# The second dual of a von Neumann algebra

Let $$A$$ be a von Neumann algebra and let $$A^{**}$$ be the second dual space of $$A$$.

Is this true that $$A^{**}=A$$ ?

No, it is in general not true. For example, $$L^{\infty}(\mathbb{R})$$ is a von-Neumann algebra, but it is not a reflexive Banach space, that is, the second dual of $$L^{\infty}(\mathbb{R})$$ is not isomorphic to $$L^{\infty}(\mathbb{R})$$.
It is probably true that being reflexive for von Neumann algebras implies being finite dimensional. Since every continuous linear map would be automatically weak-$$\ast$$ continuous, the weak-$$\ast$$ and norm topologies would have the same closed compact sets, but then the unit ball would be compact.