Sequence spaces – "Math for Non-Geeks"

Fehler: Aktuelle Seite wurde in der Sitemap nicht gefunden. Deswegen kann keine Navigation angezeigt werden The sequence space is a vector space consisting of infinitely long tuples $(x_{1},x_{2},x_{3},\ldots )$. The operations on the sequence space are component-wise addition and scalar multiplication.

Motivation

We have already learned about the coordinate spaces $K^{n}$ for a field $K$ as an example for vector spaces. Here, each element consists of $n$ distinct, that is, finitely many, entries from $K$. For example, $(1,4,2,-1)$ is an element of $\mathbb {R} ^{4}$. We can also consider infinite tuples. For example, $(1,4,9,16,\dots )=(n^{2})_{n\in \mathbb {N} }$ is such an "infinite tuple". A better name for infinite tuples is "sequence". If $K$ is the field of real or complex numbers, these are exactly the already known sequences from calculus.

How do we define the vector space operations on the sequences? On $K^{n}$ we have defined the operations component-wise. We already know that we can also add and scale sequences component-wise. Therefore, we can also define addition and scalar multiplication on infinite sequences over arbitrary fields. This leads us to the conjecture that the set of all sequences with entries in $K$ should form a vector space. We call it the sequence space over $K$.

We will first define the sequence space precisely and then prove that it is indeed a vector space. Then, in the section Subspaces of the sequence space, we will consider examples of subspaces of the sequence spaces over the real and complex numbers, which are important for advanced calculus.

Notation

Let $K$ be a field.

We always write $(x_{i})_{i}$ instead of $(x_{i})_{i\in \mathbb {N} }$ in this article for sequences with elements from $K$.

Definition of a sequence space

Definition (The sequence space as a set)

We define the set $\omega :=\left\{\,(x_{i})_{i}\,{\bigg \vert }\,x_{i}\in K{\text{ for all }}i\in \mathbb {N} \,\right\}$ We call it the set of all sequences over $K$, or the sequence space over $K$.

In analogy to the coordinate space, we can also define an addition and a scalar multiplication on $\omega$:

Definition (Vector space operations on $\omega$)

The addition $\boxplus \colon \omega \times \omega \to \omega$ is defined by ${\begin{aligned}(x_{i})_{i}\boxplus (y_{i})_{i}:=(x_{i}+y_{i})_{i}.\end{aligned}}$

Similarly we define scalar multiplication $\boxdot \colon K\times \omega \to \omega$ by $\lambda \boxdot (x_{i})_{i}=(\lambda \cdot x_{i})_{i}.$

The sequence space is a vector space

Theorem ($\omega$ is a vector space)

$(\omega ,\boxplus ,\boxdot )$ is a $K$-vector space.

How to get to the proof? ($\omega$ is a vector space)

We proceed as in the article Proofs for vector spaces. Because the sequence space is defined similarly to the coordinate space, we use the same strategy as for the proof that the coordinate Space is a vector space.

Proof ($\omega$ is a vector space)

We have to check the eight vector space axioms.

Proof step: Associativity of addition

Let $x=(x_{i})_{i},y=(y_{i})_{i},z=(z_{i})_{i}\in \omega$. Then: ${\begin{aligned}(x\boxplus y)\boxplus z&=((x_{i})_{i}\boxplus (y_{i})_{i})\boxplus (z_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=(x_{i}+y_{i})_{i}\boxplus (z_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=((x_{i}+y_{i})+z_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{associativity of the addition in }}K\right.}\\[0.3em]&=(x_{i}+(y_{i}+z_{i}))_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=(x_{i})_{i}\boxplus (y_{i}+z_{i})_{i}=x\boxplus (y\boxplus z).\end{aligned}}$ This shows the associativity of the addition.

Proof step: Commutativity of addition

Let $x=(x_{i})_{i},y=(y_{i})_{i}\in \omega$. Then: ${\begin{aligned}x\boxplus y&=(x_{i})_{i}\boxplus (y_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=(x_{i}+y_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{commutativity of the addition }}K\right.}\\[0.3em]&=(y_{i}+x_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=(y_{i})_{i}\boxplus (x_{i})_{i}=y\boxplus x.\end{aligned}}$ This establishes the commutativity of the addition.

