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Math Help - An Equality ?

  1. #1
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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(x)\,dx\leqslant\frac{1}{4}|f'(c)|.
    Last edited by bigli; June 6th 2010 at 03:01 AM. Reason: I made some mistakes in writing the question and NOW the question is correct.
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  2. #2
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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}) . If f be differentiable over (0,1) prove that there exist a  c\in (0,1) such that \Big|\int_0^1\!f(x)\,dx\Big|\leqslant\frac{1}{4}|f  '(c)|.
    I think you made some mistake writing the question. The way you wrote, it doesn't make sense. Take f to be a constant 1 function. Then f'(c)=0 for any c, but the integral is positive
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  3. #3
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    Try the mean value inequality
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  4. #4
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    Yes, rebghb was right and I fixed it.NOW, How does I solve it?
    Last edited by bigli; June 6th 2010 at 03:07 AM.
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  5. #5
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    Please HINT me.
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  6. #6
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    Quote Originally Posted by bigli View Post
    Please HINT me.
    is there f(0)=f(1/2)=0 or f(0)=f(1)=0?
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  7. #7
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    ah, ok, i think f(0)=f(1/2)=0 is the right condition.

    i am assuming that f is actually a positive on [0,1] function, since if not then we can essentially take |f| instead of f.

    denote M=\int_0^1 f(x)dx. assume for any c\in (0,1), |f'(c)|\le 4M

    first consider interval x\in[0,1/4]. by the mean value theorem there:  f(x)=f'(\theta_x) (x-0) (since f(0)=0), where \theta_x\in[0,x]. therefore f(x)=|f(x)|\le 4Mx for x\in[0,1/4]. therefore \int_0^{1/4}f(x)dx\le 4M\int_0^{1/4}xdx=M/8.

    same arguments for x\in[1/4,1/2] (using mean value theorem at the point 1/2, where f is equal to zero) show \int_{1/4}^{1/2}f(x)dx\le M/8.

    finally consider x\in[1/2,1]. again, by the mean value theorem:  f(x)=f'(\theta_x) (x-1/2) (since f(1/2)=0), where \theta_x\in[1/2,x]. therefore f(x)=|f(x)|\le  4M(x-1/2) for x\in[1/2,1]. therefore \int_{1/2}^{1}f(x)dx\le 4M\int_{1/2}^{1}(x-1/2)dx=M/2.

    combining, we get M=\int_{0}^{1}f(x)dx\le M/8+M/8+M/2=3M/4, contradiction.
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