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 $1$? – Davide Giraudo May 8 '12 at 15:21
I have edited the question. i hope it makes sense :D the question came just as a thought . – Theorem May 8 '12 at 15:24
Just take Davide's suggestion and introduce a very small bump. – David Mitra May 8 '12 at 15:26
$$f(x)=a+bx$$ where $a \in [0,1]$ and $b$ is very very small. – N. S. May 8 '12 at 15:28
Perhaps you meant to say $f$ is onto (i.e. surjective)? – AD. May 8 '12 at 15:30

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. – Theorem May 8 '12 at 15:57
@Vedananda I filled in steps. – Sasha May 8 '12 at 16:52
i was missing a number in my calculation , stupid of me. thanks:) – Theorem May 8 '12 at 16:59