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? 



Rolle's theorem in n dimensions
Here is a solution in the $C^1$ case [but see upd]. Suppose the vectors $F'(t\_1),\ldots, F'(t\_n)$ are linearly independent for all $0\leq t\_1< \cdots < t\_n\leq 1$. Let $L(t\_1,\ldots,t\_{n-1})$ be vector space spanned by the first $n-1$ of these. Let $$t^i=(t\_1^i,\ldots,t\_{n-1}^i),0\leq t^i\_1< \cdots < t^i\_{n-1}\leq 1$$ be a sequence such that $t^i\_{n-1}\to 0$ as $i\to \infty$. Since the space of all vector hyperplanes in $\mathbf{R}^n$ is compact, we can assume the sequence $L(t^i)$ has a limit $L$.


Claim: for any $s,t$ such that $0< s < t\leq 1$ the hyperplane $L$ does not separate $F'(s)$ and $F'(t)$.


Proof of the claim: if $L$ does, then for all sufficiently large $i$ the hyperplane $L(t^i)$ also separates $F'(s)$ and $F'(t)$. Choose $i$ so that moreover $t^i\_{n-1} < s$. Then the determinants of the matrices $(F'(t^i\_1),\ldots, F'(t^i\_{n-1}),F'(s))$ and $(F'(t^i\_1),\ldots, F'(t^i\_{n-1}),F'(t))$ have different signs. This is impossible, since $F'$ is continuous and the determinant is never zero. Claim is proven.


Now one of the two things can happen: either all $F'(t),0 \leq t\leq 1$ are in $L$, in which case $F'(t\_1),\ldots, F'(t\_n)$ are linearly dependent for all $t\_1,\ldots, t\_n$, or there is a $t$ such that for $l(F'(t))\neq 0$ where $l$ is a linear equation defining $L$. Say, $l(F'(t)) > 0$. Then $l(F(0)) < l(F(1))$, so we can't have $F(0)=F(1)$.


upd: here is how one can take care of the case when $F'$ is not assumed continuous. Basically, the only thing that changes is the proof of the claim; the claim itself remains the same except that we assume $t < 1$. Choose $i$ as above and set $g(x),s\leq x\leq t$ to be the determinant of $(F'(t^i\_1),\ldots,F'(t^i\_{n-1}),F(x))$. This function is differentiable and we have $g'(x)=det(F'(t^i\_1),\ldots,F'(t^i\_{n-1}),F'(x))$. So $g'(s)$ and $g'(t)$ have different signs. The claim follows now from the following statement, which is a consequence of the classical Rolle's theorem: if $f:[a,b]\to\mathbf{R}$ is differentiable at each point of $[a,b]$ and $f'(a)$ and $f'(b)$ have different signs, then there is an $x\in (a,b)$ such that $f'(x)=0$.


Then we deduce from the claim that all $F'(t),0 < t < 1$ are in the same half-space with respect to $L$. This suffices.


