Putnam and Beyond Problem 21 I am having a hard time solving this rather interesting question below:
Prove that any function defined on the entire real axis can be written as the sum of two functions whose graphs admit centers of symmetry.
I know we have to set the two functions that are mentioned in the question as an odd function (because it is a function with a graph that is symmetrical with respect to the origin) but I am having a hard time trying to come up with the proof as there are some cases that we need to consider but I don't know what are the appropriate cases we should imply. Thanks!
 A: A method that almost works is letting 
$$
\begin{aligned}
f_e&= \dfrac{f(x) + f(-x)}{2} \\
f_0&= \dfrac{f(x) - f(-x)}{2}
\end{aligned}
$$
Then $f(x)= f_e(x) + f_o(x)$ and $f_e,f_o$ are even and odd, respectively. Then $f_o$ has a point of symmetry $(0,0)$ because it is odd. But $f_e$ has an axis of symmetry rather than necessarily a point of symmetry. So we will need something else. I will outline the method suggested by the book, and leave you to fill in the gaps and be sure you understand it.
$1$. Let $g$ be an odd function on $[-1,1]$ with $g(1)=f(1)$. [Be sure you can give at least one concrete example for any give point $(1,f(1))$. It shouldn't be too hard. For example, try with line segments! But it may be beneficial to try to create some 'fancier' examples.]
$2$. Now $g$ is odd (thus far) so it will have $(0,0)$ as a point of symmetry. We need to define another function, which along with $g$, will sum to $f$. An easy way is to define $h(x)= f(x) -  g(x)$ on $[-1,1]$. Check that $f= g + h$. 
$3$. Now we need to extend $g,h$ so that they are defined everywhere. For $n \geq 1$, define $h(x):= -h(2-x)$ and then extend $g$ via $g(x):= f(x) - h(x)$ for $x \in (2n-1,2n+1]$. Now we need to extend the functions 'on the other side.' Define $g(x):= -g(-x)$ and $h(x):= f(x) - g(x)$ for $x \in [-2n-1,-2n+1)$. 
Try this for $n=1,2$ with an explicit function, something simple like $f(x)=x^2$ to understand the process. Then go to the 'generic' $f(x)$. For $n=1,2$, check that $g$ is odd and $h(x)$ has symmetry point $(1,0)$, then check that $f(x)= g(x)+h(x)$. Once you have this done. You should be able to finish the induction argument to show that $f= g+h$ and each have a point of symmetry. Remember, an explicit example always helps!
