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EDIT: IncompleteThe following solution is incomplete. We need to make sure that if $F^{\prime}\left(t\right)$, $F^{\prime\prime}\left(t\right)$, ..., $F^{\left(n-1\right)}\left(t\right)$ are linearly dependent vectors for every $t$, then the coordinate functions of $F^{\prime}$ are linearly dependent on a sufficiently small interval. This reminds me of the follows from Wronskian considerations if the coordinate functions of $F^{\prime}$ are sufficiently nice (i. e., but is there such a result?

My analysis background is laughablelocally real-analytic), so please forgive me some more spamthis solves the problem for this nice class of functions, but I can't use this ansatz further.

"SOLUTION".

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EDIT: Incomplete. We need to make sure that if $F^{\prime}\left(t\right)$, $F^{\prime\prime}\left(t\right)$, ..., $F^{\left(n-1\right)}\left(t\right)$ are linearly dependent vectors for every $t$, then the coordinate functions of $F^{\prime}$ are linearly dependent on a sufficiently small interval. This reminds me of the Wronskian, but is there such a result?

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My analysis background is laughable, so please forgive me some more spam.

IMPORTANT: I consider $F$ to be a map from $S^1$ to $\mathbb R^n$, because a map from an interval with equal values at the ends is the same as a map from the circle. I will assume continuity of $F^{\prime}$ (yes, this includes the two endpoints of the interval which I have glued together). So I don't claim I have 100% solved the original problem.

I will say that an $n$-tuple of distinct points $\left(t_1,t_2,...,t_n\right)\in \left(S^1\right)^n$ is in counterclockwise position if there is an orientation-preserving map $\Phi:S^1\to \left[0,1\right]$, continuous except at one point, such that $\Phi\left(t_1\right)<\Phi\left(t_2\right)<...<\Phi\left(t_n\right)$. I need the following intuitively obvious fact:

(1) If an $n$-tuple of distinct points $\left(t_1,t_2,...,t_n\right)\in \left(S^1\right)^n$ is in counterclockwise position, and an $n$-tuple of distinct points $\left(s_1,s_2,...,s_n\right)\in \left(S^1\right)^n$ is in counterclockwise position as well, then there exists a smooth way to move the points $t_1$, $t_2$, ..., $t_n$ along $S^1$ such that they occupy the places of $s_1$, $s_2$, ..., $s_n$ at the end, and such that they stay distinct at any time during the process.

I could formalize this if anyone asks me to.

As a consequence of (1) and the intermediate value theorem (the usual one, for functions $\mathbb R\to \mathbb R$), we have:

(2) If an $n$-tuple of distinct points $\left(t_1,t_2,...,t_n\right)\in \left(S^1\right)^n$ is in counterclockwise position, and an $n$-tuple of distinct points $\left(s_1,s_2,...,s_n\right)\in \left(S^1\right)^n$ is in counterclockwise position as well, and $R:\left(S^1\right)^n\to \mathbb R$ is a continuous map that never takes the value $0$ on $n$-tuples of distinct points, then the reals $R\left(t_1,t_2,...,t_n\right)$ and $R\left(s_1,s_2,...,s_n\right)$ have the same sign.

In other words,

(3) If $R:\left(S^1\right)^n\to \mathbb R$ is a continuous map that never takes the value $0$ on $n$-tuples of distinct points, then for any $n$-tuple of distinct points $\left(t_1,t_2,...,t_n\right)\in \left(S^1\right)^n$ in counterclockwise position, the real $R\left(t_1,t_2,...,t_n\right)$ has the same sign.

Now, let's solve the problem: Assume that the assertion is wrong, and thus $F^{\prime}\left(t_1\right)$, $F^{\prime}\left(t_2\right)$, ..., $F^{\prime}\left(t_n\right)$ are linearly independent for any $n$-tuple of distinct points $\left(t_1,t_2,...,t_n\right)\in \left(S^1\right)^n$. Then, applying (3) to the continuous map

$R:\left(S^1\right)^n\to \mathbb R,$ $\left(t_1,t_2,...,t_n\right)\mapsto \det\left(F^{\prime}\left(t_1\right),F^{\prime}\left(t_2\right),...,F^{\prime}\left(t_n\right)\right)$,

we obtain that:

(4) For any $n$-tuple of distinct points $\left(t_1,t_2,...,t_n\right)\in \left(S^1\right)^n$ in counterclockwise position, the real $\det\left(F^{\prime}\left(t_1\right),F^{\prime}\left(t_2\right),...,F^{\prime}\left(t_n\right)\right)$ has the same sign.

