Is there a differentiable map surjective from low to high dimension？

Question

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?

Answer 1

$\DeclareMathOperator\R{\mathbf{R}}$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 $$(*)\qquad F_f(x)=\limsup_{y\to x,\;y\neq x}\frac{\|f(y)-f(x)\|}{\|y-x\|}<\infty.$$

Define, for $p$ positive integer $$X_p=\{x\in\mathbf{R}^n:\forall y\in\mathbf{R}^n:\|y-x\|\le 1/p \Rightarrow \|f(y)-f(x)\|\le p\|y-x\|\}.$$ 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.

Answer 2

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, 129-135.

Answer 3

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 $$\Vert \mathrm d_x f\Vert^n r_x^n o(r_x^{m-n})~.$$ 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 $\epsilon >0$.