Can a Continuous Curve Only Intersect a Differentiable Curve at the Origin?

[caffeinemachine]

Hello MHB,

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. (Forgot to add this earlier.)

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

 

[HallsofIvy]

What's wrong with defining, say, g(x)= f(x)+ (x², x²)?

 

[caffeinemachine]

What's wrong with defining, say, g(x)= f(x)+ (x², x²)?

It might so happen that $g(x_1)=f(x_1)+(x_1^2,x_1^2)=f(x_2)\neq (0,0)$.

 

[Opalg]

[From my phone by Tapatalk, so this has to be short. ]

Suppose $ f (t)=t^3 \sin (1/t) e^{i\pi/t} $, with $ f (0) =0$. This is differentiable, and I don't see how a function $ g $ can get away from the origin without crossing the graph of $ f $.

 

[blue_raver22]

Dear [Name],

Thank you for sharing your conjecture with me. It is an interesting problem and I can see why you are having difficulty settling it. I have looked into it and unfortunately, I do not believe your conjecture is true.

To disprove your conjecture, I have found a counterexample. Consider the curve $f:[0,1]\to\mathbb R^2$ defined by $f(x)=(x^2,x^3)$ which is differentiable and passes through the origin. However, there does not exist a continuous function $g:[0,1]\to\mathbb R^2$ that satisfies all three conditions in your conjecture. This is because $f([0,1])$ is the parabola $y=x^2$ which intersects the curve $g([0,1])$ at the points $(0,0)$ and $(1,0)$. Therefore, your conjecture is not true in general.

However, I would encourage you to continue exploring this problem and see if you can modify your conjecture to make it true. Perhaps by adding additional conditions or restrictions, you can find a subset of differentiable curves for which your conjecture holds. Keep up the good work and don't be discouraged by this counterexample.

Best regards,