# Bicommutant theorem for commutative operator algebras

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

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).