Let $A$ be a $C^*$-algebra which is not $W^*$ (i.e. it has no predual Banach space). Is it possible that there exists a Banach space $X$ such that $X^{**}=A^*?$ This is the next best thing and, as shown by David Gao, this would correspond to a minimal closed, weakly$^*$-dense subspace of $A^*.$ As of right now, it is not clear whether a minimal closed, weakly$^*$-dense subspace exists for any such $A$ so this is of great interest to me.
For example, if $A=c_0,$ then it has no predual. Moreover, $\ell^1=c_0^*$ has no bipredual, by Onur Oktay. One example I would be very interested in is $A=K(H)$ for separable $H.$ Then is $A^*=L^1(H)$ the bidual of some Banach $X?$
Update: If $A^*=X^{**},$ then $A^{**}=X^{***}$ is a W$^*$-algebra which is the third dual of a Banach space. The article Onur provided shows that, in this case, $X$ is the predual of a W$^*$-algebra. Thus, there exists some W$^*$-algebra $B$ s.t. $B=X^*,$ so $B^*=A^*,$ but it is then not clear that $B$ is $*$-isomorphic to $A.$
The question can be rephrased (equivalently): Let $A$ be C$^*$, $B$ be W$^*$, and suppose $A^*$ is isometrically isomorphic to $B^*.$ Is it necessary that $A$ is W$^*$? If it helps, you also have that $A^{**}$ and $B^{**}$ are $*$-isomorphic W$^*$'s.
Some ideas: By Sakai, $B^*\cong B_*\oplus B^*_s$, thus $A^*\cong X\oplus B^*_s.$ Therefore, we have that $X$ is complemented in $A^*$.
