Multivariable Analysis: Another Question Re: D&K Lemma 2.2.7

  • Context: Undergrad 
  • Thread starter Thread starter Math Amateur
  • Start date Start date
  • Tags Tags
    Analysis Multivariable
Math Amateur
Gold Member
MHB
Messages
3,920
Reaction score
48
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 Lemma 2.2.7 (Hadamard...) ... ...

Duistermaat and Kolk's Lemma 2.2.7 and its proof read as follows:
D&K - 1 -  Lemma 2.2.7 ... ... PART 1 ... .png

D&K - 2 -  Lemma 2.2.7 ... ... PART 2 ... .png


Near to the end of the above text D&K write the following:

" ... ... A direct computation gives ##\| \epsilon_a(h) h^t \|_{ Eucl } = \| \epsilon_a(h) \| \| h \|## , hence##\lim_{ h \rightarrow 0 } \frac{ \| \epsilon_a(h) h^t \|_{ Eucl } }{ \| h \|^2 } = \lim_{ h \rightarrow 0 } \frac{ \| \epsilon_a(h) \| }{ \| h \| } = 0## This shows that ##\phi_a## is continuous at ##a##. ... ... "

My questions are as follows:

Question 1

... how/why does the above show that ##\phi_a## is continuous at ##a##. ... ...?

Can someone please demonstrate explicitly, formally and rigorously that ##\phi_a## is continuous at ##a##. ... ...?Question 2

How/why does the proof of Hadamard's Lemma 2.2.7 imply that ##f## is continuous at ##a## if ##f## is differentiable at ##a## ... ?
Help will be much appreciated ... ...

Peter==========================================================================================

NOTE:

The start of D&K's section on differentiable mappings may help readers of the above post understand the context and notation of the post ... so I am providing the same as follows:
D&K - 1 - Start of Section 2.2 on Differentiable Mappings ... PART 1 ... .png

D&K - 2 - Start of Section 2.2 on Differentiable Mappings ... PART 2 ... .png

The start of D&K's section on linear mappings may also help readers of the above post understand the context and notation of the post ... so I am providing the same as follows:
D&K - 1 -  Linear Mappings ... Start of Section - PART 1.png

D&K - 2 -  Linear Mappings ... Start of Section - PART 2 ... ... .png

D&K - 3 -  Linear Mappings ... Start of Section - PART 3 ... ... .png

Hope the above helps readers understand the context and notation of the post ...

Peter
 

Attachments

  • D&K - 1 -  Lemma 2.2.7 ... ... PART 1 ... .png
    D&K - 1 - Lemma 2.2.7 ... ... PART 1 ... .png
    13.5 KB · Views: 584
  • D&K - 2 -  Lemma 2.2.7 ... ... PART 2 ... .png
    D&K - 2 - Lemma 2.2.7 ... ... PART 2 ... .png
    24 KB · Views: 665
  • D&K - 1 - Start of Section 2.2 on Differentiable Mappings ... PART 1 ... .png
    D&K - 1 - Start of Section 2.2 on Differentiable Mappings ... PART 1 ... .png
    29.9 KB · Views: 484
  • D&K - 2 - Start of Section 2.2 on Differentiable Mappings ... PART 2 ... .png
    D&K - 2 - Start of Section 2.2 on Differentiable Mappings ... PART 2 ... .png
    26.9 KB · Views: 469
  • D&K - 1 -  Linear Mappings ... Start of Section - PART 1.png
    D&K - 1 - Linear Mappings ... Start of Section - PART 1.png
    8.6 KB · Views: 495
  • D&K - 2 -  Linear Mappings ... Start of Section - PART 2 ... ... .png
    D&K - 2 - Linear Mappings ... Start of Section - PART 2 ... ... .png
    33.7 KB · Views: 532
  • D&K - 3 -  Linear Mappings ... Start of Section - PART 3 ... ... .png
    D&K - 3 - Linear Mappings ... Start of Section - PART 3 ... ... .png
    40.3 KB · Views: 520
Physics news on Phys.org
.
Math Amateur said:
" ... ... A direct computation gives##\| \epsilon_a(h) h^t \|_{ Eucl } = \| \epsilon_a(h) \| \| h \|## , (1)

hence

##\lim_{ h \rightarrow 0 } \frac{ \| \epsilon_a(h) h^t \|_{ Eucl } }{ \| h \|^2 } = \lim_{ h \rightarrow 0 } \frac{ \| \epsilon_a(h) \| }{ \| h \| } = 0## (2)This shows that ##\phi_a## is continuous at ##a##. ... ... "

My questions are as follows:

Question 1

... how/why does the above show that ##\phi_a## is continuous at ##a##. ... ...?
Let ##h=x-a##. With that substitution we have

$$\frac{ \| \epsilon_a(h) h^t \|_{ Eucl } }{ \| h \|^2 }
= \frac{ \| \epsilon_a(x-a) (x-a)^t \|_{ Eucl } }{ \|x-a \|^2 }
=\|\phi_a(x)-\phi_a(a)\|
$$
per the above definition of ##\phi_a(x)##

So what you have quoted shows that the limit as ##x\to a## of ##\|\phi_a(x)-\phi_a(a)\|## is zero, which is one way of defining continuity of ##\phi_a## at ##a##..
 
  • Like
Likes   Reactions: Math Amateur
Thanks Andrew ...

Appreciate your help ...

Peter
 

Similar threads

Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K