Idempotent multipliers on the space of analytic singular measures (Q1058094)

Let G be a LCA group with dual group \(\hat G.\) Let \(\psi:\hat H\to {\mathbb{R}}\) (the reals) be a nontrivial continuous homomorphism. Let M(G) be the usual measure algebra, and let \(M_ s(G)\) be the closed subspace of M(G) consisting of singular measures. We say that \(\mu\in M(G)\) is of analytic type, if \({\hat \mu}(\gamma)=0\) whenever \(\psi(\gamma)<0\), where \({\hat \mu}\) means the Fourier-Stieltjes transform of \(\mu\). The space of all such measures is denoted by \(M^ a(G)\), and put \(M^ a(G)_ s=M^ a(G)\cap M_ s(G)\). We consider only the case that \(M^ a(G)_ s\neq \{0\}.\) A function \(\Phi\) on \(\hat G\) is called a multiplier on \(M^ a(G)_ s\), if \(\Phi{\hat \mu}\in \{{\hat \nu}:\nu \in M^ a(G)_ s\}\) for all \(\mu\in M^ a(G)_ s\). In this case, there exists a unique bounded linear operator S on \(M^ a(G)_ s\) such tha \(S(\mu\hat)=\Phi{\hat\mu}\). The operator S is called a multiplier operator induced by \(\Phi\). We say that \(\Phi\) is an idempotent multiplier if \(S^ 2=S.\) Convolution operators on \(M^ a(G)_ s\) induced by bounded regular measures on G become multiplier operators on \(M^ a(G)_ s\). However it is known that there exists a multiplier \(\Phi\) on \(M^ a(G)_ s\) such that S is not given by convolution with a bounded regular measure on G. The author characterizes convolution operators on \(M^ a(G)_ s\) induced by regular measures on G. Moreover, if \(\psi(\hat G)\) is dense in \({\mathbb{R}}\), he proves that idempotent multiplier operators on \(M^ a(G)_ s\) are necessarily given by convolution with bounded regular measures on G of certain type. (If \(\psi(\hat G)\) is not dense in \({\mathbb{R}}\), there exists an idempotent multiplier operator on \(M^ a(G)_ s\) which is not given by convolution with a bounded regular measure on G.)