Example of operator that commutes with a given multiplication that is not itself a multiplication.

Question

I am aware that given a diagonal matrix $D$ whose diagonal entries are all distinct, any matrix $A$ that commutes with $D$ must be itself diagonal. I am also aware that this result does not generally extend to linear operators on Banach spaces, where in place of the given diagonal matrix we use a multiplication operator. What I am after is a counterexample.

To be more precise, let $\mathcal{F}$ be the Banach space of essentially bounded real functions on a subset the real line. Define $D_d$ to be the multiplication operator associated with the essentially bounded function $d$ that is not constant on any set of positive measure, that is, $D_d[f](x)=d(x)f(x)$. I am looking for an example of a function $d$ and a linear operator $A$, so that $A$ commutes with $D_d$ but $A$ is not itself a multiplication operator. Any help would be greatly appreciated.

Edit: What if I strengthen the restrictions so that $d$ is one-to-one, which more closely aligns with the matrix example?

Answer 1

Suppose that $d$ is constant on a set of positive measure. To make it concrete, let $\mathcal F=L^\infty(\mathbb R)$, let $E=[0,1]$ and $d=1_E$. Then $D_d$ commutes with any $A$ that acts on $1_E\,\mathcal F$.

For example, let $A$ be given by $$ (Ag)(t)=1_E(t)\,g(1-t). $$ Then $$ (AD_dg)(t)=(A1_Eg)(t)=1_E(1-t)\,g(1-t)=1_E(t)\,g(1-t)=(D_dAg)(t). $$ So $A$ commutes with $D_d$.

Answer 2

Consider $L^\infty(\mathbb R)$ with $d(x) = x^2$ and $A f(x) = f(-x)$. Then $A D_g f(x) = x^2 f(-x) = D_g A f(x)$.