Let $H$ be a Hilbert space and $T : H \supseteq \mathrm{dom}(T) \to H$ be a self-adjoint linear operator on $H$. Define a one-parameter group $(\tau_t)_{t \in \mathbb{R}}$ of $\ast$-automorphisms on the $C^\ast$-algebra $\mathscr{B}(H)$ of bounded operators on $H$ by setting $$ \tau_t(A) = e^{i t T} A \, e^{- i t T} $$ for all $A \in \mathscr{B}(H)$, $t \in \mathbb{R}$. I came across the following statement in some book: the group $(\tau_t)$ is strongly continuous on $\mathscr{B}(H)$ if and only if the operator $T$ is bounded on $H$.
Question: How can one prove this equivalence, especially the implication that strong continuity implies boundedness of $T$?
Since $T$ is self-adjoint, it would suffice to show that $\mathrm{dom}(T) = H$. I thought about mimicking the well-known proof of the fact that a strongly continuous one-parameter group of operators is uniformly continuous iff its generator is defined everywhere, but this did not work out as I hoped.
Context of the question: If $T$ is interpreted as the Hamiltonian of a quantum system, then the group $(\tau_t)$ describes the time evolution of (bounded) observables of the system. Hence, the above statement implies that the quantum dynamical system $(\mathscr{B}(H), \tau_t)$ is a $C^\ast$-dynamical system iff $T$ is bounded.
$\def\abajo{\\[0.2cm]}\def\e{\varepsilon}$ Suppose that $T$ is unbounded. Note first that if $\|T-T_0\|<\e$ and $TT_0=T_0T$, then \begin{align} \|e^{itT}Ae^{-itT}-e^{itT_0}Ae^{-itT_0}\| &\leq\|e^{itT}Ae^{-itT}-e^{itT_0}Ae^{-itT}\| + \|e^{itT_0}Ae^{-itT}-e^{itT_0}Ae^{-itT_0}\|\abajo &=\|e^{itT}A-e^{itT_0}A\| + \|Ae^{-itT}-Ae^{-itT_0}\|\abajo &\leq\|e^{itT}-e^{itT_0}\|\,\|A\| + \|e^{-itT}-e^{-itT_0}\|\,\|A\|\abajo &=\|e^{it(T-T_0)}-I\|\,\|A\| + \|I-e^{-it(T-T_0)}\|\,\|A\|. \end{align} The above allows us to replace, without loss of generality, $T$ with a discretized version (via the Spectral Theorem, see details at the end). That is, we assume $$\tag1 T=\sum_n t_n\,E_{nn}, $$ where the $E_{nn}$ are pairwise orthogonal rank-one projections, and $\{t_n\}$ is an unbounded sequence of real numbers. This gives us $$ e^{ihT}=\sum_ne^{iht_n}E_{nn}. $$
By fixing an appropriate orthonormal basis, we extend the $E_{nn}$ to a system of matrix units $\{E_{kj}\}$. Using that $\{t_n\}$ is unbounded, choose a subsequence $\{t_{n_k}\}$ such that $$ |t_{k}-t_{n_k}|>k. $$ Let $$ A=\sum_k E_{k,n_k}. $$ It is straighforward to see that $A^*A\leq I$, and so $A$ is bounded. We have \begin{align} \|A-e^{ihT}Ae^{-ihT}\| &=\Big\|\sum_k E_{k,n_k}-\sum_{n,m}\sum_k e^{ih(t_n-t_m)}E_{nn}E_{k,n_k}E_{mm}\Big\|\abajo &=\Big\|\sum_k E_{k,n_k}-\sum_k E_{k,n_k}e^{ih(t_k-t_{n_k})}\,E_{k,n_k}\Big\|\abajo &=\Big\|\sum_k (1-e^{ih(t_k-t_{n_k})})\,E_{k,n_k}\Big\|\abajo &=\max\{|1-e^{ih(t_k-t_{n_k})}|:\ k\} \end{align} If we now consider the sequence $h_r=\frac\pi{t_r-t_{n_r}}$ then $h_r\to0$ and $$ \|A-e^{ih_rT}Ae^{-ih_rT}\|\geq|1-e^{ih_r(t_r-t_{n_r})}| =1-e^{i\pi}=2. $$ So $\tau_t(A)$ is not continuous at $t=0$.
How to get $T_0$.
By the Spectral Theorem, $$ T=\int_{\sigma(T)}\lambda\,dE(\lambda). $$ Given $\e>0$, choose $\{a_n\}_{n\in\mathbb Z}$ to be a partition of $\mathbb R$ with $|a_{n+1}-a_n|<\e$. Form $I_n=[a_n,a_{n+1})$. Define $$f=\sum_n a_n\,1_{I_n}$$ and $$ T_0=\int_{\sigma(T)}f(\lambda)\,dE(\lambda)=\sum_na_n\,E(I_n). $$ By writing each $E(I_n)$ as a sum of pairwise orthogonal rank-one projections and repeating the corresponding $a_n$ accordingly, we get $(1)$. And from $\|\operatorname{id}-f\|_\infty<\e$, $$ T-T_0=\sum_nTE(I_n)-a_nE(I_n), $$ and $$ a_nE(I_n)\leq\int_{I_n}\lambda\,dE(\lambda)=TE(I_n)\leq a_{n+1}E(I_n), $$ we get $$ \|T-T_0\|=\max\{\|TE(I_n)-a_nE(I_n)\|: n\}<\e. $$