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show/hide this revision's text 3 fixed typos

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 $$t^i=(t_1^i,\ldots,t_{n-1}^i),0\leq t^i_1< \cdots < t^i_{n-1}$$ 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 one a $F'(t)$ 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.
show/hide this revision's text 2 added the case when the derivative is not assumed continuous

Here is a solution in the $C^1$ case. 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 $$t^i=(t_1^i,\ldots,t_{n-1}^i),0\leq t^i_1< \cdots < t^i_{n-1}$$ 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 one $F'(t)$ such that for $l(F'(t))\neq 0$ where $l$ is a linear eqation 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.
show/hide this revision's text 1

Here is a solution in the $C^1$ case. 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 $$t^i=(t_1^i,\ldots,t_{n-1}^i),0\leq t^i_1< \cdots < t^i_{n-1}$$ 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 one $F'(t)$ such that for $l(F'(t))\neq 0$ where $l$ is a linear eqation defining $L$. Say, $l(F'(t)) > 0$. Then $l(F(0)) > l(F(1))$, so we can't have $F(0)=F(1)$.