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Consider $ f$ from $[0,1]$to $[0,1] $ be continuous and non constant .

Then, is there $c\in[0,1]$ such that $f(c) =\int^1 _0 f^2(t) dt $ ?

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    In fact if $f$ is onto the function need not be continuous, integrable would be sufficient.2012-05-08

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Counterexample
Consider $f(x) = \frac{1}{2} (1-x) + \frac{1}{4} x = \frac{1}{4} \left(2-x\right)$. Then $ \int_0^1 f^2(x) \mathrm{d} x = \frac{1}{16} \int_0^1 \left(2-x\right)^2 \mathrm{d} x = \frac{1}{16} \int_1^2 y^2 \mathrm{d} y = \frac{1}{16} \cdot \frac{2^3-1^3}{3} = \frac{7}{48} $ But the equation $ \frac{1}{4} \left(2-x\right) = \frac{7}{48} \quad x = 2 - \frac{7}{12} = \frac{24-7}{12} = \frac{17}{12} $ but the solution $ x= \frac{17}{12}$ lies outside $[0,1]$ interval.

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    i was missing a number in my calculation , stupid of me. thanks:)2012-05-08