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I've been trying to prove that $\sin(\lambda x)$ only has a single fixed point (over all the real number) for $0< \lambda < 1$. I've thought of using the fixed point theorem, since it's obvious that $|sin'(\lambda x)| < 1$. But, the other criteria (that, for $g(x) = \sin(\lambda x)$, $g([a,b]) \subset [a,b] $), aren't met.

I know, intuitively the result is true and that the only fixed point is at $x = 0$, but I'm struggling to prove that.

Rodrigo
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If $x\ne0$ then $|\sin(\lambda x)|<|\lambda x|<|x|$.

David
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