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.

  • 2
    $\begingroup$ Try sketching f(x) on the domain and ask yourself what the maximum of $|f(x) - f(y)|/(x - y)|$ could be. $\endgroup$
    – Simon S
    Oct 31, 2014 at 15:38
  • 6
    $\begingroup$ $|1/x-1/y|=|x-y|/|xy|$. What is the bound of $1/|xy|$? $\endgroup$
    – Hanul Jeon
    Oct 31, 2014 at 15:39

2 Answers 2


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.


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!!!


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