I was thinking about Lipschitz constants and their dependence on the domain. This led me to the following question:

Let $f$ a function that is Lipschitz continuous over the domain $A\subset\mathbb{R}^{n}$ be given. Suppose $K(A)$ is the set of compact subsets of $A$. This has a topology induced by the Hausdorff metric (See here for the definition).

Let $L_{f}(B)\colon K(A)\to\mathbb{R}$ be the (best) Lipschitz constant for $f$ on the compact set $B\in K(A)$. Is this continous with respect to the standard topology of euclidean space and the Hausdorff metric topology of $K(A)$?

I suspect the answer is probably not. A function with a sufficient discontinuous derivative should break things, but I'm having trouble engineering an example.

I'm also wondering when it might be true? For $C^{\infty}$ functions? For polynomials? Maybe this is true for a certain class of bounds for $L_{f}$ (e.g bounds on the gradient) for a given class of functions $f$ (e.g $C^{1}$ functions)?

Another thought: What is the relation between $L_{f}$ and the idea of a local Lipschitz function?


For an easy counterexample to continuity of $L_f(B)$, take $$f(x) = \begin{cases} x \,\text{if $x \ge 0$} \\ 2x \,\text{if $x \le 0$} \end{cases} $$ So $L_f[0,1]=1$ and $L_f[-\epsilon,1]=2$ for all $\epsilon > 0$, whereas the Hausdorff distance between $[0,1]$ and $[-\epsilon,1]$ equals $\epsilon$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.