March 4, 2002
Inequalities that Imply the
Isoperimetric Inequality
Andrejs Treibergs
University of Utah
Abstract.
The isoperimetric inequality says that the area of any region in the plane
bounded by a curve of a fixed length can never exceed the area of a circle whose
boundary has that length. Moreover, if some region has the same length and area as
some circle, then it must be the circle. There are dozens of proofs. We give several
arguments which depend on more primitive geometric and analytic inequalities.
âThe cicrle is the most simple, and the most perfect figure. â
Poculus.
Commentary on the first book of Euclidâs Elements.
âLo cerchio `e perfettissima figura.â
Dante.
In these notes, I will present a few of my favorite proofs of the isoperimetric
inequality. It is amusing and very instructive to see that many different ideas can
be used to establish the same statement. I will concentrate on proofs based on more
primitive inequalities. Several important proofs are omitted (using the calculus of
variations, integral geometry or Steiner symmetrization), due to the fact that they
have already been discussed by others at previous occasions [B], [BL], [S].
Among all closed curve of length
L
in the plane, how large can the enclosed
area be? Since the regions are enclosed by the curves with the same perimeter,
we are asking for the largest area amongst isoperiometric regions. Which curve
(or curves) encloses the largest possible area? We could ask the dual question:
Among all regions in the plane with prescribed area
A
, at least how long should the
perimiter be? Since the regions all have the same area, we are asking for the smallest
length amongst isopiphenic regions. Which figures realize the least perimeter? The
answer to both questions is the circle. A way to formulate a statement equivalent
to both is in the form of the Isoperimetric Inequality.
1
2
Theorem. Isoperimetric Inequality.
Among all regions in the plane, enclosed
by a piecewise
C
1
boundary curve, with area
A
and perimeter
L
,
4
ĎA
â¤
L
2
.
If equality holds, then the region is a circle.
One doesnât need to assume this much smoothness on the boundary for the
isoperimetric inequality to hold, rectifiability suffices [B],[G]. This is for simplicity.
The student may be more familiar with rigorous arguments under such hypotheses.
1. Some applications of the isoperimetric inequality.
This is a useful inequality. Following Polya[P] letâs illustrate how it may be used
to prove things that may be much trickier to prove using calculus alone.
The largest quadrilateral problem.
Suppose four side lengths
a, b, c, d
are
given. Which quadrilateral in the plane with these side lengths maximizes the
area? The answer is the cyclic quadrilateral, the one whose vertices can be placed
in order around a circle,
i.e.
the one inscribable in some circle.
First of all, can a quadrilateral whose side lengths are four numbers be con-
srtructed at all? The answer is yes provided that the side lengths satisfy the in-
equalities that require that the sum of the lengths of any three of the sides exceeds
the length the remaining one. (See Lemma 1. of the Appendix.) One can also
prove that there is a cyclic quadrilateral with side lengths
a, b, c, d
provided that
the inequalities are strict [B]. (See Lemma 2. of the Appendix.)
Fig. 1.
Suppose we wished to show that among quadrilaterals in the plane with given
four side lengths
a, b, c, d
, that the cyclic quadrilateral has the largest area. This
problem can be handled by calculus, but it may become messy depending on your
formulation. We will show this fact independently of the isoperimetric inequality
later in two ways, using trigonometry (Brahmaguptaâs Inequality) and as a conse-
quence of Ptolemyâs inequality. After all we will wish to deduce the isoperimetric
3
inequality from these! But assuming the isoperimetric inequality gives an easiest
argument that cyclic quadrilaterals have the largest area.
By Lemma 2. we can construct a cyclic quadrilateral with the given side lengths.
Let
ABCD
be its vertices in order on
Z
, the circle containing the vertices. Now
imagine the regions between the sides of the quadrilateral and the circle are made of
a rigid material such as a stiff plastic. Furthermore, imagine that the quadrilateral
with its plastic flanges attached has hinges at the vertices, so that the entire figure
can flex in the plane. The problem of finding the largest area enclosed amongst
these flexings is the same as finding the largest area enclosed by the quadrilateral
is the same as finding the largest area enclosed by the the circular arcs, because the
additional plastic includes a fixed area. By the isoperimetric inequality, the circular
figure has the largest area, thus the cyclic quadrilateral contains the largest area.
This argument applies to polygons with any number of sides. Hence, the polygon
with given side length has largest area if and only if the polygon is inscribed in a
circle.
2.
Princess Didoâs Problem.
Here is another application we shall motivate
by a tale from antiquity [P]. Princess Dido, daughter of a Tyrian king and future
founder of Carthage purchased from the North African natives an amount of land
along the coastline ânot larger than what an oxhide can surround.â She cut the
oxhide into strips and made a very long string of length
L
. And then she faced
the geometrical problem of finding the region of maximal area enclosed by a curve,
given that she is allowed to use the shoreline as part of the region boundary. In the
interior of the continent the answer would be the circle, but on the seashore the
problem is different. Assuming that the seashore is a straight line (the
x
-axis), then
the maximum area enclosed is a semicircle with its diameter on the shoreline and
area
A
=
L
2
/Ď
. This easily follows from the isoperimetric inequality. We regard
the shoreline as a mirror and reflect the curve Î whose points are (
x
(
s
)
, y
(
s
)) to Î
0
whose corresponding points are (
x
(
s
)
,
â
y
(
s
)) on the other side of the axis. Then
the composite curve Î
âŞ
Î
0
has total length 2
L
. The maximal area enclosed is
the circle. Centering the circle on the shoreline means that the upper and lower
semicircles coincide under reflection, thus the lower semicircle is the maximal area
region.
Exercise.
Suppose Dido bargained to buy land along a cape (in
{
(
x, y
) :
y
â¤
â
c
|
x
|}
, for
c >
0 some constant.) What now would be the shape of the largest land
she could surround with a string of length
L
? (see [P].)
3. Ptolemyâs Inequality, Complex Numbers and the Quadrilateral In-
equality.
We envision a theorem about the four vertices of a quadrilateral, but
the statement holds for any four points.
Theorem. Ptolemyâs Inequality.
Let
ABCD
be four points in the plane. Let
a, b, c, d
be the lengths of the sides
AB
,
BC
,
CD
,
DA
and
p, q
be the lengths of the
diagonals
AC
,
BD
, resp. Then
(1)
pq
â¤
ac
+
bd.
4
If equality holds, then
ABCD
is cyclic (its vertices lie in order on a circle) or itâs
contained in a line segment (the length of one side equals the sum of the other
three.)
A
B
C
D
a
d
b
c
β
δ
p
q
Îą
Îł
Fig. 2. Quadrilateral for Ptolemyâs Theorem.
It may happen that the quadrilateral degenerates to a triangle or bigon or a point,
all of which we regard as cyclic figures. If equality holds in any of these cases, then
cyclicity implies that the figures are convex. Under equality, the quadrilateral may
also degenerate to a line segment in which the length of one side equals the sum
of the other three. Then the vertices are not contained in a circle, but since the
proof uses projective transformations, the line may be regarded as a circle through
infinity.
Proof.
We consider two cases: that one of the points is distinct from the other three
or not.
We use a little bit of complex arithmetic for the first case where
A
is distinct
from all
BCD
. If the vertices
ABCD
are assigned to complex numbers 0,
z
1
,
z
2
and
z
3
with
z
i
6
= 0 then
a
=
|
z
1
|
,
b
=
|
z
1
â
z
2
|
,
c
=
|
z
2
â
z
3
|
,
d
=
|
z
3
|
,
p
=
|
z
2
|
and
q
=
|
z
1
â
z
3
|
. Now we consider triangle inequality for the inversions
1
z
1
â
1
z
3
â¤
1
z
1
â
1
z
2
+
1
z
2
â
1
z
3
.
or
|
z
1
â
z
3
|
|
z
1
||
z
3
|
â¤
|
z
1
â
z
2
|
|
z
1
||
z
2
|
+
|
z
2
â
z
3
|
|
z
2
||
z
3
|
.
