Function spaces – Serlo

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

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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

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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

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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

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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  .