Four-vertex theorem

Remarks

The idea is clearest for a convex curve with nonconstant $k$. Since $k$ is periodic, it has a maximum and a minimum, which are two vertices. Assume there were no others. Then the line through the two corresponding points of the curve meets it only there, and $k'$ has a fixed sign on each of the two arcs. Choose a linear function $f(x) = \ip{x}{v} + c$ that vanishes on that line; then $k'(s) f(\gamma(s))$ has constant sign. But the plane Frenet formulas give $\int_0^L k'(s) \, \gamma(s) \, ds = 0$ and $\int_0^L k'(s) \, ds = 0$, so the integral of $k' \cdot (f \circ \gamma)$ is $0$, a contradiction. The general case of simple closed curves needs more work. Example: the ellipse $x^2/a^2 + y^2/b^2 = 1$ with $a > b > 0$ has exactly four vertices, the ends of its two axes. The hypothesis that the curve is simple is needed: a limaçon with an inner loop is a closed regular curve with only two vertices. The theorem was proved for convex curves by Mukhopadhyaya in 1909 and for all simple closed curves by Kneser in 1912.

Going deeper

For a student

When $k$ never vanishes, the vertices of $\gamma$ are exactly the points where its evolute has zero velocity. Typically these are cusps of the evolute, as for the four cusps of the evolute of an ellipse. The proof above uses only the plane Frenet equations $T' = kN$ and $N' = -kT$, integrated over a period. The theorem is an early example of a global result in differential geometry, in which a local quantity is constrained by the shape of the whole curve.