
Problem 1. The function $f(x)=\sqrt{1-x}(x \leq 1)$ is given. Let $F(x)=f(f(x))$.

a) Solve the equations $f(x)=x$ and $F(x)=x$.

b) Solve the inequality $F(x)>x$.

c) If $a_{0} \in(0,1)$, prove that the sequence $\left\{a_{n}\right\}_{n=0}^{\infty}$, determined by $a_{n}=f\left(a_{n-1}\right)$ for $n=$ $1,2, \ldots$, is convergent and find its limit.


Solution. The functions $f(x), F(x)$ and $F(x)-x$ are defined for all $x \in[0,1]$.

a) The equation $f(x)=x$, i.e. $\sqrt{1-x}=x$ has only one root $\alpha=\frac{-1+\sqrt{5}}{2}$. It is clear that $\alpha \in(0,1)$ and the roots of the equation $F(x)=x$ are $x_{1}=0, x_{2}=1, x_{3}=\alpha$.

b) In $[0, \alpha]$ the function $F(x)-x$ has a constant sign. The contrary would imply that there is $\beta \in(0, \alpha)$ such that $F(\beta)=\beta$ and this contradicts the result from a). Analogously in $[\alpha, 1]$ the function $F(x)-x$ has a constant sign. On the other hand $\frac{1}{4} \in(0, \alpha), \frac{3}{4} \in(\alpha, 1)$ and $F\left(\frac{1}{4}\right)-\frac{1}{4}>0, F\left(\frac{3}{4}\right)-\frac{3}{4}x$ iff $x \in\left(0, \frac{-1+\sqrt{5}}{2}\right)$.

c) Let $a_{0}=\alpha$. It follows from a) that $a_{n}=\alpha$ for $n=0,1,2, \ldots$ and hence the sequence is convergent and its limit is $\alpha$. Let now $a_{0}\operatorname{frac}(\alpha)=\alpha$, i.e. $a_{1}>\alpha$. By induction $a_{2 n} \in(0, \alpha)$ and $a_{2 n+1} \in(\alpha, 1)$ for all $n=0,1,2, \ldots$. On the other hand it follows from the result of b) that for all $x \in(0, \alpha)$ we have $F(x)>x$, while for $x \in(\alpha, 1)$ we have $F(x)0$, i.e. $F(x)$ is increasing and hence $F(x) \in(x, \alpha)$ if $x \in(0, \alpha)$ and $F(x) \in(\alpha, x)$ if $x \in(\alpha, 1)$. By induction we get

$$
\begin{aligned}
& a_{0}a_{3}>\cdots>a_{2 n+1}>\cdots>\alpha
\end{aligned}
$$

Both sequences are convergent and let their limits be $\alpha_{1}$ and $\alpha_{2}$, respectively. We have $F\left(a_{2 n}\right)=a_{2 n+2}$ and $F\left(a_{2 n+1}\right)=a_{2 n+3}$ for $n=0,1,2, \ldots$. The function $F(x)$ is continuous and thus $F\left(\alpha_{1}\right)=\alpha_{1}$ and $F\left(\alpha_{2}\right)=\alpha_{2}$. We get $\alpha_{1}=\alpha_{2}=\alpha$ because the only solution of $F(x)=x$ in $(0,1)$ is $\alpha$. Therefore the sequence $\left\{a_{n}\right\}_{n=0}^{\infty}$ is convergent and its limit is $\alpha$. The case $\alpha<a_{0}$ is analogous.
