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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).

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    $\begingroup$ 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
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    $\begingroup$ @Mohammer M.Zerrak check space filling curve $\endgroup$
    – Cloud JR K
    Feb 19, 2020 at 15:46
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    $\begingroup$ 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
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    $\begingroup$ You should look into Sards theorem. $\endgroup$ Feb 19, 2020 at 17:47
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    $\begingroup$ @Stephen: Are you sure Werner does not assume $C^1$? $\endgroup$ Feb 20, 2020 at 15:02

1 Answer 1

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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.

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  • $\begingroup$ 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
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    $\begingroup$ @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

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