Sequence spaces – Serlo

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The sequence space is a vector space consisting of infinitely long tuples . The operations on the sequence space are component-wise addition and scalar multiplication.

Motivation[Bearbeiten]

We have already learned about the coordinate spaces for a field as an example for vector spaces. Here, each element consists of distinct, that is, finitely many, entries from . For example, is an element of . We can also consider infinite tuples. For example, is such an "infinite tuple". A better name for infinite tuples is "sequence". If 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 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 should form a vector space. We call it the sequence space over .

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[Bearbeiten]

Let be a field.

We always write instead of in this article for sequences with elements from .

Definition of a sequence space[Bearbeiten]

Definition (The sequence space as a set)

We define the set

We call it the set of all sequences over , or the sequence space over .

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

Definition (Vector space operations on )

The addition is defined by

Similarly we define scalar multiplication by

The sequence space is a vector space[Bearbeiten]

Theorem ( is a vector space)

is a -vector space.

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

We have to check the eight vector space axioms.

Proof step: Associativity of addition

Let . Then:

This shows the associativity of the addition.

Proof step: Commutativity of addition

Let . Then:

This establishes the commutativity of the addition.

Proof step: Neutral element of addition

We now have to show that there is a neutral element , that is, for all . Since we trace all properties back to the properties in , we choose

as an approach for the neutral element.

Let . Then:

Thus we have shown that is the neutral element of addition.

Proof step: Inverse with respect to addition

Let . We need to show that there is a such that . As with the neutral element of addition, we use the corresponding counterpart from as a starting point. That is, we choose for the sequence . Then

Thus we have shown that for any there is a with .

Proof step: Scalar distributive law

Let and . Then:

Thus the scalar distributive law is shown.

Proof step: Vector Distributive Law

Let and . Then:

Thus the vectorial distributive law is shown.

Proof step: Associativity with respect to multiplication

Let and . Then:

This establishes the associative law for multiplication.

Proof step: Existence of a unit

Let . Then:

Thus we have shown the unitary law.

Thus we have established all eight vector space axioms and is a vector space.

Subspaces of the sequence space[Bearbeiten]

The sequence space has some frequently used subspaces. Most of these subspaces can be defined only over the fields and . 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 [Bearbeiten]

Definition (Set of sequences with finite support)

We define

.

Theorem (The sequences with finite support form a subspace)

is a subspace.

Proof (The sequences with finite support form a subspace)

We check the three subspace criteria.

Proof step:

The zero element of is the sequence which is constant . Thus, for , for all , it holds that , viz. .

Proof step: is closed under addition.

Let . By assumption, there exist with the property that for all and for all . We set . Then, for all we have that . This shows .

Proof step: is closed under scalar multiplication.

Let and . By assumption, there exists with the property that for all . Then for all , it holds that . Hence .

Thus we have shown that is a subspace of .

For example, the notation 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 or . The latter subspace is usually denoted by . The stands for convergence and the 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 is of course significantly stronger than the condition to converge against . Therefore the space of sequences with finite support gets an additional zero into the index.

Subspaces from calculus[Bearbeiten]

In the following, we assume that .

Exercise (Vector space of bounded sequences)

The subset

is a subspace of the sequence space. The space is called vector space of bounded sequences.

Solution (Vector space of bounded sequences)

We check the three subspace criteria.

Proof step:

The zero element of is the sequence which is constant . This is of course bounded (e.g. by ) and thus lies in .

Proof step: is closed under addition

Let . Then . From the triangle inequality for all we obtain

Thus .

Proof step: is closed under scalar multiplication.

Let and . Then . Since for all we obtain

Hence .

Thus we have shown that is a subspace of .

Exercise (Vector space of convergent sequences)

The subset

is a subspace of . It is called vector space of convergent sequences.

Solution (Vector space of convergent sequences)

We check the three subspace criteria.

Proof step:

The zero element of is the sequence which is constant . This obviously converges to and thus lies in .

Proof step: is closed under addition

Let . Then, . We have . The two limits on the right hand side exist by assumption, so the limit on the left hand side exists. In particular, converges, so it is true that .

Proof step: is closed under scalar multiplication.

Let and . Then, . We have . The two limits on the right hand side exist by assumption, so the limit on the left hand side also exists. Hence .

Thus we have shown that is a subspace of .

Exercise (Vector space of zero sequences)

The subset

is a subspace of . This subspace is called vector space of zero sequences.

Solution (Vector space of zero sequences)

We check the three subspace criteria.

Proof step:

The zero element of is the sequence which is constant . This naturally converges to and thus lies in .

Proof step: is closed under addition

Let . Then . We have . Consequently, converges to . Thus .

Proof step: is closed under scalar multiplication

Let and . Then . We have . Hence .

Thus we have shown that the space of zero sequences is a subspace of .

Exercise (Vector space of absolutely summable sequences)

The subset

is a subspace of . It is called the vector space of absolutely summable sequences.

Solution (Vector space of absolutely summable sequences)

We check the three subspace criteria.

Proof step:

The zero element of is the sequence which is constant . For this sequence holds. So .

Proof step: is closed under addition

Let . Then . Because of the triangle inequality for all we obtain . With the sum rule for series we get . Hence .

Proof step: is closed under scalar multiplication

Let and . Then . Since for all it follows that . Hence .

Thus we have shown that is a subspace of .

Relationships between the subspaces[Bearbeiten]

We have now learned about some subspaces of the sequence space for . 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:

Exercise (Inclusions between the subspaces)

It is true that .

Solution (Inclusions between the subspaces)

We will now show the four inclusions one after the other. Thereby we also prove that in each case no equality is valid.

Proof step:

Proof step:

Let . By definition there exists some , such that for all . Thus and we obtain . This shows .

Proof step:

Consider the sequence . We have , since . But , since for all . This shows .

Proof step:

Proof step:

Let . This means, . Then, by means of the term test, we have that . That is, and thus . This shows .

Proof step:

Consider the sequence . We have , since . But , since diverges by means of the theorem on the divergence of the harmonic series. This shows .

Proof step:

Proof step:

This is clear since every sequence converging to zero converges.

Proof step:

Consider the sequence . We have , since . But , since . This shows .

Proof step:

Proof step:

From calculus we know, that every convergent sequence is bounded. So the inclusion holds true.

Proof step:

Consider the sequence . We have , since for all . But , since does not exist. This shows .

Proof step:

Proof step:

We already know this because is a subspace of .

Proof step:

Consider the sequence . We have , since . This shows .