Lemma 14

Let $u$ be a self-adjoint partial operator on E and let $\varrho (t) =e^{itu}$ be the unitary representation of $\mathbf{R}$ defined by $u$. The infinitesimal generator of $\varrho$ is equal to $u$.

Let $x\in$ dom($u$). Put $\psi_x(t) =\varrho (t)x$ for all $t\in \mathbf{R}$. For every nonzero real number $h$, we have

When $h$ tends to 0, we have

1 $_{iht}$

for all $t\in \mathbf{R}$. Moreover,

It then follows from Proposition 6 of IV, p. 276 that the function $\psi_x$ is differentiable on $\mathbf{R}$ and satisfies $\psi '_x(t) =iu(\varrho (t)x)$ for all $t\in \mathbf{R}$. Consequently, the domain of $u$ is included in the set of $x\in E$ such that $\psi_x$ is differentiable at 0 and one then has $\psi '_x(0) =iu(x)$. This means by definition that the infinitesimal generator of $\varrho$ is an extension of $u$. These two operators are therefore equal since they are symmetric and $u$ is self-adjoint (IV, p. 238, Remark 5).