Let $G$ be a finite discrete group. We denote by $A(G)$ the Fourier algebra of $G$ and $M_{cb}A(G)$ the space of completely bounded multipliers of $A(G)$.
Is is true that $A(G)=M_{cb}A(G)$ isometrically?
2026-04-12 11:36:20.1775993780
Fourier algebra and multipliers of a finite discrete group
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