4.10. Sequence Spaces#

We shall assume the field of scalars \(\FF\) to be either \(\RR\) or \(\CC\).

4.10.1. The Space of all Sequences#

Recall that a sequence is a map \(\bx : \Nat \to \FF\) and is written as \(\{ x_n \}\). The set of all sequences of \(\FF\) is denoted by \(\FF^{\Nat}\) or just \(\FF^{\infty}\) in Cartesian product notation.

Definition 4.109 (Zero sequence)

The zero sequence is defined as:

\[ \bzero = (0, 0, 0, \dots). \]

Definition 4.110 (Vector addition of sequences)

Let \(\bx = \{ x_n \}\) and \(\by = \{ y_n \}\) be any two sequences in \(\FF^{\infty}\).

Their vector addition is defined as:

\[ \bx + \by \triangleq \{ x_n + y_n \}. \]

Definition 4.111 (Scalar multiplication of sequence)

Let \(\bx = \{ x_n \}\) be any sequence in \(\FF^{\infty}\) and let \(\alpha \in \FF\).

The scalar multiplication of \(\alpha\) with \(\bx\) is defined as:

\[ \alpha \bx \triangleq \{ \alpha x_n\}. \]

Theorem 4.127

The set of sequences \(\FF^{\infty}\) is closed under vector addition and scalar multiplication defined above.

This is obvious from definition.

Definition 4.112 (Vector space of all sequences)

The set \(\FF^{\infty}\) equipped with the vector addition and scalar multiplication defined above is a vector space. It is known as the space of all sequences.

Definition 4.113 (Sequence space)

Any linear subspace of the space of all sequences \(\FF^{\infty}\) is known as a sequence space.

4.10.2. The Space of Absolutely Summable Sequences#

Definition 4.114 (Absolutely summable sequence)

A sequence \(\{x_n\}\) of \(\FF\) is called absolute summable if

\[ \sum_{n=1}^{\infty} |x_n| < \infty. \]

Theorem 4.128 (Closure under addition)

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

\[ \sum_{n=1}^{\infty} |x_n + y_n| \leq \sum_{n=1}^{\infty} |x_n| + \sum_{n=1}^{\infty} |y_n|. \]

Proof. Consider the partial sum:

\[ S_n = \sum_{k=1}^{n} |x_k + y_k| \leq \sum_{k=1}^{n} (|x_k| + |y_k|) = \sum_{k=1}^{n} |x_k| + \sum_{k=1}^{n} |y_k|. \]

Taking the limit

\[ \lim_{n \to \infty} S_n \leq \lim_{n \to \infty}\sum_{k=1}^{n} |x_k| + \lim_{n \to \infty} \sum_{k=1}^{n} |y_k| = \sum_{n=1}^{\infty} |x_n| + \sum_{n=1}^{\infty} |y_n|. \]

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

Theorem 4.129 (Closure under scalar multiplication)

If the sequence \(\{x_n \}\) is absolutely summable, then for any \(\alpha \in \FF\), the sequence \(\{ \alpha x_n \}\) is absolutely summable with:

\[ \sum_{n=1}^{\infty} | \alpha x_n| = | \alpha| \sum_{n=1}^{\infty} | x_n|. \]

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:

\[ \lim_{m \to \infty} S_m = | \alpha | \lim_{m \to \infty} \sum_{n=1}^{m} | x_n| = | \alpha | \sum_{n=1}^{\infty} | x_n|. \]

Hence \(\{ \alpha x_n \}\) is absolutely summable.

Definition 4.115 (\(\ell^1\) The space of absolutely summable sequences)

Let \(\ell^1\) denote the set of all absolutely summable sequences of \(\FF\). 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 4.116 (Norm for the \(\ell^1\) space)

The standard norm for the \(\ell^1\) space is defined for any \(\bx \in \ell^1\) as:

\[ \| \bx \|_1 = \sum_{n=1}^{\infty} |x_n|. \]

The \(\ell^1\) space equipped with the norm \(\| \cdot \|_1\) is a normed linear space.

Theorem 4.130

The norm defined for \(\ell^1\) space in Definition 4.116 is indeed a norm.

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

[Positive homogeneity] Let \(\bx = \{ x_n \}\) be absolutely summable. From Theorem 4.129, we have:

\[ \| \alpha \bx \|_1 = \sum_{n-1}^{\infty} | \alpha x_n | = |\alpha | \sum_{n-1}^{\infty} |x_n| = | \alpha | \| \bx \|_1. \]

[Triangle inequality] Let \(\bx = \{ x_n \}\) and \(\by = \{ y_n \}\) be absolutely summable. From Theorem 4.128, we have:

\[ \| \bx + \by \|_1 = \sum_{n=1}^{\infty} | x_n + y_n | \leq \sum_{n=1}^{\infty} |x_n| + \sum_{n=1}^{\infty} |y_n| = \| \bx \|_1 + \| \by \|_1. \]

Theorem 4.131

\(\ell^1\) 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.