Consider $l^\infty(\mathbb{Z})$ the Banach space of bounded complex valued functions on the abelian group $\mathbb{Z}$ with the supremum norm. It has a natural action by $\mathbb{Z}$ given by $(zf)(g):=f(z^{-1}g)$. Then let $\mathcal{B}^{\mathbb{Z}}(l^\infty(\mathbb{Z}))$ be the algebra (multiplication by composition) consisting of all bounded $\mathbb{Z}$-equivariant operators $l^\infty(\mathbb{Z})\rightarrow l^\infty(\mathbb{Z})$.


Question: Is this a commutative algebra?


If $l^\infty$ is replaced by $l^2$ the algebra is the von Neumann algebra $L(\mathbb{Z})$ and in this cases $\mathbb{C}(\mathbb{Z})$ lies densely (wrt WOT or SOT) in $L(\mathbb{Z})$ as a subalgebra and therefore $L(\mathbb{Z})$ is commutative. But I think $\mathbb{C}(\mathbb{Z})$ does not lie densely in $\mathcal{B}^{\mathbb{Z}}(l^\infty(\mathbb{Z}))$, so the situation is a bit different.


Question: What happens if $\mathbb{Z}$ is replaced by an arbitrary abelian group?



Is $\mathcal{B}^{\mathbb{Z}}(l^\infty(\mathbb{Z}))$ a commutative algebra?
The following is an abstract (Banach) algebraic take on Werner's construction. Let $A=\ell^1(\mathbb Z)$ with convolution (but in general $A$ is any Banach algebra). We turn the dual space $A^\*$ into an $A$-bimodule (though in our example, $A$ is commutative) by dualising the actions:
$$ (a\cdot f)(b) = f(ba), \quad (f\cdot a)(b) = f(ab) \qquad
(f\in A^\*, a,b\in A). $$
Then you can check that $T:A^\*\rightarrow A^\*$ is $\mathbb Z$-equivariant if and only if $T$ is a module homomorphism: I'll write $T\in\hom\_A(A^\*)$ to mean $T(f\cdot a) = T(f)\cdot a$ (the right action is more appropriate in the non-commutative group setting). If $e\in A$ is the unit then
$$ T(f)(a) = T(f)(ae) = (T(f)\cdot a)(e) = T(f\cdot a)(e)
\qquad (T\in \hom\_A(A^\*), f\in A^\*, a\in A). $$
So again $T$ is determined uniquely by $m\in A^{\*\*}, m(f) = T(f)(e)$. Conversely, given $m\in A^{\*\*}$ we can define $T\in\hom\_A(A^\*)$ by this relation. So $\hom\_A(A^\*)\cong A^{\*\*}$. A similar result holds if $A$ only has a bounded approximate identity.


So $A^{\*\*}$ becomes an algebra for the product induced from $\hom\_A(A^\*)$. Define $$ (m\cdot f)(a) = m(f\cdot a) \qquad (m\in A^{\*\*}, f\in A^\*,a\in A).$$
Then if $m\_i$ is associated to $T\_i$ for $i=1,2$,
$$ (m\_1m\_2)(f) = (T\_1\circ T\_2)(f)(e) = T\_1(T\_2(f))(e)
= m\_1(T\_2(f)). $$
However, for $a\in A$,
$$ T\_2(f)(a) = m\_2(f\cdot a) = (m\_2\cdot f)(a) $$
and so
$$ (m\_1m\_2)(f) = m\_1(m\_2\cdot f) $$
This is precisely the "1st Arens product", say $\Box$, as studied in Banach algebra theory. By making the symmetric choices we get the "2nd Arens product", say $\diamond$, which might differ. If $A$ is commutative then $m\_1\Box m\_2 = m\_2\diamond m\_1$ and so $\hom\_A(A^\*)$ is commutative precisely when the Arens products agree.


However, Young showed in "The irregularity of multiplication in group algebras"
Quart J. Math. Oxford Ser. (2) 24 (1973), 59–62, [MathSciNet](http://www.ams.org/mathscinet-getitem?mr=320756) that for any locally compact group $G$, we have that $L^1(G)$ is not Arens regular. So Werner's question has a negative answer for all Abelian groups (with the axiom of choice!)


