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

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

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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 $L^{\infty}(\mathbb{R})$.

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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.

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