Proving that a function is local but not global Lipschitz continuous

Question

How would one go about proving that e.g. the function $f(x)=x^{2}$ is locally Lipschitz-continuous but not globally?

Answer 1

First show that if $M > 0$ and $|x|,|y| \le M$, then the expression $\dfrac{|f(x) - f(y)|}{|x-y|}$ is bounded. Hint: it does not exceed $2M$.

Next show that expressions of the form $\dfrac{|f(x) - f(y)|}{|x-y|}$ are unbounded for $x,y \in \mathbb R$. Hint: use $y=2x$.

Answer 2

$f$ is not globally lipschitz since you can check that $$ \lim_{y\to \infty}\frac{|f(x)-f(y)|}{|x-y|^\alpha}=+\infty $$ for some fixed $x\in \mathbb{R}$. Therefore, $$ \sup_{x,y\in \mathbb{R}}\frac{|f(x)-f(y)|}{|x-y|^\alpha}=+\infty. $$ You can also check that $$ \lim_{y\to x}\frac{|f(x)-f(y)|}{|x-y|^\alpha}<\infty. $$ How do you conclude from here that $f$ is locally lipschitz?