Multiplying by
|
z
1
||
z
2
||
z
3
|
gives
|
z
2
||
z
1
â
z
3
| ⤠|
z
3
||
z
1
â
z
2
|
+
|
z
1
||
z
2
â
z
3
|
which is the same as (1). Equality in the triangle inequality holds if the points
z
â
1
1
,
z
â
1
2
,
z
â
1
3
,
â
lie (in order) on a line.
5
If that line contains the origin, then
ABCD
lie on a line through the origin.
Since
z
â
1
2
lies between
z
â
1
1
and
z
â
1
3
then, depending on where 0 lies relative to the
other points, the figure
ABCD
has the property that one of the sides has a length
equal to the sum of the other three. Unless the three points coincide
B
=
C
=
D
,
the figure is not cyclic.
However, if the line does not pass through the origin, inversion maps it to a circle
through the origin. Thus
z
1
,
z
2
,
z
3
, 0 lie in order on a circle through the origin.
It may happen that two points
B
=
C
or
C
=
D
coincide yielding a triangle. But,
since the points occur in order around the circle,
B
=
D
implies
B
=
C
=
D
or the
figure is a bigon.
The last case is if none of the points is distinct from the other three. If all points
coincide
A
=
B
=
C
=
D
and (1) is 0 = 0. There three remaining degenerate
subcases. If
A
=
B
6
=
C
=
D
then
a
=
c
and
b
=
d
=
p
=
q >
0 and equality
holds in (1). If
A
=
C
6
=
B
=
D
then
p
=
q
= 0 and
a
=
b
=
c
=
d >
0 and
strict inequality holds trivially in (1). If
A
=
D
6
=
B
=
C
then
b
=
d
= 0 and
a
=
c
=
p
=
q >
0 and equality holds in (1).
An isoperimetric inequality for quadrilaterals is based on a sharp upper bound
for area of the quadrilateral. By the triangle inequality, the length of each side is
less than the sum of the other three sides. Conversely, this condition on lengths
is sufficient to construct a quadrilateral in the plane with those given side lengths.
If one side length equals the sum of the other three sides then the quadrilateral
degenerates to a line segment.
In order to avoid fussing about signed areas or worrying about cases in which
opposite sides cross, we shall insist that by quadrilateral we mean a figure which
is bounded by a simple (non-selfintersecting) polygonal curve consisting of four
line segments connected end to end. Area then is the measure of the bounded
region enclosed by the polygon. We allow some of the side lengths to be zero, thus
triangles, bigons and points are degenerate examples. In any case, these degenrerate
figures may be approximated by nondegenerate ones, so that both the length and
area of a quadrilateral approach those of the limiting degenerate figure and so that
the inequality for nondegenerate figures may be passed to the limit. The maximum
property for cyclic quadrilaterals was first observed by Steiner.
Maximum Property for Cyclic Quadrilaterals.
Let
ABCD
be the vertices of
a quadrilateral in the plane. Let
a, b, c, d
be the lengths of the sides
AB
,
BC
,
CD
,
DA
, resp., and
F
the area. Then
(2)
16
F
2
â¤
(
a
+
b
+
c
â
d
)(
a
+
b
â
c
+
d
)(
a
â
b
+
c
+
d
)(
â
a
+
b
+
c
+
d
)
.
If equality holds, then
ABCD
is cyclic (its vertices lie on a circle) or itâs contained
in a line segment (with one side as long as the sum of the other three lengths.)
Proof.
A proof of this statement may be based on Calculus or on the observation
that the maximum area of a triangle for which two sides are given occurs for the
right triangle. (see Polya [P].)
6
We prefer to deduce it from Ptolemyâs theorem. If the quadrilateral degenerates
to a line segment, then
F
is zero and (2) holds. If the right side of (2) is also zero,
then one of the terms is zero and one side is the sum of the other three.
If
ABCD
is a convex quadrilateral, then let
E
be the intersection of
AC
and
BD
and
p
1
, q
1
, p
2
, q
2
the lengths of
EC
,
ED
,
EA
,
EB
resp. Note that
θ
=
CED
=
AEB
and
Ď
â
θ
=
DEA
=
BEC
. Then the area of the quadrilateral is the
sum of the areas of the triangles
2
F
=
p
1
q
1
sin
θ
+
p
2
q
1
sin(
Ď
â
θ
) +
p
2
q
2
sin
θ
+
p
1
q
2
sin(
Ď
â
θ
)
=(
p
1
q
1
+
p
2
q
1
+
p
2
q
2
+
p
1
q
2
) sin
θ
=(
p
1
+
p
2
)(
q
1
+
q
2
) sin
θ
=
pq
sin
θ.
On the other hand, the lengths of the four sides are expressed by the cosine law
a
2
=
p
2
2
+
q
2
2
â
2
p
2
q
2
cos
θ
b
2
=
p
2
1
+
q
2
2
â
2
p
1
q
2
cos(
Ď
â
θ
)
c
2
=
p
2
1
+
q
2
1
â
2
p
1
q
1
cos
θ
d
2
=
p
2
2
+
q
2
1
â
2
p
2
q
1
cos(
Ď
â
θ
)
Thus
(3)
a
2
â
b
2
+
c
2
â
d
2
=
â
2(
p
2
q
2
+
p
1
q
2
+
p
1
q
1
+
p
2
q
1
) cos
θ
=
â
2(
p
1
+
p
2
)(
q
1
+
q
2
) cos
θ
=
â
2
pq
cos
θ
Thus, rewriting the area, we get using Ptolemyâs inequality, and (3)
16
F
2
=4
p
2
q
2
sin
2
θ
=4
p
2
q
2
â
4
p
2
q
2
cos
2
θ
â¤
4(
ac
+
bd
)
2
â
(
a
2
â
b
2
+
c
2
â
d
2
)
2
=[2
ac
+ 2
bd
+
a
2
â
b
2
+
c
2
â
d
2
][2
ac
+ 2
bd
â
a
2
+
b
2
â
c
2
+
d
2
]
=[(
a
+
c
)
2
â
(
b
â
d
)
2
][(
b
+
d
)
2
â
(
a
â
c
)
2
]
=(
a
+
b
+
c
â
d
)(
a
â
b
+
c
+
d
)(
a
+
b
â
c
+
d
)(
â
a
+
b
+
c
+
d
)
.
thus (2) holds. If equality in (2) holds, then there is equality in Ptolemyâs inequality
and the points
ABCD
lie in order on a circle.
Note that if one of the sides of the quadrilateral degenerates (e.g.
d
= 0), then
the resulting figure is a triangle, which is cyclic. It follows that (2) is equality,
which is nothing more than
Heronâs formula
for the area of a triangle:
4 Area(
ABC
) =
p
(
a
+
b
+
c
)(
a
+
b
â
c
)(
a
â
b
+
c
)(
â
a
+
b
+
c
)
.
7
4. Brahmagupta Inequality.
Brahmagupta was an mathematician from Ujjain, India, living in the early sev-
enth century. Brahmagupta was primarily concerned with number theory and in-
tegral solutions of equations. He found many integral triangles and quadrilarerals
which motivated his generalization of Heronâs formula for the area of a triangle in
terms of its side lengths to the case of the quadrilateral. The basic formula he gave
applied to cyclic quadrilaterals, those whose vertices occur in order around a circle.
By giving a trigonometric proof, it is easy to adjust his formula for all quadrilaterals
in the plane. An immediate corollary is the maximum property for quadrilaterals
obtained by Steiner. In fact, Brahmaguptaâs formula gives an expression for the
error term in Steinerâs inequality. There are several equivalent formulations of this
formula.
Brahmaguptasâs Quadrilateral Formula.
Suppose a quadrilateral
P QRS
is
given in the plane whose side lengths are
a
=
P Q
,
b
=
QR
,
c
=
RS
, and
d
=
SP
order around the quadrilateral and whose interior angles are
Îą
=
P QR
,
β
=
QRSZ
,
Îł
=
RSP
,
δ
=
SP Q
then the area of the quadrilateral
F
is given by
16
F
2
= (
a
+
b
+
c
â
d
)(
a
+
b
â
c
+
d
)(
a
â
b
+
c
+
d
)(
â
a
+
b
+
c
+
d
)
â
16
abcd
cos
2
Îą
+
Îł
2
Equality implies that that the quadrilateral is cyclic (the vertices occur in order
around some circle) or that the vertices occur on a line segment such that one side
length is the sum of the other three side lengths.
