
93.1. Let $F$ be an increasing real function defined for all $x, 0 \leq x \leq 1$, satisfying the conditions

$$
\begin{aligned}
& F\left(\frac{x}{3}\right)=\frac{F(x)}{2} \\
& F(1-x)=1-F(x)
\end{aligned}
$$

Determine $F\left(\frac{173}{1993}\right)$ and $F\left(\frac{1}{13}\right)$.


Solution. Condition (i) implies $F(0)=\frac{1}{2} F(0)$, so $F(0)=0$. Because of condition (ii), $F(1)=1-F(0)=1$. Also $F\left(\frac{1}{3}\right)=\frac{1}{2}$ and $F\left(\frac{2}{3}\right)=1-F\left(\frac{1}{3}\right)=\frac{1}{2}$. Since $F$ is an increasing function, this is possible only if $F(x)=\frac{1}{2}$ for all $x \in\left[\frac{1}{3}, \frac{2}{3}\right]$. To determine the first of the required values of $F$, we use (i) and (ii) to transform the argument into the middle third of $[0,1]$ :

$$
F\left(\frac{173}{1993}\right)=\frac{1}{2} F\left(\frac{519}{1993}\right)=\frac{1}{4} F\left(\frac{1557}{1993}\right)
$$

$$
\begin{gathered}
=\frac{1}{4}\left(1-F\left(\frac{436}{1993}\right)\right)=\frac{1}{4}\left(1-\frac{1}{2} F\left(\frac{1308}{1993}\right)\right) \\
=\frac{1}{4}\left(1-\frac{1}{4}\right)=\frac{3}{16}
\end{gathered}
$$

To find the second value of $F$, we use (i) and (ii) to form an equation fron which the value can be solved. Indeed,

$$
\begin{gathered}
F\left(\frac{1}{13}\right)=1-F\left(\frac{12}{13}\right)=1-2 F\left(\frac{4}{13}\right) \\
=1-2\left(1-F\left(\frac{9}{13}\right)\right)=2 F\left(\frac{9}{13}\right)-1=4 F\left(\frac{3}{13}\right)-1 \\
=8 F\left(\frac{1}{13}\right)-1
\end{gathered}
$$

From this we solve

$$
F\left(\frac{1}{13}\right)=\frac{1}{7}
$$
