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

Sufficiency. Assume conditions 1, 2 and 3, let $(x_k)$ converge to $L$, and let $\epsilon >0$ be an arbitrary real; fix $M\ge 0$ as in condition 3, and fix $D\ge 0$ with $\mathrm{\mid }{x}_k-L\mathrm{\mid }\le D$ for every $k$, which exists because a convergent sequence is bounded.

Choose $K\in \mathbb{N}$ with $\mathrm{\mid }{x}_k-L\mathrm{\mid }<\epsilon (3(M+1){)}^{-1}$ for every $k\ge K$.

For every $n$, choosing an admissible bound $R\ge K$ for row $n$, one has $y_n-L={\sum }_{k=0}^R{c}_{n,k}(x_k-L)+({\sum }_k{c}_{n,k}-1)L$, since ${\sum }_{k=0}^R{c}_{n,k}L=L{\sum }_k{c}_{n,k}$.

Necessity. The remaining steps, apart from 2.1, 2.2, 2.3 and 3.1 which finish the sufficiency argument above, assume instead that $c$ is regular.

Suppose, towards a contradiction, that the row absolute sums $(r_n)$ are not bounded above, that is, for every $T\in \mathbb{R}$ there is $n$ with $r_n>T$.

For each $n$ the set of admissible bounds for row $n$ is a nonempty subset of $\mathbb{N}$, so it has a least element $\rho (n)$; thus $c_{n,k}=0$ for every $k>\rho (n)$, and every $R\ge \rho (n)$ is admissible for row $n$.

For every $n$: $\mid {\sum }_{k=0}^R{c}_{n,k}(x_k-L)\mid \le {\sum }_{k<K}\mathrm{\mid }{c}_{n,k}\mathrm{\mid }\mathrm{\mid }{x}_k-L\mathrm{\mid }+{\sum }_{k=K}^R\mathrm{\mid }{c}_{n,k}\mathrm{\mid }\mathrm{\mid }{x}_k-L\mathrm{\mid }\le D{\sum }_{k<K}\mathrm{\mid }{c}_{n,k}\mathrm{\mid }+\epsilon (3(M+1){)}^{-1}{r}_n\le D{\sum }_{k<K}\mathrm{\mid }{c}_{n,k}\mathrm{\mid }+\epsilon \mathrm{/}3$, the last step because $r_n\le M<M+1$.

By condition 1 each of the finitely many columns $k<K$ satisfies $c_{n,k}\to 0$, hence $\mathrm{\mid }{c}_{n,k}\mathrm{\mid }\to 0$ since $\mid \mathrm{\mid }{c}_{n,k}\mathrm{\mid }-0\mid =\mathrm{\mid }{c}_{n,k}-0\mathrm{\mid }$; a sum of finitely many null sequences is null, by induction on the number of summands, so ${\sum }_{k<K}\mathrm{\mid }{c}_{n,k}\mathrm{\mid }\to 0$ in $n$ and there is $N_1$ with ${\sum }_{k<K}\mathrm{\mid }{c}_{n,k}\mathrm{\mid }<\epsilon (3(D+1){)}^{-1}$ for every $n\ge N_1$.

By condition 2 there is $N_2$ with $\mid {\sum }_k{c}_{n,k}-1\mid <\epsilon (3(\mathrm{\mid }L\mathrm{\mid }+1){)}^{-1}$ for every $n\ge N_2$, so that $\mid ({\sum }_k{c}_{n,k}-1)L\mid <\epsilon \mathrm{/}3$ for such $n$.

Condition 1 holds. Fix $k$ and let $e$ be the sequence with $e_k=1$ and $e_j=0$ for $j\mathrm{\ne }k$; it is eventually $0$, so it converges to $0$, and its transform at row $n$ is $c_{n,k}$ because every other term of the row sum vanishes. Regularity gives ${\lim }_n{c}_{n,k}=0$.

Condition 2 holds. The constant sequence with value $1$ converges to $1$ and its transform at row $n$ is the row sum ${\sum }_k{c}_{n,k}$, so regularity gives ${\lim }_n{\sum }_k{c}_{n,k}=1$.

