Let $f$ be periodic function with fundamental period $\tau$. Prove that function $f(nx)$ have fundamental period $\frac{\tau}{n}$.

Question

Let $f$ be periodic function with fundamental period $\tau$, prove that function $f(nx)$ have fundamental period $\frac{\tau}{n}$.
My idea:
First I need to show that $\frac{\tau}{n}$ is a period. $$f(n(x+\frac{\tau}{n}))= f(nx + \tau) = f(nx).$$ But how to show that this is the smallest such? Any idea?

Answer 1

Let $g(x) = f(nx)$. You have shown that if $f$ has a period $\tau$ then $g$ has period ${\tau \over n}$.

The same reasoning shows that if $g$ has period $T$ then $f$ has period $nT$.

If $\tau$ is a fundamental period of $f$ then any $T$ period of $g$ must therefore satisfy $nT \ge \tau$, or in other words, $T \ge { \tau \over n}$. Hence $ { \tau \over n}$ is a fundamental period of $g$.