Are all $1$-periodic functions writable as functions of $\exp(2i\pi z)$?

Question

All is essentially in the title. Let $f$ be a function defined over $\mathbb{C}$ such that $f(z+1)=f(z)$ for every $z$. I have a good feeling of why it is so but I have no argument to say properly that there is a function $f_0$ such that $f(z) = f_0(\exp(2i\pi z))$. How can it be proven properly?

Answer 1

$f_0(\exp(2 i \pi z)) = f(z)$ is actually a definition of $f_0$, since it constrains the value of $f_0$ at every complex number (except 0) as long as there is no ambiguity. (And choose $f_0(0)$ arbitrarily).

Suppose there was an ambiguity: $\exp(2 i \pi z_1) = \exp(2 i \pi z_2)$ and yet $f(z_1) \ne f(z_2)$

$$\exp(2 i \pi z_1 - 2 i \pi z_2) = 1$$ $$\exp(2 i \pi (z_1 -z_2)) = 1$$

Then $z_1 - z_2$ must be integer real. But then $f(z_1) = f(z_2)$ so the definition is unambiguous.

Answer 2

Let $g(z) = \exp(2 \pi i z)$, and consider the equivalence relation $\sim$ on $\mathbb{C}$ such that $z \sim w \iff z - w \in \mathbb{Z}$. Further, let $X = \mathbb{C} / \sim$.

Define $h : X \to \mathbb{C} \setminus \lbrace 0 \rbrace$ by $h([z]) = g(z)$. This is well-defined, because $$g(z) = g(w) \iff \frac{g(z)}{g(w)} = 1 \iff g(z - w) = 1 \iff z \sim w.$$ The above argument also shows that $g$ is injective. It's also not difficult to see that $h$ is surjective, as $g$ maps onto $\mathbb{C} \setminus \lbrace 0 \rbrace$.

Let $f_1 : \mathbb{C} \setminus \lbrace 0 \rbrace \to \mathbb{C}$ be defined by $f_1(z) = f(w)$, where $w \in h^{-1}(z) \in X$. Again, this is well-defined, because $f$ is $1$-periodic. Further, let $f_0$ be an extension of $f$ to $\mathbb{C}$ where $f_0(0) = 0$ (or indeed anything else). I claim that $f_0 \circ g = f$.

To prove this, suppose $z \in \mathbb{C}$. Then $$(f_0 \circ g)(z) = f_1(g(z)) = f(w)$$ where $w \in h^{-1}(g(z))$, since $g(z) \neq 0$. So, it suffices to show $z \in h^{-1}(g(z))$. But this follows from the definition of $h$: we know that $h([z]) = g(z)$, so $h^{-1}(g(z)) = [z] \ni z$. This confirms that $f_0 \circ g = f$, as required.

Answer 3

Here is an answer in terms of "covers":

The $z$-plane ${\mathbb C}$ is the universal cover of the punctured $w$-plane $\dot{\mathbb C}$. Hereby the covering map is $$\psi:\quad{\mathbb C}\to\dot{\mathbb C},\qquad z\mapsto w:=e^{2\pi i z}\ .$$ Since $\psi(z+k)\equiv \psi(z)$ when $k\in{\mathbb Z}$ the group $G$ of deck transformations consists of the horizontal translations $$T_k:\quad {\mathbb C}\to{\mathbb C},\qquad z\mapsto z+k\qquad\qquad(k\in{\mathbb Z})\ .$$ In the case at hand we have a function $z\mapsto f(z)$ which is defined on the universal cover, and at the same time is periodic with period $1$, hence is invariant under the group $G$. It follows that this $f$ defines a pushdown $f_0$ on the punctured $w$-pane. This $f_0$ is characterized by $$f(z)=f_0\bigl(\psi(z)\bigr)\qquad(z\in{\mathbb C})\ .$$ An explicit formula for $f_0$ is given by $$f_0(w)=f\left({1\over2\pi i}\log w\right)\qquad(w\ne0)\ .\tag{1}$$ As $f$ is periodic with period $1$ the RHS in $(1)$ is well defined, even though the $\log$ is only defined up to an additive constant $2k\pi i$.

If the given $f$ is an analytic function of $z$ then $(1)$ defines $f_0$ as an analytic function in the punctured $w$-plane. For a proof consider a $w_0\ne0$. As $\psi$ is surjective there is a $z_0$ with $w_0=\psi(z_0)$. Since $\psi'(z_0)\ne0$ we know that $\psi$ maps a small neighborhood $U$ of $z_0$ conformally onto a neighborhood $V$ of $w_0$, and we have a local inverse $\psi^{-1}_{\rm loc}:\>V\to U$ which is analytic as well. It follows that $$f_0(w)=f\bigl(\psi^{-1}_{\rm loc}(w)\bigr)\qquad(w\in V)\ ,$$ hence $f_0$ is analytic in $V$.