A Reverse Isoperimetric Inequality For Closed
Strictly Convex Plane Curves ∗
Shengliang Pan
Hong Zhang
Department of Mathematics,
East China Normal University,
Shanghai, 200062, P. R. China
email: [email protected]
May 26, 2006
Abstract In this note we present a reverse isoperimetric inequality for closed convex curves,
which states that if γ is a closed strictly convex plane curve with length L and enclosing an area
A, then one gets
L2 ≤ 4π(A + |Ã|),
where |Ã| denotes the absolute area of the domain enclosed by the locus of curvature centers of γ.
The equality holds if and only if γ is a circle.
M athematics Subject Classif ication : 52A38, 52A40
Key words convex curves, Minkowski’s support function, locus of centers of curvature, integral
of radius of curvature, reverse isoperimetric inequality.
1
Introduction
The classical isoperimetric inequality in the Euclidean plane R2 states that for a simple closed
curve γ of length L, enclosing a region of area A, one gets
L2 − 4πA ≥ 0,
(1.1)
and the equality holds if and only if γ is a circle. This fact was known to the ancient Greeks, the
first mathematical proofs were only given in the 19th century by Steiner [13]. There are many
proofs, sharpened forms, generalizations, and applications of this inequality, see for instance [1],
[2], [3], [4], [7], [8], [9], [10], [11], [12], [14], [15], etc., and the literature therein.
In [5], there is a reverse isoperimetric inequality for the plane curves under some assumption
on curvature. Now, in the present note, we establish a reverse isoperimetric inequality for convex
curves, which states that if γ is a closed strictly convex curve in the plane R2 with length L and
enclosing an area A, then we get
L2 ≤ 4π(A + |Ã|),
(1.2)
∗ This work is supported in part by the National Science Foundation of China (10371039), the Shanghai Science
and Technology Committee Program and the Shanghai Priority Academic Discipline
1
where |Ã| denotes the absolute area of the domain enclosed by the locus of curvature centers of γ,
and the equality holds if and only if γ is a circle.
In this note, we first recall some facts about Minkowski’s support function of closed convex
plane curves and then give some properties of the locus of curvature centers of closed strictly convex plane curves, and finally present the above reverse isoperimetric inequality.
Remarks. Since we get the above reverse isoperimetric inequality by the integral of radius of
curvature, our curves must be strictly convex. We wonder if this reverse isoperimetric inequality
holds for any simple closed curves in the plane. And furthermore, it would be interesting to generalize this reverse isoperimetric inequality to the higher dimensional cases.
Acknowledgements. The authors are grateful to Professors Yu Zheng and Xuecheng Pang,
and other colleagues of our geometric inequalities seminar for many useful and stimulating discussions. And especially, the authors would like to thank Professors Xingbin Pan and Yibing Shen
for reading the manuscript of this note and giving us some suggestion and invaluable comments.
2
Minkowski’s Support function for Convex Plane Curves
From now on, without loss of generality, suppose that γ is a smooth regular positively oriented and
closed strictly convex curve in the plane, and take a point O inside γ as the origin of our frame.
Let p be the oriented perpendicular distance from O to the tangent line Γ at a point on γ, and θ be
the oriented angle from the positive x1 -axis to this perpendicular ray. Clearly, p is a single-valued
periodic function of θ with period 2π. The equation of the tangent line Γ can be written in the
form
x1 cos θ + x2 sin θ = p(θ).
(2.1)
γ can be thought of as the envelope of all its tangent lines. To parameterize γ, it suffices to
determine the point (γ1 , γ2 ) of tangency of tangent line (2.1) with γ. According to the method of
determining the envelope we differentiate (2.1) with respect to θ to get −x1 sin θ + x2 cos θ = p0 (θ),
which, together with (2.1), gives us the parametrization of γ in terms of θ as follows
¡
¢ ¡
¢
γ(θ) = γ1 (θ), γ2 (θ) = p(θ) cos θ − p0 (θ) sin θ, p(θ) sin θ + p0 (θ) cos θ .
