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Let's consider a linear transformation $T$ of sequences:

Suppose $\{c_{jk}\}$ is a matrix of complex numbers. For each sequence $\alpha=\{a_n\}$ made up of complex numbers, $\beta=T\alpha$ is a sequence, as well, such that $b_n=\sum_{k=1}^\infty c_{nk}a_k$.

I need a necessary and sufficient condition that $\{b_n\}$ converges whenever $\{a_n\}$ converges.

For example, according to Silverman-Toeplitz theorem, if $\lim_{j\to\infty}c_{jk}=0$, $\lim_{j\to\infty}\sum_{k=1}^\infty c_{jk}=1$ and $\sum_{k=1}^\infty\lvert c_{jk}\rvert$ is uniformly bounded according to different $j$'s, $\{b_n\}$ converges to the same number where $\{a_n\}$ converges.

However, I don't need such a condition that both sequences converges to a same number. For example, $$\sum_{n=1}^\infty d_na_n$$ converges whenever $\{a_n\}$ converges, by virtue of Cauchy's convergence test, so set $c_{jk}=d_k$, we've got a transformation, not of Toeplitz-type.

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I would say that the matrix $(c_{jk})$ does what you want if and only if (i) $\sum_{k=1}^\infty \vert c_{jk}\vert$ is uniformly bounded with respect to $j$, (ii) $\lim_{j\to\infty} c_{jk}$ exists for every $k$ and (iii) $\lim_{j\to\infty} \sum_{k=1}^\infty c_{jk}$ exists.

Condition (ii) and (iii) are obviously necessary: to check (ii), consider for fixed $k$ the sequence $a=(0,0,\dots ,1,0\dots)$ where the $1$ is at place $k$); and for (iii) consider the constant sequence $a_n\equiv 1$. That (i) is also necessary follows from the Banach-Steinhaus theorem (as in the proof of Silverman-Toeplitz). Conversely, if (i) and (ii) holds, then it is not hard to prove that the matrix $(c_{jk})$ maps sequences $(a_n)$ tending to $0$ to Cauchy (and hence convergent) sequences; and if moreover (iii) holds this gives the result for all convergent sequences (by substracting the limit).

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  • $\begingroup$ It seems good. For (i): the space $V$ of convergent sequence is a closed and bounded subspace of $\mathcal l^\infty$, therefore complete. Let $\Lambda_n\alpha=\sum_{k=1}^\infty c_{nk}a_k$. Since $\{\Lambda_1\alpha,\Lambda_2\alpha,\dotsb\}$ is bounded for each $\alpha$, by virtue of Banach-Steinhaus theorem, it's uniformly bounded. Am I right? Since I'm a novice to real analysis, I'm not sure whether my inference is right. $\endgroup$ – Yai0Phah Jun 1 '13 at 15:01
  • $\begingroup$ In addition, such a matrix could be decomposed as the summation of a Silverman-Toeplitz matrix and copies of a sequence whose sum absolutely converges. $\endgroup$ – Yai0Phah Jun 1 '13 at 15:19

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