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(D1) A function $f$ is called Lipschitz continuous on a set $A$ if there exists $K \ge 0$ such that, for $x, y \in A$,

$|f(x_1) - f(x_2)| \le K |x_1 - x_2|$

(D2) A function is called locally Lipschitz continuous on $A$ if for each $x \in A$, there is a neighborhood $U_x$ such that $f$ is Lipschitz on $U_x \cap A$.

For my application, I need a condition in between the two:

(D3?) A function is uniformy locally Lipschitz continuous on $A$ if there is a $K \ge 0$ and a $\delta >0$ such that for $x, y \in A$, $|x_1 - x_2| < \delta$ implies

$|f(x_1) - f(x_2)| \le K |x_1 - x_2|$

This definition is especially useful for non-convex sets, in which case we may have $D3$ true but $D1$ untrue.

Is there anything of the sort in the literature?

Guillaume F.
  • 854
  • Note that what is more important if you want D3 true but D1 false is to have $A$ unbounded. If $A$ is in $B(0,M)$ for some $M$, given $K$ in D3 for $\delta=1$, $K+M$ is a proper Lipschitz constant (in D1 sense). – charmd May 26 '18 at 08:49
  • Conversely, if $A$ is unbounded, you can have D3 but not D1: consider $f : x \in \mathbb{R} \mapsto x^2$ – charmd May 26 '18 at 08:50

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