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Showing that $ \int_0^1 x^{2n}f(x) dx = 0 $ implies $f = 0$
Hi, tried looking at this link.
I did not understand how to answer regarding odd function, I receive the answer is that it is not correct, The integral is indeed $0$, but only because the integral of multiplication of odd functions is $0$, it does not say that $h(x)$ must be constant 0.

Fourier_Asker
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1 Answers1

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Let $g(x)=xh(x)$. Then $\int_0^{1} x^{2n}g(x)dx=0$ for $n=0,1,2,\cdots$. Hence, $\int_0^{1}\phi (x)g(x)dx=0$ for any $\phi \in span \{1,x^{2},x^{4},\cdots\}$. By Stone-Wierstrass Theorem $span \{1,x^{2},x^{4},\cdots\}$ is dense in $C[0,1]$ (w.r.t. the sup norm). It follows now that $\int_0^{1}[g(x)]^{2}dx=0$ and so $g \equiv 0$ and hence, $h\equiv 0$.

Kavi Rama Murthy
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  • Hi, thanks for the answer, I can not understand why you are allowed to choose h(x) as you wish? sorry for being a begineer, I can not understand the proof. – Fourier_Asker Dec 06 '22 at 08:47
  • and just a question, I have to prove $h(x) = 0$, not $g(x)=0$, because I see you had g(x) and h(x). so it doesnt mean that $h(x)=0$, the statement is false, am I right? – Fourier_Asker Dec 06 '22 at 08:48
  • You have different notations in the title and the question. I am using the one in the question. My $h$ is not the given funciton, $f$ is the given function. Now $\int hg =0$ for every continuous function $h$, so I can take $h=g$ to get $\int g^{2}=0$. @Fourier_Asker – Kavi Rama Murthy Dec 06 '22 at 08:50
  • Oh my, sorry, I will change the title, it has to be $2n+1$, sorry!! one moment!! ( as should be odd moment, i accidently wrote even ), I will fix!! – Fourier_Asker Dec 06 '22 at 08:51
  • Okay, if I understand it:------------ f(x) is like h(x) on mine---------- h(x) is a "new function" ------- and g(x) is also a"new function"------, problem is, you are saying $g(x)=0$ and not $f(x)=0$, so it doesnt imply that $f(x)=0$ – Fourier_Asker Dec 06 '22 at 08:53
  • @Fourier_Asker Anyway, I have edited my answer so tha you can understand it better. – Kavi Rama Murthy Dec 06 '22 at 08:55
  • Oh I see now, now I understand it, when you said the one in the question, you meant the linked one. Thank you very much! – Fourier_Asker Dec 06 '22 at 08:56