Example. Direct sum of even and odd functions

Let us define the sets \begin{eqnarray*} F &=& \{ f\, |\, f\,\mbox{ is a function from}\, \sR\, \mbox{ to}\, \sR \}, \\ F_+ &=& \{ f\in F \,|\, f(x)=f(-x) \,\mbox{for all}\, x\in \sR\}, \\ F_- &=& \{ f\in F \,|\, f(x)=-f(-x)\,\mbox{for all}\, x\in \sR\}. \end{eqnarray*} In other words, $F$ contain all functions from $\sR$ to $\sR$ , $F_+\subset F$ contain all even functions, and $F_-\subset F$ contain all odd functions. All of these spaces have a natural vector space structure: for functions $f$ and $g$ we define $f+g$ as the function $x\mapsto f(x)+g(x)$ . Similarly, if $c$ is a real constant, then $cf$ is the function $x\mapsto cf(x)$ . With these operations, the zero vector is the mapping $x\mapsto 0$ .

We claim that $F$ is the direct sum of $F_+$ and $F_-$ , i.e., that \begin{eqnarray} \label{eq10} F &=& F_+ \oplus F_-. \end{eqnarray} To prove this claim, let us first note that $F_\pm$ are vector subspaces of $F$ . Second, given an arbitrary function $f$ in $F$ , we can define \begin{eqnarray*} f_+(x) &=& \frac{1}{2}\big( f(x) + f(-x) \big), \\ f_-(x) &=& \frac{1}{2}\big( f(x) - f(-x) \big). \end{eqnarray*}Now $f_+$ and $f_-$ are even and odd functions and $f=f_+ + f_-$ . Thus any function in $F$ can be split into two components $f_+$ and $f_-$ , such that $f_+ \in F_+$ and $f_-\in F_-$ . To show that the sum is direct, suppose $f$ is an element in $F_+\cap F_-$ . Then we have that $f(x)=-f(-x)=-f(x)$ , so $f(x)=0$ for all $x$ , i.e., $f$ is the zero vector in $F$ . We have established equation .