Lipschitz continuity of $\frac{1}{x}$ and $x^2$

Question

Show that function $f(x)=\frac{1}{x}$ fulfills lipschitz continuity on all rays $(\epsilon, + \infty), \epsilon>0$ whereas does not fulfill lipschitz continuity on ray $(0, +\infty)$

Show that function $f(x)=x^2$ fulfills lipschitz continuity on all bounded intervals whereas does not fulfill on any ray.

How to approach those two more or less simmilar questions as i have utterly no idea.

Answer 1

Almost alawys to recall the definition is a good way to start. So:

A function $f$ is Lipschitz continuous on a set $S$ if there exists some constant $C_S$ such that $|f(x)-f(y)|\le C_S |x-y|$ for all $x, y \in S$.

So let us look at $|f(x)- f(y)|$ for $f(x)= x^2$. We have $|x^2 - y^2| = |x+y||x-y|$.

Now if we consider this on a bounded interval $I$ then we can bound $|x+y|$ for all $x,y\in I$ by some constant $C$ and we are done. By contrast if the set of definition is not bounded, so for example if it is a ray, we cannot find a bound for $|x+y|$ that holds for all $x,y$ in that set.

Now, for the other function
$$| \frac{1}{x} - \frac{1}{y}| = |x-y|\frac{1}{|xy|}$$ On $(\epsilon, \infty)$ one can say $\frac{1}{|xy|} < \epsilon^{-2}$ and setting $C= \epsilon^{-2}$ we are done.

On the other hand on $(0, \infty)$ the quantity $\frac{1}{|xy|}$ is not bounded.