# Does a periodic function have to be bounded?

Let a function $f$ satisfy the relation $f(x)=f(x+1)$ for all $x\in \Bbb{R}$. Should this function always be bounded?

I think so, but the book doesn't. Any help will be greatly appreciated. Please note that the function need not be continuous.

• Is f continuous? Commented Nov 3, 2014 at 17:12
• What about $\tan (\pi x)$? Commented Nov 3, 2014 at 17:19
• It's not defined on $\mathbb{R}$, technically. Need to extend it. Commented Nov 3, 2014 at 17:19
• sigh extend $\tan(\pi x)$ to $\mathbb{R}$ by setting $f(x) = \tan(\pi x) for x \notin \mathbb{Z}$ and to $0$ for $x \in \mathbb{Z}$. There you go. Commented Nov 4, 2014 at 5:10
• @Alqatrkapa, $\tan(\pi z)$ ($z \in \mathbb{Z}$) is already zero without redefinition... You should be concerned with $\pi/2 + \pi z$ with $z\in\mathbb{Z}$.
– Joel
Commented Nov 4, 2014 at 17:30

Nope: $f(x) = 1/(1-x)$ for $x \in [0,1)$. Now extend periodically $f(x + n) = f(x)$ for integers $n$.

• But $f(x)=f(x+1)$ for all $x$! We cannot say that $f(1)=f(2)$ here as $f(1)$ (or $f(2)$) is not defined! Commented Nov 3, 2014 at 17:15
• Read the solution more carefully. $f(0) = 1$, and thus $f(n) = 1$ for all integers $n$. Commented Nov 3, 2014 at 17:16
• Look again at the definition, $f(n) =1$ for all integers $n$. Commented Nov 3, 2014 at 17:16

If you take the function $$f(x) = \left\{ \begin{array}{ll} \cot(x) & x \neq k \pi, k \text{ integer}\\0 & \text{otherwise}\end{array}\right.$$ This function is defined for all $x \in \mathbb{R}$ and is periodic with period $\pi$. This function is not bounded.

Of course, if you require $f$ to be continuous, then if the function has (WLOG) period 1, it is bounded on $[0,1]$ because it is continuous. It follows that $f$ is bounded on all of $\mathbb{R}$ since it is periodic.

Just to add another example that has period $1$ and is unbounded on every open interval: $$f(x)=\begin{cases}\min\{\,n\in\mathbb N:nx\in\mathbb Z\,\}&\text{if x\in\mathbb Q}\\0&\text{otherwise}\end{cases}$$

Take the function $g:[0,1)\to\mathbb{R}^+$ whose value is $1$ on the irrational numbers and $q$ for any rational number of the form $\frac{p}{q}$ with $\gcd(p,q)=1$, then define $f(x)$ as $g(\{x\})$. Since $g$ is unbounded, $f$ is unbounded, too.

• Apart from the definition for $x\notin\mathbb Q$, this is exactly the same function as @HagenvonEitzen s. Commented Nov 3, 2014 at 17:43
• @AlexR: oh, you're right. I am sorry, I didn't notice Hagen von Eitzen's answer before posting mine. Commented Nov 3, 2014 at 17:56
• Yours is likely the definition that makes periodicity easier to show. Commented Nov 3, 2014 at 18:00
• @hardmath: I don't find showing $xn\in\Bbb Z\iff(x+k)n\in\Bbb Z$ (with $k,n\in\Bbb Z$) all that hard. Commented Nov 4, 2014 at 14:33