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Let $f\colon\mathbb R\to\mathbb R$ defined by

$f(x)=a_1\sin x+a_2\sin2x+a_3\sin3x+\cdots+a_n\sin nx,$ for some values $a_1,a_2,a_3,\cdots,a_n\in\mathbb R.$ Prove that $ |f(x)|\le|\sin x|\quad \forall x\in \mathbb{R}$ implies: $|a_1+2a_2+3a_3+\cdots+na_n|\le1$

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    Should that be $3a_3$ in your last displayed equation?2011-12-20

2 Answers 2

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For $x$ such that $\sin x\neq 0$, we have $\left|\sum_{j=1}^na_j\frac{\sin(jx)}{\sin x}\right|\leq 1$ so $\left|\sum_{j=1}^nja_j\frac{\sin(jx)}{jx}\frac x{\sin x}\right|\leq 1,$ and taking the limit $x\to 0$ we get $\left|\sum_{j=1}^nja_j\right|\leq 1,$ which is the wanted result.

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Here's another perspective on Davide's answer:

$f$ is a finite sum of differentiable functions and therefore differentiable, and $|f(x)|\le|\sin x|$ everywhere implies |f'(0)|\le |\sin'(0)|=1. Now the sought inequality follows simply by evaluating f'(0).