P
Q
R
S
a
d
b
c
β
δ
p
Îą
Îł
Quadrilateral used in the proof of Brahmaguptaâs Formula
Corollary. Maximum Property for Cyclic Quadrilaterals.
Among quadri-
lateral given in the plane with side lengths
a, b, c, d
(satisfying the triangle relations,
each side length is less than the sum of the other three, e.g.
a
â¤
b
+
c
+
d
) the largest
area is attained for a cyclic quadrilateral or a degenerate one whose verticies lie on
a line such that one side length is the sum of the other three.
8
Proof.
Take the angles as in Brahmaguptaâs formula. Since the angles may be taken
so 0
â¤
Îą
+
Îł
= 2
Ď
â
β
â
δ
â¤
2
Ď
in Brahmaguptaâ formula, the max occurs when
Îą
+
Îł
=
Ď
so
β
+
δ
=
Ď
or one of the sides, say
d
= 0. If a side has length zero, the
figure degenerates to a triangle, biangle or point, which are all cyclic quadrilaterals.
If all lengths are nonzero and one of the angles, say
Îą
= 0 is zero, then the other
Îł
=
Ď
and
S
is an interior point of the line segment
RP
. It follows that the four
points are collinear and
a
=
b
+
c
+
d
or
b
=
c
+
d
+
a
. Similarly if
Îą
=
Ď
then
Îł
= 0
and so either
a
+
b
+
c
=
d
or
d
+
a
+
b
=
c
. Let
p
=
P R
. If 0
< Îą < Ď
then the chord
P R
of the circle through
P QR
subtends an angle 2
Îą
. Similarly, the chord
P R
of
the circle through
RSP
subtends an angle 2
Îł
. But since 2
Îą
+ 2
Îł
= 2
Ď
the circles
agree. and the two triangles
P QR
and
RSP
are on opposite sides of the chord
P R
since the subtended angles are on opposite sides of the circle. If
Îą
=
Îł
=
Ď/
2 then
In other words, the quadrilateral is cyclic.
Proof of Brahmagupataâs formula.
The idea is that the area of a quadrilateral is
the sum of the areas of the triangles on opposite sides of a diagonal
P R
Thus
2
F
= 2 Area(
P QR
) + 2 Area(
RSP
) =
ab
sin
Îą
+
cd
sin
Îł.
Squaring,
(4)
16
F
2
=4(
ab
sin
Îą
+
cd
sin
Îł
)
2
=4(
a
2
b
2
sin
2
Îą
+ 2
abcd
sin
Îą
sin
Îł
+
c
2
d
2
sin
2
Îł
)
=4(
a
2
b
2
â
a
2
b
2
cos
2
Îą
+ 2
abcd
sin
Îą
sin
Îł
+
c
2
d
2
â
c
2
d
2
cos
2
Îł
â
2
abcd
cos
Îą
cos
Îł
+ 2
abcd
cos
Îą
cos
Îł
)
=4
a
2
b
2
+ 4
c
2
d
2
â
(2
ab
cos
Îą
â
2
cd
cos
Îł
)
2
â
8
abcd
cos (
Îą
+
Îł
)
.
Using the fact that the diagonal
p
=
P R
can be expressed using the cosine law
from either triangle,
p
2
=
a
2
+
b
2
â
2
ab
cos
Îą
=
c
2
+
d
2
â
2
cd
cos
Îł,
we get
(5)
2
ab
cos
Îą
â
2
cd
cos
Îł
=
a
2
+
b
2
â
c
2
â
d
2
.
The double angle formula implies the identity
(6)
cos(
Îą
+
Îł
) = 2 cos
2
Îą
+
Îł
2
â
1
9
Substituting (5) and (6) into (4) yields
16
F
2
= 4
a
2
b
2
+ 4
c
2
d
2
â
a
2
+
b
2
â
c
2
â
d
2
2
â
16
abcd
cos
2
Îą
+
Îł
2
+ 8
abcd
=(2
ab
+ 2
cd
)
2
â
a
2
+
b
2
â
c
2
â
d
2
2
â
16
abcd
cos
2
Îą
+
Îł
2
=(2
ab
+ 2
cd
+
a
2
+
b
2
â
c
2
â
d
2
)(2
ab
+ 2
cd
â
a
2
â
b
2
+
c
2
+
d
2
)
â
16
abcd
cos
2
Îą
+
Îł
2
= (
a
+
b
)
2
â
(
c
â
d
)
2
(
c
+
d
)
2
â
(
a
â
b
)
2
â
16
abcd
cos
2
Îą
+
Îł
2
=(
a
+
b
+
c
â
d
)(
a
+
b
â
c
+
d
)(
a
â
b
+
c
+
d
)(
â
a
+
b
+
c
+
d
)
â
16
abcd
cos
2
Îą
+
Îł
2
as to be proved.
Exercise.
Give another proof of Brahmaguptaâs inequality based on Ptolemyâs
inequality.
Mechanical Argument.
Consider the following mechanical âproofâ of the quad-
rilateral inequality. Imagine that the quadrilateral is allowed to flex in three space.
Attach the ends of four sufficiently long but equally long new edges to the existing
vertices and bind the four opposite ends together to make a new vertex. Assume
that the whole structure is allowed to flex at the vertices in three space. There is still
a degree of freedom but the quadrilateral is no longer necessarily planar. However,
if the whole apparatus is dropped onto a planar surface (table top), with the new
vertex held above the quadrilateral base, it settles to a planar quadrilateral, with
the four new edges forming a pyramid above the base. Orthogonally projecting the
four new edges to the table top give four equally long segments in the plane of the
quadrilateral connecting at an interior vertex, showing that the mechanically stable
pyramid configuration has a cyclic quadrilateral base. One could build a model by
tying drinking straws together with string.
Exercise (Solution Unknown to the author!)
Show that there is only
one physically stable configuration (one with the lowest center of gravity) and this
one has the largest area of the base. Perhaps one should look among pyramids that
have the property that the new vertex must have a projection inside the projection
of the base.
5. Steinerâs Four Hinge âProof â.
Jakob Steiner(1796â1863), a self made Swiss farmerâs son and contemporary of
Gauss was the foremost âsynthetic geometer.â He hated the use of algebra and
analysis and distrusted figures and once wrote âCalculating replaces thinking while
geometry stimulates itâ[St]. He proposed several arguments to prove that the circle
is the largest figure with given boundary length. One very important method
is Steiner symmetrization [B]. The other is his four-hinge method that has great
intuitive appeal, but is limited to two dimensions.
10
If the curve bounding the region is not a circle, then there are four points on the
boundary which are not cyclic. Consider the quadrilateral with these vertices, and
regard as the rest of the domain rigidly attached to the sides of the quadrilateral
as before. Suppose that the boundary can be articulated at these points. Then
flexing the quadrilateral so to move the points to a circle results in a larger area by
the quadrilateral inequality, with the same boundary length. Call this the âFour
Hinge Maneuver.â Steiner argued that for noncircular figures, the area could be
increased, and therefore the circle has the largest area.
The fact that this is not a proof because there is no demonstration of the exis-
tence of a maximizing figure was pointed out to Steiner by Dirichlet. WeierstraĂ
eventually gave the first rigorous demonstration. If we denote the application of
a Four Hinge Maneuver to a noncircular domain ⌠bounded by a piecewise
C
1
curve by
F
(âŚ), then we can envision a sequence âŚ
0
= âŚ, âŚ
n
+1
=
F
(âŚ
n
), for
n
= 0
,
1
,
2
, . . .
such that Area(âŚ
n
+1
)
>
Area(âŚ
n
) and L(
â
âŚ
n
+1
) = L(âŚ
n
). If it
could be proved that for a (subsequence taken from a) sequence of Four Hinge Ma-
neuvers, âŚ
n
â
B
an
n
â â
where
B
is a closed unit disk, using the semicontinuity
(see Appendix), then the proof of the isoperimetric inequality would be given by
Area(âŚ)
<
lim
n
ââ
Area âŚ
n
= Area
lim
n
ââ
âŚ
n
= Area(
B
)
,
L(
â
âŚ) = lim inf
n
ââ
L(
â
âŚ
n
)
âĽ
L
â
lim
n
ââ
âŚ
n
= L(
âB
)
.
