I am reading literature on self-adjoint extensions of Hamiltonians (particle interaction) and I came across the following statement (in context of separating total momentum $P$):
Operator $H$ defined by $(Hf)(P)=\frac{P^2}{m+2}f(P)$ is a self-adjoint operator in $L_2(\mathbb{R}^{3})$, i.e. $f \in L_2(\mathbb R ^3)$.
So from this I figure that $H$ is defined on $L_2(\mathbb R^3)$, but we can obviously find some $f \in L_2(\mathbb R ^3)$ s.t. $\frac{P^2}{m+2}f(P)\notin L_2(\mathbb R ^3)$, in other words $H(L_2(\mathbb R ^3))$ is larger than $L_2(\mathbb R ^3)$. So the operator is not defined on the whole domain (since it has to be closed w.r.t. Hilbert space it is acting on in order to be well-defined), and its domain $D(H)$ has to be restricted. However, the paper just states that $H$ is self-adjoint on $L_2(\mathbb R ^3)$ and goes ahead.
So my question is, am I missing something here? Should the domain really be restricted, or is this operator indeed self-adjoint as it is?
Thank you in advance!