Let $\alpha: I \to \mathbb{R}^n$, $n\geq 1$ be a differentiable regular curve such that all of it's tangent lines pass through origin. This means that for every $t\in I$ there exists $k(t) \in \mathbb{R}$ such that $\alpha(t)+k(t)\alpha'(t)=0$. So each coordinate $x$ of $\alpha$ satisfies the differential equation $x+kx'=0$. If $k$ was integrable and non-zero, the solutions to this equation would be $x=Ce^F$ where $F$ is an antiderivative of $-\frac{1}{k}$. And then it follows that the whole curve is a multiple of a constant and therefore contained in a line. However i don't see how to show if $k$ is necessarily integrable/continuous and non-zero.

Note: I am aware there is an easier way if the curve is twice differentiable. My question is more general.

  • $\begingroup$ Is your $\alpha$ merely differentiable or is it $C^1$? $\endgroup$ – Arctic Char Jun 11 at 8:41
  • $\begingroup$ @ArcticChar Just differentiable, but an answer about the $C^1$ case would still be interesting. $\endgroup$ – Carla is my name Jun 11 at 9:00

Partial answer: In the following we assume that $|\alpha'|$ is locally bounded. This is the case when $\alpha$ is $C^1$.

Let $\epsilon >0$ and consider the curve $$ \alpha_\epsilon =\alpha |_{I_\epsilon}, $$ where $I_\epsilon = \alpha^{-1} \{y\in \mathbb R^n: | y| > \epsilon\}$ (that is, consider only the portion of $\alpha$ which is as least distance $\epsilon$ away from the origin). Then $I_\epsilon$ is open and is a disjoint union of connected components (which are intervals). It suffices to show that $\alpha_\epsilon$ is a straight line when restricted to each such intervals.

The condition gives the equality $$\tag{1}\alpha + k \alpha ' = 0,$$ -taking dot product with $\alpha'$ in (1) gives $$k = -\frac{\alpha \cdot \alpha '}{\|\alpha'\|^2}.$$ In particular, since $\alpha'$ is measureable (see here), $k$ is also measurable;

-also, (1) implies $$|k| =\frac{ \|\alpha'\|}{\|\alpha\|}.$$ Hence on $\alpha_\epsilon$ we have $|1/k|\le \|\alpha'\|\epsilon^{-1}$. Since $\|\alpha'\|$ is assumed to be locally bounded, $1/k$ is locally bounded, thus integrable.

As a result, as you did in the post, one obtains that $\alpha = C e^{F}$ on each connected component of $I_\epsilon$.

  • $\begingroup$ Can $|\alpha'|$ be non locally bounded except for at most 2 points? (since the derivative is defined everywhere, feels like can only be unbounded at the extremes of the interval, but i'm not sure) Wouldn't the argument then extend for the whole curve since derivative is locally bounded inside? $\endgroup$ – Carla is my name Jun 12 at 7:49

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