All-Russian Olympiad question (sum of symmetrical functions)

Question

(All-Russian Olympiad, $1995$, $11^{th}$ Graders, Final Round)

Prove that every real function, defined on all of $\mathbb R$, can be represented as a sum of two functions each of which has a vertical axis of symmetry (presumably the line $x=$ something).

There's a "solution" on Art of Problem Solving but it's hardly complete and gives me little help. In essence, it says that arbitrary functions $g,h$ such that $g+h=f$ on $[0,1]$ can be extended so that $g$ is symmetrical around $x=0, h$ around $x=1$. How can such an extension be done?

Answer 1

Initially, we have $g$ and $h$ defined on $[0,1]$ and they sum up to $f$. This determines $g$ on $[-1,0]$ and $h$ on $[1,2]$, so $h$ needs to be defined on $[-1,0]$ as $f-g$, and $g$ needs to be defined on $[1,2]$ as $f-h$. This determines $g$ on $[-2,-1]$ and $h$ on $[2,3]$. Continuing this will define them on the whole $\mathbb R$.

Answer 2

We will write out given function $f(x)$ as the sum of $g(x)$, a function symmetric across the axis $x = 0$, and $h(x)$, a function symmetric across the axis $x = a$ for some fixed (but arbitrary) $a >0$. We will define the function successively on the intervals$$[-a, a], \text{ }[a, 3a], \text{ }[-3a, -a], \text{ } [3a, 5a], \text{ etc}.$$On $[-a, a]$, put$$g(x) = 0, \quad h(x) = f(x).$$We are forced to put$$h(x) = f(2a - x)$$for $x \in [a, 3a]$, so we put$$g(x) = f(x) - f(2a - x).$$Now we are forced to put$$g(x) = g(-x)$$for $x \in [-3a, -a]$, so we put$$h(x) = f(x) - g(-x)$$there, and so forth.