Show that $U_1 \oplus U_2=V$

Question

Let $V=\mathbb{R}^\mathbb{R}$ be a $\mathbb{R} $ vector space of all mappings from $\mathbb{R}$ to $\mathbb{R}$

$$U_1=\{f \in V:f(-x)=f(x), \forall x \in\mathbb{R} \}$$

$$U_1=\{f \in V:f(-x)=-f(x), \forall x \in\mathbb{R} \}$$

Show that $U_1 \oplus U_2=V$.

Can someone give me a hint on how to start with it?

My initial idea was to show that $U_1 \cap U_2 = {0}$ and $\dim_\mathbb{R}(U_1)+\dim_\mathbb{R}(U_2)=\dim_\mathbb{R}(V)$

Answer 1

$U_1\cap U_2=\{0\}$ is easy and I will let you handle it. For the second property use the fact that $f=g+h$ where $g(x)=\frac {f(x)+f(-x)} 2$ and $h(x)=\frac {f(x)-f(-x)} 2$

Answer 2

Hint: Every function can be written as a sum of an odd and an even function . For example: for any $g\in V$, note that $g(x) =Even +Odd=\frac{g(x) +g(-x)} {2}+\frac{g(x)-g(-x)}{2}\in U_1+U_2$.
For $U_1\cap U_2$, think about the function which is odd and even both!

Answer 3

$f(x)={1\over 2}(f(x)+f(-x))+{1\over 2}(f(x)-f(-x))$