Here area is in the sense of Lebesgue measure, length is in the sense of one di-
mensional Hausdorff measure (mass of a rectifiable 1-current) and convergence of
domains is in the sense of Hausdorff distance (see the definitions and discussion in
the Appendix about Hausdorff convergence and the continuity of areas and lengths.)
This program has been carried out for Steiner Symmetrization.
One of the issues is whether a Four Hinge Maneuver sequence can be found that
actually forces the domains to converge to the circle. Consider the following sugges-
tive reasoning. Starting with the domain ⌠with sufficiently nice boundary, choose
a point
X
0
â
â
âŚ
,
and some
Îľ >
0. Starting from
X
0
, go both directions around
â
âŚ
and mark off points
X
0
1
, X
0
2
, . . . , X
0
m
going one direction
X
00
1
, X
00
2
, . . . , X
00
m
in the
other whose
R
2
distances apart are
Îľ
,
i.e.
Îľ
= dist(
X
0
m
, X
0
m
â
1
) = dist(
X
0
m
â
1
, X
0
m
â
2
) =
¡ ¡ ¡
= dist(
X
0
1
, X
0
) = dist(
X
0
, X
00
1
) =
¡ ¡ ¡
= dist(
X
00
m
â
1
, X
00
m
)
.
Then construct
m
â
1 quadrilaterals
Q
k
=
X
0
k
X
0
k
+1
X
00
k
+1
X
00
k
and apply the Four
Hinge Maneuver
m
â
1 times based on the
Q
k
âs. Since the two sides
Îľ
= dist(
X
0
k
, X
0
k
+1
) = dist(
X
00
k
, X
00
k
+1
)
the result of the maneuver is a trapezoid whose sides
F
(
X
0
k
+1
X
00
k
+1
) and
F
(
X
0
k
X
00
k
)
are parallel an have a common perpendicular bisector. Lining up
F
m
â
1
(âŚ) on this
11
bisector results in a domain that is nearly symmetric about its axis, has the same
boundary length as ⌠but has larger area. If
Îľ
is small, then the polygon length
approximates the length of
â
⌠and so is Hausdorff close to the polygon which is
the convex hull of the new vertices, which is symmetric along a line. Now
Îľ
can
be made to decrease and the choice of starting point, hence direction of symmetry,
can be made arbitrarily, thus the sequence of Four Hinge Maneuvers converges to
a domain such that every line is a line of symmetry,
i.e.
a circle. One expects
that a proof of convergence of such a sequence will follow the lines of the proofs
of convergence of Steiner Symmetrization sequences, although the author doesnât
know if this has been done. (The circle in the figure is a radius
Îľ
-circle.)
6"
11"
1'-3"
1'-5"
1'-4"
1'-2"
1'-1"
1'-2"
1'-1"
1'-0"
8"
6"
8"
9"
8"
7"
7"
7"
7"
6"
6"
8"
9"
8"
7"
7"
7"
7"
6"
Several Four Hinge Maneuvers
Fig. 3. Left region is converted by Four Hinge Maneuvers to the nearly symmetric one on right.
6. Hurwitzâs proof using the Wirtinger inequality.
The Wirtinger inequality bounds the
L
2
norm of a function by the
L
2
norm of its
derivative. In more general settings, the inequality is also known as the
Poincar´e
Inequality.
[GT,PS]. The best constant in the Poincar´e Inequality is known as
the first eigenvalue of the Laplace operator, and has been the inspiration of much
geometric study (see, e.g. [LT].) We state stronger hypotheses than necessary.
Theorem. Wirtingerâs inequality.
Let
f
(
θ
)
be a piecewise
C
1
(
R
)
function with
period
2
Ď
(for all
θ
,
f
(
θ
+ 2
Ď
) =
f
(
θ
))
. Let
ÂŻ
f
denote the mean falue of
f
ÂŻ
f
=
1
2
Ď
Z
2
Ď
0
f
(
θ
)
dθ.
Then
Z
2
Ď
0
f
(
θ
)
â
ÂŻ
f
2
dθ
â¤
Z
2
Ď
0
(
f
0
(
θ
))
2
dθ.
12
Equality holds if and only if
(7)
f
(
θ
) = ÂŻ
f
+
a
cos
θ
+
b
sin
θ
for some constants
a, b
.
Proof.
The idea is to express
f
and
f
0
in Fourier series. Thus, since the derivative
is bounded and
f
is continuous, the Fourier series converges at all
θ
[W]
f
(
θ
) =
a
0
2
+
â
X
k
=1
{
a
k
cos
kθ
+
b
k
sin
kθ
}
where the Fourier coefficients are determined by formally multiplying by sin
mθ
or
cos
mθ
and integrating to get
a
m
=
1
Ď
Z
2
Ď
0
f
(
θ
) cos
mθ dθ,
b
m
=
1
Ď
Z
2
Ď
0
f
(
θ
) sin
mθ dθ,
hence 2 ÂŻ
f
=
a
0
. Since the sines and cosines are complete, the Parseval equation
holds
(8)
Z
2
Ď
0
f
â
ÂŻ
f
2
=
Ď
â
X
k
=1
a
2
k
+
b
2
k
.
Formally, this is the integral of the square of the series, where after multiplying out
and integrating, terms like
R
cos
mθ
sin
kθ
= 0 or
R
cos
mθ
cos
kθ
= 0 if
m
6
=
k
drop
out and terms like
R
sin
2
kθ
=
Ď
contribute
Ď
to the sum. The Fourier Series for
the derivative is given by
f
0
(
θ
)
âź
â
X
k
=1
{â
ka
k
sin
kθ
+
kb
k
cos
kθ
}
Since
f
0
is square integrable, Besselâs inequality gives
(9)
Ď
â
X
k
=1
k
2
a
2
k
+
b
2
k
â¤
Z
2
Ď
0
(
f
0
)
2
.
Wirtingers inequality is deduced form (8) and (9) since
Z
2
Ď
0
(
f
0
)
2
â
Z
2
Ď
0
f
â
ÂŻ
f
2
âĽ
Ď
â
X
k
=2
(
k
2
â
1)
a
2
k
+
b
2
k
âĽ
0
.
Equality implies that for
k
âĽ
2, (
k
2
â
1)
a
2
k
+
b
2
k
= 0 so
a
k
=
b
k
= 0, thus
f
takes
the form (7).
13
Recall Greenâs theorem. If
p
and
q
are differentiable functions on the plane and
Î is a piecwise
C
1
curve bounding the region ⌠then
I
Î
p dx
+
q dy
=
Z Z
âŚ
(
q
x
â
p
y
)
dx dy.
If we take
q
=
x
and
p
= 0 then Greenâs theorem says
(10)
I
Î
x dy
= Area(âŚ)
.
The same formula can be used to make sense of area even for curves that are merely
rectifiable, namely, those whose length is the limit of lengths of approximating
polygonal curves. (see e.g. [B].)
Hurwitzâs proof of the Isoperimetric Inequality.
We suppose that the boundary
curve has length
L
is parameterized by arclength, thus given by two piecewise
C
1
and
L
periodic functions
x
(
s
),
y
(
s
) that satisfy
dx
ds
2
+
dy
ds
2
= 1
.
We convert to 2
Ď
periodic functions
f
(
θ
) =
x
Lθ
2
Ď
,
g
(
θ
) =
y
Lθ
2
Ď
so that writing â
0
â =
d/dθ
gives
(11)
(
f
0
)
2
+ (
g
0
)
2
=
L
2
4
Ď
2
.
