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Possible Duplicates:
Nonzero $f \in C([0, 1])$ for which $\int_0^1 f(x)x^n dx = 0$ for all $n$

Slight generalization of an exercise in (blue) Rudin

What can we say about $f$ if $\int_0^1 f(x)p(x)dx=0$ for all polynomials $p$?

I found a nice problem I would like to share.

Problem: If $f$ is continuous on $[0,1]$, and if

$\int_0^1 f(x)x^n \ dx =0$

for every positive integer $n$, prove that $f(x)=0$ on $[0,1]$.

Source: W. Rudin, Principles of Mathematical Analysis, Chapter 7, Exercise 20.

I have posted a proposed solution in the answers.

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    Sorry, I'm horrible at searching for duplicates apparently. I have marked it for deletion, but it will take two days (says the site).2012-07-02

1 Answers 1

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We will show that the integral of $f^2$ over $[0,1]$ is 0. This will show that $f$ is zero, because if $f$ were not identically equal to zero, $f^2$ would be positive on some interval (by continuity) and have a nonzero integral.

Using the Stone-Weierstrass theorem, we can approximate $f$ uniformly by polynomials $P_n$ so that $||P_n-f||<1/n$ in the $\sup$ norm. The given condition obviously implies that the integral of $f(x)P(x)$ is zero for any polynomial $P$, by linearity. Note that

$\left|\int_0^1 f(x)f(x) \ dx \right|=\left|\int_0^1 f(x)f(x) \ dx - \int_0^1 f(x)P_n(x)\right|\le \int_0^1 |f(x)| |f(x)-P_n(x)|\ dx \\\le \frac{1}{n}\int_0^1 |f(x)| \ dx.$

The last integral is constant, so taking $n$ arbitrarily large completes the proof.

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    @lsy Yes it matters. For $\int_{0}^{1} f(x) P_n(x) dx$ to be $0$, you need the condition with $x^n$.2016-12-10