# Function spaces – Serlo

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In this article we consider the space of functions, that is, the vector space of all maps ${\displaystyle f\colon X\to V}$ of a set ${\displaystyle X}$ into a vector space ${\displaystyle V}$.

## Definition of function spaces

Let ${\displaystyle K}$ be a field, ${\displaystyle (V,+_{V},\cdot _{V})}$ a ${\displaystyle K}$-vector space and ${\displaystyle X}$ some set.

Then we can define the set of maps of ${\displaystyle X}$ to ${\displaystyle V}$:

Definition (Set of maps from ${\displaystyle X}$ to ${\displaystyle V}$)

We denote the set of all maps from ${\displaystyle X}$ to ${\displaystyle V}$ by ${\displaystyle \operatorname {Fun} (X,V)}$. This means formally ${\displaystyle \operatorname {Fun} (X,V):=\{f\,\vert \,f\colon X\to V\,\}}$.

Hint

For this definition we have not yet used that ${\displaystyle V}$ is a vector space. It is sufficient if ${\displaystyle V}$ is just any set.

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

Definition (Vector space operations on ${\displaystyle \operatorname {Fun} (X,V)}$)

The addition ${\displaystyle \boxplus \colon \operatorname {Fun} (X,V)\times \operatorname {Fun} (X,V)\to \operatorname {Fun} (X,V)}$ is defined by

${\displaystyle (f\boxplus g)(x):=f(x)+_{V}g(x)}$

for all ${\displaystyle f,g\in \operatorname {Fun} (X,V)}$ and ${\displaystyle x\in X}$.

Similarly, we define the scalar multiplication ${\displaystyle \boxdot \colon K\times \operatorname {Fun} (X,V)\to \operatorname {Fun} (X,V)}$ by

${\displaystyle (\lambda \boxdot f)(x):=\lambda \cdot _{V}f(x)}$

for all ${\displaystyle f\in \operatorname {Fun} (X,V),\lambda \in K}$ and ${\displaystyle x\in X}$.

Hint

${\displaystyle f\boxplus g}$ and ${\displaystyle \lambda \boxdot f}$ as in the definition above are in fact maps ${\displaystyle X\to V}$ again, since we have specified them on every element ${\displaystyle x\in X}$ (and ${\displaystyle V}$ is closed under ${\displaystyle +_{V}}$ and ${\displaystyle \cdot _{V}}$).

Hint

For the definition we only need that ${\displaystyle V}$ is a vector space. ${\displaystyle X}$ can actually be an arbitrary set (i.e. without an algebraic structure).

## The function space is a vector space

Theorem (${\displaystyle V}$ is a vector space)

${\displaystyle (\operatorname {Fun} (X,V),\boxplus ,\boxdot )}$ is a ${\displaystyle K}$-vector space.

How to get to the proof? (${\displaystyle V}$ is a vector space)

We proceed as in the article proofs for vector spaces.

Proof (${\displaystyle V}$ is a vector space)

We establish the eight vector space axioms. In the following, we consider ${\displaystyle x\in X}$.

Let ${\displaystyle f,g,h\in \operatorname {Fun} (X,V)}$. Then, we have:

{\displaystyle {\begin{aligned}&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]((f\boxplus g)\boxplus h)(x)&=(f\boxplus g)(x)+_{V}h(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=(f(x)+_{V}g(x))+_{V}h(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{associativity of addition in }}V\right.}\\[0.3em]&=f(x)+_{V}(g(x)+_{V}h(x))\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=f(x)+_{V}(g\boxplus h)(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=(f\boxplus (g\boxplus h))(x)\\[0.3em]\end{aligned}}}

This shows the associativity of addition.

Let ${\displaystyle f,g\in \operatorname {Fun} (X,V)}$. Then, we have:

{\displaystyle {\begin{aligned}&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em](f\boxplus g)(x)&=f(x)+_{V}g(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{commutativity of addition in }}V\right.}\\[0.3em]&=g(x)+_{V}f(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=(g\boxplus f)(x)\\[0.3em]\end{aligned}}}

This shows the commutativity of addition.

Proof step: Neutral element of addition

We now need to show that there is a neutral element ${\displaystyle 0_{Fun}\in \operatorname {Fun} (X,V)}$ exists.

That is, ${\displaystyle (f\boxplus 0_{Fun})(x)=f(x)}$ should hold for all ${\displaystyle f\in \operatorname {Fun} (X,V)}$. It is obvious that the zero mapping ${\displaystyle 0_{Fun}\colon X\to V\ ;x\mapsto 0_{V}}$ has this property.

Let ${\displaystyle f\in \operatorname {Fun} (X,V)}$. Then, we have:

{\displaystyle {\begin{aligned}&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em](f\boxplus 0_{Fun})(x)&=f(x)+_{V}0_{Fun}(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}0_{Fun}\right.}\\[0.3em]&=f(x)+_{V}0_{V}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{neutral element of addition in }}V\right.}\\[0.3em]&=f(x)\end{aligned}}}

This shows that ${\displaystyle \operatorname {Fun} (X,V)}$ has a neutral element with respect to addition.

