Let$ v$ : $\mathbb R \rightarrow \mathbb R^2$ be a differentiable function such that the velocity vector $\cfrac{dv}{ dt}\neq 0$ at all $t\in \mathbb R$. Prove that $v$ is not surjective.

I tried to work with definition and tried to prove contrapositive statement but It get me nowhere. Could anyone give me a hint to start with (not solution).

  • 1
    $\begingroup$ Interesting problem. Maybe you can show that the image of $v$ has measure 0 (or even just finite measure) by bounding it within a thin region (that gets thinner when $t$ is farther from 0 so that its total area is a convergent sum). $\endgroup$
    – Karl
    Feb 19, 2020 at 7:56
  • 2
    $\begingroup$ @Mohammer M.Zerrak check space filling curve $\endgroup$
    – Cloud JR K
    Feb 19, 2020 at 15:46
  • 1
    $\begingroup$ You don't need the condition on the velocity vector. There is no surjective differentiable function from $\mathbf{R}^m$ to $\mathbf{R}^n$ when $m<n$. This appears e.g. as an exercise (with substantial hints!) in chapter 1 of Warner's excellent book Foundations of differentiable manifolds and Lie groups. $\endgroup$
    – Stephen
    Feb 19, 2020 at 15:59
  • 1
    $\begingroup$ You should look into Sards theorem. $\endgroup$ Feb 19, 2020 at 17:47
  • 1
    $\begingroup$ @Stephen: Are you sure Werner does not assume $C^1$? $\endgroup$ Feb 20, 2020 at 15:02

1 Answer 1


Let $f: {\mathbb R}^m\to {\mathbb R}^n$ be a map differentiable everywhere, $m< n$. I will identify ${\mathbb R}^m$ with a linear subspace $E$ in ${\mathbb R}^n$. Then $mes(E)=0$ where $mes$ is the Lebesgue measure on ${\mathbb R}^n$. Now, apply Lemma 7.25 from W.Rudin, "Real and Complex analysis" (3rd edition). Since $f$ is differentiable everywhere on $E$, the assumption of Lemma 7.25 holds: One only needs $$ \forall x\in E, \limsup_{y\to x} \frac{|f(x)-f(y)|}{|x-y|}<\infty. $$ Hence, $mes (f(E))=0$. Thus, $f$ cannot be surjective. qed

No idea, why do they assume nonzero derivative (maybe just to confuse you).

The bottom line: If you want to get to TIFR, make sure you know Rudin's book really well.

Edit. 1. Robert Israel gives a proof of the relevant result from Rudin (for $m=1$) here. The proof is quite short and is essentially the same as the one in Rudin's book.

  1. While the original space-filling curves were nowhere differentiable, Lebesgue constructed a space-filling curve which is almost everywhere differentiable (with Holder exponent arbitrarily close to $1/2$), see Theorem 5.4.2 in

H.Sagan, "Space-filling curves," Springer-Verlag, 1984.

  1. Here is an alternative (longer!) proof which uses the assumption that $v$ is $C^1$and $v'(t)\ne 0$ for all $t$. I will give this proof as a sequence of steps which you should be able to manage yourself (in fact, if you want to get to TIFR, you should not need these hints).

a. Argue that $v$ is locally injective.

b. Prove that for every finite subinterval $I\subset {\mathbb R}$, $v(I)$ has empty interior.

c. Conclude by using Baire's Theorem.

  • $\begingroup$ That means derivative not equal 0 is not necessary. But with that hypothesis does it makes proof any simpler. I ask this because this is one of the top institute for mathematics and if they ask then there might be something in that $\endgroup$
    – Cloud JR K
    Feb 20, 2020 at 22:43
  • $\begingroup$ @CloudJR: No, Rudin's proof does not get simpler (and it is already quite simple if you know what you are doing). It is possible that they had in mind a totally different proof. $\endgroup$ Feb 20, 2020 at 23:55
  • $\begingroup$ which rudin u mention ? Baby rudin or real and complex analysis $\endgroup$
    – Cloud JR K
    Feb 21, 2020 at 6:14
  • $\begingroup$ @CloudJR It is written in my answer. $\endgroup$ Feb 21, 2020 at 13:23
  • 1
    $\begingroup$ @CloudJR: No, sorry: I prefer to keep my anonymity. One thing I can say, I am not based in India. $\endgroup$ Mar 2, 2020 at 18:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.