Von Neumann algebra generated by a subalgebra

Question

Let A be a C*-algebra of operators on a Hilbert space H. Show that if $A\subset K(H)$, then $\{A'\cap K(H)\}'\cap K(H) = A$
I do not have any idea about it. Please give me a hint. Thanks.

Answer 1

As stated this result is false.

Take $H=\mathbb{C^n}$. Then, $K(H)=B(H)$. Let $A=\{0\}$.

Then, $A'=B(H)$. and we have:

$\{A' \cap K(H)\}' \cap K(H) = \mathbb{C}I \neq A$.

Note that $A$ is contained in LHS.

EDIT: I think the result is false in general as well. Here's the counterexample.

Let $e_1,e_2, \cdots$ be a basis of $H$. Let $A$ consist of operators $T$ such that $Te_1=ce_1$ and $Te_n=0, n \ge2$.

Then all $2$ by $2$ diagonal matrices belong to the LHS, but do not belong to $A$.