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Twice differentiability

  1. Dec 6, 2008 #1
    a. Suppose f is twice differentiable on (0, infinity). Suppose that |f(x)|< or equal A0 for all x>0 and that the second derivative satisfies |f''(x)|< or equal A2 for all x>0.
    Prove that for all x>0 and all h>0
    |f'(x)| < or equal 2A0/h + A2h/2
    This is sometimes called Landau's inequality.

    b. Use part a to show that for all x>0
    |f'(x)| < or equal 2 sqrt(A0A2)


    I have no idea how to go about this problem.
    Should I just try to do this problem backwards by trying to find what f(x) by using the f'(x)?

    Thanks
     
  2. jcsd
  3. Dec 6, 2008 #2

    HallsofIvy

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    Looks to me like the mean value theorem should work. If f is twice differentiable on the positive numbers, then you can apply the mean value theorem to f' on any interval of positive numbers: (f'(x)- f'(x0))/(x- x0)= f"(c) where c is between x0 and x. Taking x- x0= h, that is f'(x)- f'(x0)= f"(c)h.
    Also use f(x)- f(x0)= f'(d)h, the mean value theorem applied to f.
     
  4. Dec 10, 2008 #3
    I'm not sure how to do this, can you elaborate how to do this esp with the mean value theorem?

    Thank you
     
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