Return to Question

This looks like a statement from a calculus textbook, which perhaps it should be.

"Rolle's theorem". Let $F\colon [a,b]\to\mathbb R^n$ be a continuous function such that F(a)=F(b) and F'(t) exists for all $a<t<b$. Then there exist numbers $a < t_1 < t_2 < \dots < t_n < b$ such that the vectors $F'(t_1),\dots,F'(t_{n})$ are linearly dependent.

We are all familiar with the case n=1. The case n=2 is not hard either: pick any $a<t^\ast<b$ and find, using the mean value theorem, numbers $a<t_1<t^\ast$ and $t^*<t_2<b$ such that $F'(t_j)$ is collinear with $F(t^\ast)-F(a)$. Note that we avoided using the parameter value $t^\ast$, which will be important in a moment. When n=3, we pick $a<t^\ast<b$ and project F onto the orthogonal complement of $F'(t^\ast)$, then apply the case n=2 to the projection ($t^\ast$ will become the third chosen parameter value). So far so good.

But I get stuck at n=4. If the above process is followed, then after F is projected down to two dimensions, we must avoid two particular parameter values. Which is not possible in general: if in two dimensions F parametrizes a triangle and F(a) is a vertex, then one of points $F(t_j)$ must be one of two other vertices. Presumably this problem can avoided by a generic choice of points of projection, but how?

This looks like a statement from a calculus textbook, which perhaps it should be.

"Rolle's theorem". Let $F\colon [a,b]\to\mathbb R^n$ be a continuous function such that $F(a)=F(b)$ and $F'(t)$ exists for all $a<t<b$. Then there exist numbers $a < t_1 < t_2 < \dots < t_n < b$ such that the vectors $F'(t_1),\dots,F'(t_{n})$ are linearly dependent.

We are all familiar with the case $n=1$. The case $n=2$ is not hard either: pick any $a<t^\ast<b$ and find, using the mean value theorem, numbers $a<t_1<t^\ast$ and $t^*<t_2<b$ such that $F'(t_j)$ is collinear with $F(t^\ast)-F(a)$. Note that we avoided using the parameter value $t^\ast$, which will be important in a moment. When $n=3$, we pick $a<t^\ast<b$ and project F onto the orthogonal complement of $F'(t^\ast)$, then apply the case $n=2$ to the projection ($t^\ast$ will become the third chosen parameter value). So far so good.

But I get stuck at $n=4$. If the above process is followed, then after $F$ is projected down to two dimensions, we must avoid two particular parameter values. Which is not possible in general: if in two dimensions $F$ parametrizes a triangle and $F(a)$ is a vertex, then one of points $F(t_j)$ must be one of two other vertices. Presumably this problem can avoided by a generic choice of points of projection, but how?