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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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    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

3 Answers 3

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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 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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    @James Fennell: thank you, I missed the case when$f$does not change sign.2012-11-29