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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.

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

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  • $\begingroup$ 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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