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:
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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 $x \in \mathbb{R}$)
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To prove the proposition for all other $f(x)$, I have found that the following statements hold:
- $f(0)=f(T)$ (or actually, $f(k)=f(k+T)$ for all $k \in \mathbb{R}$)
- There exists at least one value $f(y)$, with $y \in [0,T]$ for which $f(y)\neq f(0)$ and $f(y) \neq f(T)$
- 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?
Best Answer
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).