Prove that the set of even function defined on the real line where $x\in \mathbb{R}$with operations of addition and scalar multiplication defined as $$(f+g)(x)=f(x)+g(x); (cf)(x)=c[f(x)]$$
forms a Vector Space
Def of Vector space
A vector space V over field F consists of a set on which 2 operations (addition, scalar multiplication) are defined so that
$$\begin{aligned}
&\forall x,y \in V; x+y\in V && \text{closed add} \\
&\forall a\in F ,x\in V ;\exists \text{ unique } ax\in V && \text{closed mult scalar and vector} \\
&\forall x,y \in V, x+y=y+x && \text{communitive addition} \\
&\forall x,y,z \in V , (x+y)+z=x+(y+z)
&& \text{associativity of addition} \\
&\forall x\in V, \exists 0_v \in V ; x+0_v=x &&\text{existence of neutral additive element} \\
&\forall x\in V , \exists y\in V ; x+y=0_v && \text{existince of additive inverse} \\
&\forall{x \in V},1x=x &&\text{existence neutral element mult} \\
&\forall a,b\in F,\forall x \in V;(ab)x=a(bx) &&\text{associtive ____} \\
&\forall a\in F, \forall x,y \in V; a(x+y)=ax+ay &&\text{Dist 1}\\
&\forall a,b \in F,\forall x\in V ; (a+b)x=ax+bx && \text{Dist 2}
\end{aligned}$$
Def of even function $f(x)=f(-x)$
Attempt
Assuming $F=\mathbb{R}$. Call $V[x]$ The set of all even functions.Let $f(x),g(x) \in V[x], \forall a\in \mathbb{R}$.
$$\begin{aligned}
\text{Closed Add} && (f+g)(x)=f(x)+g(x)=f(-x)+g(-x)=(f+g)(-x)
\\\text{closed Mult scalar, vector} && (af)(x)=af(x)=af(-x)=(af)(-x) \\ \text{comm. add} && (f+g)(x)=f(x) +g(x)=g(x)+f(x)=(g+f)(x)
\end{aligned}$$
Asso. Add
$$\begin{aligned} (f+(g+h))(x)&=f(x)+(g+h)(x)=f(x)+(g(x)+h(x)) =(f(x)+g(x))+h(x)=(f+g)(x)+h(x)\\&=((f+g)+h)(x)
\end{aligned}$$
Let $0_v(x)=0 $ be function where $\forall x \in R$ $0_v(x)=0$.$$ (f+0_v)(x)=f(x)+0_v(x)=f(x)+0_v=f(x)=(f)(x)$$
$$\begin{align}
\text{existence additive inver.} && (f+-1f)(x)=f(x)+-1f(x)=f(x)*(1-1)=f(x)*(0_v)=0_v \\
\text{Neutral elem. mult.} && (1f)(x)=1f(x)=f(x)=(f)(x) \\
\text{ass.__} && ((ab)f)(x)=(ab)(f(x))=a(bf(x))=(a(bf))x \\
\text{Dist 1} && (a(f+g))(x)=a(f(x)+g(x))=af(x)+bg(x)=(af)(x)+(ag)(x) \\
\text{Dist 2} && ((a+b)f)(x)=(a+b)f(x)=af(x)+bf(x)=(af)(x)+(bf)(x)
\end{align}$$
Just reversed engineered the answer. Not sure if it actually holds. The middle part formatting of the attempt is a little chaotic. I'll try to make it prettier.
Appreciate constructive critique.
Best Answer
This is correct, except for a minor point under existence of additive inverse: I think $f(x)â(0_\nu)$ should just be $f(x)â0$. You haven't defined $fâg$, where $f$ and $g$ are functions. Also, your expression leads to the number $0$, not the zero function. It's indeed true based on your definitions that $0*f(x) = 0_\nu$.