real analysis - Rolle's theorem in n dimensions

EDIT: The following solution is incomplete. We need to make sure that if $F^{prime}left(tright)$, $F^{primeprime}left(tright)$, ..., $F^{left(n-1right)}left(tright)$ are linearly dependent vectors for every $t$, then the coordinate functions of $F^{prime}$ are linearly dependent on a sufficiently small interval. This follows from Wronskian considerations if the coordinate functions of $F^{prime}$ are sufficiently nice (i. e., locally real-analytic), so this solves the problem for this nice class of functions, but I can't use this ansatz further.

"SOLUTION".

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 $left(t_1,t_2,...,t_nright)in left(S^1right)^n$ is in counterclockwise position if there is an orientation-preserving map $Phi:S^1to left[0,1right]$, continuous except at one point, such that $Phileft(t_1right)<Phileft(t_2right)<...<Phileft(t_nright)$. I need the following intuitively obvious fact:

(1) If an $n$-tuple of distinct points $left(t_1,t_2,...,t_nright)in left(S^1right)^n$ is in counterclockwise position, and an $n$-tuple of distinct points $left(s_1,s_2,...,s_nright)in left(S^1right)^n$ 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 $mathbb Rto mathbb R$), we have:

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

In other words,

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

Now, let's solve the problem: Assume that the assertion is wrong, and thus $F^{prime}left(t_1right)$, $F^{prime}left(t_2right)$, ..., $F^{prime}left(t_nright)$ are linearly independent for any $n$-tuple of distinct points $left(t_1,t_2,...,t_nright)in left(S^1right)^n$. Then, applying (3) to the continuous map

$R:left(S^1right)^nto mathbb R,$
$left(t_1,t_2,...,t_nright)mapsto detleft(F^{prime}left(t_1right),F^{prime}left(t_2right),...,F^{prime}left(t_nright)right)$,

we obtain that:

(4) For any $n$-tuple of distinct points $left(t_1,t_2,...,t_nright)in left(S^1right)^n$ in counterclockwise position, the real $detleft(F^{prime}left(t_1right),F^{prime}left(t_2right),...,F^{prime}left(t_nright)right)$ 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 $left(t_1,t_2,...,t_nright)in left(S^1right)^n$ in counterclockwise position, the real $detleft(F^{prime}left(t_1right),F^{prime}left(t_2right),...,F^{prime}left(t_nright)right)$ 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, $detleft(F^{prime}left(t_1right),F^{prime}left(t_2right),...,F^{prime}left(t_nright)right)$ tends to $detleft(F^{prime}left(t_1right),F^{primeprime}left(t_1right),...,F^{left(n-1right)}left(t_1right),F^{prime}left(t_nright)right)$. So, we get:

(5) For any pair of distinct points $left(t_1,t_nright)in left(S^1right)^2$ in counterclockwise position, the real $detleft(F^{prime}left(t_1right),F^{primeprime}left(t_1right),...,F^{left(n-1right)}left(t_1right),F^{prime}left(t_nright)right)$ 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 $left(t,sright)in left(S^1right)^2$, the real $detleft(F^{prime}left(tright),F^{primeprime}left(tright),...,F^{left(n-1right)}left(tright),F^{prime}left(sright)right)$ 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 $left(t,sright)in left(S^1right)^2$, the real $detleft(F^{prime}left(tright),F^{primeprime}left(tright),...,F^{left(n-1right)}left(tright),F^{prime}left(sright)right)$ is nonnegative.

But if we fix $t$ and integrate $detleft(F^{prime}left(tright),F^{primeprime}left(tright),...,F^{left(n-1right)}left(tright),F^{prime}left(sright)right)$ over $sin S^1$, we must get zero (because $detleft(F^{prime}left(tright),F^{primeprime}left(tright),...,F^{left(n-1right)}left(tright),F^{prime}left(sright)right)$ is linear in $F^{prime}left(sright)$, and the integral of $F^{prime}left(sright)$ over $S^1$ is zero). The integral of a continuous nonnegative function is zero only if the function itself is identically zero. Thus, $detleft(F^{prime}left(tright),F^{primeprime}left(tright),...,F^{left(n-1right)}left(tright),F^{prime}left(sright)right)$ for all $sin S^1$. This means that $F^{prime}left(sright)$ lies in a fixed hyperplane for all $sin S^1$. Now, taking ANY $n$ points $t_1$, $t_2$, ..., $t_n$ on $S^1$, we get linearly dependent vectors $F^{prime}left(t_1right)$, $F^{prime}left(t_2right)$, ..., $F^{prime}left(t_nright)$, and this is of course a contradiction!

Or do we?

• Next ag.algebraic geometry - Why does the algebraic condition of flatness on the structure sheaves give a good definition of family?
• Previous alexandrov geometry - Are isometries the only geodesic preserving maps in a CAT(0)-space?