Page 223 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes Theorem 4 (Four-vertex Theorem.). A simple closed convex curve has at least four vertices.
Remark 4. This theorem was first given by Mukhopadhyaya (1909). The following proof was
due to G. Herglotz. It is also true for non-convex curves, but the proof will be more difficult.

18.9 Isoperimetric Inequality in the Plane

Among all simple closed curves having a given length the circle bounds the largest area, and is
the only curve with this property. We shall state the theorem as follows:
Theorem 5. Let L be the length of a simple closed curve C and A be the area it bounds. Then
L – 4A  0 . ...(14)
2
Moreover, the equality sign holds only when C is a circle.

The proof given below is due to E. Schmidt (1939).
We enclose C between two parallel lines g, g’, such that C lies between g, g’ and is tangent to
them at the points P, Q respectively. Let s = 0, s be the parameters of P, Q. Construct a circle C
0
tangent to g, g’ at P, Q respectively. Denote its radius by r and take its center to be the origin
of a coordinate system. Let x(s) = (x (s), x (s)) be the position vector of C, so that
1
2
(x (0), x (0)) = (x (L), x (L)) .
1 2 1 2
As the position vector of C we take (x 1(s), x ), such that
2
x (s) = x (s),
1
1
2
2
x (s) =  r  x (s),0 s  s 0

1
2
2
s
2
=  r  x (s),s   L.
0
1
Denote by A the area bounded by C . Now the area bounded by a closed curve can be expressed
by the line integral
L L 1 L
'
'
'
'
2 
1 
A  x x ds   x x ds   (x x  x x )ds.
2
1
2
1
1
2
0 0 2 0
Applying this to our two curves C and C , we get
L
'
A   x1x ds,
2
0
L L
'
2
'
2 
2 
A   r   x x ds   x x ds.
1
1
0 0
Adding these two equations, we have
L L
'
'
2
'
'
A + r = (x x  x x )ds   (x x  x x ) ds

2
2
2
2
1
2
1
1
1
0 0
L
2
'2
2
'2
  (x  x )(x  x )ds ...(15)
1
2
1
2
0
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