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Homework Help: Analysis Question-Schwarz Derivative

  1. Nov 21, 2009 #1
    1. The problem statement, all variables and given/known data
    Def. f is Schwarz Differentiable at a pt c in its domain if

    lim(h->0) [f(c+h)-f(c-h)]/2h exists as a finite limit.

    1.)Prove or disprove: f is differentiable at c => f is Schwarz Differentiable at c
    2.)Prove or disprove: f is Schwarz Differentiable at c => f is differentiable at c

    2. Relevant equations

    a function f is differentiable at c if lim(x->c) [f(x)-f(c)]/[x-c] exists

    3. The attempt at a solution

    My suspicion, from picturing each derivative, is that 1 is true and 2 is false. To prove 1, I've tried to set h=|x-c| and evaluate the derivative for x>c, and x<c and then use some linear combination of those limits to derive the schwarzian derivative, but I keep running into problems with extra terms.
    Last edited: Nov 21, 2009
  2. jcsd
  3. Nov 21, 2009 #2
    [tex]f'(c) = \lim_{h \rightarrow 0}\frac{f(c + h) - f(c)}{h} = \lim_{h \rightarrow 0}\frac{f(c - h) - f(c)}{-h} = \lim_{h \rightarrow 0}\frac{f(c) - f(c + h) }{h}[/tex]
  4. Nov 21, 2009 #3
    Big Help. Thank you snipez but, as the third part of the equality, did you mean to write f(c-h) instead of f(c+h)? As in

    f'(c)=~=~=lim(h->0)[f(c)-f(c-h)]/h ?

    I believe that's correct, and gives me what I need to proceed to part 1.

    For part two I'm trying to think of a function that will provide me with a counterexample. Maybe f=|x| a x=0?
    Last edited: Nov 21, 2009
  5. Nov 21, 2009 #4


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    If you think that's a good counterexample, why don't you show us why it is?
  6. Nov 21, 2009 #5
    Sorry, I still had some figuring to do on my own.

    Clearly, for f=|x|, f is not differentiable at c=0.

    However, fs(c)=lim(h->0)[f(c+h)-f(c-h)]/2h=0, as |x| is symmetric about the y-axis. That is, f(c+h)-f(c-h) is always zero.

    Seems right to me. What do you think?
  7. Nov 22, 2009 #6


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    I think that's excellent.

    (I would write it specifically as fs(0)= lim(h->0) (|h|- |-h|)/2h= 0, however.)
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