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The following problem was proposed by Pietro Majer as an extension of an earlier question of mine on Lipschitz functions.

For $f$ a Lipschitz function on $\mathbb R^n$, we denote by

$$\text{Lip}(f, U) := \sup_{y,z \in U} \frac{f(y) - f(z)}{|y- z|}$$

the Lipschitz constant of $f$ on the set $U \in \mathbb R^n$.

Question: Suppose $\mathbb R^n = \bigcup_{i=1}^N A_i$ is a partition of $\mathbb R^n$ into finitely many pairwise disjoint measurable sets $A_1, \dots, A_N$. Is it true that

$$\text{Lip}(f, \mathbb R^n) = \max_{1 \leq i \leq N} \text{Lip}(f, A_i)?$$

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    $\begingroup$ Ah it's true indeed. The idea is that you can start from a in A, and go to b in B, along a chain a, x, y, z,... b, in such a way that every consecutive pair is in the ckosure of the same set. $\endgroup$
    – Pietro Majer
    Commented Jul 12 at 10:52
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    $\begingroup$ Try a covering by three closed sets $A, B, C$, and take $a \in A$ and $b \in B$. Now starting from $a$ along the segment $ [a,b]$ you meet a point $x$ which is in $A$ and in some $X\neq A$. If $X=B$ you are done with the chain $(a,x,b)$, if $X=C$, do the same on the segment $[x,b]$ starting from $b$; you'll find a point $y$ in $B$ and in some $Y\neq B$. If $Y=A$, the chain $(a,y,b)$ is ok, if $Y=C$, the chain $(a,x,y,b)$ is ok $\endgroup$
    – Pietro Majer
    Commented Jul 12 at 11:31
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    $\begingroup$ Ahh, and I think I see the argument now for countably many sets too - you want to go to the element of $\partial A_1 \cap [a, b]$ that is closest to $b$. And then repeat for the next set $A_2$ you obtain and so on, until the countably many sets are exhausted and you eventually hit $B$, so you are done. $\endgroup$
    – Nate River
    Commented Jul 12 at 11:44
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    $\begingroup$ Yes, it also works with no measurability assumption on the $A_i$ and with the constraint that $f$ is Lipschitz weakened to $f$ continuous. $\endgroup$
    – Will Sawin
    Commented Jul 12 at 13:44
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    $\begingroup$ I'd say it holds for X a length metric space. Completeness (in any sense) does not seem necessary $\endgroup$
    – Pietro Majer
    Commented Jul 12 at 19:39

1 Answer 1

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Here is a statement in a slightly more general setting (since it is for free).

Let $X$ be a length metric space, $Y$ a metric space, $f:X\to Y$ continuous, $(A_i)_{i\in \mathbb N}$ a countable covering of $X$, such that $f$ is $L$-Lipschitz on each set $A_i$. Then $f$ is $L$-Lipschitz on $X$. In other words $$\text{Lip}(f,X)=\sup_{i\in\mathbb N} \text{Lip}(f,A_i).$$

Rmk. Since $X$ is a length metric space, it is sufficient to assume $[0,1]$ as a domain of $f$. We can also assume $Y:=\mathbb R$: the case of normed-space valued functions follows from Hahn-Banach by composing with linear functionals. By the Fréchet-Kuratowski embedding this includes the case of general metric space $Y$.

Proof 1. Consider $$R:=\{(x,y)\in X\times X: |f(x)-f(y)|\le L|x-y|\}.$$ As a relation on $X$, $R$ is a closed equivalence relation, with at most countably many classes. So by Sierpiński‘s theorem it is trivial, meaning that $f$ is $L$-Lipschitz.

[edit] Also

Proof 2. Let $Z_i$ be the (countable) set of isolated points of $A_i$, and $Z:=\cup_{i\in\mathbb N}Z_i$. Every $x\in[0,1]\setminus Z$ is an accumulation point for some $A_{i_*}$, therefore $$\displaystyle D_*f(x)\le\liminf_{y\to x\atop y\in A_{i_*}}\frac{f(x)-f(y)}{x-y}\le L$$ holds for all points but a countable set, which implies (together with the same bound for $-f$) that $f$ is $L$-Lipschitz.

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  • $\begingroup$ (transitivity of the relation $xRy, yRz \implies xRz$ is less obvious if $x<z<y$ and needs a proof) $\endgroup$
    – Pietro Majer
    Commented Jul 13 at 9:03
  • $\begingroup$ Sierpinski’s theorem is very pretty - I am surprised I have not come across it before. By the way, I wonder under what conditions on the $A_i$ the sup is actually achieved by some $A_i$. $\endgroup$
    – Nate River
    Commented Jul 15 at 8:28
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    $\begingroup$ If $X$ is compact and the $A_i$ are open (or more generally if the interior of the $A_i$ still covers X), then yes, because the Lipschitz constant of $f$ can be achieved by a sequence $(x_n, y_n) $ with $d(x_n,y_n)\to0$, thus eventually in the same $A_i$. Outside this case, I guess one can find $f$ for which $ \text{Lip}(f,A_i)<\text{Lip}(f)$ for all $i$ $\endgroup$
    – Pietro Majer
    Commented Jul 15 at 10:44

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