# direct sum of even/odd functions (example)

Let us define the sets

 $\displaystyle F$ $\displaystyle=$ $\displaystyle\{f\,|\,f\,\mbox{ is a function from}\,\mathbb{R}\,\mbox{ to}\,% \mathbb{R}\},$ $\displaystyle F_{+}$ $\displaystyle=$ $\displaystyle\{f\in F\,|\,f(x)=f(-x)\,\mbox{for all}\,x\in\mathbb{R}\},$ $\displaystyle F_{-}$ $\displaystyle=$ $\displaystyle\{f\in F\,|\,f(x)=-f(-x)\,\mbox{for all}\,x\in\mathbb{R}\}.$

In other words, $F$ contain all functions from $\mathbb{R}$ to $\mathbb{R}$, $F_{+}\subset F$ 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

 $\displaystyle F$ $\displaystyle=$ $\displaystyle F_{+}\oplus F_{-}.$ (1)

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

 $\displaystyle f_{+}(x)$ $\displaystyle=$ $\displaystyle\frac{1}{2}\big{(}f(x)+f(-x)\big{)},$ $\displaystyle f_{-}(x)$ $\displaystyle=$ $\displaystyle\frac{1}{2}\big{(}f(x)-f(-x)\big{)}.$

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 1.

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