# On functions with strict Lipschitz constant

We say a measurable subset $$S$$ of $$\mathbb R^n$$ is measure dense if for every open set $$U \subset \mathbb R^n$$, $$U \cap S$$ is of positive Lebesgue measure.

Let $$n \geq 2$$, and let $$f: \mathbb R^n \to \mathbb R$$ be a Lipschitz continuous function with strict Lipschitz constant $$L > 0$$.

That is, $$|f(x) - f(y)| < L|x - y|$$ for all $$x \neq y$$ in $$\mathbb R^n$$.

Question: Is it possible that $$|Df| = L$$ on a measure dense set?

Note: Here $$Df$$ denotes the total derivative of $$f$$, and $$|\cdot|$$ the operator norm of a linear map.

• I'd rewrite a bit. You say "Let $f$...", thus fixing $f$. Then you ask "can there exist a function $f$". Apr 7, 2022 at 0:46
• Ah true, one second. Apr 7, 2022 at 0:47

I guess it suffices to give an example for $$n = 1$$. If $$f: \mathbb{R} \to \mathbb{R}$$ is an example then $$g(x_1, \ldots, x_n) = f(x_1)$$ will be an example for any $$n \geq 1$$.
All we need is a measurable set $$A \subseteq \mathbb{R}$$ such that both $$A$$ and its complement have positive measure in every interval. See here, for example. Then define $$f(x) = \int_0^x 1_A = \begin{cases} m(A \cap [0,x])&x \geq 0\cr -m(A\cap [x,0])&x < 0 \end{cases}.$$ It should be clear that 1 is a strict Lipschitz constant, but $$f'(x) = 1$$ at every Lebesgue point of $$A$$, so the derivative is $$1$$ on a measure dense set.
• I have been working on an answer along the same lines, but your presentation is much better than the one I was going to provide (which would include a construction of the kind of measurable set $A$ that you describe.) Upvoted. Apr 7, 2022 at 4:10
• @Nik Weaver By the way, by taking a set $S$ such that both it and its complement are measure dense, and integrating the function $2(\mathbf 1_S - 1)$, we get a function $f$ with $|Df| = L$ a.e.! Apr 8, 2022 at 4:47