I have found in some book the following :
$\int^{1}_{0} f(t)t^{n}dt$ = 0 for all $n$ in $N$ iff $f(t)$ = 0 where $f(t)$ is a real valued continuous function on [0,1].
I don't understand the proof of this. How to prove this ?
I have found in some book the following :
$\int^{1}_{0} f(t)t^{n}dt$ = 0 for all $n$ in $N$ iff $f(t)$ = 0 where $f(t)$ is a real valued continuous function on [0,1].
I don't understand the proof of this. How to prove this ?
$f$ is a continuous function on a compact set [0,1], by weierstrass approximation theorem, we know there is a polynomial that can approximate $f$ arbitrarily close in that interval.
Let $M=max\{f(x): x\in [0,1]\}$ and an $\epsilon>0$ be given.
Let $p(x)$ be the polynomial such that $|f(x)-p(x)|<\epsilon/M$ on $[0,1]$.
Observed that
the hypothesis gives $\int_0^1 f(x)p(x)dx=0$ since $\int_0^1 f(x)x^n dx=0 , n=0,1,2, \cdots$ hence
$|\int_0^1 f^2 dx|\leq \int_0^1 |f(x)||(f(x)-p(x))|dx \leq M\epsilon/M=\epsilon$
This gives $\int_0^1 f^2=0$ and since $f$ continuous, we must conclude that $f=0$ on $[0,1]$.