(2.2)
Thus, if θ and p(θ) are given, one can determine a unique point (γ1 (θ), γ2 (θ)) on the curve γ, and
vice versa. The couple (θ, p(θ)) is usually called the polar tangential coordinate on γ, and p(θ)
Minkowski’s support function of γ.
Differentiating (2.2) with respect to θ yields
¡
¢
γ 0 (θ) = p(θ) + p00 (θ) (− sin θ, cos θ).
(2.3)
Now, the unit tangent vector of γ can be expressed as T = (− sin θ, cos θ) = (dγ1 /ds, dγ2 /ds),
and one also has k = dθ/ds > 0. Thus, one gets
− sin θ =
h
i
dγ1 dθ
dγ1
=
= − (p + p00 ) sin θ k,
ds
dθ ds
cos θ =
2
i
dγ2
dγ2 dθ h
=
= (p + p00 ) cos θ k.
ds
dθ ds
Therefore, the curvature k and the radius of curvature ρ of γ can be calculated by
k(θ) =
dθ
1
=
> 0,
ds
p(θ) + p00 (θ)
ρ(θ) =
ds
= p(θ) + p00 (θ) > 0.
dθ
(2.4)
Now, denote L and A the length of γ and the area it bounds, respectively. Then one can get
Z
Z 2π
Z 2π h
Z 2π
i
00
L=
ds =
ρ(θ)dθ =
p(θ) + p (θ) dθ =
p(θ)dθ,
(2.5)
γ
and
A=
1
2
0
Z
p(θ)ds =
γ
1
2
0
Z
2π
0
0
Z
h
i
i
1 2π h 2
2
p(θ) p(θ) + p00 (θ) dθ =
p (θ) − p0 (θ) dθ.
2 0
(2.6)
(2.5) and (2.6) are known as Cauchy’s formula and Blaschke’s formula, respectively.
3
Some Properties of the Locus of Curvature Centers
Now, we turn to studying the properties of the locus of curvature centers of a closed strictly convex
plane curve γ which is given by (2.2). Let β represent the locus of centers of curvature of γ. Then
¡
¢
β(θ) = β1 (θ), β2 (θ) can be given by
¡
¢
β(θ) = γ(θ) + ρ(θ)N(θ) = − p0 (θ) sin θ − p00 (θ) cos θ, p0 (θ) cos θ − p00 (θ) sin θ ,
(3.1)
where N(θ) = (− cos θ, − sin θ) is the unit inward normal vector field along γ. Thus we calculate
¡
¢
β 0 (θ) = − p0 (θ) + p000 (θ) (cos θ, sin θ)
(3.2)
and
β 00 (θ) =
=
−(p00 + p(4) )(cos θ, sin θ) − (p0 + p000 )(− sin θ, cos θ)
¡
¢
− (p00 + p(4) ) cos θ + (p0 + p000 ) sin θ, −(p00 + p(4) ) sin θ − (p0 + p000 ) cos θ . (3.3)
From the above calculations we get
Proposition 3.1. The locus of curvature centers of a closed strictly convex plane curve has at
least four cusps at which the curvature is infinite.
Proof. From the well-known four-vertex theorem, a closed convex plane curve has at least four
vertices, each of which is defined to be a point where the curvature k of the curve has a relative
extremum; that is to say, at which one gets k 0 = 0. Now, the curvature, denoted by k̃, of β can be
calculated by
k2
β 0 β 00 − β100 β20
1
=
,
k̃ = ¡ 1 2
=
¢
3/2
|p0 + p000 |
|k 0 |
β10 2 + β20 2
(3.4)
from which one can easily arrive at the claimed conclusion.
2
Proposition 3.2 The oriented area of the domain enclosed by β is nonpositive. And moreover, if
β is simple, then orientation of β is the reverse direction against that of the original curve γ and
the total curvature of β is equal to −2π.
