Fourier series of periodic function with functional equation

Question

Given $\alpha$ and $\beta$. let $f(z)$ be the function satisfying $f(\alpha z+ \beta) = f(z)$ for all $z$. What is the Fourier expansion of $f$ ?

Answer 1

The goal is to determine a period of the function $f$. Let $g:\mathbb{R}\rightarrow\mathbb{R}$ denote the linear function $x\mapsto \alpha x + \beta$. By induction, one shows $$\forall n \in \mathbb{N}, \qquad f(x) = (f\circ g^n)(x) = f\left(\alpha^n x + \beta \sum_{k=0}^{n-1} \alpha^{k}\right) . $$ If $\alpha = 0$, then the function $f$ is constant, which is not of big interest to write its Fourier series. Else, $g$ is a bijection, and one shows $$\forall n \in \mathbb{N}, \qquad f(x) = (f\circ g^{-n})(x) = f\left( \alpha^{-n}x - \beta \sum_{k=1}^n \alpha^{-k} \right) . $$ Let us denote by $T$ the fundamental period of $f$. For all $x$ and all $n$, \begin{equation} f(x+T) = f\left(g^n(x) + \alpha^n T\right) = f\left(g^{-n}(x) + \alpha^{-n} T \right) . \end{equation} Since $g^n$ and $g^{-n}$ are bijections, one can write $$ f\left(x + \alpha^n T\right) = f(x) \qquad\text{and}\qquad f\left(y + \alpha^{-n} T\right) = f(y) $$ for all $(x,y)$ in $\mathbb{R}^2$. Due to the $T$-periodicity of $f$, these equalities rewrite as \begin{align} &f\left(x + \alpha^{n} T\right) = f\left(x + \left\lbrace\alpha^{n}\right\rbrace T + \left\lfloor\alpha^{n} \right\rfloor T\right) = f\left(x + \left\lbrace\alpha^{n} \right\rbrace T\right) = f(x)\, ,\\ &f\left(y + \alpha^{-n} T\right) = f\left(y + \left\lbrace\alpha^{-n}\right\rbrace T + \left\lfloor\alpha^{-n} \right\rfloor T\right) = f\left(y + \left\lbrace\alpha^{-n} \right\rbrace T\right) = f(y) \, , \end{align} where $\lfloor\cdot\rfloor$ denotes the floor function and $\left\lbrace\cdot\right\rbrace$ denotes the fractional part. If $|\alpha|\neq 1$, one can always find $n$ such that $\left\lbrace\alpha^{\pm n} \right\rbrace$ is nonzero. In this case, $f$ is $\left\lbrace\alpha^{\pm n} \right\rbrace T$-periodic, which contradicts the fact that $T$ is the fundamental period of $f$. Therefore, $|\alpha|=1$.

• If $\alpha=1$, then one has in particular $f(x) = f(x + |\beta|)$ for all $x$. The function $f$ is $|\beta|$-periodic.

• If $\alpha=-1$, then for all $x$, $$ f(x + \beta/2) = f(-x + \beta/2) \, . $$ The function $f$ is symmetric with respect to $x = \beta/2$. It seems that more information is needed to determine a periodicity in $f$.

Note: this answer concerns real functions of real variables. Maybe, the case of complex analysis should be addressed.