 2.2.5 read as follows:View attachment 7825
View attachment 7826In the above text by D&K we read the following:

" ... ... Indeed \(\displaystyle A(a+h) - A(a) = A(h)\), for every \(\displaystyle h \in \mathbb{R}^n\); and there is no remainder term. ... ... "Now I can see that

\(\displaystyle A(a + h) = A(a) + A(h)\) ... ... (1) from the definition of A ...

and in (2.10) we have ...

\(\displaystyle A(a +h) - A(a) = DA(a)h + \epsilon_a(h)\) ... ... (2)

So ... from (1) and (2) we get

\(\displaystyle A(h) = DA(a)h + \epsilon_a(h)\)

... BUT ... why, in D&K's terms is "there no remainder term" ...

... in other words ... why is \(\displaystyle \epsilon_a(h) = 0\) ...
Hope someone can help ...

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

The above post refers to equation (2.10) which occurs in Definition 2.2.2 ... so I am providing Definition 2.2.2 and the accompanying text ... as follows:View attachment 7827
View attachment 7828I hope that helps readers understand the context and notation of the above post ...

Peter

 

[S.G. Janssens]

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 an aspect of Example 2.2.5 ... ...

Duistermaat and Kolk's Example 2.2.5 read as follows:
In the above text by D&K we read the following:

" ... ... Indeed \(\displaystyle A(a+h) - A(a) = A(h)\), for every \(\displaystyle h \in \mathbb{R}^n\); and there is no remainder term. ... ... "Now I can see that

\(\displaystyle A(a + h) = A(a) + A(h)\) ... ... (1) from the definition of A ...

and in (2.10) we have ...

\(\displaystyle A(a +h) - A(a) = DA(a)h + \epsilon_a(h)\) ... ... (2)

So ... from (1) and (2) we get

\(\displaystyle A(h) = DA(a)h + \epsilon_a(h)\)

... BUT ... why, in D&K's terms is "there no remainder term" ...

... in other words ... why is \(\displaystyle \epsilon_a(h) = 0\) ...

By (2.10), the derivative of $A$ at $a$ is the (unique, remember the lemma on uniqueness we discussed) linear mapping $DA(a)$ satisfying
\[
A(a + h) = A(a) + DA(a)h + \epsilon_a(h)
\]
with $\epsilon_a(h) = o(\|h\|)$. Now, as follows from what you wrote yourself, the above equality is satisfied for $DA(a) = A$, because in that case $\epsilon_a(h) \equiv 0$ identically, and clearly $0 = o(\|h\|)$. Since derivatives are unique, it follows that $DA(a) = A$ and the remainder vanishes identically. The latter is what they mean by saying that there is no remainder term.

 

[Math Amateur]

By (2.10), the derivative of $A$ at $a$ is the (unique, remember the lemma on uniqueness we discussed) linear mapping $DA(a)$ satisfying
\[
A(a + h) = A(a) + DA(a)h + \epsilon_a(h)
\]
with $\epsilon_a(h) = o(\|h\|)$. Now, as follows from what you wrote yourself, the above equality is satisfied for $DA(a) = A$, because in that case $\epsilon_a(h) \equiv 0$ identically, and clearly $0 = o(\|h\|)$. Since derivatives are unique, it follows that $DA(a) = A$ and the remainder vanishes identically. The latter is what they mean by saying that there is no remainder term.

Oh! OK ... get the idea ...

Thanks ...

Peter