# Unitary representation is strictly continuous

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.

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.