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Proof of derivative uniqueness

  1. May 26, 2015 #1
    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: May 26, 2015
  2. jcsd
  3. May 26, 2015 #2

    Mark44

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    Can you post a slightly larger image? That one is difficult for me to read.
     
  4. May 27, 2015 #3
    Proof.png I'm sorry. Here you go.
     
  5. May 27, 2015 #4

    HallsofIvy

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    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]
     
  6. May 27, 2015 #5

    micromass

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    [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]
     
  7. May 27, 2015 #6
    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.
     
  8. May 27, 2015 #7
    Thanks alot for answers.
     
  9. May 28, 2015 #8

    mathwonk

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    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: May 28, 2015
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