9
$\begingroup$

Let $ f $ be a continuous function defined on $ [0,\pi] $. Suppose that

$$ \int_{0}^{\pi}f(x)\sin {x} dx=0, \int_{0}^{\pi}f(x)\cos {x} dx=0 $$

Prove that $ f(x) $ has at least two real roots in $ (0,\pi) $

  • 0
    I have the feeling that the Fourier expansion $f(x) = a_0 + \sum_{n\geq 1} (a_n \sin 2n x + b_n \cos 2n x)$ might help, but can't see how exactly.2012-11-29
  • 0
    @LiorB-S Since $f$ is "just@ continuous you can't be sure that Fourier series of $f$ converges to $f$.2012-11-29
  • 0
    @Norbert: You are right, but I have no proof for smooth $f$'s either. Moreover it would be nice, I think, to have if a proof using Fourier expansion exists...2012-11-29
  • 1
    It's worth remarking that the function $f(x)=\sin(3x)$ satisfies the conditions and has exactly 2 roots in $(0,\pi)$, so 2 is the best possible integer in this problem.2012-11-29

2 Answers 2