Solve the functional equation $f(x) + f(x^2) = 2$

Question

What are the solutions of the functional equation $f(x) + f(x^2) = 2$? Will they be one to one or many to one? Will they be periodic or not?

Answer 1

Select some positive $q\ne 1$ and $g(t)=f(q^{2^t})-1$. Then $$ g(t+1)=f((q^{2^t})^2)-1=-g(t)=g(t-1) $$ Thus $g$ is $2$-periodic. But note that $q$ may be arbitrarily close to $1$.

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In the end this means that you can divide the positive numbers in equivalence classes $\{q^{4^n}:n\in\Bbb Z\}$ where $f$ is necessarily constant, $f(q^{4^n})=f(q)$. The classes of $q$ and $q^2$ are connected by the functional equation and in general the function values of every equivalence class are determined by the values over $(1/4,1/2]$ and $[2,4)$ for the positive axis and thus for the full real line, adding the trivial values $f(\pm1)=1=f(0)$.

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Only by demanding some continuity condition do you get the unique solution which is the constant function $1$.

Answer 2

Let $x=2^{2^t}$ and $f(x)=g(t)$.

Then,

$$g(t)+g(t+1)=2,$$ which is solved by $$g(t)=1+C(-1)^t.$$

As only values one unit apart are related to each other, we can write

$$g(t)=1+c(t-\lfloor t\rfloor)(-1)^{\lfloor t\rfloor}$$ where $c(t)$ is arbitrary in $[0,1)$.

Thus $$f(x)=1+h(\sqrt[\lfloor \text{lg}(\text{lg}(x))\rfloor]x)(-1)^{\lfloor \text{lg}(\text{lg}(x))\rfloor},$$

where $h(x)$ is arbitrary in $[2,4)$.