# When does a matrix define a convolution operator on a hypergroup?

Let $H$ be a discrete hypergroup. Suppose I have a matrix $A=(A_{x,y})$ indexed over $H$ with nonnegative entries which defines a bounded operator on $\ell^2(H)$. When does there exist $f\in\ell^1(H)$ such that $A_{x,y}=\langle f*\delta_x,\delta_y\rangle$, i.e., $A$ is the matrix of transition probabilities for a random walk given by convolution with $f$?

A necessary condition is that $A$ commutes with $\ell^1(H)$ convolution on the right. Is this sufficient?

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Presumably then you have an answer to my question as to when and why hypergroups appear in mathematics mathoverflow.net/questions/30288/… –  David Corfield Jul 12 '10 at 10:44

In case of discrete groups, it requires amenability of $H$. Indeed, $H$ is amenable if and only if $f\in\ell^1(H)$ for all $f\geq0$ such that $[f(xy^{-1})]_{x,y} \in B(\ell^2H)$. I just don't know what are hypergroups.