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Let $f: \mathbb R^n \to \mathbb R$ be a Lipschitz function with $\nabla f$ nonzero almost everywhere with respect to Lebesgue measure.

What is the supremal Hausdorff dimension of the set on which $f$ is differentiable with $\nabla f = 0$?

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2 Answers 2

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Checking this in perfect detail might be a bit technical, but the following looks like it could work for any dimension up to n:

For 1d: Take an S curve $\phi: [0,1] \to [0,1]$ such that $\phi(0) = 0$, $\phi(1) = 1$ and $\phi'(0) = \phi'(1) = 0$ and nonzero derivative elsewhere. Now for a (non-fat) Cantor-set C on any dimension, replace the removed intervals with a rescaled copy, lined up to be continuous. Specifically scale it in such a way that replacing an interval of length $l$ is done by a shifted version of $l^2 \phi(x/l)$. Call the resulting function $u$. Then $u' \neq 0$ on the interior of all removed intervals, which is a set of full measure.

What is left to show is that $u' = 0$ on the Cantor-set. Call $u_k$ the approximating function, where only intervals of order up to $k$ are replaced. Certainly $u_k' = 0$ on C. Additionally, as the maximum derivative gets exponentially smaller the smaller the intervals get, we should have $u_k \to u$ in $C^1$, so the same has to hold for the limit.

For n dimensions, just do the same in one direction and keep it constant in the others.

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  • $\begingroup$ Maybe as an additional afterthought, if $\phi$ is chosen in such a way that all derivatives are zero at the ends and the prefactor $l^2$ is replaced with something exponential, then I think that $u$ should even be smooth by the same argument. $\endgroup$
    – mlk
    Commented Jun 21 at 8:25
  • $\begingroup$ I think this checks out, nice. So this Cantor set would have to have to have full Hausdorff dimension, but I don’t think that affects the argument. $\endgroup$
    – Nate River
    Commented Jun 21 at 8:29
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    $\begingroup$ @NateRiver The construction should work with any Cantor set, even with the fat ones. Only with the latter the set of zero derivative will obviously have non-zero measure. As there is a Cantor-set for any dimension below $1$, that proves the supremum to be $n$. I guess there should also be a "skimmed" fat-Cantor set of full dimension but zero measure, but I leave that as an exercise for the reader. $\endgroup$
    – mlk
    Commented Jun 21 at 8:38
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    $\begingroup$ @NateRiver In general the set $\{f =0\}$ for a smooth function can be quite ugly. This is kind of the same, only for the derivative. I guess if you do the same in each component and play around a bit, you can even find a bijection $f: \mathbb{R}^n \to \mathbb{R}^n$ that has $\nabla f = 0$ on a set of dimension $n$, which also indirectly links this to half of all the unsolved open problems in nonlinear elasticity. $\endgroup$
    – mlk
    Commented Jun 21 at 8:44
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    $\begingroup$ @NateRiver In non-linear elasticity one wants to study deformations which are (almost everywhere) bijections. In a variational context if $\det \nabla f >0$ uniformly, then any small enough smooth variation gives me again a (local) bijection. Isolated points where this is not true, one could cut out and treat seperately. But the example mentioned in my last comment would have $\det \nabla f = 0$ on a dense set. $\endgroup$
    – mlk
    Commented Jun 21 at 9:00
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This example is essentially the same in mlk’s answer, so I just give it as a variation on their construction for $n=1$.

The zero-set of a $C^\infty$ function $g: \mathbb R \to \mathbb R$ can be every closed set $C\subset \mathbb R$ (a well-known result by Whitney in $\mathbb R^n$, which is easy for $n=1$; if we are OK with a Lipschitz function, we can just take $g(x):=\text{dist}_C(x) $). A primitive $f$ of $g$ is a differentiable function such that $\{f’=0\}=C$.

So, since $C$ can have zero Lebesgue measure and any Hausdorff dimension less than $1$, one concludes as in mlk’s answer.

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  • $\begingroup$ Nice! I will ask this as a question later, but I am thinking about the following now - what if we require instead $|\nabla f| = 1$ a.e. (and $f$ Lipschitz)? In one dimension this reduces to the following problem: Let $A \subset \mathbb R$ be measurable. What is the maximal Hausdorff dimension of the set at which both the upper and lower density of $A$ are equal to $\frac{1}{2}$? $\endgroup$
    – Nate River
    Commented Jun 21 at 9:07
  • $\begingroup$ Do you mean the same question as in the original post but now with $|\nabla f|=1$ a.e.? $\endgroup$
    – Pietro Majer
    Commented Jun 21 at 9:39
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    $\begingroup$ Indeed, and $f$ still Lipschitz. $\endgroup$
    – Nate River
    Commented Jun 21 at 9:43

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