Show 1/x is Lipschitz continuous Show that $\space f:(c,∞)\rightarrow\mathbb{R}$ for some $c>0$, and defined by $f(x)= \frac{1}{x}$, is Lipschitz continuous. 
I'm not quite certain how to go about this.
 A: We must show that there exists a constant $C > 0$ such that for all $x, y \in (c, \infty)$ 
$|f(x) - f(y)| \leq C|x-y|$.
Now, $|f(x) - f(y)| = |\frac{1}{x} - \frac{1}{y}| = |\frac{|y-x|}{xy}| = \frac{|x-y|}{|xy|} \leq \frac{1}{c^2}|x-y|$ as needed.
A: Differentiable functions such as $f(x) = x^{-1}$ are always Lipschitz continuous, at least locally.  This notion works itself out in what follows:
$f(x)$ is continuously differentiable on $(c, \infty)$, and we have $\vert f´(x) \vert = \vert -x^{-2} \vert < c^{-2}$ for all $x \in (c, \infty)$.  Thus for $x_1, x_2 \in (c, \infty)$,
$\vert f(x_2) - f(x_1) \vert = \vert \int_{x_1}^{x_2} f'(s) ds \vert  = \vert (\vert \int_{x_1}^{x_2} f'(s) ds \vert) \vert$
$\le \vert  \int_{x_1}^{x_2} \vert f'(s) \vert ds \vert \le \vert \int_{x_1}^{x_2} c^{-2} ds \vert = c^{-2} \vert \int_{x_1}^{x_2} ds \vert = c^{-2}\vert x_2 - x_1 \vert; \tag{1}$
this shows that $c^{-2}$ is in fact a global Lipschitz constant for $f(x) = x^{-1}$ on $(c, \infty)$; (1) applies for all $x_1, x_2 \in (c, \infty)$.This choice of Lipschitz constant is in fact sharp; it is easy to see, by taking $x_1$, $x_2$ sufficiently close to $c$, that no lesser positive constant will suffice; this since $c^{-2}$ is the supremum of the values $\vert f'(x) \vert$ may take on $(c, \infty)$.
Hope this helps.  Cheerio,
and as ever,
Fiat Lux!!!
