# Prove $v$ is not surjective

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).

• 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).
– Karl
Feb 19, 2020 at 7:56
• @Mohammer M.Zerrak check space filling curve Feb 19, 2020 at 15:46
• 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. Feb 19, 2020 at 15:59
• You should look into Sards theorem. Feb 19, 2020 at 17:47
• @Stephen: Are you sure Werner does not assume $C^1$? Feb 20, 2020 at 15:02

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.

• 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 Feb 20, 2020 at 22:43
• @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. Feb 20, 2020 at 23:55
• which rudin u mention ? Baby rudin or real and complex analysis Feb 21, 2020 at 6:14
• @CloudJR It is written in my answer. Feb 21, 2020 at 13:23
• @CloudJR: No, sorry: I prefer to keep my anonymity. One thing I can say, I am not based in India. Mar 2, 2020 at 18:14