Proof. To get the claimed results, we calculate the oriented area, denoted by Ã, of β by Green’s
formula. Since
¡
¢
β1 dβ2 − β2 dβ1 = p0 (θ) p0 (θ) + p000 (θ) dθ,
3
the oriented area of the domain enclosed by β is given by
Z
Z
Z
¢
1
1 2π 0 ¡ 0
1 2π ¡ 0 2
2¢
à =
p (θ) p (θ) + p000 dθ =
p (θ) − p00 dθ.
β1 dβ2 − β2 dβ1 =
2 γ
2 0
2 0
(3.5)
Using the Wirtinger inequality for 2π−periodic C 2 real functions gives us à ≤ 0. If β is simple,
then, from Green’s formula and the fact that à ≤ 0, it follows that orientation of β is the reverse
direction against that of γ and the total curvature of β is equal to −2π..
2
The following result is essential to the proof of the main result of this note.
Proposition 3.3. Let γ be a C 2 closed and strictly convex curve in the plane, ρ the radius of
curvature of γ, A the area enclosed by γ and |Ã| the area enclosed by β. Then we have
Z
2π
ρ2 dθ = 2(A + |Ã|).
(3.6)
0
Proof. From (2.4), we have p00 = ρ − p, and thus,
2
p00 = ρ2 − 2pρ + p2 = ρ2 − 2p(p + p00 ) + p2 = ρ2 − 2pp00 − p2 .
Now, according to (3.5), the absolute area |Ã| can be rewritten as
|Ã|
=
1
2
=
1
2
=
1
2
=
1
2
=
1
2
Z
2π
2
(ρ2 − 2pp00 − p2 − p0 )dθ
0
Z
2π
Z
0
Z
2π
ρ2 −
0
2π
2
ρ dθ −
0
Z
2π
0
Z
pp00 dθ −
2π
Z
pp0 |2π
0
1
ρ2 dθ +
2
Z
1
2
2π
+
2π
0
Z
2π
2
(p2 + p0 )dθ
0
1
p dθ −
2
Z
02
2π
2
(p2 + p0 )dθ
0
02
(p − p2 )dθ
0
ρ2 dθ − A,
0
which completes the proof.
4
2
A Reverse Isoperimetric Inequality
Lemma 4.1. Let γ be a smooth closed and strictly convex curve in the plane, ρ be the radius of
curvature of γ, and L be the length of γ. We have
Z
L2
ρds ≥
.
(4.1)
2π
γ
Furthermore, the equality in (4.1) holds if and only if γ is a circle.
R
R 2π
Proof. Note that γ ρds = 0 ρ2 (θ)dθ. Now, from the Cauchy-Schwartz inequality, we get
Z
2π
2π
0
ρ2 (θ)dθ ≥
³Z
2π
ρ(θ)dθ
0
´2
³Z
´2
ds = L2 .
=
γ
4
(4.2)
And furthermore, the equality in (4.2) holds if and only if ρ is a constant which means that γ is a
circle, because a simple closed plane curve with constant curvature must be a circle.
2
Now, from the above lemma and Proposition 3.3, one can easily get our main result.
Theorem 4.2. (A Reverse Isoperimetric Inequality) If γ is a closed strictly convex plane
curve with length L and enclosing an area A, let |Ã| denote the absolute area of its locus of centers
of curvature, then we get
L2 ≤ 4π(A + |Ã|),
(4.3)
where the equality holds if and only if γ is a circle.
2
Corollary 4.3. Let β be the locus of curvature centers of a closed strictly convex plane curve
γ. Then the area à of β is zero if and only if γ is a circle and thus β is a point which is the center
of γ.
Proof. It is obvious that if γ is a circle, then β is a point, the center of γ, and thus its area
à = 0. Conversely, if à = 0, then, from the classical isoperimetric inequality (1.1) and the reverse
isoperimetric inequality (4.3), it follows that the area A and the length L of γ satisfies L2 = 4πA,
which implies that γ is a circle and, therefore, β is a point.
2
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