Given $f, g : \mathbb{R} \to \mathbb{R}$, $f$ a periodic function, $g(x) = f(ax+b)$ and $a, b$ are real numbers with $a \ne 0$, I want to prove that $g$ is a periodic function.
The first logical step in my opinion would be:
Suppose $\exists\ T \ne0$ and $g(x) = g(x+T), \forall\ x \in \mathbb{R}$. We would have $f(ax+b)= f(a(x+T)+b)$ $\forall\ x \in \mathbb{R}$. I would start giving $x$ values such as $0$: $f(b) = f(aT + b)$. If $x$ is $1$: $f(a+b) = f(a + aT + b)$. We see that $aT$ is repeating itself. How should I continue from here?
When I was writing this post, a suggested post (which I will link) had appeared. It's indeed really similar but the conclusions differ a bit and I don't know how should I connect the "pieces" to solve the problem so to speak. Any hints or methods of continuing?