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Rudin 5.15

  1. Apr 30, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose [itex]a \in \mathbb{R}[/itex], f is a twice-differentiable real function on (a, \infinty) and M_0,M_1,M_2 are the least upper bound of [itex]|f(x)|,|f'(x)|,|f''(x)|[/itex], respectively on (a,\infinity). Prove that

    [tex]M_1^2\leq 4 M_0 M_2[/tex]

    2. Relevant equations

    3. The attempt at a solution
    That is equivalent to showing that M_2 x^2 +M_1 x +M_0=0 has a real solution.

    I was trying to use Taylor's Theorem which says that if \alpha and \beta are distinct points in (a,\infinity) then there exists x between \alpha and \beta that makes the following equation true:

    [tex]f(\beta) = f(\alpha) + f'(\alpha)(\beta-\alpha) + f''(x) (\beta-\alpha)^2/2[/tex]

    I could take the absolute value of both sides and then use triangle inequality but I did not see how to get anywhere with that.
    Last edited: Apr 30, 2008
  2. jcsd
  3. Apr 30, 2008 #2
    Did you look at the hint given in rudin?
  4. Apr 30, 2008 #3
    How was I supposed to know there was a hint on the next page? That is a serious question.
  5. Apr 30, 2008 #4
    OK. The hint says:

    If h>0, Taylor's theorem show that

    [tex]f'(x) = \frac{1}{2h}[f(x+2h)-f(x)]-hf''(\xi)[/tex]

    for some [itex]\xi \in (x,x+2h)[/itex]. Hence

    [tex]|f'(x)| \leq hM_2+M_0/h[/tex]

    I don't see how the last inequality is useful (even if you square it).
  6. Apr 30, 2008 #5
    Square it and remember that it holds for ANY h > 0. There is a particular choice of h that will yield the result.
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