Steinhaus theorem on regular methods

Remarks

Proof idea: a "gliding hump" argument. By the Silverman–Toeplitz theorem, each fixed column of $A$ fades as $n$ grows, while row sums tend to $1$ and the absolute row sums stay bounded. So one can pick rows $n_1 < n_2 < \cdots$ and disjoint blocks of columns such that row $n_j$ has almost all its weight on block $j$. Put $s_k = 1$ on the blocks with odd $j$ and $s_k = 0$ elsewhere. Then $(As)_{n_j}$ is close to $1$ for odd $j$ and close to $0$ for even $j$, so $(As)_n$ does not converge. Example: for Cesàro summation, the sequence equal to $1$ for $2^{2j} \le k < 2^{2j+1}$ and $0$ otherwise has Cesàro means that oscillate between values near $\frac13$ and near $\frac23$.

Going deeper

For a student

The theorem shows that the bounded sequences are never all summable by one regular matrix. Extensions of the limit to all bounded sequences do exist, such as Banach limits, but they are not given by matrices, and their existence uses the Hahn–Banach theorem.