TITLE: Rolle's theorem in n dimensions
QUESTION [27 upvotes]: 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?

REPLY [7 votes]: Let's add a bit of Anton Petrunin's answer into algori's: namely, I'm going to spell out the 'little  bit of work' from Anton's answer, which is a version of algori's Claim. I am not proving the stronger statement made by Anton though, only the original statement. (This was supposed to be a comment: my attempt to write algori's proof without taking limits, but it turned out to be too long for comments.)
Claim 1 (Petrunin): $0\in Conv(f'(a,b))$.
Proof: Otherwise there is a hyperplane separating $0$ from the convex hull.
By Caratheodory's Theorem, we get 
$$a < t_1 < \dots < t_{n+1} < b$$ such that 
$$0\in Conv(f'(t_1),\dots,f'(t_{n+1}).$$
Claim 2 (algori): There is $s\in[t_n,t_{n+1}]$ such that 
$$0\in Span(f'(t_1),\dots,f'(t_{n-1}),f'(s)).$$
Proof: Set $$l(x)=\det(f'(t_1),\dots,f'(t_n),f(x)).$$ The condition on $s$ is that $l'(s)=0$. By Claim 1, either $$0\in Conv(f'(t_1),\dots,f'(t_{n-1}))$$ (in which case the claim is obvious) or $0$ lies between $l'(t_n)$ and $l'(t_{n-1})$. In the latter case, such $s$ exists by the usual Rolle's theorem.