I am posting my question of mathstack exchange here. (see: My post on MSE)
Let $G$ be a locally compact group with Haar measure $\mu$, and $B(G),A(G),C_r^*(G),L(G)$ be its Fourier-Stieltjes algebra, Fourier algebra, group $C^* $-algerba and von Neumann algebra respectively. For $\varphi \in L^\infty(G)$ consider the maps $$ M_\varphi: A(G) \rightarrow A(G), \, \psi \mapsto \varphi\psi \quad \text{ for all } \psi \in A(G), $$ $$ m_\varphi: C_r^*(G) \rightarrow C_r^*(G), \, f \mapsto \varphi f \quad \text{ for all } f \in C_r^*(G). $$ $$ \overline{m_\varphi}: L(G) \rightarrow L(G), \, f \mapsto \varphi f \quad \text{ for all } f \in L(G). $$ They are related in the following ways:
Proposition 1 ([1, Proposition 3.2]) Let $\varphi \in L^\infty(G)$, then $M_\varphi$ is well defined and completely bounded if and only if $\overline{m_\varphi}$ is well defined and completely bounded. Moreover, their c.b. norms coincide.
If $\varphi \in L^\infty(G)$ satisfies these properties, it is said to be a completely bounded multiplier and the set of all such functions is denoted by $M_0A(G)$.
Proposition 2 ([1, Theorem 2.2]) Let $\varphi \in L^\infty(G)$, the following are equivalent:
- $\varphi \in B(G)$;
- $m_\varphi: C_r^*(G) \rightarrow C_r^*(G)$ is bounded;
- $m_\varphi: C_r^*(G) \rightarrow C_r^*(G)$ is completely bounded;
From my understanding, this two results would imply $M_0A(G) = B(G)$, which known to be false (see for example [2]).
Fake Proof: It is well-known that $B(G) \subseteq M_0A(G)$, so the deal is to "prove" the converse. Let $\varphi \in M_0A(G)$, then $\overline{m_\varphi}: L(G) \rightarrow L(G)$ is well-defined and c.b. by Proposition 1. We can see that $C_r^*(G) \subseteq L(G)$ is invariant for $\overline{m_\varphi}$, indeed if $f \in C_c(G)$, then $\overline{m_\varphi}(f) = \varphi f \in C_c(G)$ and since all $f \in C_r^*(G)$ are norm limits of elements in $C_c(G)$, then $\overline{m_\varphi}(f) \in C_r^*(G)$. We then have that $$ \overline{m_\varphi}|_{C_r^*(G)} = m_\varphi : C_r^*(G) \rightarrow C_r^*(G) $$ is (completely) bounded which by Proposition 2 gives $\varphi \in B(G)$.
The most important step was the proof that $\overline{m_\varphi}|_{C_r^*(G)} = m_\varphi$, which seems correct to me. However, it has to be the flawed part. Can anyone point out the mistake in my proof?
1 Renault, Jean, The Fourier algebra of a measured groupoid and its multipliers, J. Funct. Anal. 145, No. 2, 455-490 (1997). ZBL0874.43003.
[2] de Cannière, Jean; Haagerup, Uffe, Multipliers of the Fourier algebras of some simple Lie groups and their discrete subgroups, Am. J. Math. 107, 455-500 (1985). ZBL0577.43002.
