$f$ 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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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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