Prove |a₁+2a₂+...+naₙ|≤1 for P(x)=∑aₖsin(kx) with |P(x)|≤|sin x|

[anemone]

Here is this week's POTW:

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Let $P(x)=a_1\sin x+a_2\sin 2x+\cdots+a_n\sin nx$ where $a_1,\,a_2,\,\cdots,\,a_n$ are real numbers. Suppose that $|P(x)|\le |\sin x|$ for all real $x$, prove that

$$|a_1+2a_2+\cdots+na_n|\le 1$$

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[anemone]

Congratulations to the following members for their correct solution::)

1. Opalg
2. lfdahl

Solution from Opalg:

Spoiler

Notice that $|P(0)| \leqslant |\sin0| = 0$, and so $P(0)=0.$

Next, $\left|\dfrac{P(x) - P(0)}x\right| = \left|\dfrac{P(x)}x\right| \leqslant \left|\dfrac{\sin x}x\right|$. Since weak inequalities are preserved by taking limits, it follows that $$\left|\lim_{x\to0}\dfrac{P(x) - P(0)}x\right| \leqslant \left|\lim_{x\to0}\dfrac{\sin x}x\right|.$$ But by the definition of a derivative, the left side is $|P'(0)| = |a_1 + 2a_2 + \ldots + na_n|$; and the limit on he right side is $1$. Therefore $|a_1 + 2a_2 + \ldots + na_n| \leqslant 1.$