Prove that weak operator topology in von Neuman Algebra implies norm topology in $C^*$ - algebra

Question

I am a beginner in Operator Theory and Functional Analysis.

On the space of bounded operators on Hilbert space $H$, We claim that " It is true that von Neumann algebras are $C^*$- algebras of operators".

It is obviously true that any von Neumann algebra is close under adjoint operation.

In von Neumann algebras, we use weak operator topology. In $C^*$ - algebra, we use norm topology. In general, weak opeator convergence does not imply norm convergence.

I wonder how can we say that they are equivalent in the space of bounded operators?

Thank you for your time.

Answer 1

We don't have equivalence in general (in the finite-dimensional case we do), but we do have one implication: a von Neumann algebra is a C$^*$-algebra.

This is because the weak operator topology is weaker in the norm topology. Hence if $A\subset B(H)$ is closed in the weak operator topology, $A^c$ is open in the weak operator topology, thus open in the norm topology, and therefore $A$ is closed in the norm topology.

An example of why the other implication does not hold is the C$^*$-algebra $K(H)$ of all compact operators on $H$ (assuming $H$ is separable and infinite-dimensional). This contains all finite-rank projections, and the identity operator (which is not compact) is the weak limit of finite-rank projections. Thus $K(H)$ is not closed in the weak operator topology.