The position operator for a system with periodic boundary condition Consider a quantum system defined on the interval $[0,L]$. Suppose we take the periodic boundary condition, 
$$ \psi(0) = \psi(L ) .  $$
It is know that for such a system, the position operator is not well defined. It is often said that even if $\psi (x)\in \mathcal{H }$,  generally $x \psi(x) \notin \mathcal{H}$. 
But are the states in the Hilbert space $\mathcal{H }$ dense in the space $L^2 (0,L)$? 
If so, $x \psi(x)$ can be approximated by some state $\phi(x)$ in $\mathcal{H }$. Then possibly we can define the position operator as 
$$ x \psi (x) = \phi(x) .$$
Can this idea work?   
 A: The issue is much easier than you suppose. If you define $X$ as multiplicative operator in the Hilbert space $L^2([0,L], dx)$, differently to what happens in $L^2(\mathbb R, dx)$,  its domain is the whole space.
If $\psi \in L^2([0,L], dx)$ then $$||X\psi||^2 = \int_0^L |x \psi(x)|^2 dx \leq L^2 \int_0^L |\psi(x)|^2 dx \leq L^2 ||\psi||^2 < +\infty\:.$$
This operator is self-adjoint, bounded and defined on the whole Hilbert space. It is possible to prove that $||X||=L$ and $\sigma(X)= \sigma_c(X)= [0,L]$.
Boundary conditions are completely irrelevant here, since the Hilbert space  $L^2(\mathbb R, dx)$ is defined independently from them. 
However, if you try to define 
$$(X_0\psi)(x) := x \psi(x)\quad \psi \in L^2([0,L], dx) \quad \psi(0) = \psi(L)$$
this definition does not make sense, as $0$ and  $L$ have zero Lebesgue measure and $\psi$ is defined up to zero-measure sets.
You may artificially add some condition to make sensible the last requirement like this
$$(X_0\psi)(x) := x \psi(x)\quad \psi \in L^2([0,L], dx)\:, \mbox{$\psi$ continuous,} \quad \psi(0) = \psi(L)\:.$$
With this definition the self-adjointness condition $X_0^* =X_0$ fails, just because it turns out that 
$$X_0^* = X$$
where $X$ is the self-adjoint operator I initially introduced which is defined in the whole Hilbert space. In other words, the adjoint of $X_0$ has a larger domain than $X_0$ which, in fact, coincides with the whole Hilbert space.
To obtain a self-adjoint operator from $X_0$ one may  maximally extend it from its initial domain, made of continuous functions, to the closure of this domain which is the whole $L^2([0,L], dx)$ since the initial domain is dense therein. This extension is possible because $X_0$ is continuous.
This way, as a matter of fact, we come back to my initial definition. 
It is however possible to prove that, since $X_0$ is symmetric and its adjoint is self-adjoint, $X$ is the only possible self-adjoint linear extension of $X_0$, obtained by continuity or not.
