Function spaces – Serlo

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In this article we consider the space of functions, that is, the vector space of all maps of a set into a vector space .

Definition of function spaces[Bearbeiten]

Let be a field, a -vector space and some set.

Then we can define the set of maps of to :

Definition (Set of maps from to )

We denote the set of all maps from to by . This means formally .

Hint

For this definition we have not yet used that is a vector space. It is sufficient if is just any set.

On this set we define an addition and a scalar multiplication:

Definition (Vector space operations on )

The addition is defined by

for all and .

Similarly, we define the scalar multiplication by

for all and .

Hint

and as in the definition above are in fact maps again, since we have specified them on every element (and is closed under and ).

Hint

For the definition we only need that is a vector space. can actually be an arbitrary set (i.e. without an algebraic structure).

The function 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.

Proof ( is a vector space)

We establish the eight vector space axioms. In the following, we consider .

Proof step: Associativity of addition

Let . Then, we have:

This shows the associativity of addition.

Proof step: Commutativity of addition

Let . Then, we have:

This shows the commutativity of addition.

Proof step: Neutral element of addition

We now need to show that there is a neutral element exists.

That is, should hold for all . It is obvious that the zero mapping has this property.

Let . Then, we have:

This shows that has a neutral element with respect to addition.

Proof step: Inverse with respect to addition

Let with . We need to show that there exists a such that holds. Since is a vector space, there exists an inverse with respect to "" with for every . We now show that is the inverse of . We have that:

Furthermore, is uniquely determined by the well-definiteness of and the uniqueness of the inverse in . Thus we have shown that for any there exists a with .

Proof step: scalar distributive law

Let and . Then, we have:

Thus the scalar distributive law is established.

Proof step: Vectorial distributive law

Let and .

Thus the vector distributive law is also established.

Proof step: Associativity of multiplication

Let and . Then, we have:

This establishes the associative law for multiplication.

Proof step: unitarity law

Let . Then, we have:

Thus we have shown the unitary law.

Hint

Some people include the completeness of addition and of scalar multiplication also within the vector space axioms. In our case, they follow from the fact that is itself a -vector space. We considered this in the hint after defining the operations.

It must also hold that is non-empty. This follows directly from the existence of a neutral element with respect to addition.

The set of differentiable functions an an -vector space[Bearbeiten]

In the previous section we showed that the set of all maps of a set into a -vector space is again a -vector space. We now consider the special case , and . We already know that is a -vector space. Hence, we know so far that the set of maps is an -vector space.

We now consider the set of differentiable functions , which is denoted (as "differentiable").


Theorem

The set of differentiable functions forms an -vector space.

Proof

The set of differentiable functions is a subset of the set of maps , i.e. . To show that also forms an -vector space, it suffices to show that is an -subspace of . To do this, we need to establish the 3 subspace axioms.

Proof step:

The function is differentiable. So: .

Proof step: For all we have that .

Let be differentiable, i.e. . We have shown in Analysis I that the function is also differentiable. Consequently we have that .

Proof step: For all and for all we have that .

Let and . We have shown in Analysis I that the map is also differentiable. Thus we have that .

Thus we have shown that is an -subspace of .

Relation to the sequence space[Bearbeiten]

We have already seen that the set of sequences over forms a vector space with respect to coordinate-wise operations. So a sequence with entries in can be seen as a function . In this sense, the sequence space is a special case of the function space by setting and .