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  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    Taylor.

    Let f be 2 times differentiable function in open interval (0,1) with following conditions:

    1. lim(x-->0)f(x) = lim(x-->1)f(x) = 0

    2.Exist M>0 so that |f ''(x)|<=M for all x in (0,1)

    Prove that |f '(x
    )|<=M/2


    Thank you all!
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Let f be 2 times differentiable function in open interval (0,1) with following conditions:

    1. lim(x-->0)f(x) = lim(x-->1)f(x) = 0

    2.Exist M>0 so that |f ''(x)|<=M for all x in (0,1)

    Prove that |f '(x
    )|<=M/2


    Thank you all!
    It looks like you have that f(0)=f(1)=0 thus by Rolle's theorem there exists some c\in (0,1) such that f'(c)=0. Form the first degree taylor polynomial around that point f'(x)=f'(c)+\frac{f''(\xi)}{2!}(x-c)^2=\frac{f''(\xi)}{2}(x-c)^2. Now, combining \left|f''(\xi)\right|\leqslant M and (x-c)^2\leqslant 1 we see that \left|f'(x)\right|=\left|\frac{f''(\xi)}{2}(x-c)^2\right|\leqslant\frac{M}{2}
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  3. #3
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Let f be 2 times differentiable function in open interval (0,1) with following conditions:

    1. lim(x-->0)f(x) = lim(x-->1)f(x) = 0

    2.Exist M>0 so that |f ''(x)|<=M for all x in (0,1)

    Prove that |f '(x
    )|<=M/2
    Use the Taylor series with remainder f(x+t) = f(x) + tf'(x) + \tfrac12t^2f''(\xi), where \xi lies between x and x+t. This implies that |tf'(x) +f(x) - f(x+t)| = |\tfrac12t^2f''(\xi)|\leqslant\tfrac12Mt^2. There are now two separate cases to consider.

    If f(x) and f'(x) have the same sign, take t = 1-x. Then |tf'(x) +f(x)| = |(1-x)f'(x) +f(x)| \geqslant |(1-x)f'(x)|. Also, f(x+t) = f(1) = 0. So the inequality in the previous paragraph tells us that |(1-x)f'(x)| \leqslant \tfrac12M(1-x)^2. Hence |f'(x)|\leqslant \tfrac12M(1-x)\leqslant \tfrac12M.

    If f(x) and f'(x) have the opposite signs, take t = -x. Then |tf'(x) +f(x)| = |-xf'(x) +f(x)| \geqslant |xf'(x)|. Also, f(x+t) = f(0) = 0. So the inequality tells us that |xf'(x)| \leqslant \tfrac12Mx^2. Hence |f'(x)|\leqslant \tfrac12Mx\leqslant \tfrac12M.

    Quote Originally Posted by Drexel28 View Post
    Form the first degree taylor polynomial around that point f'(x)=f'(c)+\frac{f''(\xi)}{2!}(x-c)^2.
    The second term in that Taylor polynomial is f''(\xi)(x-c), not \frac{f''(\xi)}{2!}(x-c)^2.
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