A function \[\text{f:(0,}\infty \text{) }\to \text{(0,}\infty \text{)}\] is called contract if, for every numbers $x,y\in \text{(0,}\infty \text{)}$ we have, $\underset{n\to \infty }{\mathop{\lim }}\,\left( {{f}^{n}}\left( x \right)-{{f}^{n}}\left( y \right) \right)=0$ where ${{f}^{n}}=\underbrace{f\circ f\circ ...\circ f}_{n\ f\text{'s}}$ a) Consider \[f:\text{(0,}\infty \text{) }\to \text{(0,}\infty \text{)}\] a function contract, continue with the property that has a fixed point, that existing ${{x}_{0}}\in \text{(0,}\infty \text{) }$ there so that $f\left( {{x}_{0}} \right)={{x}_{0}}.$ Show that $f\left( x \right)>x,$ for every $x\in \text{(0,}{{x}_{0}}\text{)}\,$ and $f\left( x \right)<x$, for every $x\in \text{(}{{x}_{0}}\text{,}\infty \text{)}\,$. b) Show that the given function \[f\text{:(0,}\infty \text{) }\to \text{(0,}\infty \text{)}\] given by $f\left( x \right)=x+\frac{1}{x}$ is contracted but has no fix number.
Problem
Source:
Tags: function, limit, inequalities, real analysis, real analysis unsolved
22.08.2013 00:29
a)The fixed point of $f$ is unique.Suppose that there exist $y_0\neq x_0$ such that $f(y_0)=y_0$ then $\lim_{n\rightarrow +\infty }(f^n(x_0)-f^n(y_0))=0$ so $x_0=y_0$. Since function $f(x)-x$ is continuous and not zero for $x\neq x_0$, must maintain sign on each of the intervals $(0,x_0)$, $(x_0,+\infty)$. Assume that $f(x)<x$ for all $x\in (0,x_0)$.Then $f(x)$ belongs to the same interval with $x$ so replacing $x$ with $f(x)$ in the previous inequality we get $f^2(x)<f(x)$ for all $x\in (0,x_0)$.Inductively we have $f^{n+1}(x)<f^n(x)<x$ for all $x\in (0,x_0)$ and for every $n$. Now $\lim_{n\rightarrow +\infty }(f^n(x)-f^n(x_0))=0$ thus $\lim_{n\rightarrow +\infty }f^n(x)=x_0$ hence $f^n(x)\geq x_0$, contradiction.Therefore $f(x)>x$ for all $x\in (0,x_0)$.Similarly we have $f(x)<x$ for all $x\in (x_0,+\infty)$. b)Because $f(x)=f\left(\frac{1}{x}\right)$ we can suppose that $x\geq 1$ and also $f(x)\geq 2$ holds. Let $x>y\geq 1$ then $f(x)-f(y)=(x-y)\left(1-\frac{1}{xy}\right)>0$.If for some $n$ have $f^n(x)-f^n(y)>0$ then $f^{n+1}(x)-f^{n+1}(y)=\left ( f^n(x)-f^n(y) \right )\left ( 1-\frac{1}{f^n(x)f^n(y)} \right )>0$. The sequence $f^n(x)$ increases because $f^{n+1}(x)=f^n(x)+\frac{1}{f^n(x)}>f^n(x)$ and it is not bounded from above, otherwise converges to $a$ with $a=a+\frac{1}{a}$. Let $x\geq y\geq1$ then there exist an $m$ such that $f^m(y)>x$.Consequently for every $n$ \begin{align*}0\leq f^{n}(x)-f^n(y)<f^n(f^m(y))-f^n(y)=f^{n+m}(y)-f^n(y)=\sum_{k=1}^{m}\left ( f^{n+k}(y)-f^{n+k-1}(y) \right )=\sum_{k=1}^{m}\frac{1}{f^{n+k-1}(y)}\end{align*} and the last sum converges to zero as $n$ tends to $+\infty$. The function $f$ has no fixed point.If such $x_0$ exist then $f(x_0)=x_0$ so $x_0=x_0+\frac{1}{x_0}$.
04.01.2025 20:37