Does there exist a map $f:\Bbb R^n \rightarrow \Bbb R^m$, where $n<m$ and $ n,m \in\Bbb N^+$ such that $f$ is surjective and differentiable?

5$\begingroup$ This is impossible when $f$ is smooth, because the set of critical values of $f$ in $\mathbb{R}^m$ will be the entire range of $f$, which has measure zero by Sard's theorem. There is a form of Sard's theorem which applies to $C^r$ functions, $r < \infty$, but the minimal $r$ to which it applies depends on $n$ and $m$. $\endgroup$– BmaOct 2, 2021 at 4:31

2$\begingroup$ The argument at the following link appears to generalize to any $C^1$ $f: \mathbb{R}^n \to \mathbb{R}^m$ where $n < m$, showing that this is impossible when the derivative is continuous. This is more or less the proof of Sard's theorem I know. See here: math.stackexchange.com/questions/1905299/… $\endgroup$– BmaOct 2, 2021 at 4:41

9$\begingroup$ sard's theorem requires the derived function of f is continuous.peano cuves and hilbert cuves only need f is continuous, before giving a strict solution,please do not underestimate the difficulty of the problem and review the questions carefully. $\endgroup$– weak solutionOct 2, 2021 at 8:07

3$\begingroup$ Sard actually called his paper "The measure of the critical values of differentiable maps", but he really does require continuous differentiability, so I can't find a reference for differentiable but not continuously differentiable functions. $\endgroup$– Ben McKayOct 2, 2021 at 8:13

3$\begingroup$ @WlodAA: as I say in my previous comment, I don't know a reference to prove such a result, and I don't see how to prove it. Would you mind giving an outline proof in an answer? $\endgroup$– Ben McKayOct 2, 2021 at 8:44
3 Answers
$\DeclareMathOperator\R{\mathbf{R}}$It's easy to check that the image of any locally Lipschitz map $f:\R^n\to\R^m$ has measure zero when $n<m$ (this encompasses the case of class$\text{C}^1$ maps, but not the case of differentiable maps).
Indeed, extend $f$ to $F:\R^m\to\R^m$ by $F(x,y)=f(x)$. This is still locally Lipschitz. So it maps the subset $\R^n$ of measure zero to a subset of measure zero, see this MathSE post (it assumes Lipschitz, but the argument is local and $\R^m$ is a countable union of subsets on which $F$ is Lipschitz).
Taking into accounts the comments: here is a setting encompassing both the cases when $f$ is locally Lipschitz, and when $f$ is differentiable.
Suppose that for every $x\in\mathbf{R}^n$, we have $$(*)\qquad F_f(x)=\limsup_{y\to x,\;y\neq x}\frac{\f(y)f(x)\}{\yx\}<\infty.$$
Define, for $p$ positive integer $$X_p=\{x\in\mathbf{R}^n:\forall y\in\mathbf{R}^n:\yx\\le 1/p \Rightarrow \f(y)f(x)\\le p\yx\\}.$$ Then $\mathbf{R}^n$ is the (countable) union of all $X_p$, and $X_p$ is a countable union of subsets $X_{p,i}$ of diameter $\le 1/p$. And $f$ is $p$Lipschitz on $X_{p,i}$ (and also on its closure, in case one wishes to get closed subsets).
So the result indeed follows, not of the Lipschitz case as strictly said, but of the same statement replacing $\mathbf{R}^n$ with a subset of $\mathbf{R}^n$ with the restriction of the Euclidean distance (namely: for $n<m$ and $Y$ subset of $\mathbf{R}^n$, every Lipschitz function $Y\to\mathbf{R}^m$ has image of measure zero). The argument for the latter seems unchanged.
PS: for a reference, it is mentioned by @Kosh that Lemma 7.25 in Rudin's Real and complex analysis (initially published in 1966) does all the job: it asserts that any map $f:\mathbf{R}^m\to\mathbf{R}$ satisfying $(*)$ maps measure zero subsets to measure zero subsets. The proof given here actually seems to roughly be the same as the one written (concisely) in Rudin's book.

3$\begingroup$ But just differentiable does not imply Lipschitz, not even locally... $\endgroup$ Oct 2, 2021 at 16:42

1$\begingroup$ @MohanRamachandran can you give me a reference for this result? You are claiming that the function the OP is looking for cannot exists. Right? $\endgroup$– KoshOct 2, 2021 at 22:08

2$\begingroup$ Yes. You can find the proof of a stronger result in Piotr Hajlasz lecture notes titled Geometric Analysis Lemma 2.8 page 18 $\endgroup$ Oct 2, 2021 at 23:55

1$\begingroup$ So in order to let your observation and Majer remark coexisting, the result is that if a function is differentiable everywhere then it is not necessarily the case that it is locally Lipschitz but still there exist countable many measurable subsets where it's restriction is Lipschitz. So a kind of local Lipschitzianity on certain measurable subsets rather than compact subsets. Am I right? $\endgroup$– KoshOct 3, 2021 at 8:18

1$\begingroup$ Yes. For approximately differentiable points this is due to Federer Surface Area II TAMS 1944 vol 55 page 453 Theorem 5.2 . $\endgroup$ Oct 3, 2021 at 14:19
I don't think that the question is "at undergraduate level", and I don't understand the downvotes. That such a map does not exist for $n=1$ is a result of Morayne, On differentiability of Peano type functions I, II. Colloq. Math. 53 (1987), no. 1, 129135.
I think the image of a differentiable map $f:\mathbb R^n \to \mathbb R^m$ with $n<m$ always has measure $0$. As mentioned my @YCor in another answer, this follows from the local Lipschitz property of $f$ when $f$ is $\mathcal C^1$. Let us try to remove this hypothesis.
First, covering $\mathbb R^n$ by countably many compact sets, we reduce to proving that $f(K)$ has measure $0$ for any compact set $K$.
Fix $\epsilon >0$. For every $x \in K$, Since $f$ is differentiable at $x$, $f(B(x,r))$ is contained in a $o(r_x)$neighbourhood of a euclidean $n$dimensional disc of radius $\Vert \mathrm d_x f\Vert r_x$, and thus has measure $$\Vert \mathrm d_x f\Vert^n r_x^n o(r_x^{mn})~.$$ We can thus choose $r_x$ sufficiently small so that $f(B(x,r_x))$ has Lebesgue measure at most $\epsilon r_x^n$.
Now, the balls $B(x,\frac{1}{3}r_x)$, $x\in K$ obviously cover $K$ and we can extract a finite cover from it. Then by Vitali's lemma, we can find $x_1,\ldots, x_k$ such that the balls $B(x_i, \frac{1}{3}r_{x_i})$ are disjoint and the balls of radius $r_{x_i}$ cover $K$.
By disjointness of the balls $B(x_i,\frac{1}{3}r_{x_i})$, we get that $$\sum_i r_{x_i}^n \leq \mathrm{Constant}\cdot \mathrm{Leb}(K)~.$$ Since the balls $B(x_i, r_{x_i})$ cover $K$ we get that $$\mathrm{Leb}(f(K)) \leq \sum_i \mathrm{Leb}(f(B(x_i,r_{x_i})) \leq \epsilon \sum_i r_{x_i}^n~.$$ Taking $\epsilon$ arbitrarily small, we conclude that $f(K)$ has measure $0$.
Remark: the only regularity we need is that for every $x$, the image of $f(B(x,r))$ has volume $o(r^n)$ when $r$ goes to $0$ (with $o(r^n)$ possibly depending on the point $x$). This is satisfied for instance if $f$ is $(\frac{n}{m}+ \epsilon)$Hölder for any $\epsilon >0$.

1
