Can someone give me a hint on the following problem? I'm not sure what to do...
Suppose $f\in L^1([0,1])$ is such that for all $n=0,1,2,...$ we have
$$\int_0^1 f(x)(\sin x)^n\,dx = 0.$$
Show that $f=0$ almost everywhere
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If $f\in L^1(0,1)$, by setting $g(x)=\frac{f(\arcsin x)}{\sqrt{1-x^2}}$, we have $g\in L^1(0,\sin 1)$ and $$ \forall n\in\mathbb{N},\qquad \int_{0}^{\sin 1} x^n\,g(x)\,dx = 0.\tag{1} $$ That implies $g\equiv 0$ a.e. on $(0,\sin 1)$, so $f\equiv 0$ a.e. on $(0,1)$.
Jack D'Aurizio
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I think you should mention why g = 0 a.e., and I think this question answers that. – Kurt Aug 21 '15 at 01:40
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The part about g = 0 a.e. just follows from the fact that g (being continuous on closed intervals) is the uniform limit of polynomials, on closed intervals of course, right? – SwifferCat Aug 21 '15 at 01:44
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@SwifferCat $g$ isn't continuous. but, you may approximate $f$ with continuous function in $L_1$ norm. – user251257 Aug 21 '15 at 01:53
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@user251257: thanks, I was just saying that. – Jack D'Aurizio Aug 21 '15 at 01:53
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@JackD'Aurizio oh sorry :( – user251257 Aug 21 '15 at 01:55
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@user251257: don't be sorry, you saved me from typing that :D – Jack D'Aurizio Aug 21 '15 at 01:59
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1@user251257 and Jack D'Aurizio. I thought it might be useful for the OP and other readers to have the reference by name to The Weierstass Approximation Theorem used herein. – Mark Viola Aug 21 '15 at 04:37