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 $\mathcal{A} = \mathcal{A}''$.



Bicommutant theorem for commutative operator algebras
The answer is negative. Consider the Hardy space $H^{\infty}(\mathbb{T})$ consisting of holomorphic functions admitting bounded extension to the unit circle and view it is a subalgebra of $B(L^{2}(\mathbb{T})$. Let us compute the commutant. Denote the operator of multiplication by $z$ by $M\_z$. If $T$ belongs to the commutant of $H^{\infty}(\mathbb{T})$, it commutes with $M\_z$. But it also has to commute with its inverse $M\_{\overline{z}}$. But these two generate $L^{\infty}(\mathbb{T})$, so the commutant is equal to $L^{\infty}(\mathbb{T})$, hence also the bicommutant, as this a maximal abelian subalgebra of $B(L^{2}(\mathbb{T}))$. It remains to see that $H^{\infty}(\mathbb{T})$ is strongly closed, but this is simple: a function $f\in L^{\infty}(\mathbb{T})$ belongs to $H^{\infty}(\mathbb{T})$ iff all of its negative Fourier coefficients vanish and this property is clearly preserved by strong convergence (even by weak convergence).


