Page 244 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Unit 20: Curves

Proof. We may assume without loss of generality that '  1. Let :[0,L]   be the continuous Notes
function given in Proposition 2 satisfying:

e = (cos , sin ),
1
and ’ = k.
Suppose that  is convex. We will show that  is weakly monotone, i.e., if t < t and (t ) = (t ) then
2
1
1
2
1
 is constant on [t , t ]. First, we note that since  is simple, we have n   by the Rotation

2
1
Theorem, and it follows that e is onto  , see Exercise 5. Thus, there is t  [0,L] such that
1
1
3
e (t ) = –e (t ) = –e (t ).
1
1
2
1
1
3
By convexity, the three parallel tangents at t , t , and t cannot be distinct, hence at least two must
3
1
2
coincide. Let p   (s ) and p   (s ),s  s denote these two points, then the line p p is
1
1
2
2
1
2
1
2
contained in . Otherwise, if q is a point on p p not on , then the line through q perpendicular
1
2
to p p intersects  in at least two points r and s, which by convexity must lie on one side of
1
2
p p . Without loss of generality, assume that r is the closer of the two to p p . Then r lies in the
1
2
1
2
interior of the triangle p p s. Regardless of the inclination of the tangent at r, the three points p ,
2
1
1
p and s, all belonging to , cannot all lie on one side of the tangent, in contradiction to convexity.
2

If p p   (s) : s 1  s s 2  , then p p   (s) : s 2  s L   (s) : 0   s s 1 . However, in that


2
2
1
1
case, we would have (s ) – (s ) = (L) – (0) = 2, a contradiction. Thus, we have
2
1
p p   (s) : s 1  s s 2   (t) : t  1  t  t 2 .

1
2
In particular (t) = (t ) = (t ).
1
2
Conversely, suppose that  is not convex. Then, there is t  [0, L] such that the function
0
     (t 0  ) e 2 changes sign. We will show that ’ also changes sign. Let t , t  [0,L] be such
+
–
that


min   (t ) 0   (t )   (t )  max .

0
[0,L]  [0,L]

Note that the three tangents at t , t and t are parallel but distinct. Since '(t )   '(t )  0, we

–

0
+
have that e (t ) and e (t ) are both equal to ±e (t ).
–
+
1
0
1
1
Thus, at least two of these vectors are equal. We may assume, after reparametrization, that there
exists 0 < s < L such that e (0) = e (s). This implies that
1
1
(s) – (0) = 2k, (L) – (s) = 2k
1.
with k, k  . By the Rotation Theorem, n  k  k   Since (0) and (s)  do not lie on a line

parallel to e (t ), it follows that is not constant on either [0, s] or [0, L]. If k = 0 then  changes sign
1
0
on [0, s], and similarly if k = 0 then changes sign on [s, L]. If kk 0, then since k + k = ±1, it
follows that kk < 0 and  changes sign on [0, L].
2
Definition 8. Let :[0,L]   be a regular plane curve. A vertex of  is a critical point of the
curvature k.
Theorem 11 (The Four Vertex Theorem). A regular simple convex closed curve has at least four
vertices.
Proof. Clearly, k has a maximum and minimum on [0,L], hence  has at least two vertices. We
will assume, without loss of generality, that  is parametrized by arclength, has its minimum at
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