Dropping continuity condition on functional equation

Question

I have the following equation that is to be satisfied by some function $f:[0,1)\rightarrow\mathbb R$:

$$f\left(\frac{2x}{1+x^2}\right)=(1+x^2)f(x\sqrt{2-x^2}).$$

I managed to prove that, given continuity at $0$, the only possible solution is

$$f(x)=\frac{f(0)}{\sqrt{1-x^2}}.$$

I wonder, however, if this still applies if we drop the continuity condition. If it's possible to prove the continuity at $0$ given the functional equation alone, for example. I thought about it for a while but haven't gotten to anything useful.

Answer 1

Idea: Let's denote $g(x) = \sqrt{1-x^2}f(x)$. From your equation, we have $$g(\frac{2x}{1+x^2}) = g(x\sqrt{2-x^2})$$ Make the change of variables $x = \frac{1 - \sqrt{1 - t^2}}{t}$ with $0<t<1$ and denote $$u(t)=u(x(t))=\frac{1 - \sqrt{1 - t^2}}{t}\sqrt{2-(\frac{1 - \sqrt{1 - t^2}}{t})^2}$$

We have then for all $0<t<1$ $$g(t)=g(u(t))$$

The function $u(t)$ have this property: $u(t)<t$ (here below is the function $t-u(t)$)

and the sequence $u^{(n)}(t)$ defined by $\{ u^{(n)}(t) = u(u^{(n-1)}(t)), u^{(0)}(t)=t \}$ converge to $0$.

Because the function $f(x)$ (and so $g(x)$) continues at $0$, we can conclude that $g(t) = g(0)$ for all $t\in (0,1)$.

PS: If we suppose the assumption that $f(x)$ is continue at $x=0$, the solution of $g(x)$ would be as follows:

• We define a set $A$ in $\mathbb{R}\cap(0,1)$ such that for each $x\in A$, $\not\exists n\ge 1$ and $y\in A$ with $x = u^{(n)}(y)$ or $y = u^{(n)}(x)$
• We define a function $k(x): A\rightarrow \mathbb{R}$, $k(x)$ is not necessary continue
• The function $g(x)$ is defined as: $g(x)=g(k(x^*))$ with $x^*$ is the unique value such that there exists $n$: $x = u^{(n)}(x^*)$ And the function $f(x)$ will be $$f(x)=\frac{g(x)}{\sqrt{1-x^2}}$$