Suppose $f: [1, \infty) \to [1, \infty]$ defined by $f(x) = x + \frac{1}{x}$ for all $x \geq 1$. I want to prove that:
\begin{equation}
|f(x)-f(y)| < |x-y|
\end{equation}
except when $x=y$, but $f$ does not have a fixed point.
By the Banach fixed point theorem we know that if a function $f: X \to X$ is a contraction of a complete metric space, then $f$ has a unique fixed point $p$ and the sequence of $(f, f \circ f, f\circ f\circ f, …)$ that is the sequence of $f$ composed with itself $n$ times at index $n$ converges to p for all $x$.
But $[1, \infty]$ is not a complete metric space. So it seems like a good idea to proceed via contradiction? Where can I go from here
All help is greatly appreciated
Best Answer
Use $$ f(x)-f(y) = x - y + \frac{1}{x} - \frac{1}{y} = (x-y) \left(1 - \frac{1}{x y}\right) $$ Thus, for $x\not= y$ $$ \left| f(x)-f(y) \right| = \left|x-y\right| \left(1 - \frac{1}{x y}\right) < \left|x-y\right| $$ since $1 - \frac{1}{x y} < 1$ for all $x>1$ and $y>1$