Why does weak continuity imply existence of self adjoint position and momentum operators in Hilbert space?

In my studies I came across Polymer Quantum Mechanics. Last line on p. 4 states:

These operators are weakly continuous in the parameters $\lambda$, $\mu$ and this ensures that the self-adjoint operators $\hat{x}$, $\hat{p}$ exist in $\mathcal{H}_{Sch}$.

By "these operators", it means the translation operators in the position and momentum space.

Could someone please provide a proof of this? I posted this here instead of math stackexchange because I would like to get a physical insight for this as well.

• This doesn't make that much sense because $X$ and $P$ do not act on all of $L^2(\Bbb R)$ anyway. It is clear that $D(X)$ and $D(P)$ are nonempty since $0\in$ both of them. So I don't know what it means for them to "exist in $\mathcal H_{Sch}$" and why this needs to be proved. – Ryan Unger Apr 11 '17 at 0:23

Asking for a proof of this is a bit too much, especially on this site. The slides allude to the Stone theorem on one-parameter unitary groups $U_t$ of operators in Hilbert spaces. Stone proved that if the group is strongly continuous in $t$ then it is generated by a self-adjoint operator, i.e. $U_t=e^{itA}$ for some self-adjoint $A$ (densely but not everywhere defined in general). Von Neumann later proved that it is enough for $U_t$ to be weakly measurable in $t$, in particular weakly continuous suffices. There is a more general result that for one-parameter semigroups in Banach spaces weak continuity always implies strong continuity, the Yosida theorem. Proofs can be found in most functional analysis texts, Reid and Simon, Fourier analysis, Self-adjointness is physicist friendly, but still not a cakewalk. Perhaps, Stone's original paper will be a shorter study, only 7 pages.
The idea is to construct the infinitesimal generator as the limit (in a certain weak sense) $\lim_{t\to0}\frac1t(U_t-I)$, which intuitively should give $iA$, and then use the properties of unitary groups and continuity to show that $A$ must be self-adjoint. Taking $U_t$ as the group of position shifts $U_t\psi(x):=\psi(x+t)$ gives us $\hat{x}$ as $A$, and similarly the group of momentum shifts $V_t$ gives us $\hat{p}$.
Later the linked slides also use the Stone–von Neumann theorem, which is stronger. It states that if groups of shifts satisfy the kind of relations that position and momentum shifts $U_t$ and $V_t$ satisfy, then their generators satisfy the canonical commutation relations. And if they are also jointly irreducible (no invariant subspaces), then they are unitarily isomorphic to the position and momentum shifts.