Proof step: Inverse with respect to addition

Let ${\displaystyle f\in \operatorname {Fun} (X,V)}$ with ${\displaystyle f\colon X\to V;x\mapsto f(x)}$. We need to show that there exists a ${\displaystyle g\in Abb(X,V)}$ such that ${\displaystyle (f\boxplus g)(x)=0_{Fun}(x)}$ holds. Since ${\displaystyle (V,+_{V},\cdot _{V})}$ is a vector space, there exists an inverse ${\displaystyle -x}$ with respect to "${\displaystyle +_{V}}$" with ${\displaystyle x+_{V}(-x)=0_{V}}$ for every ${\displaystyle x\in V}$. We now show that ${\displaystyle g\colon X\to V,x\mapsto -f(x)}$ is the inverse of ${\displaystyle f}$. We have that:

{\displaystyle {\begin{aligned}&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em](f\boxplus g)(x)&=f(x)+_{V}g(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}g\right.}\\[0.3em]&=f(x)+_{V}(-f(x))\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{inverse map in V w.r.to }}+_{V}\right.}\\[0.3em]&=0_{V}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}0_{Fun}\right.}\\[0.3em]&=0_{Fun}(x)\end{aligned}}}

Furthermore, ${\displaystyle g\in \operatorname {Fun} (X,V)}$ is uniquely determined by the well-definiteness of ${\displaystyle f}$ and the uniqueness of the inverse in ${\displaystyle V}$. Thus we have shown that for any ${\displaystyle f\in \operatorname {Fun} (X,V)}$ there exists a ${\displaystyle g\in \operatorname {Fun} (X,V)}$ with ${\displaystyle (f\boxplus g)(x)=0_{Fun}}$.

Proof step: scalar distributive law

Let ${\displaystyle \lambda ,\mu \in K}$ and ${\displaystyle f\in \operatorname {Fun} (X,V)}$. Then, we have:

{\displaystyle {\begin{aligned}&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em](\lambda +_{K}\mu )\boxdot f(x)&=(\lambda +_{K}\mu )\cdot _{V}f(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{distributive law in }}V\right.}\\[0.3em]&=(\lambda \cdot _{V}f(x))+_{V}(\mu \cdot _{V}f(x))\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=((\lambda \cdot _{V}f)\boxplus (\mu \cdot _{V}f))(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=((\lambda \boxdot f)\boxplus (\mu \boxdot f))(x)\\[0.3em]\end{aligned}}}

Thus the scalar distributive law is established.

Proof step: Vectorial distributive law

Let ${\displaystyle \lambda \in K}$ and ${\displaystyle f,g\in Fun(X,V)}$.

{\displaystyle {\begin{aligned}&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]\lambda \boxdot (f\boxplus g)(x)&=\lambda \boxdot (f(x)+_{V}g(x))\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=\lambda \cdot _{V}(f(x)+_{V}g(x))\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{distributive law in }}V\right.}\\[0.3em]&=(\lambda \cdot _{V}f(x))+_{V}(\lambda \cdot _{V}g(x))\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=(\lambda \boxdot f)(x)+_{V}(\lambda \boxdot g)(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus \right.}\\[0.3em]&=((\lambda \boxdot f)\boxplus (\lambda \boxdot g))(x)\\[0.3em]\end{aligned}}}

Thus the vector distributive law is also established.

Proof step: Associativity of multiplication

Let ${\displaystyle \lambda ,\mu \in K}$ and ${\displaystyle f\in \operatorname {Fun} (X,V)}$. Then, we have:

{\displaystyle {\begin{aligned}&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]((\lambda \cdot _{K}\mu )\boxdot f)(x)&=(\lambda \cdot _{K}\mu )\cdot _{V}f(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{associativity of multiplication in }}V\right.}\\[0.3em]&=\lambda \cdot _{V}(\mu \cdot _{V}f(x))\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=\lambda \cdot _{V}(\mu \boxdot f)(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em]&=(\lambda \boxdot (\mu \boxdot f))(x)\\[0.3em]\end{aligned}}}

This establishes the associative law for multiplication.

Proof step: unitarity law

Let ${\displaystyle f\in \operatorname {Fun} (X,V)}$. Then, we have:

{\displaystyle {\begin{aligned}&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot \right.}\\[0.3em](1_{K}\boxdot f)(x)&=1_{K}\cdot _{V}f(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{neutral element of scalar multiplication in }}V\right.}\\[0.3em]&=f(x).\end{aligned}}}

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 ${\displaystyle V}$ is itself a ${\displaystyle K}$-vector space. We considered this in the hint after defining the operations.

