9
$\begingroup$

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.

  • 1
    ... This is not how posting on this site works Potato. Do you want people to review your proof? Then ask them to. It seems to me you jsut wanted to share a proof of a classical exercise, a blog would be better suited for this.2012-07-02
  • 3
    @Olivier: [It is](http://meta.math.stackexchange.com/questions/4286/). See also [here](http://blog.stackoverflow.com/2012/05/encyclopedia-stack-exchange/)2012-07-02
  • 0
    @t.b. seems I was wrong.2012-07-02
  • 0
    See also [this thread](http://math.stackexchange.com/questions/16831/nonzero-f-in-c0-1-for-which-int-01-fxxn-dx-0-for-all-n) and [this thread](http://math.stackexchange.com/q/83813)2012-07-02
  • 0
    @OlivierBégassat: See [here](http://math.stackexchange.com/q/164827) for some context.2012-07-02
  • 0
    @Potato Sorry for my earlier comment, t.b. has shed some light on the issue.2012-07-02
  • 0
    @did: I'm not too fond of this either. However, no one objected on meta (see my comment to Olivier). In fact I edited out a "(unfortunately)" in my first comment.2012-07-02
  • 0
    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

13

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.

  • 0
    Does the $x^n$ matter in this case? Will the proof be the same if it's just another constant? Thanks.2016-04-21
  • 0
    @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