How can I find an interval where $f(x)=\frac 12(x+\frac 3x)$ is contractive mapping? I want to find out an interval where $f(x)=\frac12(x+\frac 3x)$ is contractive mapping.
How can I find this interval where $f(x)$ becomes contractive?
 A: We need a constanct $k$ with $0<k<1$ and an interval $I$ such that for arbitrary $x,y\in I$ the inequality $|f(x)-f(y)|\le k|x-y|$ be verified. Let's see.
$$\frac12|(x+\frac3x)-(y+\frac 3y)|\le k|x-y|$$ implies 
$$|(x-y)(1-\frac{3}{xy})|\le2k|x-y|\Rightarrow|\frac{x-y}{|x-y|}(1-\frac {3}{xy})|\le 2k$$
 Hence $$|(1-\frac{3}{xy})|\le 2k\iff-2k\le1-\frac {3}{xy}\le2k$$
 It follows $$\frac{3}{2k+1}\le xy\le\frac{3}{1-2k}$$ 
For all $k$ such that $1-2k\ge0$, i. e. $0\le k\le\frac12$ we can choose as endpoints of the searched interval $I$ the numbers $\sqrt {\frac {3}{2k+1}}$ and $\sqrt{\frac {3}{1-2k}}$ in order to ensure that if $x,y\in I$ then $\frac {3}{2k+1}\le xy \le \frac  {3}{1-2k}$.
Thus, for instance, $f(x)=\frac 12(x+\frac{3}{x})$ is contractive on $I=[\sqrt{\frac{15}{7}}, \sqrt 5]$ with  Lipschitz constant $k=\frac15$. 
A: We know by mean value theorem that, $|f(x)-f(y)|\leq |f'||x-y|$, where $|f'|=\sup_x|f'(x)|$. Note that $f'(x)=\frac{1}{2}-\frac{3}{2x^2}$, and if $x\geq \frac{\sqrt{6}}{2}$ we have $|f'|\leq \frac{1}{2}$. In particular, $f$ is a contraction on $[\frac{\sqrt{6}}{2},+\infty)$.
