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Many posts on this site address the following problem in the affirmative:

If $f:[0,1]\to \mathbb R$ is continuous and $\int_0^1 x^n f(x)dx=0$ for all positive integers $n$ then must $f\equiv 0$?

The standard proof is to first note that the product of $f$ and any polynomial integrates to zero, so that approximating $f$ by a polynomial using Weierstrass gives $\int f^2 = 0$.

My question is: what happens if we change the domain of $f$ to the open interval $(0,1)$ while still requiring continuity? Does there then exist a counterexample to the above statement?

RDL
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  • See my answer here, and couple with continuity of $f$ at the very end to deduce $f=0$ on $(0,1)$, not just a.e. The point is you shouldn’t look at $\int f^2$, but rather apply Weierstrass more carefully. – peek-a-boo Jan 23 '25 at 05:23
  • what do you mean by the integral? in other words does $f$ need to be absolutely integrable (in $L^1(0,1)$) or just integrable and all $x^nf$ integrable too? – Conrad Jan 23 '25 at 13:48
  • Is this the Riemann or Lebesgue integral? If $f$ is Lebesgue integrable then it is straightforward to show that $f(x)=0$ ae. – copper.hat Jan 23 '25 at 18:36

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