Why Is the Derivative Uniqueness Proof Important?

Click For Summary

Discussion Overview

The discussion centers on the proof of uniqueness of the multi-variable derivative as presented in Rudin's text. Participants explore the reasoning behind a specific inequality used in the proof and share alternative resources for understanding the topic better.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • One participant expresses confusion regarding the validity of an inequality in the uniqueness proof.
  • Another participant provides a detailed breakdown of the inequality, using specific definitions for variables involved in the proof.
  • A different participant reiterates the inequality and suggests using the absolute value properties to clarify the argument.
  • One participant introduces the concept of "little oh" functions and discusses their relevance to the uniqueness of the derivative, suggesting that the only linear function that satisfies certain conditions is zero.
  • Several participants recommend alternative texts for better understanding, critiquing the clarity of Rudin's book on this topic.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the clarity of the proof or the effectiveness of the resources discussed. There are multiple viewpoints regarding the understanding of the inequality and the suggested alternative texts.

Contextual Notes

Some participants note that the proof's complexity may depend on the definitions and assumptions used, and there is an acknowledgment of the potential difficulty in interpreting Rudin's exposition.

NanoMath
Messages
11
Reaction score
0
Proof.png
Hello.
In the proof of uniqueness of ( multi-variable ) derivative from Rudin, I am a little stuck on why the inequality holds. Rest of the proof after that is clear .
 
Last edited by a moderator:
Physics news on Phys.org
Can you post a slightly larger image? That one is difficult for me to read.
 
Proof.png
I'm sorry. Here you go.
 
It follows from the basic fact that, for any two numbers, u and v, [itex]|u- v|\le |u|+ |v|[/itex].

Here, [itex]u= f(x+ h)- f(x)- A_1h[/itex] and [itex]v= f(x+ h)- f(x)- A_2h[/itex].

[itex]u- v= A_2h- A_1h= (A_2- A_1)h= Bh[/itex] so [itex]|u- v|= |Bh|[/itex]
Of course, |u|= |f(x+ h)- f(x)- A_1h| and |v|= |f(x+ h)- f(x)- A_2h| so that [itex]|u- v|\le |u|+ |v|[/itex] becomes
[itex]|Bh|\le |f(x+ h)- f(x)- A_1h|+ |f(x+ h)- f(x)- A_2h|[/itex]
 
[tex]|B\mathbf{h}| = |A_2\mathbf{h} - A_1\mathbf{h}| = |\mathbf{f}(\mathbf{x} + \mathbf{h}) - \mathbf{f}(\mathbf{x}) - A_1\mathbf{h} - \mathbf{f}(\mathbf{x} + \mathbf{h}) +\mathbf{f}(\mathbf{x}) +A_2\mathbf{h}| \leq |\mathbf{f}(\mathbf{x} + \mathbf{h}) - \mathbf{f}(\mathbf{x}) - A_1\mathbf{h}| + |\mathbf{f}(\mathbf{x} + \mathbf{h}) -\mathbf{f}(\mathbf{x}) -A_2\mathbf{h}|[/tex]
 
Utilize ##|b-a|\leq|b|+|a|## (Note equality holds when ##b## is positive and ##a## is negative or when ##a##is positive and ##b## is negative or when either of them or both of them are 0.) with ##a=|f(x+h)-f(x)-A_1h|## and ##b=|f(x+h)-f(x)-A_2h|##
EDIT:- dang, micromass beat me to it.
 
Thanks a lot for answers.
 
a function s is "little oh" if the limit of |s(x)|/|x| is zero as x-->0. then the derivative at a of a function f is a linear function L such that (f(a+x) -f(a) - L(x)) is little oh as a function of x. so just prove that the only little oh and linear function is zero. that does it by subtraction.

oh yes, and it is possible the reason you are confused is that you are reading the worst book in the world for understanding. For this particular topic I suggest you try (pages 138 and 142 of) Loomis and Sternberg: Advanced Calculus, (free on Sternberg's website).

Or Spivak, Calculus on manifolds, page 16, or Dieudonne': Foundations of modern analysis page 143, or Lang, Analysis I, pages 302-303. or Fleming: Functions of several variables (corollary of) Prop. 12, page 156.

Best to do the proof yourself as suggested above.
 
Last edited:

Similar threads

  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
3
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K