Given the metric space $M := ([1, \infty), |.|)$, is the function $\phi(x) = x + \frac{1}{x}$ a contraction? I know the answer must be "no" because then $\phi$ should have a fixed point by the Banach fixed point theorem and it clearly doesn't have one.
But isn't it true that:
$$\left| \frac{\phi(x_2) - \phi(x_1)}{x_2 - x_2} \right| = \left| 1 - \frac{1}{x_1 x_2} \right| < 1$$
implying that $\phi$ is a contraction after all? Where am I going wrong here?
On
$$\left| \frac{\phi(x_2) - \phi(x_1)}{x_2 - x_2} \right| = \left| \frac{1}{x_1 x_2} \right| \le 1$$ but this does not say that $\phi$ is a contraction. For $\phi$ to be a contraction there must exist a number $c \in (0,1)$ (independent of $x_1,x_2$) such that $ | {\phi(x_2) - \phi(x_1)}| \le c|x_1-x_2|$ for all $x_1,x_2$ and there is no such $c$ as seen by taking $x_1=1,x_2=1+\frac 1 n$ and letting $n \to \infty$.
On
A contraction is a function for which a fixed number $C\in(0,1)$ exists such that $\vert f(x_2)-f(x_1)\vert\leq C\vert x_2-x_1\vert$. You showed that $\vert \phi(x_2)-\phi(x_1)\vert\leq\vert x_2-x_1\vert=1\cdot\vert x_2-x_1\vert$. You're basically claiming $C=1$, but $C$ needs to be smaller than 1. Basically, the difference quotient of a contraction needs to be bounded not by 1, but by a number smaller than 1. Which is not the case here.
But there's a fixed point anyway: $\phi(1)=2>1$, but $\phi(2)=\frac32<2$, so by the intermediate value theorem, a fixed point exists. As pointed out in the comments, it's the golden ratio, the solution of $1+\frac1x=x$, which can be transformed to $x^2-x-1=0$, the solutions of which can be explicitly calculated. One of them is in $[1,\infty)$.
The definition of a contracting map $\Phi$ is that it exists $0 \le k \lt 1$ such that
$$\Vert \Phi(x) - \Phi(y) \Vert \le k \Vert x- y \Vert$$ for any $x,y$ not that
$$\Vert \Phi(x) - \Phi(y) \Vert \lt \Vert x- y \Vert$$ for any $x,y$.
Your $\phi$ is not a contracting map as
$$\lim\limits_{x \to 1} \left\vert \frac{\phi(x) - \phi(1)}{x - 1} \right\vert = 1$$