Taken from Random Walks on Finite Quantum Groups by Franz & Gohm.

## Theorem

Let $\phi$ be a state on a finite quantum group $A$. Then the Cesaro mean $\displaystyle \phi_n=\frac{1}{n}\sum_{k=1}^n\phi^{\star n}$ $n\in\mathbb{N}$

converges to an idempotent state on $A$, i.e. to a state $\pi$ such that $\pi\star\pi=\pi$.

Proof : Let $\phi'$ be an accumulation point of $\{\phi_n\}_{n\geq0}$, this exists since the states on $A$ form a compact set. We have $\|\phi_n-\phi\star \phi_n\|=\frac{1}{n}\|\phi-\phi^{n+1}\|\leq \frac{2}{n}$.

I have no idea where the equality comes from.

Choose sequence $\{n_k\}_{k\geq 0}$ such that $\phi_{n_k}\rightarrow \phi'$, we get $\phi\star\phi'=\phi'$ and similarly $\phi'\star \phi=\phi'$. By linearity this implies $\phi_n\star\phi'=\phi'=\phi'\star \phi_n$. If $\phi''$ is another accumulation point of $\{\phi_n\}_{n\geq 0}$ and $\{m_{\ell}\}_{\ell\geq 0}$ a sequence such that $\phi_{m_\ell}\rightarrow\phi''$, then we get $\phi''\star\phi'=\phi'=\phi'\star\phi''$ and thus $\phi'=\phi''$ by symmetry (??). Therefore the sequence $\{\phi_n\}_{n\geq0}$ has a unique accumulation point, i.e. it converges $\bullet$

### Remark

If $\phi$ is faithful, then the Cesaro limit $\pi$ is the Haar state on $A$ (prove this).

### Remark

Due to cyclicity the sequence $\{\phi^{\star n}\}_{n\geq 0}$ does not converge in general. Take, for example, the state $\phi=\eta_2$ (p.28) on the Kac-Paljutkin quantum group $A$, then we have $\eta_2^{\star n}=\left\{\begin{array}{ccc}\eta_2&\text{if}& n\text{ is odd}\\ \varepsilon&\text{if}&n\text{ is even}\end{array}\right.$,

but $\displaystyle \lim_{n\rightarrow\infty}\frac1{n}\sum_{k=1}^n\eta_2^{\star k}=\frac{\varepsilon+\eta_2}2$.