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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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    What for $f$ constant, with a constant different from $0$ and $1$?2012-05-08
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    I have edited the question. i hope it makes sense :D the question came just as a thought .2012-05-08
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    Just take Davide's suggestion and introduce a very small bump.2012-05-08
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    $$f(x)=a+bx$$ where $a \in [0,1]$ and $b$ is very very small.2012-05-08
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    Perhaps you meant to say $f$ is onto (i.e. surjective)?2012-05-08
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    @AD how would that change , i am just trying various things !2012-05-08
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    If $f$ is onto the problem becomes trivial, since $f$ takes all values in $[0,1]$.2012-05-08
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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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    @ Sasha, i think there is something wrong with ur calculations , can u check once.2012-05-08
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    @Vedananda I filled in steps.2012-05-08
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    i was missing a number in my calculation , stupid of me. thanks:)2012-05-08