TITLE: Extension of $C^*$ isomorphism to $W^*$ isomorphism
QUESTION [5 upvotes]: Let $\mathfrak{A}$ be $C^*$algebra, and $\pi$ its faithful representation on Hilbert space $\mathcal{H}$. Bicommutant $\mathfrak{B}=\pi(\mathfrak{A})''$ is the von Neumann algebra generated by $\pi(\mathfrak{A})$. Let $\phi$ be ${}^*$-automorphism of $\mathfrak{A}$. Under what conditions the ${}^*$-automorphism $\gamma: \pi(\mathfrak{A})\rightarrow \pi(\mathfrak{A})$, $\gamma = \pi\circ\phi\circ\pi^{-1}$ of the algebra $\pi(\mathfrak{A})$ might be extended to ${}^*$-automorphism of $\mathfrak{B}$.
EDIT:
Let us assume in addition that $\forall_{A\in\mathfrak{A}}\gamma(\pi(A)) = \lim_{n\rightarrow\infty}\gamma_n(\pi(A))$, where $\gamma_n(B)=U_n B U_n^{-1}$, $U_n$ - sequence of unitary operators in $\mathcal{H}$, but $\gamma$ is not implementable by unitary operator.

REPLY [4 votes]: Here is a partial answer to your question, in the paper:
Derivations and automorphisms of operator algebras by Richard V. Kadison and John R.Ringrose, Commun.math.Phys. 4,32-63(1967)
It is proved that each derivation on your represented algebra extends to its weak closure provided the representation is faithful.
Furthur, they use this result to conclude that if an automorphism is within norm distance 2 to the identity automorphism, then it is in the connected component of the group of automorphism and therefore also extends to an ( actually inner) automorphism of the weak closure.
You should be able to find the necessary details there.