Let $U : \mathbb{R} \rightarrow H^{-1}(\mathbb{R})$ a strongly continous unitary group on $H^{-1}(\mathbb{R})$ In particular the map $t \in \mathbb{R} \rightarrow U(t)\delta_{0} \in H^{-1}$ is continuous.
Does the map : $$ t \in \mathbb{R} \rightarrow \langle U(t)\delta_{0}, \varphi \rangle $$ is continuous for $\varphi \in C_{c}^{\infty}(\mathbb{R}) $ given ?
I think it's enough to show that if $\varphi \in C_{c}^{\infty}(\mathbb{R}) $ then $L_{\varphi} : T \in H^{-1} \rightarrow \langle T, \varphi \rangle $ is continuous.
Thanks and regards.
Assume that $T_n\to 0$ in $H^{-1}$, this means that
$$\|\mathcal F^{-1}([1+\xi^2]^{-1/2}\mathcal FT_n)\|_{L^2}\to0.$$ Since the Fourier transform is unitary on $L^2$ this is the same as $\int_{\Bbb R}\frac1{1+\xi^2}|\mathcal FT_n|^2\to0$ (with a big root symbol over the integral). In particular we can multiply with $\int_{\Bbb R}(1+\xi^2)|\mathcal F\varphi|$ ($\mathcal F\varphi$ is Schwartz) and what you get will still go to $0$. Now here you have a product of the form $\|X_n\|_{L^2}\cdot\|Y\|_{L^2}$, which by Cauchy inequality bounds $|\langle X_n, Y\rangle|$. This thing that is bounded is however equal to:
$$\int \frac1{(1+\xi^2)^{1/2}}(1+\xi^2)^{1/2} \overline{\mathcal FT_n}\,\mathcal F\varphi = \langle\mathcal FT_n,\mathcal F\varphi\rangle = \langle T_n,\varphi\rangle.$$ This gives you continuity.