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I am reading Runde's Lectures on Amenability. In the proof of Johnson's theorem where he proves "$L^{1}(G)$ is amenable Banach algebra implies $G$ is amenable" : he defines a $L^1(G)$ bimodule action on $L^\infty(G)$. With the help of this action, he gets finally a $\tilde{m} \in L^\infty(G)^*$ such that $$ \langle \delta_g * \phi,\tilde{m} \rangle = \langle \phi,\tilde{m} \rangle \quad (g \in G,\phi \in L^\infty(G)).$$

Then he claims $|\tilde{m}| \neq 0$ is left invariant and $\langle 1,|\tilde{m}|\rangle^{-1}|\tilde{m}|$ is a left invariant mean. I don't understand this part.

I posted this question in MSE. The post was receiving downvotes but no response. So I deleted it.

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  • $\begingroup$ I think this should be asked in MSE. $\endgroup$
    – Uri Bader
    Commented Apr 28, 2016 at 6:40
  • $\begingroup$ @user89334 People hardly answers question related to amenability in MSE nowadays $\endgroup$
    – Mambo
    Commented Apr 28, 2016 at 11:45
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    $\begingroup$ @user89334 I just posted this question in MSE providing this page MO link. I have posted some other questions related to amenability which didn't receive any answers. That's why. Sorry if I am wasting your time. $\endgroup$
    – Mambo
    Commented Apr 28, 2016 at 13:17
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    $\begingroup$ Do ou undertand that $\langle \delta_g*\phi,\tilde{m}\rangle=\langle \phi,g^{-1}\tilde{m}\rangle$? $\endgroup$
    – Uri Bader
    Commented Apr 28, 2016 at 19:39
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    $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – Mambo
    Commented Apr 28, 2016 at 20:35

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