If $f$ belongs to $C[-\pi, \pi]$ & $\int_{-\pi}^{\pi}\ x^nf(x) =0$ for all non negative integer n. Then $f$ will be identically $0$.
How to do it by stone-weierstrass theorem?
Can anyone help me?
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It is not true. Take $f = 1_{[-\pi,0]} \cdot \sin$. – copper.hat Jan 09 '18 at 22:33
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Did you mean the integral to start at zero? You need to clarify the question. – copper.hat Jan 09 '18 at 22:41
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Can you clarify the question please? – copper.hat Jan 09 '18 at 22:59
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@m_t_: Note the space and the limits on the integration. – copper.hat Jan 09 '18 at 23:06
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Extremely sorry... I just edited the question @copper.hat m_t_ – cmi Jan 10 '18 at 03:33
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Using Stone-Weierstrass theorem, you know that it exists $\left(P_n\right)_{n \in \mathbb{N}}$ a sequence of polynomial that uniformly converges to $f$. Hence you can write that for all $\ n \in \mathbb{N}$ and for all $\ x \in \left[-\pi,\pi\right]$ $$ \left|f(x)P_n(x)-f^2(x)\right|=\left|f(x)\right|\left|P_n(x)-f(x)\right| \leq \left\|f\right\|_{\infty, \left[-\pi,\pi\right]}\left\|P_n-f\right\|_{\infty, \left[-\pi,\pi\right]} $$ You know that second term tends to $0$ as it is UNIFORMLY convergent. So $$\left\|fP_n-f^2\right\|_{\infty, \left[-\pi,\pi\right]} \underset{n \rightarrow +\infty}{\rightarrow}0$$ So we have shown that the sequence $\left(fP_n\right)_{n \in \mathbb{N}}$ converges uniformy to $f^2$. I let you conclude ( on $\left[0, \pi\right]$ )starting from there. ( that's not obvious to think to do that )
Atmos
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I proved it was the case on $\left[0,\pi\right]$, you are right. But i think he wants to show it is $0$ only on $\left[0,\pi\right]$, if not its proposition if obviously false – Atmos Jan 09 '18 at 22:44
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