Lipschitz Continuous Diffeomorphism Denote the real line by $\mathcal{R}$. I look for non-trivial examples of functions $f:\mathcal{R}\longrightarrow\mathcal{R}$ that are Lipschitz continuous, differentiable and with non-vanishing derivative. 
By trivial examples I mean the ones of the form $f(x):=ax+b$ for $a,b$ real numbers. 
 A: If you need
$f:\Bbb R \to \Bbb R \tag 1$
to be a diffeomorphism 'twixt $\Bbb R$ and itself, you can take
$f(x) = x^3 + x; \tag 2$
we have
$f'(x) = 3x^2 + 1 > 0 \tag 3$
for all $x \in \Bbb R$; this allows us to conclude that $f$ is injective; $f(x)$ is also surjective; since
$f(x) \to \infty \; \text{as} \; x \to \infty, \tag 4$
and
$f(x) \to -\infty \; \text{as} \; x \to -\infty, \tag 5$
for any $y \in \Bbb R$ we can find $M$ such that
$y \in [f(-M), f(M)]; \tag 6$
thus $f(x)$ is onto $\Bbb R$ as well.
Since $f(x)$ is differentiable, it is locally Lipschitz:  if $x \in I_L = (-L, L)$ where $L > 0$, then
$\sup_{x \in I_L} \vert f'(x) \vert \le 3L^2 + 1; \tag 7$
thus for $y, z \in I_L$ we have
$\vert f(y) - f(z) \vert = \vert \displaystyle \int_z^y f'(x) dx \vert \le \int_z^y \vert f'(x) \vert dx \le (3L^2 + 1) \vert y - z \vert, \tag 8$
which shows $f(x)$ is Lipschitz continuous on $I_L$ with constant $3L^2 + 1$.
A: Try $f(x)=\tan^{-1}x$, $x\in{\bf{R}}$, because $f'(x)=\dfrac{1}{x^{2}+1}$.
A: $f(x)= \sqrt{x^2+a}$ for $a >0$ is $K-$Lipschitz for $K=1$. Indeed, $-1 < f'(x)=\dfrac{x}{\sqrt{x^2+a}} < 1$
