A summability matrix failing exactly one Silverman-Toeplitz condition and transforming a convergent sequence to a divergent one

Refuted claim: a summability matrix whose columns tend to $0$ and whose row sums tend to $1$ is regular; equivalently, the uniform bound on the row absolute sums in A summability matrix with only finitely many nonzero entries per row is regular iff each column tends to $0$, the row sums tend to $1$, and the row absolute sums are uniformly bounded is redundant.

The witness is the matrix with exactly two nonzero entries in each row,

$c_{n,n}:=-(n+1),c_{n,n+1}:=n+2,c_{n,k}:=0\text{ for }k\mathrm{\notin }\{n,n+1\},$

together with the null sequence $x_k:=s_k\mathrm{/}(k+1)$, where $(s_k)$ is the alternating sequence of The even and odd index maps and the alternating sequence: strictly increasing $e,o$ with $\mathbb{N}$ their disjoint union, and the unique $(s_k)$ with $s_0=1$, $s_{\sigma (k)}=-s_k$, which satisfies $\mathrm{\mid }{s}_k\mathrm{\mid }=1$, $s\circ e\equiv 1$ and $s\circ o\equiv -1$. Every column of $c$ is eventually $0$; every row sum is exactly $1$; the row absolute sums are $2n+3$ and are unbounded. The transform of $(x_k)$ is

$y_n=-(n+1)\frac{s_n}{n+1}+(n+2)\frac{s_{n+1}}{n+2}=-s_n+s_{n+1}=-2s_n,$

which does not converge although $x_k\to 0$. So $c$ is not regular, and the third condition of A summability matrix with only finitely many nonzero entries per row is regular iff each column tends to $0$, the row sums tend to $1$, and the row absolute sums are uniformly bounded is not redundant.

• Exactly one condition fails, and the counterexample is arranged so. The columns are eventually $0$ and the row sums are constantly $1$, so conditions 1 and 2 of A summability matrix with only finitely many nonzero entries per row is regular iff each column tends to $0$, the row sums tend to $1$, and the row absolute sums are uniformly bounded hold outright; only the uniform bound fails, and the failure is visible in a single line, $2n+3$ being unbounded.

• How the failure is exploited. The two entries of row $n$ are large and of opposite sign, so they nearly cancel on a slowly varying input and do not cancel at all on an alternating one. The input $x_k=s_k\mathrm{/}(k+1)$ is chosen so that the large factors $n+1$ and $n+2$ exactly cancel the small factors $1\mathrm{/}(n+1)$ and $1\mathrm{/}(n+2)$, leaving the undamped oscillation $-2s_n$. That is the gliding hump of the necessity proof in A summability matrix with only finitely many nonzero entries per row is regular iff each column tends to $0$, the row sums tend to $1$, and the row absolute sums are uniformly bounded, in its simplest possible instance.

• The failure is the stronger of the two possible ones. A matrix can be irregular by changing a limit, for instance $c_{n,n}=2$ and all other entries $0$, whose row sums tend to $2$ rather than $1$ and which sends $(x_k)$ to $(2x_k)$. The matrix above destroys convergence altogether.

• Contrast with the Cesaro matrix, whose rows are nonnegative and sum to $1$, so its row absolute sums are constantly $1$ and it is regular (The Cesaro matrix satisfies the Silverman-Toeplitz conditions, giving a second proof of the Cesaro mean theorem). Uniform boundedness of the row absolute sums is what stops a weighting from amplifying, and it is the only one of the three conditions that is not tested by a single fixed input.