Prove $v$ is not surjective

Question

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).

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.

2. 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.

3. 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.