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Proof Error Q. Conway's Functions of One Complex Variable VI

  1. Aug 30, 2012 #1
    I ask this question only to those who read or have this book:

    If you have Baby Rudin, it would be even better.

    On the page 34 of the text Conway's Functions of One Complex Variable Vol 1, it proves the Chain Rule

    but it seems the proof is not valid:

    It uses sequences to show the limit is satisfied to be the differentiation i. e., f'(g(z_0))g'(z_0)

    We all know for the limit of function to hold, sequence should be chosen arbitrary

    For example, to show limf(x) = f(p) with x to p, choose arbitary x_n such that x_n to p and then show limf(x_n)=f(p).

    But in some parts of the proof

    it uses constructed sequences which, obviously, not arbitrary.

    For example, in Case 1 of the proof, when deriving f'(g(z_0)), it uses a sequence f(x+h_n) which is not arbitrary, obvious, though h_n is arbitrary.

    And in Case 2, it separate h_n to l_n and k_n which are also not arbitrary.

    So I think the proof is not really valid.

    Am I the only one who think like this?

    Btw can I use just Rudin's Chain Rule proof though this proof is on real-domain functions?
    Last edited: Aug 30, 2012
  2. jcsd
  3. Aug 30, 2012 #2

    My Baby Rudin has the chain rule in page 105 and uses no sequences. Can you be more specific about the edition?


    Pd Besides, proof with sequences is fine as we a have a little lemma that says the limit exists iff the sequential limit exists
  4. Aug 30, 2012 #3

    Ah of course yes, baby Rudin does not use sequences in his proof (in his latest edition).

    And Rudin's proof is very clear whereas Conway's seems not valid.

    As for the lemma, doesn't that lemma state that

    'limit exists iff for an "arbitrary" sequence the sequential limit exists'?

    Conway does not prove for arbitrary sequence but only for some sequence (at least I think so).
  5. Aug 30, 2012 #4

    Ok, sorry about that: I missed the Conway part and looked into Rudin. Anyway, there's no problem as [itex]\,z_0+h_n\,[/itex] covers all the open ball

    of radius r around [itex]\,z_0\,[/itex] , thus making the limit as generalized as possible.

  6. Aug 30, 2012 #5


    I think by your notion it seems very plausible to conclude that

    [tex] \lim_{n\to\infty}\frac{f(z_{0}+h_{n})-f(z_{0})}{h_{n}}=\lim_{h\to0}\frac{f(z_{0}+h)-f(z_{0})}{h} [/tex]

    though I'm not using Thm 4.2 in Baby Rudin

    But what about

    [tex] \lim_{n\to\infty}\frac{g(f(z_{0}+h_{n}))-g(f(z_{0}))}{f(z_{0}+h_{n})-f(z_{0})}=\lim_{k\to0}\frac{g(f(z_{0})+k)-g(f(z_{0}))}{k} [/tex]

    ? On what ground can we conclude this? (I have no idea indeed.)

    By the way the major concern is that

    it seems the author does not intend to use your notion namely epsilon-delta logic

    (though you use r rather than delta and any ball instead of arbitrary epsilon)

    but rather limit law namely Thm 4.2

    If the proof is right, it's just that I don't get this part
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