Relation between local lipschitz constant and global lipschitz constant

Question

Let $(X,d)$ be a metric space, $f:X \to \mathbb R$ a Lipschitz function. One can define the local lipschitz constant of $f$ as $$ \DeclareMathOperator{\lip}{lip} (\lip f) (x) = \begin{cases} 0 & \text{ if } x \text{ is isolated,}\\ \displaystyle\limsup_{y \to x} \dfrac{|f(y)-f(x)|}{d(x,y)} & \text{ otherwise}. \end{cases} $$ Let $L$ be the minimum constant s.t. $$ |f(y)-f(x)| \leq Ld(x,y), \quad \forall x,y \in X. $$ It is immediate to show that $\lip f \leq L$. I was wondering if actually it is true that $$ \sup_{x \in X} \lip f(x) = L .$$ Any idea to prove or disprove this fact?

Answer 1

$\DeclareMathOperator{\lip}{lip}$So there's a simple obstruction to this in general being true, which David Ullrich mentions in a comment, more generally if we take $X = A \sqcup B$ a disconnected metric space, then $\lip f$ only ever depends on the values of $f$ in the piece of the partition (not the same component though if the component isn't open, consider the rationals) So if we shift $f$ by different constants on $A, B$ we don't change local Lipschitz estimates but we do change global ones.

Now there are also less trivial counterexamples to the claim. Consider the set $A \subset \mathbb{R}^2$, $A = \{(\cos\theta,\sin \theta) : \theta \in [0, \pi ]\}$ with metric given by the restriction of the Euclidean metric. Now the coordinate function $\theta$ is locally Lipschitz with constant $1$, but between the two endpoints it changes by $\pi$ while the distance between them is $2$.

The essential idea behind the above construction is that even in nice spaces we may well have global metrics that are "not convex" in a sense. For a higher brow example we might consider nice distance functions on an arbitrary manifold $M$ such that they induce a Riemannian metric, and then ask if $d'$, the distance given by minimal arc lengths with respect to the Riemannian metric, is the same as $d$ (we should always have $d' \geq d$) and if not then $x \mapsto d'(x,y)$ will itself be a counterexample for at least one $x,y$.