Lemma 13

Let $\varrho$ be a unitary representation of $\mathbf{R}$ in E and let $u$ be its infinitesimal generator.

a) The domain of $u$ is the set of $x\in E$ such that the mapping $\psi_x:t\mapsto \varrho (t)x$ is differentiable on $\mathbf{R}$;

b) For every $t\in \mathbf{R}$ and every $x\in$ dom($u$), one has $\varrho (t)x\in$ dom($u$) and $\psi '_x(t) =iu(\varrho (t)x)$;

c) The partial operator $u$ is essentially self-adjoint.

For every $x\in E$, let us denote by $\psi_x$ the mapping of $\mathbf{R}$ into E defined by $\psi_x(t) =\varrho (t)x$. For every $t\in \mathbf{R}$ and every $h\in \mathbf{R}$, one has

which proves that dom($u$) is the space of elements $x\in E$ such that $\psi_x$ is differentiable on $\mathbf{R}$ and establishes that $\psi '_x(t) =\varrho (t)\psi '_x(0) =i\varrho (t)(u(x))$ for every $t\in \mathbf{R}$.

Let $x\in$ E and $t\in \mathbf{R}$. One has $\psi_{\varrho(t)x}(s) =\psi_x(s+t)$ for every $s\in \mathbf{R}$. Consequently, one has $\varrho (t)x\in$ dom($u$) if $x\in$ dom($u$), and moreover $u(\varrho (t)x) =i^{-1}\psi '_x(t) =\varrho (t)u(x)$ by a). One then obtains assertion b).

Let us prove c). Let $x\in$ Ker($u^*-i1_E$). Let us prove that $x= 0$. Let $y\in$ dom($u$), and let $f$ be the function on $\mathbf{R}$ defined by $f(t) =\langle \psi_y(t)|x\rangle$ for $t\in \mathbf{R}$. The function $f$ is bounded since $\|\psi_y(t)\|=\|y\|$ for every $t\in \mathbf{R}$; it is differentiable in $\mathbf{R}$ and, for every $t\in \mathbf{R}$, assertion b) implies

since $u^*(x) =ix$. Hence $f(t) =f(0)e^t$ for all $t\in \mathbf{R}$ (FVR, IV, p. 27). Since $f$ is bounded, the function $f$ is zero and, in particular, we have $\langle y|x\rangle =f(0) = 0$. As the space dom($u$) is dense in E, we conclude that $x= 0$. Analogously, one proves that Ker($u^*+i1_E$) is reduced to 0. By Corollary 3 of IV, p. 261, the partial operator $u$ is therefore essentially self-adjoint.