I have seen other answers to this question but they all seem to have small issues. For example in A curve where all tangent lines are concurrent must be straight line it is never clear (at least to me) why we suppose that the $\lambda (s)$ is differentiable. I was wondering if that answer could be completed or if there is another approach to it.
(Just to be clear, the curve is supposed to be in any $R^n$).
If all tangent lines to a curve meet in one point, prove the curve is a straight line
calculuscurvesdifferential-geometry
Best Answer
Good question. And you're right that we authors of elementary differential geometry texts don't always include these details.
Without loss of generality, you can assume the point in question is the origin, and then you have $$\alpha(s)=-\lambda(s)T(s),$$ where $T(s)$ is the unit tangent vector. (We may always parametrize by arclength, but in this case it's not really relevant.) Now $\alpha$ and $T$ are smooth functions, so therefore $$\lambda = -\langle\alpha,T\rangle$$ is smooth as well. (If you want to be entirely pedantic, you can prove by the product rule and induction that if $f,g$ are $C^k$ functions, then $\langle f,g\rangle$ is also $C^k$. Or you can write it out in terms of component functions and then it's just a sum of products of $C^k$ functions.)
Similar arguments pertain when you write linear combinations of all the vectors in the Frenet frame. E.g., if $\alpha$ is a smooth curve with $\kappa\ne 0$ and you write $$\alpha = \lambda T+\mu N + \nu B,$$ then the coefficient functions will similarly all be smooth.