# Statement
Let $S$ be a Set and $V$ some Vector Space over Field $F$. Then $\mathcal{F} = {f : S \to V}$ is a Vector Space with
- $(f +_\mathcal{F} g)(x) = f(x) +_V g(x)$ for $x \in S$, $f,g \in \mathcal{F}$.
- $(c *\mathcal{F} f)(x) = c *{V} f(x)$ for $x \in S$, $c \in F$, $f \in \mathcal{F}$.
# Proof
We verify the definition of a Vector Space. Let $\mathbf{0}(x) = 0$ for all $x \in S$.
- Abelian Group:
- $(\mathbf{0} +\mathcal{F} f)(x) = 0 +{V} f(x) = f(x)$ for $x \in S$. $\checkmark$
- Let $f, g, h \in \mathcal{F}$. Then $$\begin{align*} ((f+\mathcal{F} g) +\mathcal{F} h)(x) &= (f +\mathcal{F} g)(x) +{V} h(x) \\ &= f(x) +{V} g(x) +{V} h(x) \\ &= f(x) +{V} (g +\mathcal{F} h)(x) \\ &= (f +\mathcal{F} (g +\mathcal{F} h))(x). \end{align*}$$ $\checkmark$
- Let $f, g \in \mathcal{F}$. Then $$\begin{align*} (f +\mathcal{F} g)(x) &= f(x) +{V} g(x) \\ &= g(x) +{V} f(x) \\ &= (g +\mathcal{F} f)(x) \end{align*}$$ for $x \in S$. $\checkmark$
- Let $f \in \mathcal{F}$. Then $$\begin{align*} (f +\mathcal{F} (-f))(x) &= f(x) -{V} f(x) \\ &= 0 \\ &= \mathbf{0}(x) \end{align*}$$ for $x \in S$. $\checkmark$
- Compatibility with Scalar Multiplication, $f, g \in \mathcal{F}$:
- $(1*\mathcal{F} f)(x) = 1 *{V} f(x) = f(x)$ for $x \in S$. $\checkmark$
- $c_{1}, c_{2} \in F$ then $$\begin{align*} ((c_{1} *F c{2}) \mathcal{F} f)(x) &= (c{1}{F} c{2}) {V}f(x) \\ &= c{1}{V}(c{2} {V} f(x))\\ &= c{1}{V}(c{2} *\mathcal{F}f)(x) \\ &= (c{1} *\mathcal{F} (c{2} _\mathcal{F} f))(x) \end{align}$$ for $x \in S$. $\checkmark$
- $c \in F$, then $$\begin{align*} (c *\mathcal{F} (f +\mathcal{F} g))(x) &= c *{V}(f+\mathcal{F} g)(x)\\ &=c *{V}(f(x)+{V} g(x)) \\ &= c *{V} f(x) +{V} c *{V}g(x) \\ &= (c *\mathcal{F} f)(x) +{V} (c *\mathcal{F} g)(x)\\ &= (c *\mathcal{F} f +\mathcal{F} c _\mathcal{F} g)(x) \end{align}$$ for $x \in S$. $\checkmark$
- $c, d \in F$, then $$\begin{align*} ((c +{F} d) *\mathcal{F} f)(x) &= (c +{F} d) *{V} f(x) \\ &= c *{V} f(x) +{V} d *{V} f(x)\\ &= (c *{\mathcal{F}} f)(x) +{V} (d *{\mathcal{F}} f)(x) \\ &= (c *\mathcal{F} f +\mathcal{F} d _\mathcal{F} f)(x) \end{align}$$ for $x \in S$. $\checkmark$
$\blacksquare$