Let ${f}$ be a bounded real function defined for all real numbers and satisfying for all real numbers ${x}$ the condition ${ f \Big(x+\frac{1}{3}\Big) + f \Big(x+\frac{1}{2}\Big)=f(x)+ f \Big(x+\frac{5}{6}\Big)}$ . Show that ${f}$ is periodic.
parmenides51 wrote:
Let ${f}$ be a bounded real function defined for all real numbers and satisfying for all real numbers ${x}$ the condition ${ f \Big(x+\frac{1}{3}\Big) + f \Big(x+\frac{1}{2}\Big)=f(x)+ f \Big(x+\frac{5}{6}\Big)}$ . Show that ${f}$ is periodic.
Let $x\in\mathbb R$ and $a_n=f(x+\frac n6)$ (for easier writing)so that equation is
$a_{n+5}=a_{n+3}+a_{n+2}-a_n$
So $a_{n+5}+a_{n+4}-a_{n+2}-a_{n+1}=a_{n+4}+a_{n+3}-a_{n+1}-a_n$ $=a_4+a_3-a_1-a_0$
And $a_{n+6}-a_n=a_{n+4}+a_{n+3}-a_{n+1}-a_n$ $=a_4+a_3-a_1-a_0$
Which means $a_{n+6k}=a_n+k(a_4+a_3-a_1-a_0)$
And so, since bounded, $a_4+a_3-a_1-a_0=0$ and $a_{n+6}=a_n$
Which is $f(x+1)=f(x)$
Q.E.D.