Any finite group G is amenable

Question

A linear functional $μ:L^∞ (G)→\mathbb{C}$ is called a mean on $L^∞ (G)$ if $ μ(1)=1$ and is positive, i.e. if $μ(f)≥0$ for all positive $f∈L^∞ (G)$.

A group G is amenable if there exists an invariant mean on $L^∞ (G)$.

A mean on $L^∞ (G)$ is called invariant if it satisfies

$$μ(f(g^{-1}h))=μ(f(h)$$

proof:
The linear functional $μ:L^∞ (G)→\mathbb{C}$ denfied by

$$ μ(f)=|G|^{-1} ∑_{g∈G}f(g)$$

$ μ(1)=|G|/|G| =1$

if $f≥0$ then $∑_{g∈G}f(g)≥0$ so $μ(f)≥0$

$μ(f(g^{-1}h))=|G|^{-1} ∑_{g∈G}f(g^{-1}h)\quad*$

My question:How (*) is equal $μ(f(h)$?

Answer 1

Notice that $$\mu(f(g^{-1}\cdot ))=\frac 1{|G|}\sum_{a\in G}f(g^{-1}a)=\frac 1{|G|}\sum_{a\in G}f(a)=\mu(f(\cdot)).$$ Indeed, we used the definition for the first equality and for the second the map $a\mapsto g^{-1}a$ from $G$ to $G$ is bijective. It's the same is if $\sigma$ is a permutation from $\{1,\dots,n\}$ to itself and $(a_j)_{1\leqslant j\leqslant n}$ a family of complex numbers, then $\sum_{j=1}^na_j=\sum_{j=1}^na_{\sigma(j)}$.