# Can every real function be represented as two shifted even functions?

I saw the theorem that every function can be represented as the sum of and even and odd function, and this made me wonder: can every function from the reals to the reals, defined on all the reals, be represented as the sum of two functions that are each symmetric about (possibly distinct) vertical axes of symmetry?

This seems false to me, but it is harder than I expected to find a counterexample. For example, when trying to decompose f(x)=x this way, one can easily see that if our functions are a(x) and b(x), and we can translate and stretch so that we need to solve the functional equations:

$a(x)+b(x)=x$

$a(-x)=a(x)$

$b(x)=b(1-x)$

but I do not see an obvious contradiction.

Is such a decomposition always possible?

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Yes. Here's a sketch: given a function $f(x)$ we'll define functions $f_0(x), f_1(x)$ such that $f(x) = f_0(x) + f_1(x)$ and $f_i$ is symmetric about $i$. We proceed by induction:

• On $[0, 1]$, define $f_0(x) = f(x), f_1(x) = 0$.
• By symmetry this uniquely defines $f_0(x)$ on $[-1, 1]$ and defines $f_1(x)$ on $[0, 2]$.
• Define $f_0(x)$ on $(1, 2]$ uniquely so that $f(x) = f_0(x) + f_1(x)$. Do the same to $f_1(x)$ on $[-1, 0)$.
• By symmetry this uniquely defines $f_0(x)$ on $[-2, 2]$ and defines $f_1(x)$ on $[-1, 3]$.
• Etc.

The point is that at each step there is nothing stopping you from continuing the construction. Also, no choices have to be made except at the first step (which shows that this decomposition, unlike the decomposition of a function into its even and odd parts, is not unique).

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Here is a very simple-minded answer for the function $f(x)=x$ mentioned in the post.

Let $a(x)=x^2+1/4$ and let $b(x)=-(x-1/2)^2$. The first function is symmetric about $x=0$, and the second about $x=1/2$.

Or more symmetrically let $a(x)=(x+1)^2/4$ and $b(x)=-(x-1)^2/4$.

The idea can be readily extended to arbitrary polynomials. Given a polynomial $P(x)$ with real coefficients, one can produce explicit polynomials $a(x)$, $b(x)$ symmetric about $0$ and $1$ respectively such that $a(x)+b(x)$ is identically equal to $P(x)$.

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