Let $H=L^2(\mathbb R)$ and $U_tf(x) = f(x-t)$. I want to prove that $t \rightarrow U_tf$ is continuous in $H$ and $t\rightarrow U_t$ is not continuous in $B(H)$.
I can't estimate $||U_tf - U_sf||_{L^2(\mathbb R)}$. In the continuity of $U_t$ I need some counterexample, I can't think of any though.
It is enough to consider the case $t=0$ (then $U_0=I=\text{identity}$.) We need to estimate $$ \|I-U_t\|_{B(H)}=\sup_{\|f\|_H\le1}\|f-U_tf\|_H. $$ For fixed $t>0$, take $f\in H$ with compact support in an interval of length smaller than $t$. The the supports of $f$ and $U_tf$ are disjoint, and $$ \|f-U_tf\|_H^2=\int_{\Bbb R}|f(x)-f(x-t)|^2\,dx=\int_{\Bbb R}|f(x)|^2\,dx+\int_{\Bbb R}|f(x-t)|^2\,dx=2\|f\|_H^2. $$ It follows that $\|I-U_t\|_{B(H)}\ge\sqrt2$, while $\|I\|_{B(H)}=1$.