# Doesn't this contradict the contraction mapping theorem

Consider this function $f(x) = x + \frac{1}{x}$ mapping from $[2, \infty)$ to $\mathbb{R}$. Then $[2, \infty)$ is a complete metric space. We also have $d(f(x),f(y)) < d(x,y)$. However, there is no fixed point for this function.

Since $d(f(x),f(y)) < d(x,y)$, I thought we could find $\alpha$ such that $0 < \alpha < 1$ by doing $\frac{d(f(x),f(y))}{ d(x,y) }$. The function is also mapping from a complete metric space. Why doesn't the contracting mapping theorem apply here?

The contracting mapping theorem applies if there is a constant $q \in [0,1)$ with $d(f(x),f(y)) \le q d(x,y)$. But no such $q$ exists !
Compute $d(f(x),f(y))/d(x,y)$ for $x \ne y$ and you will get $d(f(x),f(y))/d(x,y) \to 1$ for $x,y \to \infty$.
The contraction mapping theorem does not apply because its' assumptions are not met. The theorem requires the existence of a constant $C<1$ such that $$d(f(x), f(y) )<Cd(x,y)$$ Your example shows that $C=1$ won't do.
• Can't I get a $C$ by doing $d(f(x),f(y))/d(x,y)$ – user1691278 Dec 8 '17 at 6:04
• @user1691278 Sure you can find $C$. You'll get $C=1$. And that is not sufficient as I tried to point out. – Thomas Dec 8 '17 at 6:09
• @Thomas You're saying that C = 1 is the sup of $d(f(x),d(y))/d(x,y)$? – user1691278 Dec 8 '17 at 6:13