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# Prove a periodic increasing function is constant

2018-03-12 20:22:50

My proof is as follows:

Let $f(x)$ be a periodic, increasing function. Then, there exists a $c$ such that $f(x) = f(x+c)$ for all $x$.

Take $a, b$ such that $a < b$. Since $f$ is increasing, $f(a) \leq f(b)$.

Since $f$ is periodic, $f(a) = f(a+kc)$ for an integer $k$. Choose $k$ such that $a + kc > b$. We have $f(a+kc) \geq f(b)$, since $f$ is increasing. However, $f(a) = f(a+kc)$, so $f(a) \geq f(b)$.

Since $f(a) \leq f(b)$ and $f(a) \geq f(b)$, we have $f(a) = f(b)$ for $a < b$. As such, $f$ is constant.

Is this proof correct?