0
$\begingroup$

We assume that $X$ is a metric space and that $A \subseteq X$ is a subset. Let $f : A \rightarrow \mathbb R$ be a Lipschitz-continuous function with Lipschitz constant $L$.

By the Kirszbraun theorem, we can extend $f$ to the whole of $X$ without increasing the Lipschitz constant: we let $F : X \rightarrow \mathbb R$ be

$F(x) = \inf_{ a \in A } \Big\{\; f(a) + L \cdot d( a, x ) \;\Big\}$.

The Lipschitz constants of $f$ and $F$ are the same but the range of $F$ is generally larger than the range of $f$.

Can we construct an extension of $f$ to all of $X$ such that not only the Lipschitz constant but also the convex hull of the range remains the same.

Improve this question
$\endgroup$
8
  • $\begingroup$ Not always, take $X=[0,1]$, $A=\{0,1\}$, $f(x)=x$ on $A$. $\endgroup$
    – Fedor Petrov
    Commented May 16, 2017 at 12:02
  • $\begingroup$ @FedorPetrov: Thank you for pointing out that example. I have edited the question: the range is convex in my application, which rules out your example. $\endgroup$
    – shuhalo
    Commented May 16, 2017 at 12:14
  • $\begingroup$ Still no. $X=[0,1]$, $A=[0,1)$, $f(x)=x$ on $A$. $\endgroup$
    – Fedor Petrov
    Commented May 16, 2017 at 12:33
  • $\begingroup$ The extension can be constructed so that the closure of convex hull of the range remains. That is about maximum one can achieve. $\endgroup$
    – Anton Petrunin
    Commented May 16, 2017 at 17:11
  • $\begingroup$ Btw, the quoted result is not Kirszbraun theorem (that deals with extending Hilbert-valued map defined on subsets of the Hilbert space). $\endgroup$
    – Pietro Majer
    Commented May 17, 2017 at 7:05

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .