Sequence Spaces
Contents
4.13. Sequence Spaces#
We shall assume the field of scalars \(\FF\) to be either \(\RR\) or \(\CC\).
4.13.1. 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.
(Zero sequence)
The zero sequence is defined as:
(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:
(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:
The set of sequences \(\FF^{\infty}\) is closed under vector addition and scalar multiplication defined above.
This is obvious from definition.
(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.
(Sequence space)
Any linear subspace of the space of all sequences \(\FF^{\infty}\) is known as a sequence space.
4.13.2. The Space of Absolutely Summable Sequences#
(Absolutely summable sequence)
A sequence \(\{x_n\}\) of \(\FF\) is called absolute summable if
(Closure under addition)
If sequences \(\{x_n \}\) and \(\{ y_n\}\) are absolutely summable, then their sum \(\{ x_n + y_n \}\) is absolutely summable with
Proof. Consider the partial sum:
Taking the limit
Thus, the sequence \(\{x_n + y_n\}\) is absolutely summable.
(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:
Proof. Consider the partial sum:
Taking the limit:
Hence \(\{ \alpha x_n \}\) is absolutely summable.
\(\ell^1\) 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 \(\ell^1\).
\(\ell^1\) space)
(Norm for theThe standard norm for the \(\ell^1\) space is defined for any \(\bx \in \ell^1\) as:
The \(\ell^1\) space equipped with the norm \(\| \cdot \|_1\) is a normed linear space.
The norm defined for \(\ell^1\) space in Definition 4.157 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 4.161, we have:
[Triangle inequality] Let \(\bx = \{ x_n \}\) and \(\by = \{ y_n \}\) be absolutely summable. From Theorem 4.160, we have:
\(\ell^1\) 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.