If $f$ is a periodic function with fundamental period $T$ and $g$ is a polynomial such that $f\circ g$ is periodic, prove that $g(x)=ax+b$ where $a,b\in\mathbb{R}$ are some constants.
My working:
Let period of $f\circ g$ be $T_1$
$\implies f(g(x+nT_1))=f(g(x))\forall x\in\mathbb{R},\forall n\in\mathbb{Z}$
$\implies g(x+nT_1)= g(x)+kT$
for some $k \in\mathbb{Z}\,\, \forall x\in\mathbb{R}, \forall n\in\mathbb{Z},n\ge n_0$
When $f$ is assumed continuous the claim can be proven as follows:
There are two points $x_1$, $x_2\in[0,T[\>$ with $f(x_i)=:y_i$ and $|y_2-y_1|=:\alpha>0$.
Assume that $t\mapsto g(t)$ is a polynomial of degree $\geq2$, and that $\phi:=f\circ g$ is periodic with period $T'>0$. Since $\phi$ is continuous on ${\mathbb R}$ it is then even uniformly continuous. It follows that there is a $\delta>0$ such that $|\phi(s)-\phi(t)|<{\alpha\over2}$ whenever $|s-t|<\delta$.
Now $g$ satisfies $\lim_{t\to\infty}g'(t)=\infty$ (or $=-\infty$). It follows that there is an interval $J'$ of length $<\delta$ far out on the $t$-axis such that $g$ maps $J'$ onto an interval $J$ of length $>T$ on the $x$-axis. The function $\phi$ then assumes in $J'$ the two values $y_1$, $y_2$ which differ by $\alpha\ $ – a contradiction.