If $f : [0,1] \to \mathbb R$ is a continuous function and $\int_{0}^1f(x)x^ndx=0 , \forall n \in \mathbb Z^+\cup\{0\}$ then is it true that
i) $\int_{0}^1(f(x))^2dx=0$ ?
ii) $\int_{0}^1f(f(x))dx=0$ ?
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Note that $f$ can be estimated to within any positive bound by finite degree polynomials. – Arthur Dec 12 '14 at 14:07
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See here: http://math.stackexchange.com/questions/16831/nonzero-f-in-c0-1-for-which-int-01-fxxn-dx-0-for-all-n/ – Hans Lundmark Dec 12 '14 at 14:16
1 Answers
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We can use Weierstrass approximation theorem see http://en.wikipedia.org/wiki/Stone–Weierstrass_theorem
To prove $f(x)=0$ Because,$\int_0^1 f^2(x)dx=\int_0^1 \left(f^2(x)-f(x)p(x)\right) dx\le \int_0^1 \left|f(x)-p(x)\right| dx<M\epsilon$. Where $M$ is the maximum of $f$ in $[0,1]$ and $p(x)$ is a polynomial choose in Weierstrass approximation theorem .
