Question 85
Let $f$ be a bounded real function defined for all real numbers and satisfying for all real numbers $x$ the condition
$$
f\left(x+\frac{1}{3}\right)+f\left(x+\frac{1}{2}\right)=f(x)+f\left(x+\frac{5}{6}\right)
$$
Show that $f$ is periodic. (A function $f$ is bounded, if there exists a number $L$ such that $|f(x)|\lt L$ for all real numbers $x$. A function $f$ is periodic, if there exists a positive number $k$ such that $f(x+k)=f(x)$ for all real numbers $x$.)
$$
f\left(x+\frac{1}{3}\right)+f\left(x+\frac{1}{2}\right)=f(x)+f\left(x+\frac{5}{6}\right)
$$
Show that $f$ is periodic. (A function $f$ is bounded, if there exists a number $L$ such that $|f(x)|\lt L$ for all real numbers $x$. A function $f$ is periodic, if there exists a positive number $k$ such that $f(x+k)=f(x)$ for all real numbers $x$.)