Problem 1095

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 $|g(y)| \leq 1$.