Solve functional equation: $\frac{F(x)}{F(1)} = 2F\left(\frac{(1+x)^2}{4a}\right)-F(x/a)$ Solve this functional equation:
$$\frac{F(x)}{F(1)} = 2F\left(\frac{(1+x)^2}{4a}\right)-F(x/a)$$
for $F(x)$ where $a > 0$ is a real parameter. Here $F(x)$ is a real function. The domain of $x$ is an open interval including 0. I do not specify the interval myself because I do not know the radius of convergence. I know there is a trivial constant solution, $F(x) = 1$. Is there a non-constant solution?
I do not know if it has an analytical solution, and have no reason to expect that it does. It showed up in some calculations I was doing. But if there is a technique I can apply, it would be nice to know about it.
 A: There is actually at least one non-constant solution for any given $a > 0$, if you're ready to accept non-continuous solutions (this here answer might thus feel unsatisfying for some since it neither solves the equation nor gives any non-constant continuous solution).
Since $x \mapsto 1$ is a solution, we can restrict our search to solutions $F$ such that $F(\mathbb{R}) \subset \{0,1\}$, or equivalently the solutions of the form $F = _A$ for a subset $A$ of $\mathbb{R}$.
(Spoiler: just taking indicator functions worked here, but generally trying to compose or multiply existing or would-be solutions by an indicator function, like $_\mathbb{Q}$, seems surprisingly useful in showing non-uniqueness of solutions of such equations, based on a few other questions I've looked at during this month.)
In order for the equation to even have a sense, we need $1 \in A$, otherwise $F(1)$ would be null. Moreover, it can be shown, through some very light brute-forcing, that we have the following condition:
$$\forall x \in \mathbb{R},\quad x \in A \,\,\Leftrightarrow\,\, \frac{x}{a} \in A \,\,\Leftrightarrow\,\, \frac{(1+x)^2}{4a} \in A \tag{1}$$
Conversely, any set $A$ satisfying this condition $(1)$ and such that $1 \in A$ will produce a solution of the equation, and so we have found exactly the solutions such that $F(\mathbb{R}) \subset \{0,1\}$. Let's show there is at least one such set that is not all of $\mathbb{R}$.
We will show something stronger: there exists a field $\mathbb{Q} \subset A \subsetneq \mathbb{R}$ for which $(1)$ holds.
To know how to find this $A$, let's first see why looking for a field could be a good idea. Assume for just this paragraph that we have such a field $A$.
Since $1 \in A$, and $ 1 = \frac{a}{a}$, we have: $a \in A$, and as such condition $(1)$ is actually equivalent to:
$$\forall x \in \mathbb{R},\quad x \in A \,\,\Leftrightarrow\,\, (1+x)^2 \in A$$
but $1 \in A$ implies $x \in A \Leftrightarrow 1 + x \in A$, therefore the above is again equivalent to, by choosing $y = 1+x$:
$$\forall y \in \mathbb{R},\quad y \in A \,\,\Leftrightarrow\,\, y^2 \in A \tag{1bis}$$
The reason why this is good is that we actually know how to, well, construct (pun intended) a field $\mathbb{Q} \subset A \subsetneq \mathbb{R}$ containing a given $a$ and satisfying $(1\text{bis})$: this is done by mimicking the creation of the field of constructible numbers (I linked the English page since we're on the mostly-English-speaking MSE but I prefer how the French page handles the matter so I'll use their notations, sorry... To anyone potentially looking at this post: do not hesitate to edit what's in this parenthesis with a link to an English source defining the constructible numbers in the same way as the French page).
Essentially boils down to: let $E_a := \{(0,0), (1,0), (a,0)\} \subset \mathbb{R^2}$ (basically we only added the point $(a,0)$ compared to the definition of the usual constructible numbers where you start with only $(0,0)$ and $(1,0)$), then define $C_n(E_a)$ as the set of points that are constructible from $E_a$ in $n$ steps for $n \in \mathbb{N}$, then take $C(E_a) := \bigcup_n C_n(E_a)$, and finally consider $A := \{x \in \mathbb{R} \mid (x,0) \in C(E_a)\}$.
By what I'll call "geometry magic" that I'll spare you the details about, $A$ is a field that contains $\mathbb{Q}$ and $a$, and for every positive element $x$ of $A$, $\sqrt{x}$ also belongs in $A$, which means that $A$ respects the condition $(1\text{bis})$.
It only remains to see that $A$ is not $\mathbb{R}$.
Since $E_a$ is finite, the $C_n(E_a)$ are actually all finite, and the reunion $C(E_a)$, which can be shown to be equal to $A^2$, is thus Lebesgue-negligeable in $\mathbb{R^2}$ as a countable union of countable sets.
In particular, $A^2 \neq \mathbb{R}^2$, granting: $A \neq \mathbb{R}$. Hence, by all the previous observations, $F := _A$ is a solution of our functional equation, and it is non-constant (and non-continuous) due to: $A \not\in \{\emptyset, \mathbb{R}\}$.
