How to decide the neighbourhood when proving a function is locally Lipschitz-continuous.

Question

I have come across a problem in one of my textbooks that I understand but could not replicate.

Show that the function $f : \mathbb{R}^+ → \mathbb{R}$ where $f(x) = {1\over x}$ is locally Lipschitz-continuous in $\mathbb{R}^+ = (0, \infty)$.

$$|f(y) − f(z)| = |{1\over y}−{1\over z}|={|y − z|\over yz}≤{|y − z|\over (x/2)^2}$$

where it chooses the neighbourhood of U $\subset$ $\mathbb{R}^+$ as U =$ ({x\over 2}, 2x)$ proving that it is locally Lipschitz-continuous in $\mathbb{R}^+ = (0, \infty)$ with $L = {4\over x^2}$

I am very confused on how they choose this value of U, if anyone could explain that would be amazing.

Answer 1

There's nothing deep. If $y, z\ge \frac x 2>0$ then $yz\ge (x/2)^2$ and so $$ \left| \frac 1 y -\frac 1 z\right| \le \frac{| y-z| }{(x/2)^2},\qquad \forall y, z\in [\tfrac x2, \infty).$$ The upper bound of $2x$ is not needed for this computation.

Answer 2

Fact: If the derivative of a function is bounded over an interval, then the function is Lip on that interval with bound equal to the supremum of the derivatie there.

Derivative of $1/x$ is $1/x^2$ which is bounded from above as long as we're away from zero. So, in fact $1/x$ is Lipschitz over any $[\delta, +\infty)$, $\delta >0$.