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Assume $f:[−1, 1] \to \mathbb{R}$ is continuous and for all $n \in \mathbb{Z}^+ $ we have $\int_{−1}^1 \sin^n(x)f(x) dx=0$. Show that $f\equiv0$.

Possibly Stone - Weierstrass and the sandwich theorem. But I don't really understand how to apply the first theorem at all. Help or hints are welcome.

Myunghyun Song
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    S.-W. says the subalgebra of $C([-1,1],\Bbb{R})$ generated by $x\mapsto\sin x$ (and $x\mapsto 1$) is a dense one. And... – metamorphy Feb 28 '19 at 17:57
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    @user439126, don't forget to select an answer of the questions you have asked. You didn't do it for any of them. Do you need help with that? It's the green tick. – Guillermo Mosse Feb 28 '19 at 18:01
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    @Guillermo 1. wait more than 1 minute before assuming the questioner forgot to accept an answer. 2. don't assume that just because there are answers, one of them needs to be accepted. 3. if those answers contain hints, expect it to take a while to work out the solution before the questioner can decide if the hint was helpful or not. – Trevor Gunn Feb 28 '19 at 18:04
  • @TrevorGunn, I agree with you in all of your points, but the user has 8 unclosed questions. For example, this one: https://math.stackexchange.com/questions/3112386/does-fx-0-for-all-x-in-0-1 I of course wan't telling him about this question in particular. – Guillermo Mosse Feb 28 '19 at 19:18
  • https://math.stackexchange.com/questions/3107293/is-there-a-point-where-the-value-of-the-function-g-is-greater-than-the-length https://math.stackexchange.com/questions/3106682/show-g0-1-rightarrow-0-1-is-continuous – Guillermo Mosse Feb 28 '19 at 19:21

1 Answers1

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Hint: Let $g(x) = \frac{f(x)}{\cos x}$ and note that by change of variable $\sin x=t$, $$ \int_{-1}^1 \sin^n (x)f(x)\mathrm dx=\int_{-\sin(1)}^{\sin 1} t^n g(\arcsin t)\mathrm dt=0. $$

Myunghyun Song
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    Here is a sketch; by invoking Weierstrass approximation theorem, we can find polynomials $p_n(t)\to g(\arcsin t)$uniformly on $[-\sin 1,\sin 1]$. With the given assumption, we have that $\int g(\arcsin t)^2 =0$, which implies $g(\arcsin t)\equiv 0$. It can be noted that $f(x)=\cos x\cdot g(x)\equiv 0$ is implied by this. The substitution is made to avoid the use of Stone-Weierstrass theorem, somewhat more advanced theorem than Weierstrass approximation theorem. – Myunghyun Song Feb 28 '19 at 23:19
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    @ViktorGlombik This would be a spoiler ... but we can see that $$\int t^k\cdot g =0,\ \forall k \implies \int p_n(t)\cdot g =0 \xrightarrow{n\to\infty} \int g^2=0.$$ That is, if $g$ is orthogonal to every monomial (thus polynomial by linearity), then it should be orthogonal to itself by denseness of polynomials. – Myunghyun Song Feb 28 '19 at 23:26