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.


2 Answers 2


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.

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

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




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

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

  • $\begingroup$ This will be the better answer once it is finished ;) $\endgroup$
    – Carsten S
    Aug 13, 2016 at 16:38

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