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 $\varepsilon > 0$ be an arbitrary real; fix $M \ge 0$ as in condition 3, and fix $D \ge 0$ with $|x_k - L| \le D$ for every $k$, which exists because a convergent sequence is bounded.

Choose $K \in \mathbb{N}$ with $|x_k - L| < \varepsilon\,(3(M+1))^{-1}$ for every $k \ge K$.

For every $n$, choosing an admissible bound $R \ge K$ for row $n$, one has $y_n - L = \sum_{k=0}^{R} c_{n,k}(x_k - L) + \big(\textstyle\sum_k c_{n,k} - 1\big)L$, 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 $n$: $\big|\sum_{k=0}^{R} c_{n,k}(x_k-L)\big| \le \sum_{k<K}|c_{n,k}|\,|x_k-L| + \sum_{k=K}^{R}|c_{n,k}|\,|x_k-L| \le D\sum_{k<K}|c_{n,k}| + \varepsilon\,(3(M+1))^{-1} r_n \le D\sum_{k<K}|c_{n,k}| + \varepsilon/3$, 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 $|c_{n,k}| \to 0$ since $\big||c_{n,k}| - 0\big| = |c_{n,k} - 0|$; a sum of finitely many null sequences is null, by induction on the number of summands, so $\sum_{k<K}|c_{n,k}| \to 0$ in $n$ and there is $N_1$ with $\sum_{k<K}|c_{n,k}| < \varepsilon\,(3(D+1))^{-1}$ for every $n \ge N_1$.

By condition 2 there is $N_2$ with $\big|\sum_k c_{n,k} - 1\big| < \varepsilon\,(3(|L|+1))^{-1}$ for every $n \ge N_2$, so that $\big|(\sum_k c_{n,k} - 1)L\big| < \varepsilon/3$ for such $n$.

Condition 1 holds. Fix $k$ and let $e$ be the sequence with $e_k = 1$ and $e_j = 0$ for $j \ne k$; 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 $N_2$: $|y_n - L| \le D\sum_{k<K}|c_{n,k}| + \varepsilon/3 + \varepsilon/3 < \varepsilon/3 + \varepsilon/3 + \varepsilon/3 = \varepsilon$; as $\varepsilon$ 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 $M_k := \sup\{\,|c_{n,k}| : n \in \mathbb{N}\,\}$ exists in $\mathbb{R}$ for every $k$; putting $B_m := \sum_{k=0}^{m} M_k$ one has $\sum_{k=0}^{m}|c_{n,k}| \le B_m$ 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, $x_k := \operatorname{sgn}(c_{n_j,k})\,j^{-1}$, where $\operatorname{sgn}(t) := 1$ for $t \ge 0$ and $\operatorname{sgn}(t) := -1$ for $t < 0$. Then $|x_k| \le 1$ for every $k$, and $x_k \to 0$: given $\varepsilon > 0$, take $J \ge 1$ with $1/J < \varepsilon$, and every $k > k_J$ lies in a block with index $j > J$, so $|x_k| = 1/j < 1/J < \varepsilon$.

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 $j^{-1}\sum_{k_{j-1}<k\le k_j}|c_{n_j,k}| = j^{-1}\big(r_{n_j} - \sum_{k \le k_{j-1}}|c_{n_j,k}|\big) \ge j^{-1}(r_{n_j} - B_{k_{j-1}})$, while the first has absolute value at most $\sum_{k\le k_{j-1}}|c_{n_j,k}| \le B_{k_{j-1}}$; 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 $|y_n| \le S$ 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.