Is $f(x)=\sqrt{(1+x^2)}$, $x \in \mathbb{R}$ a contraction mapping?

Question

We take $x, y \in \mathbb{R}$.

Then I say :

$\mid \sqrt{(1+x^2)} - \sqrt{(1+y^2)} \mid$ $\le$ $\mid 1+\sqrt{x^2} - 1-\sqrt{y^2} \mid$

$\Leftrightarrow$ $\mid \sqrt{(1+x^2)} - \sqrt{(1+y^2)} \mid$ $\le$ $\mid \sqrt{x^2} - \sqrt{y^2} \mid$

$\Leftrightarrow$ $\mid \sqrt{(1+x^2)} - \sqrt{(1+y^2)} \mid$ $\le$ $\mid \mid{x} \mid - \mid{y} \mid \mid$

$\Leftrightarrow$ $\mid \sqrt{(1+x^2)} - \sqrt{(1+y^2)} \mid$ $\le$ $\mid \mid{x} \mid - \mid{y} \mid \mid$ $\le$ $\mid x-y \mid$

Which means that $k=1$?

So $f$ is not a contraction mapping. Does it work ? Thanks in advance.

Answer 1

If it were a contraction mapping, it would have a fixed point. What are the solutions of $\sqrt{1+x^2}=x$?

Answer 2

You can also use $\frac{|\sqrt{1+x^2}-1|}{|x-0|}=\frac{|x|}{|\sqrt{1+x^2}+1|}\to 1$ as $x \to \infty$ to argue there doesn't exist $0\leq k<1$ satisfying the requirement.

Answer 3

It's not too hard to show that $$\lim_{n\to\infty} (f(n+1)-f(n))=1$$

Thus there can be no $k<1$ such that $|f(x)-f(y)|\le k|x-y|$ for all $x,y\in\mathbb{R}$.