Question

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

Answer 1

Consider $F\left(x\right)=f\left(x\right)+f(-x)...\left(1\right)$
$F(-x)=f(-x)+f\left(x\right)=F\left(x\right)\therefore Even$ $G\left(x\right)=f\left(x\right)-f(-x)...\left(2\right)$
$G(-x)=f(-x)-f\left(x\right)=-G\left(x\right)\therefore Odd.$
Now $F\left(x\right)+G\left(x\right)=2f\left(x\right)by\left(1\right)and\left(2\right)$ $\therefore f\left(x\right)=\frac{1}{2}[F\left(x\right)+G\left(x\right)]...\left(3\right)$ where F(x) is even G(x) is odd function of x.
For uniqueness : Let,if possible, there exist $F_1\left(x\right)\left(even\right)and{G}_1\left(x\right)\left(odd\right)$ 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 (3) and (4),we get $0=\frac{1}{2}[\{F\left(x\right)-F_1\left(x\right)\}+\{G\left(x\right)-G_1\left(x\right)\}]$
$\therefore F\left(x\right)-F_1\left(x\right)=0andG\left(x\right)-G_1\left(x\right)=0$
$\therefore F_1\left(x\right)=F\left(x\right)and{G}_1\left(x\right)=G\left(x\right)$
Hence the expression is unique.