# problem on definite integral

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 ?

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This can be proved by weierstrass approximation theorem. – KWO Oct 28 '12 at 13:41
The property you mention implies that $\int_0^1 f(x)P(x)dx=0$ for any polynomial $P$. Since $f$ can be approximated uniformly by a sequence of polynomials it follows that $\int_0^1 f^2(x)dx=0$ and therefore $f=0$. – Beni Bogosel Oct 28 '12 at 13:41

$f$ 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 $[0,1]$.

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If you are satisfied with this answer then you can mark this as answered, otherwise this question will still be considered unanswered. – KWO Oct 29 '12 at 9:08