A periodic chain has Cesaro but not ordinary convergence

The matrix $p$ is a transition matrix: both entries of each row are $0$ or $1$ and each row sums to one. Multiplying once, $p^2=I$, so by [F2] an induction gives $p^{(2m)}=I$ and $p^{(2m+1)}=p$ for every $m\ge 0$; in particular $p^{(n)}(0,0)=1$ for even $n$ and $p^{(n)}(0,0)=0$ for odd $n\ge 1$, while $p^{(n)}(0,\cdot )={\delta }_0$ for even $n$ and ${\delta }_1$ for odd $n$.

The chain is irreducible: $p(0,1)=1>0$ and $p(1,0)=1>0$ by step 1.1, so $0\to 1$ and $1\to 0$, and each state is accessible from itself with a zero-step path; by [F3] every pair communicates.

The law $\pi$ is invariant: ${\sum }_x\pi (x)p(x,0)=\pi (1)p(1,0)=\frac{1}{2}=\pi (0)$ and ${\sum }_x\pi (x)p(x,1)=\pi (0)p(0,1)=\frac{1}{2}=\pi (1)$, which is the criterion of [F4].

Ordinary convergence fails at the level of a single transition probability: by step 1.1 the diagonal sequence $p^{(n)}(0,0)$ equals $1$ at even $n$ and $0$ at odd $n$, so it alternates and does not converge as $n\to \mathrm{\infty }$; correspondingly the laws $p^{(n)}(0,\cdot )$ alternate between the two point masses ${\delta }_0$ and ${\delta }_1$ and do not converge.

Positive recurrence: the chain is irreducible by step 2.1 and has the invariant probability $\pi$ by step 2.2, so [F6] gives that every state is positive recurrent; in particular the hypotheses of the Cesàro supplier [F8] are met.

The chain is not aperiodic: by step 1.1 the positive return set of [F5] is $R_0=\{2,4,6,\dots \}$, whose greatest common divisor is $2$, so $d(0)=2$; the periods agree on the irreducible chain by [F7], so $\mathrm{per}(p)=2\mathrm{\ne }1$ and $p$ is not aperiodic.

Cesàro convergence: step 1.1 gives $p^{(k)}=I$ for even $k$ and $p^{(k)}=p$ for odd $k$, so for even $n=2m$ the average is $\frac{1}{2m}{\sum }_{k=0}^{2m-1}{p}^{(k)}=\frac{1}{2m}m(I+p)=\frac{1}{2}(I+p)$, and for odd $n=2m+1$ it is $\frac{m(I+p)+I}{2m+1}\to \frac{1}{2}(I+p)$; the matrix $\frac{1}{2}(I+p)$ has both rows equal to $(1\mathrm{/}2,1\mathrm{/}2)=\pi$. Hence $\frac{1}{n}{\sum }_{k=0}^{n-1}{p}^{(k)}(x,\cdot )\to \pi$ for each starting state $x$, which is the displayed Cesàro assertion; since the chain is irreducible and positive recurrent with invariant $\pi$ by steps 2.1, 2.2 and 3.1, this is exactly the conclusion of the general supplier [F8].

Boundary and scope cases: the identity at $n=0$ is included and is consistent with $p^{(0)}(0,0)=1$, so the alternation starts with the value $1$; the two-state chain is the smallest deterministic cycle and the period is exactly two, so the example exhibits the necessity of aperiodicity rather than a failure of irreducibility or of existence of $\pi$; the Cesàro average equals $\pi$ exactly for every even $n$ and converges otherwise, so no aperiodicity is needed for the averaged statement, and the example claims no converse implication in the other direction; the state space is finite, all sums are finite, and no limit is interchanged with an infinite sum; the two hypotheses needed by [F8], irreducibility and positive recurrence, are verified at steps 2.1 and 3.1 and the invariant law at step 2.2, while the matrices themselves are determined by the fixed data; and AC [A1] is spent exactly on the general suppliers [F6] and [F8], the direct computations of steps 1.1–4.1 being choice-free.