Question

A function $f\left(x\right)$ can be uniquely expressed as the sum of an even and odd function.

Answer 1

Let $F\left(x\right)=f\left(x\right)+f(-x)...\left(1\right)$
$F(-x)=f(-x)+f\left(x\right)=F\left(x\right)$

$\therefore F$ is even

$G\left(x\right)=f\left(x\right)-f(-x)...\left(2\right)$
$G(-x)=f(-x)-f\left(x\right)=-G\left(x\right)$

$\therefore G$ is odd.

Now, adding (1) and (2),

$F\left(x\right)+G\left(x\right)=2f\left(x\right)$

$\therefore f\left(x\right)=\frac{1}{2}[F\left(x\right)+G\left(x\right)]...\left(3\right)$

where $F\left(x\right)$ is even . $G\left(x\right)$ is odd function .

For uniqueness : Let, if possible, there exist $F_1\left(x\right)$, an even function and $G_1\left(x\right)$, an odd function of $x$ such that

$f\left(x\right)=\frac{1}{2}[F_1\left(x\right)+G_1\left(x\right)]...\left(4\right)$

Subtracting (4) from (3),we get

$0=\frac{1}{2}[\{F\left(x\right)-F_1\left(x\right)\}+\{G\left(x\right)-G_1\left(x\right)\}]$
$\therefore F\left(x\right)-F_1\left(x\right)=0$ and $G\left(x\right)-G_1\left(x\right)=0$
$\therefore F_1\left(x\right)=F\left(x\right)$ and $G_1\left(x\right)=G\left(x\right)$
Hence the expression is unique.