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Math Help - Derivative

  1. #1
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    Derivative

    Let f be defined on an interval I of length at least 2 and suppose that f^{''} exists there. If \vert f(x) \vert \leq 1 and \vert f^{''}(x) \vert \leq 1 for all x \in I show that \vert f^{'}(x) \vert \leq 2 on the interval.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Markeur View Post
    Let f be defined on an interval I of length at least 2 and suppose that f^{''} exists there. If \vert f(x) \vert \leq 1 and \vert f^{''}(x) \vert \leq 1 for all x \in I show that \vert f^{'}(x) \vert \leq 2 on the interval.
    What do you think? It's pretty clear which theorem we should use, no?
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    Mean Value Theorem?How do I start from there?
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    The only way I can see to do this is by contradiction. Suppose that there is a point \xi\in I at which f'(\xi)>2. Start by using the mean value theorem, and the fact that |f''|\leqslant1, to show that if \eta\in I then |f'(\eta) - f'(\xi)|\leqslant|\eta-\xi|, and therefore f'(\eta)\geqslant f'(\xi) - |\eta-\xi|.

    Now let s\in I with s\geqslant\xi. Then

    . . . . . . . \begin{aligned}f(s) &= f(\xi) + \int_\xi^sf'(\eta)\,d\eta \\ &\geqslant f(\xi) + \int_\xi^s\bigl(f'(\xi) - (\eta-\xi)\bigr)\,d\eta \\ &= f(\xi) + (s-\xi)f'(\xi) -\tfrac12(s-\xi)^2.\end{aligned}

    A similar calculation shows that if r\leqslant\xi then f(r)\leqslant f(\xi) - (\xi-r)f'(\xi) + \tfrac12(\xi-r)^2. Subtract that inequality from the previous one to get f(s) - f(r) \geqslant (s-r)f'(\xi) - \tfrac12\bigl((s-\xi)^2 + (\xi-r)^2\bigr).

    Now the interval  I has length at least 2, so we can choose r and s so that s-r=2. Notice that (s-\xi)^2 \leqslant (s-\xi)(s-r) = 2(s-\xi), and similarly (\xi-r)^2\leqslant 2(\xi-r). Therefore (s-\xi)^2 + (\xi-r)^2 \leqslant 2(s-r) = 4. It follows from the inequality in the previous paragraph that f(s) - f(r)\geqslant 2f'(\xi) - 2 > 2 (because f'(\xi)>2). But that is impossible if |f(r)|\leqslant1 and |f(s)|\leqslant1.

    That contradiction shows that f'(x)\leqslant2 for all x\in I. A similar argument shows that f'(x)\geqslant-2 (or you can save work by applying the above argument to the function -f).
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