Lemma 11

Let $u$ be a self-adjoint partial operator on E. For $t\in \mathbf{R}$, put $\varrho (t) =e^{itu}\in \mathscr{L}(E)$. Then the mapping $\varrho$ is a unitary representation of $\mathbf{R}$.

The operator $\varrho (t)$ is defined by the universally measurable functional calculus (def. 5 of IV, p. 272); it is an endomorphism of E since the function $x\mapsto e^{itx}$ is bounded on Sp($u$) (prop. 5, a) of IV, p. 275).

By Prop. 5 of IV, p. 275, we have $\varrho (0) = 1_E,\varrho (t)^*=\varrho (-t)$ for every $t\in \mathbf{R}$ and $\varrho (t_1+t_2) =\varrho (t_1)\varrho (t_2)$ for all $t_1$ and $t_2$ in $\mathbf{R}$. In particular, the endomorphism $\varrho (t)$ is unitary for every $t\in \mathbf{R}$. For every $x\in E$, the mapping of $\mathbf{R}$ into E defined by $x\mapsto \varrho (t)x$ is continuous at $t= 0$ by Prop. 6 of IV, p. 276, hence $\varrho$ is a unitary representation of $\mathbf{R}$ in E (V, p. 380, Lemma 4).