# Prove/Disprove set of even Function with scalar mult and addition forms a Vector space

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.

• Looks fine. Maybe you should have observed that $0_v$ is even. And maybe (?) many of the verifications can be omitted, since an awful lot of properties are inherited from the mother vector space of all functions from the reals to the reals. Perhaps that has already been proved to be a vector space. – André Nicolas Jan 24 '16 at 18:14
• @AndréNicolas, yes, I think you only need evenness for the closure axioms. The remaining properties are inherited from the more general space of functions on the reals. – Sherif F. Jan 24 '16 at 18:57
• I do understand to check for 4 prop closed add, closed scalar mult, existence of zero, existinence of additive inverse. Like subring test. It will save time but it is on the next section thats why. – Tiger Blood Jan 24 '16 at 19:05

This is correct, except for a minor point under existence of additive inverse: I think $f(x)∗(0_\nu)$ should just be $f(x)∗0$. You haven't defined $f∗g$, 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 $0*f(x) = 0_\nu$.