About the proof of the Four Vertex Theorem of Do Carmo.

Question

I have problems understanding this part of the proof given by Manfredo Do Carmo in Differential Geometry of Curves for the theorem of The Four-Vertex Theorem. I understand that in the first part he considers that exists a maximum and a minimum just because the parametrization of the curve is a mapping from $\mathbb{R}$ to $\mathbb{R}^2$. The next part is the one that I don't understand; he considers a line $L$ through those vertices where are the max and min of the curve, then he says this:

Let $Ax + By + C = 0$ be the equation of $L$. If there are no further vertices, $k'(s)$ keeps a constant sign on each of the arcs $\beta$ and $\gamma$ (until here, I understand it). We can then arrange the sign of all the coefficients $A, B, C$ so that the integral in Eq. (5) is positive. This contradiction shows that there is a third vertex and that $k'(s)$ changes sign on $\beta$ or $\gamma$, say, on $\beta$. Since $p$ and $q$ are points of maximum and minimum, $k'(s)$ changes sign twice on $p$. Thus, there is a fourth vertex.

The integral he refferes as Eq.(5) is:

$$\int_0^l (Ax+By+C)\frac{dk}{ds}ds=0$$

So I don't know how It would become a positive integral if it's zero.

Answer 1

The point here is that we're trying to arrive at a contradiction when we assume that there are just two vertices (critical points of $k$). On one of the arcs we have $Ax+By+C>0$ and $k'(s)>0$ except at the endpoints. On the other arc, $Ax+By+C<0$ and $k'(x)<0$ except at the endpoints. Thus, the integral (5) is positive. On the other hand, by integration by parts, we see that the integral (5) is zero. This contradiction shows that there must be at least one more critical point of $k$, hence at least two more. (If you want to see the proof written out a little bit differently, feel free to look at my differential geometry text.)