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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) $

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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. –  Lior B-S Nov 29 '12 at 17:20
    
@LiorB-S Since $f$ is "just@ continuous you can't be sure that Fourier series of $f$ converges to $f$. –  userNaN Nov 29 '12 at 17:27
    
@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... –  Lior B-S Nov 29 '12 at 17:32
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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. –  James Fennell Nov 29 '12 at 21:04

2 Answers 2

Here is one root: Let $F(x) = \int_{0}^x f(t) \sin t dt$. Then $F(0)=0$ and $F(\pi)=\int_{0}^\pi f(t)\sin tdt=0$. So by the intermediate value theorem, there exists $0<c<\pi$ such that $$ 0=F'(c) = f(c)\sin c. $$
But since $\sin c\neq 0$, we get that $f(c)=0$.

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If f has only one real root on $ (0,\pi)$, say $ a \in (0,\pi) $, then define $ g(x) = f(x) \sin(x-a) = f(x) (\sin(x)\cos(a) - \cos(x)\sin(a))$, then $ g(x) $ is either non-positive or non-negative, not identically zero, and has integral $ 0 $. Contradiction.

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Why is $g(x)$ either non-positive or non-negative? –  TonyK Nov 29 '12 at 19:32
    
@Lev - I think you assume that $f$ changes sign at its root, which doesn't necessarily happen (the function could be tangent to the $x$-axis). However, if $f$ doesn't change sign and is not identically zero, the integral $\int_0^\pi f(x) \sin(x)dx$ would be strictly positive (or strictly negative, if you take $f\leq 0$). This is again a contradiction. –  James Fennell Nov 29 '12 at 19:53
    
@TonyK: Note that $\sin (x-a)>0$ for $a<x<\pi$ and $\sin(x-a) <0$ for $0<x<a$. –  Lior B-S Nov 29 '12 at 19:57
    
Yes, OK -- if $f$ doesn't change sign at $a$, then the $\sin$ integral can't be zero. So $f$ does change sign at $a$, and therefore $f(x)\sin(x-a)$ doesn't. –  TonyK Nov 29 '12 at 20:01
    
@James Fennell: thank you, I missed the case when f does not change sign. –  Lev Nov 29 '12 at 20:36

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