# Proof of Toeplitz Transformation.

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 $$|a_k| $$\ \ \ \ \ (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?

• Remark: It's the way to define in a rigorous manner infinite dimensional lower triangular matrices. Oct 16, 2020 at 20:03