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Suppose that $s,t \in (\frac{1}{2},1)$ with $t \ne s$. Does there exist a continuous bijection $f \colon [0,1] \to [0,1]$ which simultaneously satisfies the functional equations $$ f(sx) = tx $$ and $$ f(sx + (1-s)) = tf(x) + (1-t) $$ for all $x \in [0,1]$?

It is fairly straightforward to show that such a function has to satisfy $f(0) = 0$ and $f(1) = 1$. Moreover, $f(s^n) = t^n$ and $ f(1-s^n) = 1-t^n$ for each $n \in \mathbb{N}$. This makes it seem like the existance of such a function $f$ is not possible, but I cannot yet disprove it.

Zorngo
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  • If $f(1)=1$, then it's pretty obvious from the first statement that $f(1) = f(sx) \implies 1 = \frac{t}{s}$, so it would make $t=s$. – Ninad Munshi Sep 22 '19 at 01:43
  • @NinadMunshi You can't put $x = \frac{1}{s}$ into any of the equations, since we require $x \in [0, 1]$ and $s < 1$. – Theo Bendit Sep 22 '19 at 01:48

2 Answers2

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$f(0)=0$ implies that

$$f(1-s) = 1-t$$

Since $s\in(\frac{1}{2},1)$, we have that $\frac{1-s}{s}\in(0,1)$ so

$$1-t = f(1-s) = f\left(s\cdot\left(\frac{1-s}{s}\right)\right) = t\cdot\left(\frac{1-s}{s}\right)$$

$$\implies s=t$$

Ninad Munshi
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$f(sx) = tx \implies f(x) = tx/s $.

$f(sx + (1-s)) = tf(x) + (1-t) \implies t(sx + (1-s))/s = t(tx/s) + (1-t) $ or $tx+t(1-s)/s =t^2x/s+1-t $ or $x(t^2/s-t) =t(1-s)/s-(1-t) =t/s-t+t-1 =t/s-1 $ or $xt(t/s-1) =t/s-1 $ so that $t=s $ and $f(x) = x$.

Note that no initial values of $f(x)$ or restrictions on $s$ and $t$ are needed.

marty cohen
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