Proof step: Neutral element of addition

We now have to show that there is a neutral element $0_{\omega }\in \omega$, that is, $x\boxplus 0_{\omega }=x$ for all $x\in \omega$. Since we trace all properties back to the properties in $K$, we choose $0_{\omega }=(0)_{i}=(0_{K},0_{K},\ldots )$ as an approach for the neutral element.

Let $x=(x_{i})_{i}\in \omega$. Then: $x\boxplus 0_{\omega }=(x_{i}+0)_{i}=(x_{i})_{i}=x.$ Thus we have shown that $0_{\omega }\in \omega$ is the neutral element of addition.

Proof step: Inverse with respect to addition

Let $x=(x_{i})_{i}\in \omega$. We need to show that there is a $y\in \omega$ such that $x\boxplus y=0_{\omega }$ . As with the neutral element of addition, we use the corresponding counterpart from $K$ as a starting point. That is, we choose for $y$ the sequence $(-x_{i})_{i}$. Then $x\boxplus y=(x_{i})_{i}\boxplus (-x_{i})_{i}=(x_{i}+(-x_{i}))_{i}=(0)_{i}=0_{\omega }.$ Thus we have shown that for any $x\in \omega$ there is a $y\in \omega$ with $x\boxplus y=0_{\omega }$.

Proof step: Scalar distributive law

Let $\lambda ,\mu \in K$ and $x=(x_{i})_{i}\in \omega$. Then: ${\begin{aligned}(\lambda +\mu )\boxdot x&=(\lambda +\mu )\boxdot (x_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=((\lambda +\mu )\cdot x_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{distributive law in }}K\right.}\\[0.3em]&=(\lambda \cdot x_{i}+\mu \cdot x_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=(\lambda \cdot x_{i})_{i}\boxplus (\mu \cdot x_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=(\lambda \boxdot (x_{i})_{i})\boxplus (\mu \boxdot (x_{i})_{i})\\[0.3em]&=(\lambda \boxdot x)\boxplus (\mu \boxdot x).\end{aligned}}$ Thus the scalar distributive law is shown.

Proof step: Vector Distributive Law

Let $\lambda \in K$ and $x=(x_{i})_{i},y=(y_{i})_{i}\in \omega$. Then: ${\begin{aligned}\lambda \boxdot (x\boxplus y)&=\lambda \boxdot ((x_{i})_{i}\boxplus (y_{i})_{i})\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=\lambda \boxdot (x_{i}+y_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=(\lambda \cdot (x_{i}+y_{i}))_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{distributive law in }}K\right.}\\[0.3em]&=(\lambda \cdot x_{i}+\lambda \cdot y_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=(\lambda \cdot x_{i})_{i}\boxplus (\lambda \cdot y_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=(\lambda \boxdot (x_{i})_{i})\boxplus (\lambda \boxdot (y_{i})_{i})\\[0.3em]&=(\lambda \boxdot x)\boxplus (\lambda \boxdot y).\end{aligned}}$ Thus the vectorial distributive law is shown.

Proof step: Associativity with respect to multiplication

Let $\lambda ,\mu \in K$ and $x=(x_{i})_{i}\in \omega$. Then: ${\begin{aligned}(\lambda \cdot \mu )\boxdot x&=(\lambda \cdot \mu )\boxdot (x_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=((\lambda \cdot \mu )\cdot x_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{Aassociative law for multiplication in }}K\right.}\\[0.3em]&=(\lambda \cdot (\mu \cdot x_{i}))_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=\lambda \boxdot (\mu \cdot x_{i})_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=\lambda \boxdot (\mu \boxdot (x_{i})_{i})\\[0.3em]&=\lambda \boxdot (\mu \boxdot x).\end{aligned}}$ This establishes the associative law for multiplication.

