In order to proof that a set A is a subspace of a Vector space V we'd need to prove the following:
- Enclosure under addition and scalar multiplication
- The presence of the 0 vector
And I've done decent when I had to prove "easy" or "determined" sets A.
Now this time I need to prove that F and G are subspaces of V where:
F = {f ∈ V : f(−x) = −f(x) for every x ∈R} (every odd function)
G = {f ∈ V : f(−x) = f(x) for every x ∈R} (every even function)
How can I prove the above stated question with sets liek F and G which are just functions?
Thanks!
Edit: V is the "Vector Space of the real-valued functions over R
Best Answer
Given $V=F(-∞,∞)$ i.e set of all real valued functions over $\mathbb{R}$.
Consider given set $F=\{f(x)\in V: f(-x)=-f(x)\}$.
Clearly if $f,g\in F$ then $(f+g)(-x)=f(-x)+g(-x)=-f(x)-g(x)=-(f(x)+g(x))$
So that $F$ is closed with respect to addition. Now for $k\in \mathbb{R}$ consider, $(kf)(-x)=k•f(-x)=-k•f(x)$
Hence $F$ is closed with respect to scalar multiplication.
So that $F$ is subspace of $V=F(-∞,∞)$
Now can you do same for $G$?