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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 non-negative 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.

Potato
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1 Answers1

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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.

Potato
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  • Does the $x^n$ matter in this case? Will the proof be the same if it's just another constant?

    Thanks.

     – lsy Apr 21 '16 at 06:41
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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 $x^n$. – Prince Kumar Dec 10 '16 at 11:39
  • In the second pair of absolute bars you subtract a quantity without adding a quantity, I don't follow this equality. Why is it not necessary to also add $f(x)P_n(x)$ within an integral? – oliverjones Nov 05 '20 at 23:54
  • I think this proof is not completely right, because you are not considering the polynomials with constant terms in the first part of your argument. For instance, if you assume $P(x) = x^2 + 1$, you have that $$\int_0^1 f(x)P(x)dx = \int_0^1 x^2f(x)dx + \int_0^1 f(x)dx,$$ and the hypothesis only guarantees that the first term will vanish. – Lord Vader Jan 23 '23 at 18:53
  • What justifies the interchange of summation and integration when showing $\int^1_0 \mathrm{d}x f(x)P_n(x) = 0$? Is there some condition which I overlooked? – Jono94 Mar 10 '24 at 09:08
  • @LordVader The integral against the constant vanishes by the $n=0$ case of the hypothesis. – Potato Mar 10 '24 at 19:49
  • @Jono94 It's a finite sum, using the fact that integration is additive in the integrand. Why do you think there would be a problem? – Potato Mar 10 '24 at 19:50
  • @Potato would letting $n$ run to infinity possibly make the sum not absolutley convergent, and so we cannot interchange integration and summation? – Jono94 Mar 11 '24 at 02:43
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    @Jono94 I agree that for an infinite sum, one must be more careful. But the sum in $P_n$ here is always finite. – Potato Mar 11 '24 at 03:27