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For a given locally compact group $G$ the Fourier-Stieltjes algebra $B(G)$ is defined as the algebra of matrix coefficients of unitary representations $\pi:G\to B(H)$. Similarly, the Fourier algebra $A(G)$ is defined as the algebra of matrix coefficients of the left regular representation $\lambda:G\to B(L_2(G))$ (see here or here). P.Eymard in his original paper ((3.6), $1^\circ$) says that when $G$ is compact, these algebras coincide: $$ A(G)=B(G). $$ Is this the only case when they coincide?

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Sergei Akbarov
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Yes. Any function in $A(G)$ vanishes at infinity, whereas $B(G)$ contains the constant functions. So whenever $G$ is non-compact, $A(G)\subsetneq B(G)$.

A related question is when $A(G) = B(G)\cap C_0(G)$. See e.g. http://arxiv.org/abs/1311.5400.

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  • Thank you! But Yemon Choi has already answered this question in MathOverflow http://mathoverflow.net/questions/181703/when-is-the-fourier-algebra-ag-enough-close-to-the-fourier-stieltjes-algebra – Sergei Akbarov Oct 14 '14 at 15:20