Proposition: Infinitesimal algebra alone need not exponentiate

There is a densely defined symmetric infinitesimal generator satisfying the one-dimensional Lie-algebra relations that has no self-adjoint extension and hence no unitary one-parameter implementation extending it.

Proof. On $L^2(0,\infty )$, let

Integration by parts proves symmetry. Its adjoint deficiency equations are

The first has the square-integrable solutions $c{e}^{-x}$, while the second has no nonzero square-integrable solution. The deficiency indices are $(1,0)$, so no self-adjoint extension exists. Stone's theorem rules out the claimed unitary group.

The Lie algebra has one generator and its only bracket is zero, so the formal bracket condition is satisfied.