I am trying to show that for f in C[0,1] , and ##n=0,1,2,... ## we have:(adsbygoogle = window.adsbygoogle || []).push({});

## \int_0^1 x^n f(x)dx =0 ## (&&) , then

##f(x)==0 ## .

I am using Weirstrass approximation, so that , for any ## \epsilon >0 ## , there is ## P_n(x) =

a_0+a_1x +..+x^n ## with : ##Sup_{x in [0,1]} | P_n(x)-f(x)| < \epsilon ##.

We then sub-in in (&&) to get, for ## i=0,1,..,n ##. :

##0= \int _0^1 a_i x^i => \int_0^1 P_n(x)f(x)dx =0 ## , so :

## \int_0^1 [f(x) \pm \epsilon] f(x)dx = \int_0^1 f(x)^2 + (\epsilon) f(x) =0 ##

We let ## \epsilon →0 ## , and then ##\int_0^1 f(x)^2 =0 → f(x)=0 ##.

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# Integral x^n *f(x) dx =0 ; f for all n, f in C[0,1], then f(x)=0

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