Proof step: Existence of a unit

Let $x=(x_{i})_{i}\in \omega$. Then: $1_{K}\boxdot x=1_{K}\boxdot (x_{i})_{i}=(1_{K}\cdot x_{i})_{i}=(x_{i})_{i}=x.$

Thus we have shown the unitary law.

Thus we have established all eight vector space axioms and $(\omega ,\boxplus ,\boxdot )$ is a $K$ vector space.

Subspaces of the sequence space

The sequence space has some frequently used subspaces. Most of these subspaces can be defined only over the fields $\mathbb {R}$ and $\mathbb {C}$. They have many applications in functional analysis, where they are part of an important class of examples. In the field of linear algebra over arbitrary fields, the space of sequences with finite support serves as an example in many places. It is the simplest example of a infinite-dimensional vector space and thus can be used as a good example where statements cease to hold, as the vector space if "too large".

The subspace of sequences with finite support

Definition (Set of sequences with finite support)

We define

$c_{00}:=\left\{\,(x_{i})_{i}\in \omega \,{\bigg \vert }\,{\text{there is an }}i_{0}\in \mathbb {N} ,{\text{ such that }}x_{i}=0{\text{ for all }}i\geq i_{0}\,\right\}$.

Theorem (The sequences with finite support form a subspace)

$c_{00}\subseteq \omega$ is a subspace.

Proof (The sequences with finite support form a subspace)

We check the three subspace criteria.

Proof step: $0_{\omega }\in c_{00}$

The zero element of $\omega$ is the sequence which is constant $0$. Thus, for $i_{0}=1$, for all $i\geq i_{0}$, it holds that $x_{i}=0$, viz. $0_{\omega }\in c_{00}$.

Proof step: $c_{00}$ is closed under addition.

Let $x=(x_{i})_{i},y=(y_{i})_{i}\in c_{00}$. By assumption, there exist $i_{1},i_{2}in\mathbb {N}$ with the property that $x_{i}=0$ for all $i\geq i_{1}$ and $y_{i}=0$ for all $i\geq i_{2}$. We set $i_{0}=\operatorname {max} (i_{1},i_{2})$. Then, for all $i\geq i_{0}$ we have that $(x\boxplus y)_{i}=x_{i}+y_{i}=0+0=0$. This shows $x\boxplus y\in c_{00}$.

Proof step: $c_{00}$ is closed under scalar multiplication.

Let $x=(x_{i})_{i}\in c_{00}$ and $\lambda \in K$. By assumption, there exists $i_{0}\in \mathbb {N}$ with the property that $x_{i}=0$ for all $i\geq i_{0}$. Then for all $i\geq i_{0}$, it holds that $(\lambda \boxdot x)_{i}=\lambda \cdot x_{i}=\lambda \cdot 0=0$. Hence $\lambda \boxdot x\in c_{00}$.

Thus we have shown that $c_{00}$ is a subspace of $\omega$.

For example, the notation $c_{00}$ for the space of sequences with finite support can be derived like this: This vector space is a subspace of the space of zero sequences over the fields $\mathbb {R}$ or $\mathbb {C}$. The latter subspace is usually denoted by $c_{0}$. The $c$ stands for convergence and the $0$ for the fact that we put only zero sequences of the convergent sequences into the vector space. When talking about convergence, the condition that the sequence eventually becomes $0$ is of course significantly stronger than the condition to converge against $0$. Therefore the space of sequences with finite support $c_{00}$ gets an additional zero into the index.

Subspaces from calculus

In the following, we assume that $K\in \{\,\mathbb {R} ,\mathbb {C} \,\}$.

Math for Non-Geeks: Template:Aufgabe

Relationships between the subspaces

We have now learned about some subspaces of the sequence space for $K\in \{\mathbb {R} ,\mathbb {C} \}$. This raises the question of what relations exist between them. Most of the conditions we used to construct the subspaces are conditions from calculus. Fortunately, there are already results in calculus that describe implications between the individual conditions. If we translate these implications into the world of sets and vector spaces, we get the following result:

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