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Jhevon · #8 November 11th, 2007, 01:10 AM

Quote:

Originally Posted by CaptainBlack

Let $P_n(x);\ n \in \mathbb(Z)_+$ be an orthonormal basis of polynomials for $L^2_{[0,1]}$ .

Then by the conditions specified in the problem:

$\langle P_n, f-g \rangle =0;\ \forall n \in \mathbb{Z}_+$

which implies that $f(x)-g(x)=0\ a.e. \in [0,1]$ which as f-g is continuous implies $f(x)-g(x)=0\ x\in [0,1]$

RonL

i always have trouble proving things that seem to me to be obvious.

would it be wrong to do this?

$\int_0^1 x^n f(x)~dx = \int_0^1 x^n g(x)~dx$

$\Rightarrow \int_0^1 x^n f(x)~dx - \int_0^1 x^n g(x)~dx = 0$

$\Rightarrow \int_0^1 x^n [f(x) - g(x)]~dx = 0$

Obviously, the only way this integral can be zero for all x and all n is if $f(x) - g(x) = 0$ . the result follows immediately.

Now i can see where the problem here would be, the "obviously" part.