I'm trying to understand the difference between global lipschitz and local lipschitz.

Let $f(x)=x^2$ while $x \in \mathbb{R}$ if we look at global lipschitz, for all $M \subset \mathbb{R} \times \mathbb{R}$

Global lipschitz applies if for every $x,y \in \mathbb{R}$

$$|f(x) - f(y)| \le M|x-y|$$

in this case global lipschitz doesn't apply.

$$|f(n) - f(0)| \le M|n-0| \rightarrow n \le M$$

So we found that no constant $M$ exists, therefore $f(x)=x^2$ is doesn't hold global lipschitz condition.

If I understood correctly if function apply local lipschitz then for every subset of $\mathbb{R^2}$ an interval of $R$ for example $[1,2],(0,5)$ we can find $M$ such that $|f(x) - f(y)| \le M|x-y|$ holds

Since $f(x)=x^2$ is continuous in $\mathbb{R}$ we know that maximum exist for every interval $D$ we take $m = max\{D\}$ and define $M=m^2$ Therefore the condition holds : $$|f(x) - f(y)| \le M|x-y|$$

Is it true to say that for every continuous function lipschitz local conditon holds?

I wonder if I understand correctly the difference between local and global lipschitz conditon, I'll be happy if someone could approve.

Any help will be appreciated, Thanks.


No, not all continuous functions are locally Lipschitz. The Weierstrass function is continuous but nowhere Lipschitz.

  • $\begingroup$ Can you explain why the method I used works on $f(x)=x^2$ (hence taking the maximum in the interval). $\endgroup$ – JaVaPG Apr 24 at 20:23
  • $\begingroup$ $f(x)=x^2$ has a continuous derivative, namely $f'(x)=2x$. A continuous function always achieves a maximum and a minimum on a compact subset of its domain. Therefore, if $D$ is any bounded subset of $\mathbb{R}$, so that $\overline{D}$ is compact, there exists $M$ such that $|f'(x)| \leq M$ on $D$. This in particular implies $f$ is Lipschitz with Lipschitz constant $M$ (if not, this would violate the mean value theorem). $\endgroup$ – kccu Apr 24 at 22:20

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