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I was recently looking again at functions like the Cantor staircase, the modified Dirichlet, etc., and something occurred to me. The modified Dirichlet is interesting because it's continuous almost everywhere (i.e. continuous on $[0, 1] \setminus \mathbb{Q}$), but discontinuous on a dense but countable set (i.e. $[0, 1] \cap \mathbb{Q}$). Now $\mathbb{Q}$ is large in the sense of being dense, but is small in most other topological notions (with which I'm familiar), e.g. it's meager, as well as countable, and by countability it has null measure and Hausdorff dimension $0$. That is, the set of discontinuity is quite small in most senses.

Similarly, the set of non-differentiable points of the Cantor staircase is small in the sense of being nowhere-dense, as well as null, though it does have one "size" over $\mathbb{Q}$ in that it has positive Hausdorff dimension. But other than dimension, the set of non-differentiability is fairly small.

My question is this: I'm aware that there are functions like Weierstrass which are always continuous and differentiable nowhere, but what I am not aware of are functions which are continuous everywhere but only differentiable on some "small" set. So my questions are as follows:

(a) Suppose $F \subseteq [0, 1]$ is a set of Hausdorff dimension $s < 1$. Does there exist a map $f: [0, 1] \to \mathbb{R}$ that is differentiable exactly on $F$?

(b) Suppose $F \subseteq [0, 1]$ is a set of Hausdorff dimension $1$ with Lebesgue measure $0$. Does there exist a map $f: [0, 1] \to \mathbb{R}$ that is differentiable exactly on $F$?

(c) Suppose $F \subseteq [0, 1]^{n}$ is a set of Hausdorff dimension $s < n$. Does there exist a map $f: [0, 1]^{n} \to \mathbb{R}^{m}$ that is differentiable exactly on $F$?

(d) Suppose $F \subseteq [0, 1]^{n}$ is a set of Hausdorff dimension $n$ with $n$-dimensional Lebesgue measure $0$. Does there exist a map $f: [0, 1]^{n} \to \mathbb{R}^{m}$ that is differentiable exactly on $F$?

Thanks.

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Just being small is not quite enough. The set of nondifferentiability of a continuous function $f$ is necessarily a $G_{\delta\sigma}$ set. Indeed, for every $r\in\mathbb{R}$ and every $n\in\mathbb{N}$ the sets $$ L_{r,n} = \left\{x: \exists y\text{ such that }|x-y|<\frac1n \text{ and } \frac{f(y)-f(x)}{y-x}<r \right\}$$ and $$ U_{r,n} = \left\{x: \exists y\text{ such that }|x-y|<\frac1n \text{ and } \frac{f(y)-f(x)}{y-x}>r \right\}$$ are open. The set of nondifferentiability is $$\bigcup_{r<s, \ r,s\in\mathbb{Q}}\bigcap_{n=1}^\infty (L_{r,n}\cap U_{s,n})$$ which is the countably union of countable intersections of open sets.

Similar argument works in higher dimensions: see Continuous functions are differentiable on a measurable set? on this site. Thus, in all (a)-(d) we should assume that the set is $G_{\delta\sigma}$.

Then the answer to (a) and (b) is "yes". For a construction, I refer to a more general result in

Also relevant: Nondifferentiability set of an arbitrary real function on MathOverflow


Things may get more involved in higher dimensions. They certainly get much more involved for Lipschitz functions. For example, in $\mathbb{R}^2$ there is a compact set $K$ of Hausdorff dimension $1$ such that every Lipschitz function $f:\mathbb{R}^2\to\mathbb{R}$ is differentiable at some point of $K$. See

  • M. Dore and O. Maleva, A compact universal differentiability set with Hausdorff dimension one, Israel Journal of Mathematics, 191 (2012), no. 2, 889–900. Also on ArXiv:1004.2151

These slides of David Preiss give an overview of the subject. In a nutshell: for Lipschitz functions, the answer to (c)-(d) is negative for $m<n$ and positive for $m\ge n$.


But you didn't assume the map to be Lipschitz... I think that in this case, every $G_{\delta\sigma}$ set of zero measure is a set of nondifferentiability of some continuous map, but don't have either a proof or a reference. I suggest looking up the papers that cite Zahorski or Piranian.

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  • $\begingroup$ What is $m$ in your answer? $\endgroup$
    – AJY
    Commented Aug 15, 2016 at 17:09

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