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Find all real number $n$ such that $|\frac{\sin(nx)}{n\sin(x)}|\le1\forall x\in\mathbb{R}-\{\pi k: k\in\mathbb{Z}\}$.

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Let $\phi_n(x) = |\frac{\sin(nx)}{n\sin(x)}|$. $\phi_0$ is undefined, so $n \neq 0$. Since $\phi_n = \phi_{-n}$, we can assume $n>0$.

Suppose $n>0$ is not an integer, and consider the sequence $x_k = \pi-\frac{1}{k}$. Then $x_k \to \pi$, but $n x_k \to n \pi \notin \pi \mathbb{Z}$. Consequently, $\lim_k \phi_n(x_k) = \infty$.

Now suppose $n>0$ is an integer, and let $z = e^{ix}$. Then $\phi_n(x) = |\frac{z^{n}-z^{-n}}{n(z-z^{-1})}| = |\frac{z^n}{k z}\frac{1-z^{-2n}}{1-z^{-2}}| = |\frac{z^n}{k z} \sum_{k=0}^{n-1} (z^2)^k| $. Since $|z| = 1$, we have $$\phi_n(x) \leq \frac{1}{k} \sum_{k=0}^{n-1} (1) = 1.$$

It follows that $\phi_n(x) \leq 1 $ for all appropriate $x$ iff $n$ is a non-zero integer.

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The inequality is true for all $n \in \mathbb{Z}^+$, which also implies immediately that it is true for $n \in \mathbb{Z}^-$. The proof follows immediately from induction.

For $n = 1$, it is true.

Assume it is true for $n=k$ i.e. we have $$\left\lvert \dfrac{\sin(kx)}{k \sin(x)} \right\rvert \leq 1$$

Then $$\sin((k+1)x) = \sin(kx) \cos(x) + \cos(kx) \sin(x)$$ $$\left \vert \sin((k+1)x) \right \vert = \left \vert \sin(kx) \cos(x) + \cos(kx) \sin(x) \right \vert \leq \left \vert \sin(kx) \cos(x) \right \vert + \left \vert \cos(kx) \sin(x) \right \vert\\ \leq k \left \vert \sin(x) \right \rvert + \left \vert \sin(x) \right \rvert$$

If $n \notin \mathbb{Z}^+$, then $\left\lvert \dfrac{\sin(kx)}{k \sin(x)} \right\rvert $ can be made arbitrarily large by choosing $x$ close to say $\pi$.

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Notice that, for $n=2^m \,,\,\, m = 0,1,2\dots\,,$ and using the identity $\sin(2x)=2\sin(x)\cos(x)\,,$ we have

$$ \sin( 2^m x ) = 2^m \sin(x) \cos(2x)\cos(4x)\dots\cos(2^{m-1}x) $$

$$ \implies \left|\frac{\sin{2^mx}}{2^m\sin(x)}\right|= |\cos(2x)\cos(4x)\dots\cos(2^{m-1}x)|\leq 1 \,. $$

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This does not seem to answer the question. – Did Oct 20 '12 at 13:24

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