We now simply to estimate the area integral (10). Using
R
g
0
dθ
= 0, the Wirtinger
inequality and (11),
2
A
=2
Z
2
Ď
0
f g
0
dθ
= 2
Z
2
Ď
0
(
f
â
ÂŻ
f
)
g
0
dθ
=
Z
2
Ď
0
(
f
â
ÂŻ
f
)
2
+ (
g
0
)
2
â
(
f
â
ÂŻ
f
â
g
0
)
2
dθ
â¤
Z
2
Ď
0
(
f
0
)
2
+ (
g
0
)
2
dθ
=
Z
2
Ď
0
L
2
4
Ď
2
dθ
=
L
2
2
Ď
,
which is the isoperimetric inequality. Equality forces equality in the Wirtinger
inequality so
f
(
θ
) = ÂŻ
f
+
a
cos
θ
+
b
sin
θ
14
for some constants
a, b
. Equality also forces the dropped term to vanish
Z
2
Ď
0
(
f
â
ÂŻ
f
â
g
0
)
2
dθ
= 0
so that
g
0
=
f
â
ÂŻ
f .
Hence
g
(
θ
) = ÂŻ
g
+
a
sin
θ
â
b
cos
θ.
Hence (11) implies
a
2
+
b
2
=
L
2
4
Ď
2
so (
x
(
s
)
, y
(
s
)) is a circle of radius
L/
2
Ď
.
7. Steinerâs Inequality for Parallel Sets.
Let âŚ
â
R
2
be a closed bounded region with piecewise
C
1
boundary. For
Ď
âĽ
0
the outer
Ď
-parallel set is
âŚ
Ď
= âŚ
ĎB
=
{
x
â
E
2
: dist(
x,
âŚ)
â¤
Ď
}
the set of points in the plane whose distance to
A
is at most
Ď
. (
B
is the closed
unit ball.)
Theorem. Steinerâs Inequality.
Let
âŚ
â
E
2
be a closed and bounded set with
piecewise
C
1
boundary whose area is
A
and whose boundary has length
L
. Let
Ď
âĽ
0
. Then the
Ď
-parallel set satisfies the inequalities
Area(âŚ
Ď
)
â¤
A
+
LĎ
+
ĎĎ
2
,
L(
â
âŚ
Ď
)
â¤
L
+ 2
ĎĎ.
If
âŚ
is convex, then the inequalities are equalities.
15
+
+
+
-
+
+
P
Pr
r
+
+
+
+
+
-
++
Fig. 4. Parallel Set for a polygon.
Proof.
Let
P
n
â
⌠be a sequence of polygons (convex if ⌠is convex) converging to
⌠whose boundary curves approach the boundary of ⌠so that Area(
P
n
)
â
A
and
L(
âP
n
)
â
L
as
n
â â
. Since Minkowski addition is continuous with respect to
Hausdorff convergence (see [H]) then (
P
n
)
Ď
â
âŚ
Ď
and furthermore Area((
P
n
)
Ď
)
â
A
and L(
âP
n
)
â
L
as
n
â â
. Thus if we can prove the Steiner inequality for
polygons then the general one follows. But for polygons, the Steiner inequality is
almost obvious (see Figure 4.) It is proved in the appendix.
8. Brunnâs Inequality and Minkowskiâs Proof.
The
Minkowski Addition
of two arbitrary sets
S, T
â
R
2
in the plane is defined
to be
aS
bT
:=
{
x
+
y
:
x
â
aS
and
y
â
bT
}
where
a, b
are nonnegative numbers and
aS
is the dilation by factor
a
aS
=
{
ax
:
x
â
S
}
.
For example, the Minkowski sum of the rectangles is a rectangle
(12)
r
([0
, a
]
Ă
[0
, b
])
s
([0
, c
]
Ă
[0
, d
]) = [0
, ra
+
sc
]
Ă
[0
, rb
+
sd
]
.
16
A
B
A + B
Minkowski Addition
Fig. 5. The Minkowski Sum of a triangle and a rectangle.
The theorem of Brunn-Minkowski says that since the Minkowski addition tends
to âround outâ the figures being added, the area of the added figure exceeds the
area of the summands.
Theorem. Brunnâs Inequality.
Let
A, B
â
R
2
be arbitrary bounded measurable
sets in the plane. Then,
(13)
p
Area(
A
B
)
âĽ
p
Area(
A
) +
p
Area(
B
)
.
Minkowski proved that equality holds if and only if
A
and
B
are homothetic.
Two figures
A
and
B
are
homothetic
if and only if they are similar and are similarly
situated, which means there is a point
x
and
r
âĽ
0 so that
A
=
rB
{
x
}
.
Proof.
We choose to present the proof of Minkowski, given in [F], using induction.
The inequality is proved for finite unions of rectangles first and then a limiting
process gives the general statement. Suppose that
A
=
âŞ
n
i
=1
R
i
and
B
=
âŞ
m
j
=1
S
j
where
R
i
and
S
j
are pairwise disjoint open rectangles, that is
R
i
âŠ
R
j
=
â
and
S
i
âŠ
S
j
=
â
if
i
6
=
j
. The proof is based on induction on
`
=
m
+
n
. For
`
= 2 there
17
are two rectangles as in (12). The area
(14)
Area((
a, b
)
Ă
(
c, d
)
(
e, f
)
Ă
(
g, h
)) = Area((
a
+
e, b
+
f
)
Ă
(
c
+
g, d
+
h
))
= (
b
â
a
+
f
â
e
)(
d
â
c
+
h
â
g
)
= (
b
â
a
)(
d
â
c
) + (
f
â
e
)(
h
â
g
) + (
b
â
a
)(
h
â
g
) + (
f
â
e
)(
d
â
c
)
âĽ
(
b
â
a
)(
d
â
c
) + (
f
â
e
)(
h
â
g
) + 2
p
(
b
â
a
)(
h
â
g
)(
f
â
e
)(
d
â
c
)
=
p
(
b
â
a
)(
d
â
c
) +
p
(
f
â
e
)(
h
â
g
)
2
=
p
Area((
a, b
)
Ă
(
c, d
)) +
p
Area((
e, f
)
Ă
(
g, h
))
2
,
where we have used the Arithmetic-Geometric Mean Inequality
|
X
|
+
|
Y
|
2
â
p
|
X
||
Y
|
=
1
2
p
|
X
| â
p
|
Y
|
2
âĽ
0
.
Equality in the Arithmetic-Geometric Mean Inequality implies
|
X
|
=
|
Y
|
, so that
equality in (14) implies (
b
â
a
)(
h
â
g
) = (
f
â
e
)(
d
â
c
) or both rectangles have the
same ratio of height to width. In other words,
R
and
S
are homothetic.
Now assume the induction hypothesis: suppose that (13) holds for
A
=
âŞ
n
i
=1
R
i
and
B
=
âŞ
m
j
=1
S
j
with
m
+
n
â¤
`
â
1. For
A
and
B
so that
m
+
n
=
`
, we
may arrange that
n
âĽ
2. Then some vertical or horizontal plane, say
x
=
x
1
, can
be placed between two rectangles. Let
R
0
i
=
R
i
⊠{
(
x, y
) :
x < x
1
}
and
R
00
i
=
R
i
⊠{
(
x, y
) :
x > x
1
}
and put
A
0
=
âŞ
i
R
0
i
and
A
00
=
âŞ
i
R
00
i
. By choice of the plane,
the number of nonempty rectangles in #
A
0
< n
and #
A
00
< n
, but both
A
0
and
A
00
are nonempty. Select a second plane
x
=
x
2
and set
S
0
i
=
S
i
⊠{
(
x, y
) :
x < x
2
}
and
S
00
i
=
S
i
⊠{
(
x, y
) :
x > x
2
}
and put
B
0
=
âŞ
i
S
0
i
and
B
00
=
âŞ
i
S
00
i
. Note that
#
B
0
â¤
m
and #
B
00
â¤
m
.
x
2
can be chosen so that the area fraction is preserved
θ
=
Area(
A
0
)
Area(
A
0
) + Area(
A
00
)
=
Area(
B
0
)
Area(
B
0
) + Area(
B
00
)
.
By definiton of Minkowski sum,
A
B
â
A
0
B
0
âŞ
A
00
B
00
. Furthermore, observe
that
A
0
B
0
is to the left and
A
00
B
00
is to the right of the plane
x
=
x
1
+
x
2
,
so they are disjoint sets. Now we may use the additivity of area and the induction
hypothesis on
A
0
B
0
and
A
00
B
00
.
