Why Is the Derivative Uniqueness Proof Important?

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SUMMARY

The discussion centers on the proof of uniqueness of the multi-variable derivative as presented in Walter Rudin's analysis. A key inequality utilized is |u - v| ≤ |u| + |v|, where u and v are defined in terms of the function f and linear approximations A1 and A2. The conversation highlights the importance of understanding the concept of "little oh" functions, which are essential for proving that the only linear function satisfying the derivative condition is zero. Recommended resources for further understanding include Loomis and Sternberg's "Advanced Calculus" and Spivak's "Calculus on Manifolds."

PREREQUISITES
  • Understanding of multi-variable calculus concepts
  • Familiarity with the definition of derivatives and linear functions
  • Knowledge of "little oh" notation in mathematical analysis
  • Basic proficiency in inequalities and limits
NEXT STEPS
  • Study the proof of uniqueness of derivatives in multi-variable calculus
  • Learn about "little oh" functions and their implications in calculus
  • Read Loomis and Sternberg's "Advanced Calculus" for deeper insights
  • Explore Spivak's "Calculus on Manifolds" for alternative explanations of derivatives
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Mathematics students, educators, and anyone interested in advanced calculus concepts, particularly those focusing on multi-variable derivatives and their proofs.

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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 .
 
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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, |u- v|\le |u|+ |v|.

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

u- v= A_2h- A_1h= (A_2- A_1)h= Bh so |u- v|= |Bh|
Of course, |u|= |f(x+ h)- f(x)- A_1h| and |v|= |f(x+ h)- f(x)- A_2h| so that |u- v|\le |u|+ |v| becomes
|Bh|\le |f(x+ h)- f(x)- A_1h|+ |f(x+ h)- f(x)- A_2h|
 
|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}|
 
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.
 
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