# how can I show this question : function

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? –  Dilip Sarwate Dec 20 '11 at 18:01

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.
$f$ 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 $f'(0)$.