Lemma 12

Let $\varrho$ be a unitary representation of $\mathbf{R}$ in E. Let D be the set of elements $x$ of E such that the mapping $\psi_x:t\mapsto \varrho (t)x$ is differentiable at 0. The mapping $u$ of D into E given by $x\mapsto i^{-1}\psi '_x(0)$ defines a symmetric partial operator on E.

Let $f\in \mathscr{D}(\mathbf{R})$ and $x\in E$. Put $y=\varrho (f)x$. For every $h\in \mathbf{R}$, we have

The function of $\mathbf{R}^2$ into $\mathbf{C}$ defined by $(t, h)\mapsto f(t-h)$ is indefinitely differentiable; its derivative with respect to $h$ is the function defined by $(t, h)\mapsto$ $-f'(t-h)$, which is bounded by $\|f'\|_{\infty}$. When $h= 0$, this derivative is zero for $t$ outside a compact set. Since $\|\psi_x(t)\|=\|x\|$ for every $t$, we deduce from Prop. 2 of IV, p. 197 that the function $\psi_y$ is differentiable at 0 and that its derivative is given by

Hence $y\in D$. Since the space $\mathscr{D}(\mathbf{R})$ is dense in $L^1(\mathbf{R})$ (Prop. 4 of IV, p. 202), we deduce from INT, VIII, p. 139, § 2, n$^o7$, Prop. 10 that the space D is dense in E.

Let $x_1$ and $x_2$ be in D. We compute

$\langle u(x_1)|x_2\rangle =$ lim $\langle (\varrho (h)-1_E)x_1|x_2\rangle$

= lim $\langle x_1|(\varrho (-h)-1_E)x_2\rangle =\langle x_1|u(x_2)\rangle$.

Consequently, the partial operator $u$ is symmetric.