Let $f: \mathbb{R} \to \mathbb{R}$ be continuous periodic function with period $T>0$

Question

Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous periodic function with period $T>0$. Show that there exists an element $x \in \mathbb{R}$ for which $f(x)=f(x+T/2)$.

This is what I came up with to prove the statement:

• For $f(x)=b$, with $b$ a constant, the proof is trivial (because in such a case $f(x)=f(x+T/2)$ for all $x \in \mathbb{R}$)

• To prove the proposition for all other $f(x)$, I have found that the following statements hold:

  1. $f(0)=f(T)$ (or actually, $f(k)=f(k+T)$ for all $k \in \mathbb{R}$)
  2. There exists at least one value $f(y)$, with $y \in [0,T]$ for which $f(y)\neq f(0)$ and $f(y) \neq f(T)$
  3. All values of $f$ between $x=0$ and $x=T$ appear at least twice, except for the maximum and minimum, which only appear once (and they must exist).

I'm guessing I have to make use of the intermediate value theorem to prove the result, but I'm unable to figure out how. Any help?

Answer 1

The simplest approach is: Consider $g(x) = f(T/2+x) - f(x).$ Now if $g(0)$ is positive, then $g(T/2)$ has to be negative (since $g(x) + g(x+T/2) = f(T+x) - f(x) =0.$) The result then follows by the intermediate value theorem (as you have guessed).

Answer 2

I want to answer the more general case posed above. Let $\alpha \in \mathbb{R}$. We want to show that there exists a $y\in\mathbb{R}$ such that $f(y+\alpha)=f(y)$. Sometimes in order to show two functions are equal, it helps to consider a function that is the two functions subtracted.

Consider $g(y)=f(y+\alpha)-f(y)$. We know that periodic continuous functions have a max and min value. Let $x_0$ be the max. Then,

1) $g(x_0)=f(x_0+\alpha)-f(x_0)\leq 0$ and

2) $g(x_0-\alpha)=f(x_0)-f(x_0-\alpha)\geq 0$.

(1) and (2) are true since $f(x_0)$ is the max of the function. Now if either (1) or (2) equal zero, we are done and have found our value $y$ such that $f(y+\alpha)=f(y)$. If neither of them are zero, then we can use the Intermediate Value Theorem to show that there exists a $y\in [x_0-\alpha,x_0]$ such that $g(y)=0$. This implies that $f(y+\alpha)=f(y)$.