For every $n$ beyond both $N_1$ and $N_2$: $\mathrm{\mid }{y}_n-L\mathrm{\mid }\le D{\sum }_{k<K}\mathrm{\mid }{c}_{n,k}\mathrm{\mid }+\epsilon \mathrm{/}3+\epsilon \mathrm{/}3<\epsilon \mathrm{/}3+\epsilon \mathrm{/}3+\epsilon \mathrm{/}3=\epsilon$; as $\epsilon$ was arbitrary, $y_n\to L$, and as $(x_k)$ was an arbitrary convergent sequence, $c$ is regular.

Each column converges, hence is bounded, so $M_k:=\sup \{\mathrm{\mid }{c}_{n,k}\mathrm{\mid }:n\in \mathbb{N}\}$ exists in $\mathbb{R}$ for every $k$; putting $B_m:={\sum }_{k=0}^m{M}_k$ one has ${\sum }_{k=0}^m\mathrm{\mid }{c}_{n,k}\mathrm{\mid }\le B_m$ for every $n$ and every $m$, and $B_m\ge 0$.

Define by recursion $k_0:=0$, $n_0:=0$ and, for $j\ge 1$, first $T_j:=$ the larger of $\max \{r_n:n\le n_{j-1}\}$ and $B_{k_{j-1}}+j(j+B_{k_{j-1}})$, then $n_j:=\min \{n:r_n>T_j\}$, which exists by step 1.5 and well-ordering, and then $k_j:=$ the larger of $k_{j-1}+1$ and $\rho (n_j)$; then $n_j>n_{j-1}$, because $r_{n_j}$ exceeds every $r_n$ with $n\le n_{j-1}$, and $k_{j-1}<k_j$ with $c_{n_j,k}=0$ for every $k>k_j$, and $r_{n_j}>B_{k_{j-1}}+j(j+B_{k_{j-1}})$.

Since $(k_j)$ is strictly increasing with $k_0=0$, every $k\ge 1$ lies in exactly one block $k_{j-1}<k\le k_j$ with $j\ge 1$; define $x_0:=0$ and, for $k$ in the $j$-th block, $x_k:=\mathrm{sgn}(c_{n_j,k})j^{-1}$, where $\mathrm{sgn}(t):=1$ for $t\ge 0$ and $\mathrm{sgn}(t):=-1$ for $t<0$. Then $\mathrm{\mid }{x}_k\mathrm{\mid }\le 1$ for every $k$, and $x_k\to 0$: given $\epsilon >0$, take $J\ge 1$ with $1\mathrm{/}J<\epsilon$, and every $k>k_J$ lies in a block with index $j>J$, so $\mathrm{\mid }{x}_k\mathrm{\mid }=1\mathrm{/}j<1\mathrm{/}J<\epsilon$.

For every $j\ge 1$: the terms of $y_{n_j}$ with $k>k_j$ vanish, so $y_{n_j}={\sum }_{k\le k_{j-1}}{c}_{n_j,k}{x}_k+{\sum }_{k_{j-1}<k\le k_j}{c}_{n_j,k}{x}_k$; the second sum equals $j^{-1}{\sum }_{k_{j-1}<k\le k_j}\mathrm{\mid }{c}_{n_j,k}\mathrm{\mid }=j^{-1}(r_{n_j}-{\sum }_{k\le k_{j-1}}\mathrm{\mid }{c}_{n_j,k}\mathrm{\mid })\ge j^{-1}(r_{n_j}-B_{k_{j-1}})$, while the first has absolute value at most ${\sum }_{k\le k_{j-1}}\mathrm{\mid }{c}_{n_j,k}\mathrm{\mid }\le B_{k_{j-1}}$; hence $y_{n_j}\ge j^{-1}(r_{n_j}-B_{k_{j-1}})-B_{k_{j-1}}>j^{-1}j(j+B_{k_{j-1}})-B_{k_{j-1}}=j$.

But $(x_k)$ converges to $0$, so regularity makes $(y_n)$ converge, hence bounded, so some $S\in \mathbb{R}$ has $\mathrm{\mid }{y}_n\mathrm{\mid }\le S$ for every $n$; taking $j$ with $j\cdot 1_{\mathbb{R}}>S$, available by the Archimedean property, step 6.1 gives $y_{n_j}>j>S$, a contradiction.

The assumption of step 1.5 is therefore untenable and condition 3 holds; with steps 2.4 and 2.5, regularity implies all three conditions.

Sufficiency is step 3.1 and necessity is step 8.1, so $c$ is regular exactly when conditions 1, 2 and 3 all hold.