Let $f:$ $\mathbb{R}$ $\to$ $\mathbb{R}$ be a continuous function such that $f(x +1) = f(x)$ for all $x \in \mathbb{R}$. Show that $f$ is bounded.

My first thought is that $f(x)$ is a constant function, so it will be clearly limited. But, I can't link the definition of continuous
( $\forall \varepsilon>0\exists\delta>0(\forall y \in \mathbb{R}, || x - y||<\delta\rightarrow ||f(x) - f(y)||<\varepsilon$) with it, because there isn't any relation that I can figure out between $x,y, \delta$. Could someone help me how to do it? Thanks

  • 1
    $\begingroup$ It does not need to be constant. Doesn't $f(x) = \sin(2\pi x)$ satisfy your definition? $\endgroup$
    – dfnu
    Commented Mar 8, 2019 at 20:54
  • $\begingroup$ What do you mean by limited? It's not necessarily true that $f$ will be constant. For example, $f$ could be the function $\sin(2\pi x)$ $\endgroup$ Commented Mar 8, 2019 at 20:54
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    $\begingroup$ @confused_wallet I guess what is meant is that the function is bounded? $\endgroup$
    – dfnu
    Commented Mar 8, 2019 at 20:58
  • $\begingroup$ I would go for a proof by contradiction. Maybe you can prove that if the function is unbounded and periodic it must be non continuous. $\endgroup$
    – dfnu
    Commented Mar 8, 2019 at 21:04
  • $\begingroup$ @confused_wallet , yes I'll correct, it's bounded. $\endgroup$
    – iaguet
    Commented Mar 8, 2019 at 21:26

2 Answers 2


Suppose the function is not bounded from above. Then for each $n>0$ there must be an $x_n\in \Bbb R$ such that $f(x_n)>n$. Since the function is periodic we can find one of such $x_n$'s in the interval $[0,1]$. In this way we can construct a sequence $(x_n)_{n\in\Bbb Z^+}$, that is bounded. By Completeness principle, therefore, $(x_n)$ must have a converging subsequence, say $(z_n)$, whose limit, say $\overline z$, must also lie in the closed interval $[0,1]$. But this contradicts continuity since $(z_n) \to \overline z$ with $(f(z_n)) \not\to f(\overline z)$. $\blacksquare$

EDIT Without Completeness the assertion becomes obviously false. Consider, for instance, in $\Bbb Q$ the function $$g(x) = \begin{cases}\frac{1}{|2x^2-1|} & (0 \leq x <1) \\ 0 & (\mbox{otherwise})\end{cases}$$ and generate from it the function $f:\Bbb Q \to \Bbb Q$ $$f(x) = \sum_{k=-\infty}^{+\infty} g(x-k).$$ This function is continuous in $\Bbb Q$ and periodic with period $1$, and yet it is unbounded.



1) Since $f$ is continuous it attains its max and min on the compact (closed and bounded) interval $[0,1]$.

Hence $f$ is bounded on $[0,1].$

2) Let $x >1$, real.

There is a $n \in \mathbb{Z^+}$ s.t.

$n \le x <n+1$, or

$0 \le x-n <1$.

3) $f(x+1)=f(x)$, i.e. is periodic with period $1$.

$f(x)=f(x-1)$, successive applications give $f(x)=f(x-n)$.

4) Hence for $x >1$ , with $(x-n) \in [0,1]$

$f(x)=f(x-n)$, bounded,

A similar argument for $x <0.$


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