Let $\mathcal{B}(H)$ denote the space of bounded linear operators on a complex Hilbert space $H$. The *von Neumann bicommutant theorem* says:

**Theorem.** Suppose that $\mathcal{A}$ is a $C^*$-subalgebra of $\mathcal{B}(H)$ and that $\mathcal{A}$ contains the identity operator. If $\mathcal{A}$ is closed with respect to the strong operator topology, then $\mathcal{A}$ coincides with its *bicommutant* $\mathcal{A}''$.

[*Note:* The *commutant* $\mathcal{S}'$ of a subset $\mathcal{S} \subseteq \mathcal{B}(H)$ is defined as the set of all operators in $\mathcal{B}(H)$ that commute with all operators in $\mathcal{S}$.]

**Question.** Does this theorem remain true if we replace the assumption "$\mathcal{A}$ is a $C^*$-subalgebra of $\mathcal{B}(H)$" with the assumption "$\mathcal{A}$ is a commutative subalgebra of $\mathcal{B}(H)$"?

The question seems to be a bit bold at first glance, but it is motivated by the following simple finite-fimensional observation:

**Motivation.** The bicommutant theorem fails if we merely assume $\mathcal{A}$ to be a subalgebra of $\mathcal{B}(H)$ that is strongly closed and contains the identity operator. As a simple counterexample we can choose $\mathcal{A}$ to be the set of all upper triangular matrices in $\mathbb{C}^{2\times 2} = \mathcal{B}(\mathbb{C}^2)$.

On the other hand, the algebra $\mathcal{A}$ is not commutative in this example, and if we try to "adjust" the example by choosing $\mathcal{A}$ as the commutative algebra \begin{align*} \mathcal{A} := \{ \begin{pmatrix} a & b \newline 0 & a \end{pmatrix}: \; a, b \in \mathbb{C} \}, \end{align*}, then we indeed have $\mathcal{A} = \mathcal{A}''$.