# $f(xy)=f(x(1-y)) \forall x,y\in (0,1)$, Find all $f:(0,1)\rightarrow \mathbb{R}$

Find all functions $f:(0,1)\rightarrow \mathbb{R}$ such that $f(xy)=f(x(1-y))$ for all $x,y\in (0,1)$.

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For any $x \in (0,\frac12)$, $$f(x) = f\left((\frac{x}{x+\frac12})(x + \frac12)\right) = f\left((1 - \frac{x}{x+\frac12})(x + \frac12)\right) = f(\frac12)\tag{*}$$ For any $x \in (\frac12,1)$, $$f(x) = f\left((\frac{2x}{1+x})(\frac{1+x}{2})\right) = f\left((1 - \frac{2x}{1+x})(\frac{1+x}{2})\right) = f\left(\frac{1-x}{2}\right) \underbrace{= f(\frac12)}_{\text{by (*)}}$$

This implies $f(x) = f(\frac12)$ is a constant over $(0,1)$.

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We have $\Delta = \{(xy,x(1-y)) | x,y \in (0,1) \} = \{(s,t) | s,t \in (0,1), s+t<1 \}$, and we have $f(a) = f(b)$ for all $(a,b) \in \Delta$.

Hence $f$ must be a constant.

To see this, note that $(0,\frac{1}{2})^2 \in \Delta$, hence if $x,y \in (0, \frac{1}{2})$, then $f(x) = f(y)$, ie, $f$ is constant on $(0, \frac{1}{2})$.

Note also that if $\alpha \in (0,1)$, then we also have $S_\alpha = (0,\alpha)\times (0,1-\alpha) \subset \Delta$. If $(a,b) \in S_\alpha$, we have $\min(a,b) < \min(\alpha, 1-\alpha) = \frac{1}{2}$, from which it follows that $f(a) = f(\min(a,b))$, hence $f$ is constant on $(0,\alpha)$. Since $\alpha \in (0,1)$ is arbitrary, we see that $f$ is constant on $(0,1)$.

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Take analytical continuation of a function $f$ to the domain $[0,1]$ and that satisfies the same functional equation for $f$, so $g(x) = y(x), \forall x \in (0,1)$. Consider $y = 0$ in $g(xy) = g(x-xy)$, you'll get that $g(x) = g(0)$ which is a constant, so is $f$ then.
You cannot take $y=0$, though. –  Andres Caicedo Mar 1 at 6:41
@copper.hat I mean extension of $f: \mathbb (0,1) \rightarrow\mathbb R$ to a $g$ s.t. $g: [0,1] \rightarrow\mathbb R$ and $g(x) = f(x)$ for $\forall x \in (0.1)$, and moreover satisfying same functional extension but for $\forall x \in [0,1]$. –  Kaster Mar 1 at 7:27