# Suppose that f:RR->RR has the properties (a) |f(x)| le 1, forall x in RR (b) f(x+13/42)+f(x)=f(x+1/6)+f(x+1/7), forall x in RR Prove that f is periodic?

Feb 4, 2017

We are given

$f \left(x + \frac{13}{42}\right) + f \left(x\right) = f \left(x + \frac{1}{6}\right) + f \left(x + \frac{1}{7}\right)$

Substituting in $x + \frac{1}{6}$ for $x$, we get

$f \left(x + \frac{20}{42}\right) + f \left(x + \frac{1}{6}\right) = f \left(x + \frac{2}{6}\right) + f \left(x + \frac{13}{42}\right)$

Adding these equations and canceling the terms which appear on each side, we get

$f \left(x + \frac{20}{42}\right) + f \left(x\right) = f \left(x + \frac{2}{6}\right) + f \left(x + \frac{1}{7}\right)$

Next, we can substitute $x + \frac{1}{6}$ for $x$ into the new equation, add the two equations, and cancel to obtain

$f \left(x + \frac{27}{42}\right) + f \left(x\right) = f \left(x + \frac{3}{6}\right) + f \left(x + \frac{1}{7}\right)$

Repeating this process three more times results in the equation

$f \left(x + \frac{48}{42}\right) + f \left(x\right) = f \left(x + 1\right) + f \left(x + \frac{1}{7}\right)$

Next we substitute in $x = \frac{1}{7}$ for $x$, add, and cancel to obtain

$f \left(x + \frac{54}{42}\right) + f \left(x\right) = f \left(x + 1\right) + f \left(x + \frac{2}{7}\right)$

Repeating this process five more times results in the equation

$f \left(x + 2\right) + f \left(x\right) = f \left(x + 1\right) + f \left(x + 1\right)$

$\implies f \left(x + 2\right) - f \left(x + 1\right) = f \left(x + 1\right) - f \left(x\right)$

The above equation implies that the difference between $f \left(x\right)$ and $f \left(x + 1\right)$ is constant for all $x$. If that difference were nonzero, then the sequence $f \left(x\right) , f \left(x + 1\right) , f \left(x + 2\right) , \ldots$ would tend to $- \infty$ or $\infty$. However, we are given the condition that $| f \left(x\right) | < 1$, a contradiction. Thus the difference must be $0$, meaning $f \left(x\right) = f \left(x + 1\right) \text{ } \forall x \in \mathbb{R}$, i.e. $f \left(x\right)$ is periodic with a period of $1$.

Feb 4, 2017

See below.

#### Explanation:

Here $f \left(x + a + b\right) + f \left(x\right) = f \left(x + a\right) + f \left(x + b\right)$

We have

$f \left(x + b + a\right) - f \left(x + a\right) = f \left(x + b\right) - f \left(x\right)$

so calling $p \left(x\right) = f \left(x + b\right) - f \left(x\right)$ we have

$p \left(x + a\right) = p \left(x\right)$ so $p \left(x\right)$ is periodic with period $a$.

Also considering

$f \left(x + b + a\right) - f \left(x + b\right) = f \left(x + a\right) - f \left(x\right)$

making $q \left(x\right) = f \left(x + a\right) - f \left(x\right)$ we have

$q \left(x + b\right) = q \left(x\right)$ which is periodic with period $b$

and now we can write

$f \left(x\right) = \alpha p \left(x\right) + \beta q \left(x\right) + \gamma p \left(x + a\right) + \delta q \left(x + b\right)$

or

$f \left(x\right) = \left(\alpha - \delta\right) f \left(x + b\right) - \left(\alpha + \beta\right) f \left(x\right) + \left(\gamma + \delta\right) f \left(x + a + b\right) + \left(\alpha - \gamma\right) f \left(x + a\right)$

but

$f \left(x\right) = f \left(x + a\right) + f \left(x + b\right) - f \left(x + a + b\right)$

so

$\left\{\begin{matrix}\alpha + \beta = 0 \\ \alpha - \gamma = 1 \\ \alpha - \delta = 1 \\ \gamma + \delta = - 1\end{matrix}\right.$

solving for $\alpha , \beta , \gamma , \delta$ we obtained

$\left(\alpha = \frac{1}{2} , \beta = - \frac{1}{2} , \gamma = - \frac{1}{2} , \delta = - \frac{1}{2}\right)$

so $f \left(x\right)$ is periodic because it can be composed as a finite sum of periodic functions and the distinct periods are relatively rational.