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I have the following conjecture which I cannot seem to settle either way:

Let $f:[0,1]\to\mathbb R^2$ be a differentiable function such that $f(0)=(0,0)$. Then there exists a continuous function $g:[0,1]\to\mathbb R^2$ such that:

1) $g(0)=(0,0)$

2) $g([0,1])\cap f([0,1])=\{(0,0)\}$.

3) $g$ is not a constant function. (Thanks to user68061 for pointing this out.)

Basically what I am trying to prove is that if we have a differentiable curve in $\mathbb R^2$ which passes through origin then we can find a continuous curve in $\mathbb R^2$ which intersects the given curve only at origin.

Does anybody know if this is an already known result or if there exists a counterexample?

Thanks in advance for your help.

caffeinemachine
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    you want $g$ to be injective, i think – user68061 Jan 30 '14 at 17:56
  • @user68061 Thanks. I don't really require $g$ to be injective but I don't want $g$ to be a constant function. Let me see if I can edit it. Thank you again for pointing this out. – caffeinemachine Jan 30 '14 at 18:08
  • @Felipe Jacob. Thanks for participating. I think your reasoning is not correct. The uniform continuity of $f$ would be there even if $f$ we merely only continuous. Now since there exist 'space filling' continuous curves, we won't be able to find $g$ if we take $f$ to be only continuous (although the reason you gave can still be given). – caffeinemachine Jan 30 '14 at 18:10

2 Answers2

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First define $F_0: [0,1] \rightarrow \mathbb{R}^2$ so that $F_0(0)=(1,0)$ and for the first half it goes around a unit circle, and then for the second half it goes along the line from $(1,0)$ to $(1/2,0)$ making sure to have the derivative vanish exponentially fast at the beginning, middle, and end to be safe.

Next define $F_1: [0,1] \rightarrow \mathbb{R}^2$ by $F_1(x)=\frac{1}{2^{n-1}}F_0(2^n(x-\frac{2^{n-1}-1}{2^{n-1}}))$ on $[\frac{2^{n-1}-1}{2^{n-1}},\frac{2^{n}-1}{2^{n}}]$ and $F(1)=(0,0)$

I'm not entirely sure I got the indices right there, but the point is taking $f(x)=F_1(1-x)$ we should have the image be a line segment from (0,0) to (1,0) along with circles of radius $\frac{1}{2^n}$.

Then the image of $g$ can't contain any points other than 0 without crossing $f$ by the Jordan curve theorem.

Nate
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  • Thank you for participating. To understand your solution I need to know if you claim that $F_1$ plays the role of $g$. Is that correct? – caffeinemachine Jan 31 '14 at 06:01
  • No, I'm saying that for $f(x)=F_1(1-x)$ than any non constant $g$ with $g(0)=(0,0)$ must intersect it's image somewhere other than $(0,0)$, so this is a counterexample. – Nate Jan 31 '14 at 17:01
  • Oh! I though you were giving a proof. Hehe. Thanks. Let me verify and vote up. – caffeinemachine Feb 02 '14 at 03:22
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A counterexample is $$ f(t)=t^2\Bigl(\sin\Bigl(\frac1t\Bigr),\cos\Bigl(\frac1t\Bigr)\Bigr),\quad t\ne0,\qquad f(0)=(0,0). $$

Edit

This is a counterexample if $g'(0)\ne(0,0)$. See Dejan Govc's comment.

Julián Aguirre
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