Prove that $f(x)=x+1/x$ is a contraction but not a strong one on $[1, \infty]$.

Question

Question: Prove that $x+1/x$ is a contraction but not a strong one on $[1, \infty]$.

Attempt: $|(x+1/x) -(y+1/y)|= |(x-y) + (\frac {1}{x} - \frac {1}{y} | \le |x-y| + \frac {|x-y|}{|xy|} \le 2 |x-y|$.

This is not even a contraction.

Could someone please point out a way. Thanks a lot.

Answer 1

You have that

\begin{align} \lvert f(x)-f(y) \rvert &= \left \lvert (x-y) \left(1-\frac{1}{xy} \right) \right \rvert \\ &= \lvert x-y \rvert \left \lvert 1-\frac{1}{xy} \right \rvert \end{align} Now, since $xy \geq 1$ you have that

$$\left \lvert 1-\frac{1}{xy} \right \rvert \leq 1$$

and the thesis follows.

Answer 2

$f'(x)=1-\frac 1 {x^{2}} \in [0,1)$ so MVT theorem shows that $f$ is a contraction.

To show that $f$ is not strong contraction (prove by contradiction and) use the points $x=n$ and $y=n+\frac 1 n$. I will leave the details to you.