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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 understand that if $m$ is a left invariant finitely additive measure on a group $G$, then $|m|$ is also a left invariant measure. But how to view this here.

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

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.

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 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 understand that if $m$ is a left invariant finitely additive measure on a group $G$, then $|m|$ is also a left invariant measure. But how to view this here.

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

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 mean $\tilde{m} \in L^\infty(G)^*$ 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 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.