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$f$ is Riemann integrable in $[a,b]$,and $\int_a^b f(x)dx>0$. If the polynomial $P(x)$ satisfies $\int_a^b P^2(x)f(x)dx=0$. Prove $P(x)=0$.

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    @EduardoSiva: Note that this is one particular $f$ and one particular $P$.2012-04-21

2 Answers 2

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If, in addition, $f$ is non-negative, here is a hint.

Hint: Polynomials that are not identically $0$ can vanish only on a finite set. Consider the open sets $ U_k=\left\{x\in(a,b):|P(x)|>\frac1k\right\}\tag{1} $ and let $ F_k=\int_{U_k}f(x)\,\mathrm{d}x\tag{2} $ Show that $ \int_a^bf(x)\,\mathrm{d}x=F_1+\sum_{k=2}^\infty(F_{k}-F_{k-1})\tag{3} $ and $ \int_a^bP^2(x)f(x)\,\mathrm{d}x\ge F_1+\sum_{k=2}^\infty(F_{k}-F_{k-1})\frac{1}{k^2}\tag{4} $ What conclusions can you draw from $(3)$ and $(4)$?

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You need more conditions on $f, P$.

Choose $f(x) = 1$ when $x \in [0,\sqrt[3]{\frac{2}{3}}]$, and $f(x)=-2$ when $x \in (\sqrt[3]{\frac{2}{3}}, 1]$. Let $P(x) = x$. Then I have $\int_0^1 f(x) dx > 0$, and $ \int_0^1 P^2(x) f(x) dx = 0$, but clearly $P \neq 0$.

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    It is true under this condition, as @robjohn pointed out earlier.2012-04-21