Let $ \mathcal{H} $ be a not-necessarily-separable Hilbert space. Let $ G $ be a locally compact Hausdorff group. It is easy to see that if $ U: G \to \mathbb{U}(\mathcal{H}) $ is a norm-continuous homomorphism from $ G $ to the group of unitary operators on $ \mathcal{H} $, then we can define a strongly continuous action $ \alpha $ of $ G $ on the $ C^{\ast} $-algebra $ \mathbb{K}(\mathcal{H}) $ of compact operators on $ \mathcal{H} $ by
$$
\forall g \in G: \qquad
\alpha\_{g}
\stackrel{\text{df}}{=}
\left\{
\begin{matrix}
\mathbb{K}(\mathcal{H}) & \to & \mathbb{K}(\mathcal{H}) \\
T & \mapsto & U(g) \circ T \circ U(g^{-1})
\end{matrix}
\right\}.
$$



> 
> **Question.** If $ \alpha $ is a **strongly continuous** action of $ G $ on $ \mathbb{K}(\mathcal{H}) $, does it necessarily arise from a **norm-continuous** homomorphism from $ G $ to $ \mathbb{U}(\mathcal{H}) $ as in the manner described above?
> 
> 
> 


I am pretty sure that the answer to this is known already, but I am frustrated by my inability to locate a reference.


Thank you so much for your assistance!




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**Clarification**


To avoid any confusion, I wish to clarify that a group acts on a $ C^{\ast} $-algebra by $ \ast $-automorphisms.



Strongly Continuous Group Actions on the $ C^{\ast} $-Algebra of Compact Operators on a Hilbert Space
The answer is no. Let $G$ be the Cartesian product of a sequence of copies of the unit circle, identified with the functions in $l^\infty$ whose modulus is constantly $1$. For $f \in G$ let $U\_f \in B(l^2)$ be multiplication by $f$. Let $G$ act on $K(l^2)$ by conjugation by these unitaries.


The action is strongly continuous because convergence in $G$ is pointwise, so if $f\_\alpha \to f$ and $T \in K(l^2)$ then $U\_{f\_\alpha}TU\_{f\_\alpha}^\* \to U\_fTU\_f^\*$ in norm, since $T$ approximately lives on finitely many coordinates. However, if we let $f\_n$ be the function which is $-1$ in the $n$th entry and $1$ everywhere else, then $f\_n \to 1$ in $G$ but $U\_{f\_n}$ does not go to $I$ in norm.


