# Questions regarding period of continuous function on $\mathbb{R}$.

Let $$f:\mathbb{R} \rightarrow \mathbb{R}$$ be a continuous fundtion such that $$f(x+1)=f(x) \forall x\in \mathbb{R}$$. Pick out the true statements from the following:

(A) $$f$$ is necessarily bounded above

(B) There exists a unique $$x_0$$ such that $$f(x_0+\pi)=f(x_0)$$

(C) There exists infinitely many $$x_0$$ such that $$f(x_0+\pi)=f(x_0)$$

(D) There exists NO $$x_0$$ such that $$f(x_0+\pi)=f(x_0)$$

I am a bit confused with options B, C, D.

(A) Is TRUE because $$f$$ is a periodic, continuous function hence it is bounded

(B, C, D) are FALSE. Because the period of the function $$f$$ is either $$n$$ or $$1/n$$ for some natural number $$n$$. Hence we cannot have $$f(x_0+\pi)=f(x_0)$$.

• It is not true that B,C,D are false. In fact exactly one of C and D must be true. You can use continuity to determine which. Sep 13, 2022 at 4:20
• $f(x_0+\pi)=f(x_0)$ for some $x_0$ doesn't imply that $f$ has period $\pi$ (or a fraction of it). For example, let $f(x)=\sin(2\pi x)$ and let $x_0=\frac74-\frac\pi2$. Then you can check $f$ is periodic with period 1 but $f(x_0)=f(x_0+\pi)$. See here: desmos.com/calculator/3iipojrvcw Sep 13, 2022 at 4:22

A continuous periodic function with period 1 has a point x which gives same value as x+π shows that there is at least one $$x_0$$ with $$f(x_0+\pi)=f(x_0)$$. But then we can replace $$x_0$$ by $$x_0+n$$ for any integer $$n$$ so (C) is true. Of cousre, this proves that (B) and (D) are false.