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Math Help - An inequality

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    An inequality

    Let f: [0,1]\longrightarrow\Bbb{R} be a continous function and f(0)=f(\frac{1}{2})=0 . If f be differentiable over (0,1), prove that there exist   c\in (0,1) such that \int_0^1\!f(t)\,dt\leqslant\frac{1}{4}|f'(c)|.
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    What ideas have you already tried?
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    MHF Contributor Unknown008's Avatar
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    Because, anyone doesn't answered me and I guess the question is probably a question in analysis.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by bigli View Post
    Let f: [0,1]\longrightarrow\Bbb{R} be a continous function and f(0)=f(\frac{1}{2})=0 . If f be differentiable over (0,1), prove that there exist   c\in (0,1) such that \int_0^1\!f(t)\,dt\leqslant\frac{1}{4}|f'(c)|.
    You're clearly going to need to use the MVT and the fact that an integral is less than the measure of the interval times the supremum of it on that interval.
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  6. #6
    Member mohammadfawaz's Avatar
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    First, let F(x)=\int_0^x f(t)dt. Apply MVT to this function on [0,1]: F(1)-F(0)=F'(c_1) where c between 0 and 1.
    Clearly, F(0)=0 and F(1)=\int_0^1 f(t)dt. Also, by the FTC, F'(c_1)=f(c_1). Hence, we have: \int_0^1 f(t)dt = f(c_1).
    Let's now apply the MVT for the function f on the interval [c_1, \frac{1}{2}] (or the other way around). We get: |f(c_1)-f(0.5)|=|f'(c_2)(c_1-0.5)|\leq \frac{1}{2}|f'(c_2)|.
    Therefore, |f(c_1)|\leq \frac{1}{2}|f'(c_2)| i.e. |\int_0^1 f(t)dt|\leq \frac{1}{2}|f'(c_2)|.
    That what I was able to reach so far...Inspire from this. The continuation must be in a similar manner.

    Hope this helps
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