Why is $f(x) = x + \frac{1}{x}$ a mapping contraction? Why is $f(x) = x + \frac{1}{x}$ a mapping contraction? The metric space in question is $[1,\infty)$. Also, if this were a contraction, wouldn't it have a fixed point by Banach's theorem?
It looks to me like it's not a contraction because for example if I take $x=5$ and $x'=6$:
$$d\left(5+\frac{1}{5},6+\frac{1}{6}\right) > 1 = d(5,6)$$
Would appreciate any help.
 A: $$d(f(x), f(y))= \left|x+\frac{1}{x}-y-\frac{1}{y} \right| = \left|x-y-\frac{x-y}{xy} \right| =\left|x-y \right|\cdot \left| 1-\frac{1}{xy} \right|< |x-y|$$
Note that to have a fixed point you need a stronger inequality:
$$d(f(x), f(y)) < C d(x,y)$$
with $0<C<1$. In this case, you can prove that such $C$ cannot exist by letting $x,y$ go to $\infty$.
A: Note that $$f'(x)=1-\frac{1}{x^2}<1$$ and therefore $|f'(x)|<1$ for all $x \in [1,+\infty)$. By the Mean Value Theorem, for any two points $x, y \in I$ there exists
a point $c$ between $x$ and $y$ such that
$$d(f(x), f(y)) = |f(x) − f(y)| = |f'(c)(x − y)| = |f'(c)|d(x, y) <1 \cdot d(x, y)$$ so $f$ is a contraction. 

However as $x\to \infty$ you have that $f'(x)\to 1$ and therefore you cannot find a constant $C<1$ so that $|f'(x)|\le C$ for all $x \in \mathbb R$. Existance of such a constant is a necessary condition (for a continuous differentiable function, such as $f$) in order to apply the Fixed Point Theorem, so there is no contradiction in this case.  
But you could think then, that by bounding the domain of $f$, say $I=[1,N]$ for a big $N$ instead of $[1, \infty]$ then you can find such a constant and then you can apply the Fixed Point Theorem to find a fixed point. However observe that $f:[1,N]\mapsto[2,N+1/N]$ so now is violated the condition that $f: I \mapsto I$ (which is the other necessary condition to apply the Fixed Point Theorem).