Area(
A
B
)
âĽ
Area(
A
0
B
0
) + Area(
A
00
B
00
)
âĽ
p
Area(
A
0
) +
p
Area(
B
0
)
2
+
p
Area(
A
00
) +
p
Area(
B
00
)
2
=
θ
p
Area(
A
) +
p
Area(
B
)
2
+ (1
â
θ
)
p
Area(
A
) +
p
Area(
B
)
2
=
p
Area(
A
) +
p
Area(
B
)
2
.
18
Thus the induction step is complete.
Finally every compact region can be realized as the intersection of a decreasing
sequence of open sets
A
n
â
A
n
+1
so that
A
=
âŠ
n
A
n
.
A
n
can be taken as the
interiors of a union of finitely many closed squares. For each
Îľ
= 2
â
n
>
0 consider
the closed squares in the grid of side
Îľ
which meet the set. Then the interior of the
union of these squares is
A
n
. Removing the edges of the squares along gridlines
A
0
n
results in a set with the same area. The result follows since Lebesgue measure
of the limit is limit of the Lebesgue measure for decreasing sequences. Since the
Minkowski sum of a decreasing set of opens is itself a decreasing set of opens, it
follows that
p
Area(
A
B
) = lim
n
ââ
p
Area(
A
n
B
n
)
âĽ
lim
n
ââ
p
Area(
A
0
n
B
0
n
)
âĽ
lim
n
ââ
p
Area(
A
0
n
) +
p
Area(
B
0
n
)
=
p
Area(
A
) +
p
Area(
B
)
and we are done.
Brunnâs inequality can be proved by other means.
Exercise.
Give a proof of Brunnâs inequality using the Wirtinger inequality.
Hint: Suppose
K
â
R
2
is strictly convex with
C
2
boundary and contains a neigh-
borhood of the origin. For each
θ
â
[0
,
2
Ď
) let
N
(
θ
) = (cos
θ,
sin
θ
) be the outward
normal vector and
h
(
θ
) be the support function in the
θ
direction. That is, the
line given by
N
(
θ
)
¡
X
=
h
(
θ
) is tangent to
K
so
h
(
θ
) is the distance from the
origin to the supporting line. Show that if the curve is parameterized by
θ
then
the arclength is
ds
= (
h
00
+
h
)
dθ
so the curvature is
Îş
= 1
/
(
h
+
h
00
), the area is
2
A
=
R
2
Ď
0
h
(
θ
)
ds
=
R
2
Ď
0
h
2
+
hh
00
dθ
=
R
2
Ď
0
h
2
â
(
h
0
)
2
dθ
and that the length is
L
=
R
âK
ds
=
R
2
Ď
0
h
+
h
00
dθ
=
R
2
Ď
0
h dθ.
Minkowskiâs proof of the isoperimetric inequality using the Brunn Inequality.
The
result follows immediately from the Brunn inequality and Steinerâs inequality. Thus,
taking
A
to be the area of a set ⌠with piecewise
C
1
boundary and
B
the closed
unit ball, we obtain for any
r
âĽ
0
A
+
Lr
+
Ďr
2
âĽ
Area(âŚ
r
) = Area(âŚ
rB
)
âĽ
p
Area(âŚ) +
p
Area(
rB
)
2
âĽ
â
A
+
r
â
Ď
2
=
A
+
r
â
4
ĎA
+
Ďr
2
,
as desired.
Exercise.
Give another proof of the isoperimetric inequality using Wirtingerâs
inequality and the hint from the previous exercise. (This was Hurwitzâs first proof.)
9.
Convex Hulls and why the Isoperimetric Inequality need only be
proved for convex sets.
For if not, then we can construct a new curve, the
convex hull of the figure, whose boundary has smaller length but encloses strictly
19
more area, and thus has stricly greater isoperimetric ratio. The convex hull is
defined to be the smallest convex set enclosing the figure. Thus if ⌠is the closed
region bounded by the curve, then
Ë
⌠=
\
{
K
0
:
K
0
is convex and âŚ
â
K
0
.
}
.
If
K
is convex then
K
= Ë
K
. The key fact is that taking the convex hull increases the
area and decreases the boundary length. Thus it suffices to prove the isoperimetric
inequality for convex domains.
âŚ
Oi
Î
i
Î
i
Yi
Zi
ÂŁ
Convex Hull.
Convexity Lemma.
Let
âŚ
â
R
2
be a closed and bounded set surrounded piecewise
C
1
Jordan boundary curve. Let
Ë
âŚ
be the convex hull. Then
Area( Ë
âŚ)
âĽ
Area(âŚ)
and
L(
â
( Ë
âŚ))
â¤
L(
â
âŚ)
.
If
âŚ
is nonconvex then both of the inequalities are strict.
Proof.
Let
K
= Ë
⌠be the convex hull.
K
must have interior points or else it is
contained in a line as is
â
⌠which therefore canât be a 1
â
1 image of a circle as a
Jordan curve must be. The interior points of
K
which are not in ⌠are denoted by
the (at most countable since
O
i
are open) disjoint union
q
i
O
i
=
K
âŚ
â
âŚ
,
where
O
i
are the open connected components. If ⌠is convex, this is the empty
set. Every point
X
in the boundary
âK
is either a point of
â
⌠or not. If not, by a
theorem of Caratheodory, it is a point of a line
ÂŁ
â
R
2
such that
K
is on one side
of the line
ÂŁ
and there are points
Y
i
, Z
i
â
â
âŚ
âŠ
ÂŁ
such that
X
is strictly between
Y
i
and
Z
i
. Let
O
i
be the âlagoonâ region between
ÂŁ
and âŚ. Since
X /
â
â
⌠then
O
i
contains part of a ball neighborhood of
X
and thus it has positive area. It follows
that Area(
K
) = Area(âŚ) +
P
i
Area(
O
i
)
>
Area(âŚ). On the other hand, we cut
the boundary curve of ⌠into two pieces at the points
Y
i
and
Y
i
. Call Î
i
the part
20
of
â
⌠from
Y
i
to
Y
i
along
âO
i
. Call Î
i
=
âO
i
â
Î
i
the line segment from
Y
i
to
Z
i
.
Since Î
i
is a straigt line in the plane from
Y
i
to
Z
i
, its length
L
(Î
i
)
< L
(Î
i
). On
the other hand the boundary of
K
= âŚ
⪠âŞ
i
O
i
so that, by semicontinuity (see
Appendix,)
L
(
âK
)
â¤
L
(
â
âŚ) +
P
i
(
L
(Î
i
)
â
L
(Î
i
))
â¤
L
(
â
âŚ).
It should be remarked that if a set has various components with various boundary
pieces, then its
IR
= 4
ĎA/L
2
can be increased. Suppose that the
K
=
K
1
âŞ
K
2
so that
âK
=
âK
1
âŞ
âK
2
. Then
L
(
âK
) =
L
1
+
L
2
where
L
i
=
L
(
K
i
). Similarly
A
=
A
1
+
A
2
. However, if both
A
1
>
0,
IR
4
Ď
=
A
1
+
A
2
(
L
1
+
L
2
)
2
=
L
2
1
(
L
1
+
L
2
)
2
A
1
L
2
1
+
L
2
2
(
L
1
+
L
2
)
2
A
2
L
2
2
<
max
A
1
L
2
1
,
A
2
L
2
2
= max
IR
1
4
Ď
,
IR
2
4
Ď
.
Thus we do better by taking one or other part of a nontrivial partition. Finally, if
K
is not simply connected, then the isoperimetric ratio is improved by filling in the
area and throwing out the boundary of any holes. Thus to find the extreme figure
for the isoperimetric ratio, it suffices to consider convex domains
K
with interior.
10. Hadwigerâs Proof using only Steinerâs Inequality.
The proof of Hadwiger depends only on Steinerâs inequality. For this purpose
we need to define the
incircle
,
incenter
and
inradius
of a compact set âŚ
â
E
2
. The
inradius is
r
I
= sup
{
r
âĽ
0 : there is
x
â
E
2
such that
x
rB
â
âŚ
}
An incenter is any such
x
I
so that the incircle
x
I
r
I
B
â
âŚ.
