Periodic Decomposition Problem
<return to the blog>
Here is the problem: do there exist two periodic functions
$f,g : \mathbb{R} \to \mathbb{R}$ such that
$\forall x \in \mathbb{R},\ (f+g)(x)=x$?
Surprisingly, the answer is yes, but only if we allow very wild,
nowhere continous non-mesurable functions.
Constructing functions
The identity
The main killing argument is to consider a basis of $\mathbb{R}$ as a $\mathbb{Q}$-vector space.
Such a basis, called a Hamel basis, are only constructible because of the axiom of choice.
Suppose we have $\alpha \not\in \mathbb{Q}$ and consider a basis containing both $\alpha$ and $1$.
Every real number $x$ can be written uniquely as
$$
x = q'\alpha + q \cdot 1 + \text{other basis terms},
$$
with $q,q' \in \mathbb{Q}$.
Then we can define $f$ and $g$ to be the projections on $\alpha$ and its supplement as follows:
$$
f(x)=q'\alpha. \
g(x)=x-f(x).
$$
By construction, $f(x)+g(x)=x$. Moreover, since $f$ is $\mathbb{Q}$-linear and a projection on $\alpha$,
we have $f(x+1) = f(x)+f(1) = f(x)$, so $f$ is $1$-periodic.
Similarly, $g$ is $\alpha$-periodic.
Therefore the identity function is indeed the sum of two periodic functions.
Notice that $f$ is more than just $1$-periodic : it is $q$-periodic for every $q \in \mathbb{Q}$
Polynomials
More generally, every polynomial $P \in \mathbb{R}_d [X]$ of degree
at most $d$ is the sum of $d+1$ periodic functions.
Choose some real numbers
$B = \{\alpha_1,\dots,\alpha_d\}$ that are linearly independent over $\mathbb{Q}$,
and for all $i \le d$, define $f_i$ the projection over $\alpha_i$ and $f_0$ the
projection over the supplement of $\text{Vect}(B)$.
We therefore have $\forall x \in \R, (f_0 + ... + f_d)(x) = x$.
Then, for $P \in \mathbb{R}_d [X]$, consider develloping $P' := P(f_0 + ... + f_d)$.
Since we have that $\deg P(Q) = \deg P \times \deg Q$, then $\deg P' = d$ (with the $(f_i)_i$
as variables), so every monom of $P(f_0 + ... + f_d)$ must have a $f_i$ that isn't present.
So every monom is $\alpha$ periodic for at least one $\alpha \in B$ (can be more than one).
By regrouping such monoms, we have a sum of periodic functions equaling the polynomial.
For example, for $f_1, f_2$ two projections over $\text{Vect}_\mathbb{Q} (\alpha_1)$ and $\text{Vect}_\mathbb{Q} (\alpha_2)$,
and $f_0$ the projection over the supplement, we have
$$
x^2 = (f_0 + f_1 + f_2)^2 = \underbracket{f_1^2 + 2f_1 f_0+ 2f_1 f_2 }_{\alpha_1-periodic}
+ \underbracket{f_2^2 + 2f_2 f_0}_{\alpha_2-periodic}
+ \underbracket{f_0^2}_{\R\backslash\mathbb{Q}-periodic}
$$
Non decomposable functions
We will say here that a function is non-decomposable if it cannot be written as a finite sum of periodic functions.
The offset $\Delta_a$ operators
For $\alpha \in \R^*$, define the difference operator $\Delta_\alpha (F)(x) = F(x) - F(\alpha-x)$. Then, notice that if $f$ is $\alpha$-periodic, then $\Delta_\alpha(f) = \overline{0}$. This lead to this easy claim: if $f$ is a function that can be written as a sum of periodic function, then there exists a $n \in \N$ and $\alpha_1,...,\alpha_n \in\R$ such than $\Delta_{\alpha_1} \circ ... \circ \Delta_{\alpha_n} (f) = \overline{0}$.
Exponential
Suppose
Final words
Conclusion
Polynomials can be built from periodic functions, but only by using highly
non-regular functions. The identity already requires a non-continuous,
non-measurable construction. The exponential, on the other hand, is too
rigid: finite differences never kill it, so it cannot be a finite sum of
periodic functions.