Sachchidanand Prasad

04-05-2023

Problem: Consider the map \[ f:[1,\infty ) \to [1,\infty ),~ x \mapsto x + \frac{1}{x}. \] Show that \[ \left\vert f(x) - f(y) \right\vert \lt \left\vert x - y \right\vert,~\text{for all } x \neq y. \] Does $f$ have any fixed points? Why does not it contradict to the contraction mapping theorem?

Solution: Note that for any $x\neq y$, \begin{align*} \left\vert f(x) - f(y) \right\vert & = \left\vert x+\frac{1}{x} - y -\frac{1}{y} \right\vert \\[1ex] & = \left\vert x-y + \frac{1}{x} - \frac{1}{y} \right\vert \\[1ex] & = \left\vert (x-y) + \left( \frac{y-x}{xy} \right) \right\vert \\[1ex] & = \left\vert (x-y)\left( 1-\frac{1}{xy} \right) \right\vert \\[1ex] & = \vert x-y \vert \left\vert 1 - \frac{1}{xy} \right\vert. \end{align*} As $x\neq y$ and $x,y\geq 1$ so we have \[ xy \gt 1 \implies 1-\frac{1}{xy} \lt 1. \] Hence, we have \[ \vert f(x) - f(y) \vert = \vert x-y \vert \left\vert 1 - \frac{1}{xy} \right\vert lt \vert x-y \vert . \]

Observe that $f$ does not have a fixed point. If \[ f(x) = x \implies x + \frac{1}{x} = x \implies \frac{1}{x} = 0, \] not possible. Hence $f$ does not have a fixed point.

We recall the contraction mapping theorem:

Let $\left( X,d_X \right) $ and $\left( Y,d_Y \right) $ be two metric spaces. Let $f:\left( X,d_x \right) \to \left( Y,d_Y \right) $ be a continuous function. Suppose that there exists a real number $k$ with $0\leq k \lt 1$ and for every distinct points $x,y\in X$ we have \[ d_Y\left( f(x), f(y) \right) \leq k d_X(x,y). \] Then $f$ has a fixed point.

Note that in the theorem $k \lt 1$, but we showed that the $|f(x)-f(y)| \lt \vert x-y \vert $, which does not satisfy the hypothesis of the contraction mapping theorem and hence, it does not contradict the contraction mapping theorem.