It must also hold that ${\displaystyle \operatorname {Fun} (X,V)}$ is non-empty. This follows directly from the existence of a neutral element with respect to addition.

## The set of differentiable functions ${\displaystyle f:(0,1)\to \mathbb {R} }$ an an ${\displaystyle \mathbb {R} }$-vector space

In the previous section we showed that the set of all maps of a set ${\displaystyle X}$ into a ${\displaystyle K}$-vector space ${\displaystyle V}$ is again a ${\displaystyle K}$-vector space. We now consider the special case ${\displaystyle X=(0,1)\subseteq \mathbb {R} }$, ${\displaystyle K=\mathbb {R} }$ and ${\displaystyle V=\mathbb {R} }$. We already know that ${\displaystyle V}$ is a ${\displaystyle K}$-vector space. Hence, we know so far that the set of maps ${\displaystyle f:(0,1)\to \mathbb {R} }$ is an ${\displaystyle \mathbb {R} }$-vector space.

We now consider the set of differentiable functions ${\displaystyle f:(0,1)\to \mathbb {R} }$, which is denoted ${\displaystyle {\mathcal {D}}((0,1),\mathbb {R} )}$ (as "differentiable").

Theorem

The set of differentiable functions ${\displaystyle f:(0,1)\to \mathbb {R} }$ forms an ${\displaystyle \mathbb {R} }$-vector space.

Proof

The set of differentiable functions ${\displaystyle f:(0,1)\to \mathbb {R} }$ is a subset of the set of maps ${\displaystyle f:(0,1)\to \mathbb {R} }$, i.e. ${\displaystyle {\mathcal {D}}((0,1),\mathbb {R} )\subseteq \operatorname {Fun} ((0,1),\mathbb {R} )}$. To show that ${\displaystyle {\mathcal {D}}((0,1),\mathbb {R} )}$ also forms an ${\displaystyle \mathbb {R} }$-vector space, it suffices to show that ${\displaystyle {\mathcal {D}}((0,1),\mathbb {R} )}$ is an ${\displaystyle \mathbb {R} }$-subspace of ${\displaystyle \operatorname {Fun} ((0,1),\mathbb {R} )}$. To do this, we need to establish the 3 subspace axioms.

Proof step: ${\displaystyle {\mathcal {D}}((0,1),\mathbb {R} )\neq \emptyset }$

The function ${\displaystyle f:(0,1)\to \mathbb {R} ,x\mapsto 0}$ is differentiable. So: ${\displaystyle {\mathcal {D}}((0,1),\mathbb {R} )\neq \emptyset }$.

Proof step: For all ${\displaystyle f,g\in {\mathcal {D}}((0,1),\mathbb {R} )}$ we have that ${\displaystyle f+g\in {\mathcal {D}}((0,1),\mathbb {R} )}$.

Let ${\displaystyle f,g:(0,1)\to \mathbb {R} }$ be differentiable, i.e. ${\displaystyle f,g\in {\mathcal {D}}((0,1),\mathbb {R} )}$. We have shown in Analysis I that the function ${\displaystyle f+g:\mathbb {R} \to \mathbb {R} ,x\mapsto f(x)+g(x)}$ is also differentiable. Consequently we have that ${\displaystyle f+g\in {\mathcal {D}}((0,1),\mathbb {R} )}$.

Proof step: For all ${\displaystyle f\in {\mathcal {D}}((0,1),\mathbb {R} )}$ and for all ${\displaystyle \lambda \in \mathbb {R} }$ we have that ${\displaystyle \lambda \cdot f\in {\mathcal {D}}((0,1),\mathbb {R} )}$.

Let ${\displaystyle f\in {\mathcal {D}}((0,1),\mathbb {R} )}$ and ${\displaystyle \lambda \in \mathbb {R} }$. We have shown in Analysis I that the map ${\displaystyle \lambda \cdot f:(0,1)\to \mathbb {R} ,x\mapsto \lambda \cdot f(x)}$ is also differentiable. Thus we have that ${\displaystyle \lambda \cdot f\in {\mathcal {D}}((0,1),\mathbb {R} )}$.

Thus we have shown that ${\displaystyle {\mathcal {D}}((0,1),\mathbb {R} )}$ is an ${\displaystyle \mathbb {R} }$-subspace of ${\displaystyle \operatorname {Fun} ((0,1),\mathbb {R} )}$.

## Relation to the sequence space

We have already seen that the set of sequences over ${\displaystyle K}$ forms a vector space with respect to coordinate-wise operations. So a sequence ${\displaystyle (a_{n})_{n}}$ with entries in ${\displaystyle K}$ can be seen as a function ${\displaystyle \mathbb {N} \to K,n\mapsto a_{n}}$. In this sense, the sequence space is a special case of the function space ${\displaystyle \operatorname {Fun} (X,V)}$ by setting ${\displaystyle X:=\mathbb {N} }$ and ${\displaystyle V:=K}$.