# Function expressed the sum of an odd and even function [duplicate]

Show that a function defined on a domain $D$ symmetric about the origin can be expressed in a unique way as the sum of an even and an odd function.

Sorry but I have no idea about how to start.

Even function: $$e(x)=\frac{f(x)+f(-x)}{2}$$ Odd function: $$o(x)=\frac{f(x)-f(-x)}{2}$$

As the domain is symmetric about the origin, $f(-x)$ is defined whenever $f(x)$ is defined.

Moreover, $$f(x)=e(x)+o(x)$$

Suppose this is not unique. Let there be an even function $E(x)\ne e(x)$ and an odd function $O(x)\ne o(x)$ such that $f(x)=E(x)+O(x)$. Then, on subtracting, $$E(x)+O(x)-e(x)-o(x)=0\tag1$$ Replace $x$ by $-x$, $$E(-x)+O(-x)-e(-x)-o(-x)=E(x)-e(x)-O(x)+o(x)=0\tag2$$ Adding $(1)$ and $(2)$, $$E(x)=e(x)$$ and $$O(x)=o(x)$$ Thus, the functions are unique.

• how about the uniqueness of the functon f Aug 13, 2016 at 16:19
• @user361163 Added that to my answer. Aug 13, 2016 at 16:24

To show existence and uniqueness, suppose that $e$ and $o$ are even and odd functions respectively such that

$$f(x)=e(x)+o(x).\tag{1}$$

Then

$$f(-x)=e(-x)+o(-x)=e(x)-o(x)\tag{2}.$$

Summing $(1)$ and $(2)$, we get

$$f(x)+f(-x)=2e(x)\iff e(x)=\frac{f(x)+f(-x)}{2}.$$

Subtracting $(2)$ from $(1)$, we get

$$f(x)-f(-x)=2o(x)\iff o(x)=\frac{f(x)-f(-x)}{2}.$$

• This will be the better answer once it is finished ;) Aug 13, 2016 at 16:38