Let's WLOG say that it is positive all the time (if its negative, just rewrite the proof with negative instead of positive...), i. e. we have:

(5) For any $n$-tuple of distinct points $\left(t_1,t_2,...,t_n\right)\in \left(S^1\right)^n$ in counterclockwise position, the real $\det\left(F^{\prime}\left(t_1\right),F^{\prime}\left(t_2\right),...,F^{\prime}\left(t_n\right)\right)$ is positive.

Now, move $t_2$, $t_3$, ..., $t_{n-1}$ (not $t_n$) closer and closer to $t_1$ (while keeping the counterclockwise position, of course), while keeping $t_1$ and $t_n$ fixed. Then, $\det\left(F^{\prime}\left(t_1\right),F^{\prime}\left(t_2\right),...,F^{\prime}\left(t_n\right)\right)$ tends to $\det\left(F^{\prime}\left(t_1\right),F^{\prime\prime}\left(t_1\right),...,F^{\left(n-1\right)}\left(t_1\right),F^{\prime}\left(t_n\right)\right)$. So, we get:

(5) For any pair of distinct points $\left(t_1,t_n\right)\in \left(S^1\right)^2$ in counterclockwise position, the real $\det\left(F^{\prime}\left(t_1\right),F^{\prime\prime}\left(t_1\right),...,F^{\left(n-1\right)}\left(t_1\right),F^{\prime}\left(t_n\right)\right)$ is nonnegative.

Notice how "positive" became "nonnegative" due to the limiting process (the limit of positive reals needs not be positive, but is always nonnegative).

Now, any pair of distinct points on $S^1$ is in counterclockwise position (and in clockwise position, too), so (5) can be simply rewritten as follows:

(6) For any pair of distinct points $\left(t,s\right)\in \left(S^1\right)^2$, the real $\det\left(F^{\prime}\left(t\right),F^{\prime\prime}\left(t\right),...,F^{\left(n-1\right)}\left(t\right),F^{\prime}\left(s\right)\right)$ is nonnegative.

We can drop the "distinct" in (6), as well, because for $s=t$, the real is simply $0$. So we have:

(7) For any pair of points $\left(t,s\right)\in \left(S^1\right)^2$, the real $\det\left(F^{\prime}\left(t\right),F^{\prime\prime}\left(t\right),...,F^{\left(n-1\right)}\left(t\right),F^{\prime}\left(s\right)\right)$ is nonnegative.

But if we fix $t$ and integrate $\det\left(F^{\prime}\left(t\right),F^{\prime\prime}\left(t\right),...,F^{\left(n-1\right)}\left(t\right),F^{\prime}\left(s\right)\right)$ over $s\in S^1$, we must get zero (because $\det\left(F^{\prime}\left(t\right),F^{\prime\prime}\left(t\right),...,F^{\left(n-1\right)}\left(t\right),F^{\prime}\left(s\right)\right)$ is linear in $F^{\prime}\left(s\right)$, and the integral of $F^{\prime}\left(s\right)$ over $S^1$ is zero). The integral of a continuous nonnegative function is zero only if the function itself is identically zero. Thus, $\det\left(F^{\prime}\left(t\right),F^{\prime\prime}\left(t\right),...,F^{\left(n-1\right)}\left(t\right),F^{\prime}\left(s\right)\right)$ for all $s\in S^1$. This means that $F^{\prime}\left(s\right)$ lies in a fixed hyperplane for all $s\in S^1$. Now, taking ANY $n$ points $t_1$, $t_2$, ..., $t_n$ on $S^1$, we get linearly dependent vectors $F^{\prime}\left(t_1\right)$, $F^{\prime}\left(t_2\right)$, ..., $F^{\prime}\left(t_n\right)$, and this is of course a contradiction!

Or do we?