Consider the unit torus $\mathbb T:=\mathbb R /\mathbb Z$. As far as I understand, the points of $\mathbb T$ are equivalence classes $[x]$ defined through the relation $x\sim x+k$, with $k \in \mathbb Z$. So for $y\in \mathbb R$ we have $y\in [x]$ if $x=y+k$ for some $k$.
I want to define functions on $\mathbb T$. Do all functions have to be periodic with period $1$? I guess so because I'd define functions by the following relation:
$$f([x]) :=f(x) \quad , \tag{1}$$
where $x\in [x] \in \mathbb T$ and $f$ on the RHS is a function defined on $\mathbb R$ (abusing the notation a bit). So the function evaluated at an equivalence class is defined through the function value of a representative of that class and thus
$$f([x])=f(x)=f([x+1])=f(x+1) \quad . \tag{2}$$
Is my analysis correct? Does the definition make sense?
Background: I study a quantum mechanical system defined on $L^2(\mathbb T)$ and in particular a linear operator $V$ defined on $L^2([0,1])$ by $(V\psi)(x):= v(x)\psi(x)$ for some real-valued function $v$. I want to 'lift' this operator to $L^2(\mathbb T)$ and I guess this is only possible if I consider some sort of periodic extension of the function $v$. I think that for $\psi\in L^2(\mathbb T)$ we have that $(V\psi)([x]) = v(x) \psi(x)$ is well defined only if $v(x)$ is periodic, no? In other words: $V:\mathcal D(V) \subset L^2(\mathbb T)\longrightarrow L^2(\mathbb T)$ only if $v$ is periodic?!
Yes, that looks right to me. Note that the function need not be continuous, but it must be periodic.
More precisely, a function $f$ on $\mathbb T$ induces a function $\tilde f$ on $\mathbb R$ in the obvious way, namely $\tilde f = f\circ q$ where $q:x\mapsto [x]$ is the quotient map corresponding to $\sim$. In order to go in the other direction - which we might loosely write as $f = \tilde f \circ q^{-1}$ despite the fact that $q$ is not injective - we must have that $\tilde f (x)= \tilde f(x+k)$ for $k\in \mathbb Z$ (otherwise it would not be a well-defined function on the quotient space).
There would be nothing wrong with having a function $\tilde f:\mathbb R\ni x \mapsto x-\lfloor x\rfloor$, which returns the fractional part of $x$. This function is discontinuous, but it is periodic, and so $\tilde f(x) = \tilde f(x+k)$ for $k\in \mathbb Z$ and our induced function $f = \tilde f \circ q^{-1}$ is well-defined.