Isoperimetric Inequality of Hadwiger.
Suppose
âŚ
is a compact set with piece-
wise
C
1
boundary of area
A
and boundary length
L
. Let
M
be a line through the
incenter
X
I
of the convex hull
K
of
âŚ
and
a
= L(
K
âŠ
âŚ)
the length of a chord
passing through the center. Then
L
2
â
4
Ď
A âĽ
Ď
2
4
(
a
â
2
r
I
)
2
.
Observe that equality in the isoperimetric inequality implies that all chords
through the incenter of
K
agree with the diameter of the incircle, hence
K
is a
circle.
Proof.
[G] Let
A
be the area of
K
and
L
the length of its boundary. Then by the
convexity lemma,
L
2
â
4
Ď
A âĽ
L
2
â
4
ĎA
so we may estimte the deficit of the convex
hull from below. Choose
R >
0 so that the interior of
B
R
(
X
I
)
â
K
. Let ÎĽ be the
annular region between the ball and
K
, ÎĽ =
B
R
(
X
I
)
â
K
âŚ
and let
F
and
G
be the
21
closures of the two halves of the annulus split by the line ÎĽ
â
M
. Let
P
and
Q
denote the disjoint line segments
P
âŞ
Q
=
M
âŠ
ÎĽ. Let
`
1
and
`
2
be the boundary
components of
âK
in
F
and
G
respectively, and
L
1
and
L
2
be the two components
of boundary
âF
âŠ
âB
R
(
X
I
) and
âG
âŠ
âB
R
(
X
I
) resp. We have
L
=
`
1
+
`
2
,
ĎR
2
= Area(
F
) + Area(
G
) +
A.
Choose a number 2
r
I
â¤
2
r
â¤
a
. Consider the parallel neighborhood of the ring
domain ÎĽ
r
. Since
r
âĽ
r
I
the entire interior region is covered so Area(ÎĽ
r
) =
Ď
(
R
+
r
)
2
.
If
p
and
q
denote the lengths of
P
and
Q
then Area(
P
r
) = 2
pr
+
Ďr
2
and Area(
Q
r
) = 2
qr
+
Ďr
2
.
âŚ
K
R
rI
F
G
P
Q
XI
l1
l2
L1
L2
M
Fig. 6. Diagram for Hadwigerâs proof.
Finally we apply Steinerâs Inequality to
F
r
and
G
r
Area(
F
r
)
â¤
Area(
F
) + (
p
+
L
1
+
q
+
ĎR
)
r
+
Ďr
2
,
Area(
G
r
)
â¤
Area(
G
) + (
p
+
L
2
+
q
+
ĎR
)
r
+
Ďr
2
.
Observe that any point of ÎĽ
r
is either a point of
F
r
or
G
r
. Also any point of
P
r
or
Q
r
are in both
F
r
and
G
r
. Hence
Area(ÎĽ
r
) + Area(
P
r
) + Area(
Q
r
)
â¤
Area(
F
r
) + Area(
G
r
)
which implies
Ď
(
R
+
r
)
2
+2(
p
+
q
)
r
+2
Ďr
2
â¤
Area(
F
)+Area(
G
)+(2
p
+
L
1
+
L
2
+2
q
+2
ĎR
)
r
+2
Ďr
2
22
or
A
+
Ďr
2
â¤
Lr
which is equivalent to
L
2
â
4
ĎA
âĽ
(
L
â
2
Ďr
)
2
.
Sunstituting
r
=
r
I
and
r
=
a/
2 implies
L
2
â
4
ĎA
âĽ
1
2
(
L
â
2
Ďr
I
)
2
+
1
2
(
L
â
Ďa
)
2
âĽ
Ď
2
4
(
a
â
2
r
I
)
2
,
using 2
A
2
+ 2
B
2
âĽ
(
A
â
B
)
2
.
11. Knotheâs Proof and Croftonâs formula for curves.
This is one of the directest proofs, it only requires one of the most basic formulas
from integral geometry. (See [S] for a thorough exposition.) From the Convexity
Lemma, it suffices to prove the isoperimetric inequality for convex domains
K
with
interior points in the plane. Let
K
have area
A >
0. By translating, we may
suppose 0
â
K
. For simplicity, we assume that the boundary curve Î of
K
is
piecewise
C
1
, is parameterized by arclength and has length
L
. Thus there are
L
-periodic functions (
x
(
s
)
, y
(
s
)) : [0
, L
]
â
R
2
that circumscribe the boundary.
Îą2
β1
β2
Ď2
Îą1
s1
s2
r1
r2
Ď1
Î
K
Fig. 7. Diagram for Knotheâs Proof.
An unoriented line
L
(
β, p
) in the plane may be parameterized by the angle
β
â
[0
,
2
Ď
) its normal makes with the
x
-axis and the distance
p
âĽ
0 to the origin.
Thus the line
L
(
β, p
) has a normal vector (cos
β,
sin
β
) and its equation is given by
x
cos
β
+
y
sin
β
=
p.
The kinematic measure on the space of lines, invariant under rigid motions of the
plane, is
dL
=
dp
â§
dβ
. Knotheâs proof will amount to an integration over all
pairs of lines that meet
K
. So let
L
=
{
L
:
K
âŠ
L
6
=
â }
denote the space of
23
lines that meet
K
. The essence of intergral geometry is to describe the space of
lines in two ways and derive different but equal expressions of the integral for these
ways, utilizing the change of variables formula for the integration. The first way is
L
=
{
L
(
p, β
) :
β
â
[0
,
2
Ď
)
, p
â
[0
, P
(
β
)]
}
where
P
(
β
)
âĽ
0 is the largest value of
p
so that the lines
L
(
p, β
) meet
K
. A simple formula, due to Crofton, for the integral
of chord lengths of lines meeting
K
is
Z
L
r
(
L
)
dL
=
Z
2
Ď
0
Z
P
(
β
)
0
r
(
p, β
)
dp
â§
dβ
=
Z
Ď
0
Z
P
(
β
)
â
P
(
β
+
Ď
)
r
(
p, β
)
dp
â§
dβ
=
Z
Ď
0
A dβ
=
ĎA
where the length of the intersection is
r
= L(
K
âŠ
L
).
The other way to describe a line is by a point (
x
(
s
)
, y
(
s
))
â
Î and by the
angle between the tangent vector to Î and the line
Îą
=
((
x
0
(
s
)
, y
0
(
s
))
, L
). Thus
L
=
{
L
(
s, Îą
) :
s
â
[0
, L
)
, Îą
â
[0
, Ď
)
}
. To express
dL
in the (
r, Îą
) variables, it is
convenient to let the tangent vector have angle
Ď
so that
x
0
(
s
) = cos
Ď,
y
0
(
s
) = sin
Ď,
x
(
s
) cos
β
+
y
(
s
) sin
β
=
p,
Ď
+
Îą
=
β
+
Ď
2
.
To compute the Jacobean for the change of variables (
s, Îą
)
â
(
p, β
) we take exterior
derivatives and exterior multiply the forms. Thus
(
x
sin
β
â
y
cos
β
)
dβ
+
dp
= (
x
0
cos
β
+
y
0
sin
β
)
ds,
dβ
=
dĎ
ds
ds
+
dÎą.
Wedging yields
dp
â§
dβ
= (
x
0
cos
β
+
y
0
sin
β
)
ds
â§
dÎą
= (cos
Ď
cos
β
+ sin
Ď
sin
β
)
ds
â§
dÎą
= cos(
Ď
â
β
)
ds
â§
dÎą
= cos
Ď
2
â
Îą
ds
â§
dÎą
= sin
Îą ds
â§
dÎą.
Thus
(15)
2
ĎA
=
Z
L
nr
(
L
)
dL
=
Z
Ď
0
Z
L
0
r
(
s, Îą
) sin
Îą ds
â§
dÎą
where
n
is the number of intersection points in
L
âŠ
Î.
n
= 2 for almost every
line,
i.e.
except for a set of lines of measure zero, because there are generically two
points (
s, Îą
) corresponding to almost every
L
â L
.
24
The second formula is for area in polar coordinates of a convex domain.
