Guys I'm stuck on this problem, I'm not even sure where to start with it. If anyone could give me a little help it would be very appreciated!
Let $F$ be any field and let $F^F$ denote the set of all functions from $F$ to $F$. The set $F^F$ is a vector space over $F$ with pointwise operations: $(f + g)(b) := f(b) + g(b), (af)(b) := af(b)$ for $f, g ∈ F^F$ and $a, b ∈ F$. A function $f ∈ F^F$ is even if $f(−b) = f(b)$ for all $b ∈ F$ and odd if $f(−b) = −f(b)$ for all $b ∈ F$. Prove that the set $U_E$ of all even functions and the set $U_O$ of all odd functions are subspaces of $F^F$.
Bonus: Show that $F^F = U_E ⊕ U_O$.
Best Answer
Investigating what the (alleged) theorem implies is often helpful. For $f:F\to F$, if $f=g+h$ where $g$ is even and $h$ is odd, then for all $x\in F$ we have $$f(x)=g(x)+h(x)$$ and $$f(-x)=g(x)-h(x).$$ Adding, we get $$f(x)+f(-x)=2 g(x).$$ Subtracting, we get $$f(x)-f(-x)=2 h(x).$$ So for $char (F)\ne 2$ if we DEFINE $g$ and $h$ by $$g(x)=(f(x)+f(-x))/2\quad \text {and }h(x)=(f(x)-f(-x))/2$$ then we see that $g$ is even, $h$ is odd, and $f=g+h$..... For $char (F)=2$, every $f:F\to F$ is already even and odd because $x=-x$ and $f(x)=-f(x)$ so $f(-x)=f(x)=-f(x).$