Bicommutant theorem for commutative operator algebras

Question

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

Answer 1

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).