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Thread: An Equality ?

  1. #1
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    An Inequality ?

    Let $\displaystyle f: [0,1]\longrightarrow\Bbb{R}$ be a continous function and $\displaystyle f(0)=f(\frac{1}{2})=0$ . If $\displaystyle f$ be differentiable over $\displaystyle (0,1)$, prove that there exist $\displaystyle c\in (0,1)$ such that $\displaystyle \int_0^1\!f(x)\,dx\leqslant\frac{1}{4}|f'(c)|$.
    Last edited by bigli; Jun 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 $\displaystyle f: [0,1]\longrightarrow\Bbb{R}$ be a continous function and $\displaystyle f(0)=f(\frac{1}{2})$ . If $\displaystyle f$ be differentiable over $\displaystyle (0,1)$ prove that there exist a $\displaystyle c\in (0,1)$ such that $\displaystyle \Big|\int_0^1\!f(x)\,dx\Big|\leqslant\frac{1}{4}|f '(c)|$.
    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; Jun 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 $\displaystyle M=\int_0^1 f(x)dx$. assume for any $\displaystyle c\in (0,1)$, $\displaystyle |f'(c)|\le 4M$

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

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

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

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