Why Is the Derivative Uniqueness Proof Important?

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The discussion centers on the proof of uniqueness of the multi-variable derivative from Rudin, with participants seeking clarification on a specific inequality within the proof. The inequality is derived from the relationship between two numbers, u and v, and involves the expressions for the differences in function values. A participant suggests that the confusion may stem from the complexity of the source material and recommends alternative texts for clearer explanations. Additionally, the concept of "little oh" is introduced, emphasizing its role in proving that the only linear function that meets certain criteria is zero. The conversation highlights the importance of understanding foundational concepts in calculus to grasp the proof effectively.
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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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