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

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

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

and in (2.10) we have ...

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

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

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

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

... in other words ... why is $$\epsilon_a(h) = 0$$ ...
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
 
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Peter said:
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 $$A(a+h) - A(a) = A(h)$$, for every $$h \in \mathbb{R}^n$$; and there is no remainder term. ... ... "Now I can see that

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

and in (2.10) we have ...

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

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

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

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

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

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.
 
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Krylov said:
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
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.
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