TITLE: Induced group action of the left regular representation strongly continuous
QUESTION [5 upvotes]: Let $G$ be a compact group and let $\lambda: G \rightarrow \mathcal{U}(L^2(G))$ be the left regular representation, i.e. $\lambda_sf(t)=f(s^{-1}t)$. Why is the  induced group action $\overline{\lambda}:G \rightarrow \mathrm{Aut}(\mathcal{K}(L^2(G)))$, $\overline{\lambda}_s(T)= \lambda_s T\lambda_{s^{-1}}$ strongly continuous?
Thank you.

REPLY [2 votes]: Let $g_j\to g$ a convergent net in $G$ and let $T\in\mathcal K$. We need to show that $\lambda_{g_j}T\lambda_{g_j^{-1}}$ converges to $\lambda_g T\lambda_{g^{-1}}$ in the strong topology. Now $\lambda_{g_j}$ converges to $\lambda_g$ strongly, i.e., $\lambda_{g_j}f$ tends to $\lambda_gf$ in the $L^2$ norm. This is seen for continuous $f$ by uniform convergence and for general $f$ by $L^2$-approximation with continuous functions. Now  write $$
\lambda_{g_j}T\lambda_{g_j^{-1}}f-\lambda_{g}T\lambda_{g^{-1}}f
=(\lambda_{g_j}T\lambda_{g_j^{-1}}f-\lambda_{g_j}T\lambda_{g^{-1}}f)
+(\lambda_{g_j}T\lambda_{g^{-1}}f-\lambda_{g}T\lambda_{g^{-1}}f)
$$
and use the triangle inequality as well as the fact that $\lambda_g$ preserves norms.