An infinite matrix A=(a_ij) maps (x_n) to (x'_n) via
An infinite matrix $A=(a_{ij})$ maps $(x_n)$ to $(x'_n)$ via $x'_n=\sum_{j=1}^{\infty}a_{nj}x_j$. Given that $\lim_{n\to\infty}a_{nk}=0$ for each fixed $k$, show that $A$ transforms every bounded sequence into a null sequence if and only if $\displaystyle\lim_{n\to\infty}\sum_{k=1}^{\infty}|a_{nk}|=0$.
Explanation of Method
Sufficiency is an immediate estimate: bounding $|x_k|\le M$ term by term gives $|x'_n|\le M\sum_k|a_{nk}|\to0$. Necessity is the delicate direction: if the row sums $\sum_k|a_{nk}|$ do not tend to $0$, a "gliding hump" construction — using the given hypothesis that each column tends to $0$ to isolate, for a subsequence of rows, disjoint blocks of column-indices carrying a fixed fraction of each row's mass — builds a single bounded sign sequence $(x_k)$ along which $x'_n$ fails to tend to $0$.
- Prove sufficiency: row sums $\to0$ forces every bounded sequence to a null sequence.
- Set up the contrapositive for necessity: assume row sums do not tend to $0$.
- Construct, via a gliding-hump argument using disjoint index blocks, a bounded sequence along which $x'_n\not\to0$.
Step 1 · Sufficiency
Suppose $\sum_k|a_{nk}|\to0$. For any bounded $(x_k)$ with $|x_k|\le M$: $$ |x'_n| = \left|\sum_{k=1}^{\infty}a_{nk}x_k\right| \le M\sum_{k=1}^{\infty}|a_{nk}| \to 0. $$ So $x'_n\to0$ for every bounded $(x_k)$.
Step 2 · Setup for necessity
Suppose $\sum_k|a_{nk}|$ does not tend to $0$; then there is $\epsilon>0$ and a subsequence $(n_j)$ with $\sum_k|a_{n_jk}|\ge\epsilon$ for every $j$. We build a single bounded $(x_k)$ (with $|x_k|\le1$) along which $|x'_{n_j}|$ stays bounded away from $0$, so $x'_n\not\to0$.
Step 3 · Gliding-hump construction
Choose $K_0=0$. Inductively, having fixed $K_{j-1}$, since each column satisfies $a_{n,k}\to0$ as $n\to\infty$, the finite sum $\sum_{k\le K_{j-1}}|a_{n,k}|$ can be made $<\epsilon/4$ by choosing $n_j$ far enough along the subsequence; this forces at least $3\epsilon/4$ of row $n_j$'s mass into $k>K_{j-1}$. Since a row's sum is (effectively) finite, choose $K_j>K_{j-1}$ large enough that $\sum_{K_{j-1}<k\le K_j}|a_{n_j,k}|\ge\epsilon/2$. Define $x_k=\mathrm{sgn}(a_{n_j,k})$ for $k$ in the block $(K_{j-1},K_j]$ (using the row active in that block). Then for each $j$: $$ x'_{n_j} = \sum_{k}a_{n_j,k}x_k \ge \sum_{K_{j-1}<k\le K_j}|a_{n_j,k}| - \left(\text{contribution from other blocks}\right) \ge \frac\epsilon2 - o(1), $$ using that the mass of row $n_j$ outside its own block is controlled by construction. So $x'_{n_j}\not\to0$, and $(x_k)$ is bounded — contradicting the assumption that $A$ sends every bounded sequence to a null sequence.
Summary Table
| Direction | Argument |
|---|---|
| Row sums $\to0$ $\Rightarrow$ null-sequence-preserving | Direct estimate $|x'_n|\le M\sum_k|a_{nk}|$ |
| Null-sequence-preserving $\Rightarrow$ row sums $\to0$ | Gliding-hump counterexample construction |
Additional Notes
- This is a close cousin of the Silverman–Toeplitz theorem characterizing regular summability matrices, whose conditions similarly require row sums of $|a_{nk}|$ to stay bounded (here, tend to $0$, a stronger requirement fitting the stronger conclusion "every bounded sequence maps to a null sequence" rather than merely "preserves limits").
- The gliding-hump technique is a standard tool in classical functional analysis (closely related to the uniform boundedness principle / Banach–Steinhaus theorem) whenever a uniform bound must be shown necessary by exhibiting a single "bad" test sequence built from an infinite family of local failures.
- The hypothesis that each column individually tends to $0$ is essential to the construction in Step 3: it is exactly what allows the mass of row $n_j$ to be pushed past any fixed finite index $K_{j-1}$, making the disjoint-block construction possible.
Result
The equivalence is proven: