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 $f(x_0+\pi)=f(x_0)$.

  • $\begingroup$ 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. $\endgroup$
    – Fishbane
    Sep 13, 2022 at 4:20
  • 1
    $\begingroup$ $f(x_0+\pi)=f(x_0)$ 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 $\endgroup$ Sep 13, 2022 at 4:22

1 Answer 1


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.


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