Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?

share|cite|improve this question
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
up vote 1 down vote accepted

$f$ 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 $[0,1]$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.