Kvant Math Problem 1501

Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 9m09s
Source on kvant.digital

Problem

V. A. Senderov

Exploration

For small $x$, the functions admit linear approximations $\sin(kx)\sim kx$ and $\sin x\sim x$. This suggests that the given inequality

$$|a_1\sin x+a_2\sin2x+\cdots+a_n\sin nx|\le|\sin x|$$

should encode a bound on the first-order behavior of the trigonometric polynomial at $x=0$, which is governed by the coefficient combination $a_1+2a_2+\cdots+na_n$.

Dividing both sides by $|\sin x|$ for $x\neq 0$ leads to a bounded expression involving $\sin(kx)/\sin x$. The central issue is whether one can pass to the limit as $x\to 0$ and identify the limit of each term as $k$. Since the sum is finite, termwise limiting is valid once each ratio is controlled near zero.

The delicate point is ensuring that the passage to the limit is legitimate despite the absolute values and the possible sign changes of $\sin x$. This is resolved by working with $x\to 0$ where $\sin x$ has fixed sign and nonzero magnitude except at the limit point.

Problem Understanding

This is a Type B problem: one must prove that a given uniform inequality for a trigonometric polynomial implies a linear inequality for its coefficients.

The structure suggests extracting the first-order Taylor behavior at $x=0$ from a global inequality valid for all $x$. The core difficulty is justifying the limiting transition from a uniform bound on a ratio of trigonometric expressions to a bound on the corresponding linear combination of coefficients.

The claim to be proved is

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

Proof Architecture

The first lemma identifies that for fixed integer $k$, the limit of $\frac{\sin(kx)}{\sin x}$ as $x\to 0$ equals $k$, justified by standard trigonometric limits.

The second lemma establishes that the given inequality implies

$$\left|\sum_{k=1}^n a_k \frac{\sin(kx)}{\sin x}\right|\le 1$$

for all sufficiently small nonzero $x$.

The third step uses the finiteness of the sum to pass the limit inside the summation and deduce convergence to $a_1+2a_2+\cdots+na_n$.

The most delicate step is the controlled passage to the limit in the presence of absolute values and a vanishing denominator, which is resolved by restricting to a neighborhood where $\sin x$ does not change sign.

Solution

For $x\neq 0$ such that $\sin x\neq 0$, the given inequality can be rewritten as

$$\left|\sum_{k=1}^n a_k \frac{\sin(kx)}{\sin x}\right|\le 1.$$

Fix $k$. Using the identity

$$\frac{\sin(kx)}{\sin x} = k \cdot \frac{\sin(kx)}{kx}\cdot \frac{x}{\sin x},$$

and the standard limit $\lim_{t\to 0}\frac{\sin t}{t}=1$, we obtain

$$\lim_{x\to 0}\frac{\sin(kx)}{\sin x} = k\cdot \lim_{x\to 0}\frac{\sin(kx)}{kx}\cdot \lim_{x\to 0}\frac{x}{\sin x} = k.$$

Since the sum is finite, it follows that

$$\lim_{x\to 0}\sum_{k=1}^n a_k \frac{\sin(kx)}{\sin x} = \sum_{k=1}^n a_k \lim_{x\to 0}\frac{\sin(kx)}{\sin x} = \sum_{k=1}^n k a_k.$$

To connect this with the inequality, consider $x$ approaching $0$ through values where $\sin x\neq 0$. The inequality implies that the absolute value of the expression is bounded by $1$ for all such $x$, hence every limit point of the expression lies in the closed interval $[-1,1]$. Since the limit exists, it follows that

$$\left|\sum_{k=1}^n k a_k\right|\le 1.$$

This completes the proof. ∎

Verification of Key Steps

The key analytic step is the evaluation of $\lim_{x\to 0}\frac{\sin(kx)}{\sin x}$. Writing it as a product of $\frac{\sin(kx)}{kx}$ and $\frac{x}{\sin x}$ isolates two standard limits, each equal to $1$ as $x\to 0$, which yields the value $k$ without ambiguity.

The second delicate point is the transition from a uniform bound on a function to a bound on its limit. The inequality ensures the expression remains in $[-1,1]$ for all sufficiently small admissible $x$, and the existence of the limit forces the limit value to remain within the same closed interval.

The finiteness of the sum prevents any interchange issues; each term converges independently, and summation preserves convergence without additional conditions.

Alternative Approaches

One alternative approach uses Fourier-analytic duality. The given inequality can be interpreted as a uniform bound on a trigonometric polynomial divided by $\sin x$, and one may test it against kernels concentrating at zero, such as rescaled Dirichlet kernels, to extract the linear functional $\sum k a_k$.

Another approach expands both sides to first order in $x$ and compares coefficients directly after proving that the inequality forces cancellation of higher-order terms, though this typically requires more careful error control than the limit argument used above.