# Prove that a specific continuous function is bounded

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

• It does not need to be constant. Doesn't $f(x) = \sin(2\pi x)$ satisfy your definition? – Matteo Mar 8 at 20:54
• What do you mean by limited? It's not necessarily true that $f$ will be constant. For example, $f$ could be the function $\sin(2\pi x)$ – confused_wallet Mar 8 at 20:54
• @confused_wallet I guess what is meant is that the function is bounded? – Matteo Mar 8 at 20:58
• 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. – Matteo Mar 8 at 21:04
• @confused_wallet , yes I'll correct, it's bounded. – iaguet Mar 8 at 21:26

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.

Attempt:

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 , 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 $$x <0.$$