Track / Stage 6 / 56 of 400 #1056 of 1964

Problem 1056

National olympiad, first round

Algebra Difficulty 6.1 Prove it

12. Let $f, g$ be real functions defined on $(-\infty,+\infty)$ , and satisfy the functional equation $f(x+y)+f(x-y)=2 f(x) g(x)$ for all $x$ and $y$ . Try to prove that if $f(x) \neq 0$ , and $|f(x)| \leqslant 1$ for all $x$ , then $|g(y)| \leqslant 1$ for all $y$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

12. Let $M=\sup |f(x)|$ , then by the given condition, $M>0$ , for any $\delta, 0 < \delta < M$ , there exists $x$ such that $|f(x)| > M-\delta$ . At this point, for any $y$ , we have
$\begin{aligned} 2 M & \geqslant|f(x+y)|+|f(x-y)| \\ & \geqslant|f(x+y)+f(x-y)| \\ & =2|f(x)||g(x)|>2(M-\delta)|g(y)|, \end{aligned}$

Thus, $|g(y)| < \frac{2M}{2(M-\delta)} = \frac{M}{M-\delta}$ . As $\delta \to 0$ , we have $|g(y)| \leq 1$ . If $|g(y)| > 1$ , then there exists $x_0$ such that $2 f(c+x_0) f(c-x_0) = f(2c) + f(2x_c) = [2 f(c) f(c) - f(0)] + [2 f(x_0) f(x_0) - f(0)] = 2[f^2(x_0) - 1] > 0$ , but $f(c+x_0) = 2 f(c) f(x_0) - f(c-x_0) = -f(x_0 - c) = -f(c-x_0)$ , thus $f(x_0 + c) f(c - x_0) = -f^2(c - x_0) \leq 0$ , which contradicts the previous statement. Therefore, for any $x \in \mathbf{R}$ , we have $|f(x)| \leq 1$ . According to the conclusion in the problem, we can deduce that $|g(y)| \leq 1$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.