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

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 $f$. – Norbert 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
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

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