Let $G$ be a compact group and $u: G \to B(H)$ be a strongly continuous unitary representation on the Hilbert space $H$. Then is $u: G \to B(H)$ strictly continuous?


That is, give $B(H)$ the topology induced by the $\*$-isomorphism $M(B\_0(H))\cong B(H)$. Explicitely, a net $(x\_i)$ in $B(H)$ converges strictly to $x$ iff $\|x\_i y -x y\|\to 0$ and $\|yx\_i \to yx\| \to 0$ for all compact operators $y \in B\_0(H)$. On bounded subsets this agrees with the $\*$-strong topology.



Unitary representation is strictly continuous
As you note, on bounded sets, the strict topology and the strong-$\ast$ topology agree on bounded sets. As the set of unitary operators *is* bounded, we can just work with the strong-$\ast$ topology. If $(u\_i)$ is a net of unitary operators converging strongly to $u$ a unitary, then for $\xi\in H$,


$$ \| u\_i^\ast(\xi) - u^\ast(\xi)\|^2 = \|u\_i^\ast(\xi)\|^2 - (u\_i^\ast(\xi)|u^\ast(\xi)) - (u^\ast(\xi)|u\_i^\ast(\xi)) + \|u^\ast(\xi)\|^2 $$


Here I write $(\cdot|\cdot)$ for the inner product. As $u\_i,u$ are unitary, this is equal to


$$ 2\|\xi\|^2 - (\xi|u\_iu^\ast(\xi)) - (u\_iu^\ast(\xi)|\xi). $$


As $u\_i(\eta)\rightarrow u(\eta)$ for any $\eta$, this converges to


$$ 2\|\xi\|^2 - (\xi|u u^\ast(\xi)) - (u u^\ast(\xi)|\xi) = 0. $$


Thus $u\_i\rightarrow u$ strong-$\ast$. This is just a proof that the strong and strong-$\ast$ topologies agree on the set of unitaries; this is surely in standard textbooks, if you look hard.


