Is there any difference in a function being locally Lipschitz on $\mathbb{R^n}$ and being globally Lipschitz?

  • $\begingroup$ Yes, for example every continuously differentiable function is locally Lipschitz. $\endgroup$ Jun 21, 2014 at 23:20
  • $\begingroup$ But say if the function was coninuously differentiable on the entirety of $\mathbb{R^n}$, would it then be globally lipschitz? $\endgroup$ Jun 22, 2014 at 9:50
  • 2
    $\begingroup$ Only if the derivative is globally bounded. Generally a globally continuously differentiable function has unbounded derivative, so it will be only locally Lipschitz (the derivative is bounded on every bounded subset of $\mathbb{R}^n$ then). $\endgroup$ Jun 22, 2014 at 11:17

1 Answer 1


A function $f$ is locally Lipschitz if for each compact subset $K \subset \mathbf{R}^n$, there is a constant $M = M_K$ such that $$|f(x) - f(y)| \le M|x-y|$$ for all $x, y \in K$. The Lipschitz constant $M$ can in general depend on the subset $K$; in particular, it could get worse as we take $K$ larger and larger. Consider for example the function $f(x) = x^2$ on $\mathbf{R}$.

We say $f$ is globally Lipschitz if there is a constant $M$ such that $$|f(x) - f(y)| \le M|x-y|$$ for all $x, y \in \mathbf{R}^n$. Equivalently, $f$ is globally Lipschitz if $f$ is locally Lipschitz but it has a Lipschitz constant $M$ that works for all compact sets $K$.

  • $\begingroup$ This definition of local Lipschitz in your answer is not correct. Here is the correct definition that I am using from Khalil's "Nonlinear Systems" book. A function $f(x)$ is said to be locally Lipschitz on a domain (open and connected set) $D\subseteq \mathbb{R}^n$ if each point of $D$ has a neighborhood $D_0$ such that $f$ satisfies the Lipschitz condition $|f(x)-f(y)|\leq L|x-y|$ for all points in $D_0$ with some Lipschitz constant $L_0$. $\endgroup$ Feb 13, 2023 at 20:39

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