Does the "differential" of a unitary representation give continuous operators on the space of smooth vectors? Let $\pi : G \rightarrow U(H)$ be a strongly continuous unitary representation of a Lie group, $G$, on a Hilbert space, $H$. Let $H_\infty$ be the space of smooth vectors in $H$, those $v$ for which $g\mapsto \pi(g)v$ is smooth.  $\pi$ induces a representation, $\pi'$, of the Lie algebra, $\frak{g}$, of $G$ on $H_\infty$ in the obvious way:  $$\pi'(X)v=\frac{d}{dt}|_{t=0}\pi(e^{tX})v$$
Is each $\pi'(X)$ continuous?
 A: If one equips $H_{\infty}$ with the topology I describe below, then yes. I don't know about the subspace topology though.
Let $U = U(\mathfrak{g})$ denote the universal enveloping algebra of $\mathfrak{g}$. Let $X_1, \dots, X_n$ be a basis of $\mathfrak{g}$. For a multi-index $\alpha = (\alpha_1, \dots, \alpha_n) \in \mathbb{Z}_{\geq 0}^n$ write $X^{\alpha} = X_1^{\alpha_1} \cdots X_n^{\alpha_n} \in U$. Equip $H_{\infty}$ with the locally convex topolgy induced by the family of semi-norms $\{||\cdot||_{\alpha}\}$, where $||v||_{\alpha}:= ||\pi(X^{\alpha}) v||$ for $v \in H_{\infty}$. Now let $Y \in \mathfrak{g}$ be arbitrary. To show that $\pi'(Y)$ (which I write as $\pi(Y)$) is continuous, it suffices to show that, given $\alpha \in \mathbb{Z}_{\geq 0}^n$, there exist finitely many multi-indices $\beta_1, \dots, \beta_m$ and a constant $c > 0$ such that for all $v \in H_{\infty}$ we have $||\pi(Y)v|| \leq c \sum_{j = 1}^m{||v||_{\beta_i}}$. To find these $\beta_j$, write $X^{\alpha}Y = \sum_{j=1}^{m}{\lambda_j X^{\beta_j}}$ in $U$, for some coefficients $\lambda_i \in \mathbb{C}$ (this is possible thanks to the Poincaré-Birkhoff-Witt-Theorem). Now for any $v \in H_{\infty}$,
$$
||\pi(Y)v||_{\alpha} = ||\pi(X^{\alpha}Y)v|| \leq \sum_{j =1}^m{|\lambda_j| ||\pi(X^{\beta_j})v||} = \sum_{j =1}^m{|\lambda_j| ||v||_{\beta_j}} \leq c \sum_{j = 1}^m{||v||_{\beta_j}}
$$
where $c:= \max_{j =1}^m{|\lambda_j|}+1$, say.
It is also true that each transformation $\pi(g): H \rightarrow H$ preserves $H_{\infty}$ and is continuous for the topology described above. I hope the above is convincing, since I came up with it by myself.
