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Toeplitz Transformation:

Let $\{c_{n,k}:1\leq k\leq n, n\geq 1\}$ be an array of real numbers such that:

i) $c_{n,k}\longrightarrow 0$ as $n\rightarrow{\infty}$ for each $k\in \mathbb{N}$.

ii) $\sum\limits_{k=1}^{n} c_{n,k} \longrightarrow 1$

iii) There exists $C>0$ such that for all positive integers n: $$\sum_{k=1}^{n}|c_{n,k}| \leq C.$$

Then for any convergent sequence $\{a_n\}$ the transformed sequence $\{b_n\}$ given by $$b_n= \sum_{k=1}^{n}c_{n,k}a_k,\ \ n\geq 1$$ is also convergent and $(b_n) \rightarrow a \leftarrow (a_n)$


Proof: Let $a_n=a$ then $$\underset{n\rightarrow \infty}{\lim} b_n= a\underset{n\rightarrow \infty}{\lim} \sum\limits_{k=1}^{n} c_{n,k} =a.\ \ \ [by (ii)]$$

Now assuming [$(a_n) \rightarrow 0 \implies (b_n) \rightarrow 0$] we can show that [$(a_n) \rightarrow a \implies (b_n) \rightarrow a$] where $a\neq 0$.

Let $d_n:=(a_n-a) \implies (d_n) \rightarrow 0$. Then from the assumption: $$ e_n:= \sum_{k=1}^{n}c_{n,k}d_k \longrightarrow 0 \\ \implies \sum_{k=1}^{n}c_{n,k}(a_k-a) \longrightarrow 0 \\ \implies \sum_{k=1}^{n}c_{n,k}(a_k) - \sum_{k=1}^{n}c_{n,k}a \longrightarrow 0 \\ \implies b_n - \sum_{k=1}^{n}c_{n,k}a \longrightarrow 0 \\ \implies b_n - a \longrightarrow 0\ \ \text{as n} \rightarrow \infty \ \ \ \textbf{USING (ii)}\\ \implies \underset{n\rightarrow \infty}{\lim} b_n=a.$$

All that remains to show is:

If $\underset{n\rightarrow \infty}{\lim} a_n=0$ then $\underset{n\rightarrow \infty}{\lim} b_n=0$.

$|b_n - 0|=|\sum\limits_{k=1}^{n}c_{n,k}a_k|\leq \sum\limits_{k=1}^{n}|c_{n,k}||a_k|\ \ \ \ \ \ \ (1)$

Now USING (iii) we have $\sum\limits_{k=1}^{n}|c_{n,k}| \leq C$, and for any $\epsilon>0$ there exists $n_1 \in \mathbb{N}$ such that for all $n \geq n_1$, $|a_n| \leq \frac{\epsilon}{2C} \ \ \ \ \ (2)$

From (2) & (1) [USING (iii)] we obtain:

For all $n\geq n_1$ we have $|b_n| \leq \sum\limits_{k=1}^{n_1-1}|c_{n,k}||a_k|+ \sum\limits_{k=n_1}^{n}|c_{n,k}||a_k| \leq \sum\limits_{k=1}^{n_1}|c_{n,k}||a_k|+C.\frac{\epsilon}{2C}(=\frac{\epsilon}{2}) \ \ \ \ \ (3)$

To get control over this $\sum\limits_{k=1}^{n_1-1}|c_{n,k}||a_k|$ one can USE (i).

As for each $k$, $(c_{n,k}) \longrightarrow 0$ as $n \rightarrow \infty$ ($1\leq k < n_1).\ \ \ \ \ \ (4)$

Also since $(a_n)$ is convergent, hence $(a_n)$ is bounded by D(say), i.e $|a_k|<D \ \ \ \text{for all} \ \ k \in \mathbb{N}$ $\ \ \ \ \ (5)$

Let $\epsilon > 0$ then there exists $n_{2,k} \in \mathbb{N}, 1\leq k < n_1$ such that $$|c_{n,k}|< \frac{\epsilon}{2(n_1-1)D} \text{for all}\ n \geq n_2:=max\{n_{2,1},n_{2,2},...,n_{2,n_1-1}\}\\ \implies \sum\limits_{k=1}^{n_1-1} |c_{n,k}| < \frac{\epsilon}{2D}\ \ \ \ \ \ \ \text{for all}\ \ n\geq n_2 \ \ \ \ \ (6)$$

Using (5) & (6) we have:

$$ \sum\limits_{k=1}^{n_1-1} |c_{n,k}||a_k| < \frac{\epsilon}{2D}.D=\frac{\epsilon}{2}\ \ \text{for all}\ \ n \geq n_2\ \ \ \ \ (7)$$

from (3) & (7) we have: for any $\epsilon > 0$ there exist $N:=\text{max}\{n_1,n_2\} \in \mathbb{N}$ such that for all $n \geq N$ we have $$|b_n - 0| \leq \sum\limits_{k=1}^{n_1}|c_{n,k}||a_k|+ \sum\limits_{k=n_1}^{n}|c_{n,k}||a_k| \leq \frac{\epsilon}{2D}.D + C.\frac{\epsilon}{2C}=\epsilon \\ \implies \underset{n\rightarrow \infty}{\lim} b_n=a.$$

Please tell me if my argument is correct.

Observation:

If $a=0$ then condition (ii) $\sum\limits_{k=1}^{n} c_{n,k} \longrightarrow 1$ can be removed. Am I correct?

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    $\begingroup$ Remark: It's the way to define in a rigorous manner infinite dimensional lower triangular matrices. $\endgroup$
    – Jean Marie
    Oct 16, 2020 at 20:03
  • $\begingroup$ Can you please tell me where can I find more about this? $\endgroup$
    – Saikat
    Oct 16, 2020 at 20:05
  • $\begingroup$ Maybe in this MSc thesis. $\endgroup$
    – Jean Marie
    Oct 16, 2020 at 20:40

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