I need to prove the following is a subspace:
Let $V$ be a set of vectors over $F=\mathbb{R}$, $V=\operatorname{Functions}(\mathbb{R} ,\mathbb{R})$ and $W$ is a subgroup of $V$ such that $$W=\{f\in V| \, f(x)=f(-x)\}$$
I'm not sure about the "close under vector addition" part.
My solution: Let be $w_1 , w_2 \in W. w_1=\{f\in V| \, f(x_1)=f(-x_1)\},\,\,\,w_2=\{f\in V| \, f(x_2)=f(-x_2)\}$. Therefore:
$$w_1+w_2=f(x_1+x_2)=f(x_1)+f(x_2)=f(-x_1)+f(-x_2)=f\big(-(x_1+x_2)\big)\\
$$
Please tell me if something is wrong here (I feel like something is wrong…). Many thanks for you time and effort.
Best Answer
Your thinking about $W$ is muddled, as evidenced when you refer to $w_1$ and $w_2$ as sets. You may as well call them $f_1$ and $f_2$ because they are functions, and they are the elements of $W$ (vectors) that you are working with.
There is no reason to index the $x$: it is just an arbitrary element of $\Bbb R$. I suspect you have fallen into a common misunderstanding among beginners. The fact is that the functions are the elements of the vector space. (Whereas beginners sometimes are overly attached to the $x$ being an element of a vector space.)
By virtue of being in $W$, both of them have the property that $f_i(x)=f_i(-x)$. Then
$$ (f_1+f_2)(x)=\dots=(f_1+f_2)(-x) $$
proving that $f_1+f_2$ is also a member of $W$. (I omitted the middle computation so you could puzzle it out :) )
See if you can do the closure under scalars now: the goal is to show that if $\lambda\in \Bbb R$, if $f\in W$, then $\lambda f\in W$. Finally, show that $0\in W$, "0" here denoting the function that is constantly 0 on $\Bbb R$.
In words, what you are doing is showing that the even functions on $\Bbb R$ form a subspace. As a followup exercise, you could additionally show that the odd functions also form a subspace of $V$.