Show that $f(C)$ has Hausdorff dimension at most zero.

Question

We say that a set $A \subset \mathbb{R}^n$ has $d$-dimensional Hausdorff measure zero if for all $\epsilon > 0$ there exists a covering of $A$ by countably many cubes $S_i$ with side lengths $s_i$ such that $\sum_i (s_i)^d < \epsilon$. The Hausdorff dimension of $A$ is then defined to be the infimum over $d \in \mathbb{R}$ such that $A$ has $d$-dimensional Hausdorff measure zero.

Suppose $f: \mathbb{R}^1 \rightarrow \mathbb{R}^1$. Let $C$ be the set of critical points of $f$. Show that $f(C)$ has Hausdorff dimension at most zero.

I know that by Sard's theorem, $f(C)$ has measure zero. But then I got stuck by proceeding to the "at most" part. Any hint will be helpful - thank you!

Answer 1

I guess that your source was Wikipedia article on Sard's theorem which states at the end

In 1965 Sard further generalized his theorem to state that if $f:M\rightarrow N$ is $C^k$ for $k\geq \max\{n-m+1, 1\}$ and if $A_r\subseteq M$ is the set of points $x\in M$ such that $df_x$ has rank less than or equal to $r$, then $f(A_r)$ has Hausdorff dimension at most $r$.

Specializing to $m=n=1$ and $r=0$, we get the statement that the set of critical values of a $C^1$ function $f:\mathbb R\to\mathbb R$ has Hausdorff dimension $0$. (By the way, it looks strange when you write "at most $0$". The Hausdorff dimension is never negative, even for the empty set, according to its common definition.)

The above statement is false. Here is a counterexample.

Let $C$ be the ternary Cantor set. Define $$f(x)=\int_0^x \operatorname{dist}(t,C)\,dt\tag1$$ where as usual, $\operatorname{dist}(t,C)=\inf\{|t-y|:y\in C\}$. You can visualize the graph of the function $t\mapsto \operatorname{dist}(t,C)$ as a collection of right isosceles triangles built on the gaps of $C$ (plus some uninteresting lines left and right of the set). Since $t\mapsto \operatorname{dist}(t,C)$ is continuous, $f\in C^1(\mathbb R)$.

Clearly, $f'(x)=0$ for every $x\in C$. I claim that $$ |f(x)-f(y)| \ge \frac{1}{36}|x-y|^2,\quad x,y\in C \tag2 $$ Assuming (2) for now, we see that $f$ is injective on $C$, and that the inverse map $f^{-1}:f(C)\to C$ is Hölder continuous with exponent $1/2$. Recall that $\alpha$-Hölder maps do not increase the Hausdorff dimension by more than the factor of $1/\alpha$. Therefore, $\dim C\le 2\dim f(C)$; put another way, $$\dim f(C)\ge \frac12 \frac{\log2}{\log 3} \tag3$$

It remains to prove (2). Take $x,y\in C$, $x<y$. Let $k$ be the smallest integer for which there is a gap of size $3^{-k}$ between $x$ and $y$. Then $x$ and $y$ are in the same component of the $(k-1)$ generation of pre-Cantor sets. Hence $y-x\le 3^{1-k}$.

Recalling the triangular shape of graph of $t\mapsto \operatorname{dist}(t,C)$ on each gap, we find that the area of the triangle is $\frac14 3^{-2k}$. Thus, $$ f(y)-f(x)=\int_x^y \operatorname{dist}(t,C)\,dt \ge \frac14 3^{-2k} \ge \frac{1}{36}(y-x)^2 $$ as claimed in (2). $\quad \Box$

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It is instructive to note that $f\notin C^2$ in this example. To make it $C^2$ smooth, I would have to replace triangles with something smoother, and also scale them differently in the vertical direction. The resulting $C^2$ map would still satisfy something like (2) but with a different power of $|x-y|$. Same goes for $C^k$ for all finite $k$; the larger $k$, the larger the exponent in (2), the smaller the lower bound in (3). But the bound will still be positive. Only for $C^\infty$ functions the counterexample disappears.

Now we are ready to appreciate the actual result that Sard proved in 1965, which was misquoted in Wikipedia. I quote from page 162 of Hausdorff Measure of Critical Images on Banach Manifolds, taking into account the errata. (Sadly, both links are paywalled.)

Define the function $c$ as follows: $$c(k,\rho )= \begin{cases} 1 &\text{ if }\ k<\rho, \\ 2+k(k-\rho)/\rho &\text{ if }\ 0<\rho\le k, \end{cases} \qquad k=0,1, \dots, \ \rho>0$$ Suppose that $f$ is a $C^q$-map of a $C^q$-manifold $M$ of dimension $m <\infty$ with countable basis into a $C^q$-Banach manifold $N$, where $1 \le q \le\infty$. Let $A_r$ denote the set of points of $M$ of rank $\le r$, $r= 0,1, \dots,m $. Put $k=m - r$.

Theorem 1. Suppose that $\rho > 0$. Then $f(A_r)$ is $(r + \rho)$-null if $$q \ge c(k, \rho).$$

In order to conclude $\dim f(A_0)=0$ for a map $f:\mathbb R\to\mathbb R$, we must be able to use arbitrarily small $\rho>0$ above, and this requires $q=\infty$. On page 169 Sard spells this out as a corollary:

Corollary. If $f$ is a $C^\infty$-map, then $f(A_r)$ is $(r+ \rho)$-null for all $\rho > O$.

and adds:

This corollary, at least in the case in which $N$ is finite dimensional, is due to Dubovickii [8].

Indeed, А. Я. Дубовицкий published this result in 1962. His paper is in free access here.

Answer 2

Here are my thoughts ("solution"), but I am very not confident with it. Anyone can point out some idea here would be greatly appreciated. Thanks!

Again without loss of generality, we assume $A$ is a cube with length $a$. So we want to find the maximum of $d$ such that $\sum_i (s_i)^d < \epsilon$.

But $(\sum_i s_i)^d < \sum_i (s_i)^d$ by triangle inequality. Since $\sum_i s_i = a$, it is obvious that the only $d$ satisfies $\forall a, a^d < \epsilon$ is that $d = 0$.

Therefore, $f(C)$ has Hausdorff dimension at most zero.

A final note: even if this proof is all correct, I still haven't justify that $d = 0$ is at most. I am aware that $d$ can only equal to 0 is sufficient to show that $d$ is at most 0. But then I prove a condition that is stronger than required - doesn't sound right to my self.