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4.14. Sequence Spaces

We shall assume the field of scalars $\mathbb{F}$ to be either $\mathbb{R}$ or $\mathbb{C}$.

4.14.1. The Space of all Sequences

Recall that a sequence is a map $\mathbf{x}:\mathbb{N}\to \mathbb{F}$ and is written as $\{x_n\}$. The set of all sequences of $\mathbb{F}$ is denoted by ${\mathbb{F}}^{\mathbb{N}}$ or just ${\mathbb{F}}^{\mathrm{\infty }}$ in Cartesian product notation.

Definition 4.167 (Zero sequence)

The zero sequence is defined as:

$$\mathbf{0}=(0,0,0,\dots ).$$

Definition 4.168 (Vector addition of sequences)

Let $\mathbf{x}=\{x_n\}$ and $\mathbf{y}=\{y_n\}$ be any two sequences in ${\mathbb{F}}^{\mathrm{\infty }}$.

Their vector addition is defined as:

$$\mathbf{x}+\mathbf{y}\triangleq \{x_n+y_n\}.$$

Definition 4.169 (Scalar multiplication of sequence)

Let $\mathbf{x}=\{x_n\}$ be any sequence in ${\mathbb{F}}^{\mathrm{\infty }}$ and let $\alpha \in \mathbb{F}$.

The scalar multiplication of $\alpha$ with $\mathbf{x}$ is defined as:

$$\alpha \mathbf{x}\triangleq \{\alpha x_n\}.$$

Theorem 4.161

The set of sequences ${\mathbb{F}}^{\mathrm{\infty }}$ is closed under vector addition and scalar multiplication defined above.

This is obvious from definition.

Definition 4.170 (Vector space of all sequences)

The set ${\mathbb{F}}^{\mathrm{\infty }}$ equipped with the vector addition and scalar multiplication defined above is a vector space. It is known as the space of all sequences.

Definition 4.171 (Sequence space)

Any linear subspace of the space of all sequences ${\mathbb{F}}^{\mathrm{\infty }}$ is known as a sequence space.

4.14.2. The Space of Absolutely Summable Sequences

Definition 4.172 (Absolutely summable sequence)

A sequence $\{x_n\}$ of $\mathbb{F}$ is called absolute summable if

$$\sum _{n=1}^{\mathrm{\infty }}|x_n|<\mathrm{\infty }.$$

Theorem 4.162 (Closure under addition)

If sequences $\{x_n\}$ and $\{y_n\}$ are absolutely summable, then their sum $\{x_n+y_n\}$ is absolutely summable with

$$\sum _{n=1}^{\mathrm{\infty }}|x_n+y_n|\le \sum _{n=1}^{\mathrm{\infty }}|x_n|+\sum _{n=1}^{\mathrm{\infty }}|y_n|.$$

Proof. Consider the partial sum:

$$S_n=\sum _{k=1}^n|x_k+y_k|\le \sum _{k=1}^n(|x_k|+|y_k|)=\sum _{k=1}^n|x_k|+\sum _{k=1}^n|y_k|.$$

Taking the limit

$$\underset{n\to \mathrm{\infty }}{lim}{S}_n\le \underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^n|x_k|+\underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^n|y_k|=\sum _{n=1}^{\mathrm{\infty }}|x_n|+\sum _{n=1}^{\mathrm{\infty }}|y_n|.$$

Thus, the sequence $\{x_n+y_n\}$ is absolutely summable.

Theorem 4.163 (Closure under scalar multiplication)

If the sequence $\{x_n\}$ is absolutely summable, then for any $\alpha \in \mathbb{F}$, the sequence $\{\alpha x_n\}$ is absolutely summable with:

$$\sum _{n=1}^{\mathrm{\infty }}|\alpha x_n|=|\alpha |\sum _{n=1}^{\mathrm{\infty }}|x_n|.$$

Proof. Consider the partial sum:

$$S_m=\sum _{n=1}^m|\alpha x_n|=\sum _{n=1}^m|\alpha ||x_n|=|\alpha |\sum _{n=1}^m|x_n|.$$

Taking the limit:

$$\underset{m\to \mathrm{\infty }}{lim}{S}_m=|\alpha |\underset{m\to \mathrm{\infty }}{lim}\sum _{n=1}^m|x_n|=|\alpha |\sum _{n=1}^{\mathrm{\infty }}|x_n|.$$

Hence $\{\alpha x_n\}$ is absolutely summable.

Definition 4.173 (${\ell }^1$ The space of absolutely summable sequences)

Let ${\ell }^1$ denote the set of all absolutely summable sequences of $\mathbb{F}$. Then ${\ell }^1$ equipped with the vector addition and scalar multiplication defined above is a vector space.

The definition is justified since:

• ${\ell }^1$ is closed under vector addition.

• ${\ell }^1$ is closed under scalar multiplication.

• The zero-sequence $(0,0,0,\dots )$ is absolutely summable and belongs to ${\ell }^1$.

Definition 4.174 (Norm for the ${\ell }^1$ space)

The standard norm for the ${\ell }^1$ space is defined for any $\mathbf{x}\in {\ell }^1$ as:

$$\Vert \mathbf{x}{\Vert }_1=\sum _{n=1}^{\mathrm{\infty }}|x_n|.$$

The ${\ell }^1$ space equipped with the norm $\Vert \cdot {\Vert }_1$ is a normed linear space.

Theorem 4.164

The norm defined for ${\ell }^1$ space in Definition 4.174 is indeed a norm.

Proof. [Positive definiteness] It is clear that the norm of the zero sequence $\Vert \mathbf{0}{\Vert }_1=0$. Now suppose that $\sum _{n=1}^{\mathrm{\infty }}|x_n|=0$. The sum of a non-negative sequence is zero only if each term is 0. Thus, $\{x_n\}=\mathbf{0}$.

[Positive homogeneity] Let $\mathbf{x}=\{x_n\}$ be absolutely summable. From Theorem 4.163, we have:

$$\Vert \alpha \mathbf{x}{\Vert }_1=\sum _{n-1}^{\mathrm{\infty }}|\alpha x_n|=|\alpha |\sum _{n-1}^{\mathrm{\infty }}|x_n|=|\alpha |\Vert \mathbf{x}{\Vert }_1.$$

[Triangle inequality] Let $\mathbf{x}=\{x_n\}$ and $\mathbf{y}=\{y_n\}$ be absolutely summable. From Theorem 4.162, we have:

$$\Vert \mathbf{x}+\mathbf{y}{\Vert }_1=\sum _{n=1}^{\mathrm{\infty }}|x_n+y_n|\le \sum _{n=1}^{\mathrm{\infty }}|x_n|+\sum _{n=1}^{\mathrm{\infty }}|y_n|=\Vert \mathbf{x}{\Vert }_1+\Vert \mathbf{y}{\Vert }_1.$$

Theorem 4.165

${\ell }^1$ is complete. In other words, every Cauchy sequence of sequences in ${\ell }^1$ converges to a sequence of ${\ell }^1$. Thus, it is a Banach space.

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