(16)
A
=
Z
Ď
0
Z
r
(
s,Îą
)
0
r dr dÎą
=
1
2
Z
Ď
0
r
(
s, Îą
)
2
dÎą.
Knotheâs Proof of the Isoperimetric Inequality.
We compute using (15) and (16),
0
â¤
Z
Î
Z
Î
Z
Ď
0
Z
Ď
0
(
r
1
sin
Îą
2
â
r
2
sin
Îą
1
)
2
dÎą
1
dÎą
2
ds
1
ds
2
=2
Z
Î
Ď
Z
0
Ď
Z
0
Z
Î
r
2
1
sin
2
Îą
2
ds
1
dÎą
1
dÎą
2
ds
2
â
2
Z
Î
Z
Î
Ď
Z
0
Ď
Z
0
r
1
sin
Îą
1
r
2
sin
Îą
2
dÎą
1
dÎą
2
ds
1
ds
2
=4
A
Z
Î
Ď
Z
0
Z
Î
sin
2
Îą
2
ds
1
dÎą
2
ds
2
â
2

ďŁ
Z
Î
Ď
Z
0
r
1
sin
Îą
1
dÎą
1
ds
1


2
=2
ĎA L
2
â
4
ĎA
.
Equality implies that for all choices
Îą
1
, Îą
2
, s
1
, s
2
, there holds
r
1
sin
Îą
2
â
r
2
sin
Îą
1
= 0
.
Thus for
s
1
, s
2
constant and
Îą
2
=
Ď/
2,
r
1
(
s
1
, Îą
1
) =
r
2
(
s
2
, Ď/
2) csc
Îą
1
so Î is a circle
with diameter
r
2
(
s
2
, Ď/
2) which is independent of
s
2
(by setting
Îą
1
=
Îą
2
=
Ď/
2).
Appendix. Definitions, Geometric Background, Various Facts.
The first two lemmata deal with the realizability of quadrilaterals and cyclic
quadrilaterals.
Lemma 1.
Suppose four nonnegative numbers
a, b, c, d
are given that satisfy
(17)
a
â¤
b
+
c
+
d,
b
â¤
c
+
d
+
a,
c
â¤
d
+
a
+
b,
d
â¤
a
+
b
+
c.
Then there is a quadrilateral in the plane whose side lengths are, in order,
a, b, c, d
.
Proof.
We cyclicly permute the lengths if necessary so that
m
=
a
+
b
= min
{
a
+
b, b
+
c, c
+
d, d
+
a
}
then in fact the three numbers
m, c, d
satisfy
m
â¤
c
+
d,
c
â¤
d
+
m,
d
â¤
m
+
c.
Thus there is a triangle with sides
m, c, d
which is also a quadrilateral whose sides
a
+
b
line up with
m
. If one of the inequalities in (17) is equality, then the quadrilateral
degenerates to a bigon.
25
Lemma 2.
Suppose there are four positive numbers
a, b, c, d
that satisfy
(18)
a < b
+
c
+
d,
b < c
+
d
+
a,
c < d
+
a
+
b,
d < a
+
b
+
c.
Then there is a cyclic quadrilateral in the plane whose side lengths are given in
order by the numbers
a, b, c, d
.
x
y
a
b
c
d
A
B
C
D
E
Fig. 8. Construction of Cyclic Quadrilateral
Proof.
If
a
=
c
then the figure is a trapezoid with
b
and
d
sides parallel. Otherwise
we may assume that
a < c
and
b
6
=
d
(by cyclically permuting the sides if needed.)
Then we may simply construct the cyclic quadrilateral whose sides are given as
follows: Imagine that there is a cyclic quadrilateral with vertices
ABCD
around
a circle with
a
=
AB
,
b
=
BC c
=
CD
and
d
=
DA
. Then extend
DA
and
CB
to their intersection
E
and suppose that the lengths of the segment
x
=
AE
and
y
=
BE
. Observe that the triangles
CED
and
AEB
are similar. To see this,
observe that the angles
CBA
and the angle
ADC
are supplementary because
they subtend opposite sides of the circle. It follows that
CDE
=
ABE
hence
the triangles are similar since all angles are the same. The similarity implies that
x
y
+
b
=
a
c
and
y
d
+
x
=
a
c
.
Since
a < c
we may solve
x
=
a
bc
+
ad
c
2
â
a
2
,
y
=
a
ab
+
cd
c
2
â
a
2
.
Finally, if we are able to construct the triangle with sides
axy
then by putting
A
=
a
âŠ
x
,
B
=
a
âŠ
y
,
E
=
x
âŠ
y
, and extending
EB
by a distance
b
to find
C
and extending
EA
a distance
d
to find
D
constructs the cyclic quadrilateral. But
c
â
a < b
+
d
, and (
c
â
a
)(
a
+
b
+
c
â
d
)
>
0 implies
c
2
â
a
2
+
bc
+
ad > ab
+
cd
,
and (
c
â
a
)(
a
â
b
+
c
+
d
)
>
0 implies
c
2
â
a
2
+
ab
+
cd > bc
+
ad
, so, resp.,
x
+
y
=
a
(
b
+
d
)
c
â
a
> a,
a
+
x
=
a
(
c
2
â
a
2
+
bc
+
ad
)
c
2
â
a
2
>
a
(
ab
+
cd
)
c
2
â
a
2
=
y,
a
+
y
=
a
(
c
2
â
a
2
+
ab
+
cd
)
c
2
â
a
2
>
a
(
bc
+
ad
)
c
2
â
a
2
=
x
26
so the triangle
axy
is constructible.
Convergence of Sets in the Hausdorff sense.
Given two compact sets in the
plane
S, T
â
R
2
(compact is equivalent to closed and bounded by the Heine-Borel
Theorem, [Wa]) their
Hausdorff Distance
, a measure of their separation is defined
by
Ρ
(
S, T
) = max
n
sup
{
dist(
x, T
) :
x
â
S
}
,
sup
{
dist(
S, y
) :
y
â
T
}
o
.
In terms of Minkowski sums this can be defined equivalently by
Ρ
(
S, T
) = inf
{
r >
0 :
S
â
T
rB
and
T
â
S
rB
}
where, as usual
B
is the closed unit ball in the plane. A sequence of compact sets
T
n
â
R
2
is said to
converge in the Hausdorff sense
to a compact set
T
â
R
2
if
Ρ
(
T
n
, T
)
â
0
as
n
â â
.
Similarly one can define a Cauchy sequence of compact sets.
{
T
n
}
is Cauchy if for
every
Îľ >
0 there is an
N
â
N
so that
m, n
âĽ
N
implies
Ρ
(
T
n
, T
m
)
< Îľ
. The
compact sets and the compact convex sets are complete: if
T
n
is a Cauchy sequence
of compact (convex) sets, then there is a compact (convex) set
T
such that
T
n
â
T
as
n
â â
.
Blaschke-Hadwiger Selection Theorem.
Let
T
n
â
R
2
be a uniformly bounded
sequence of compact sets:
T
n
â
B
for some all
n
where
B
is some fixed closed disk.
Then there is a compact set
T
â
B
and a subsequence
{
n
0
} â
Z
so that
T
n
0
â
T
as
n
0
â â
[B]
,
[F]
,
[H]
. If the
T
n
are convex then so is
T
.
Definitions of Area and Length.
For nice sets with piecewise
C
1
boundary
or which are convex, the various notions of area and length coincide. However,
the classes of sets for which these notions are defined differ. Moreover they have
different convergence properties. See [C], [H] or [R]. For us we shall take area to
mean Lebesgue measure and length to mean Jordan length, the total variation of
rectifiable curves. Then the usual line integral gives the area [G].
It is an important fact that both the area and the length of the boundary
are semicontinuous with respect to convergence in the appropriate sense. If the
boundary curves are parameterized, then the parameterizations have to be con-
trolled. The Frechet distance ÎŚ(
Îł
n
, Îł
) between two parameterized curves
Îł
n
and
Îł
is the smallest
Îľ
âĽ
0 such that there is a homeomorphism
h
:
S
1
â
S
1
such that
max
x
â
S
1
|
Îł
n
(
x
)
â
Îł
(
h
(
x
))
| â¤
Îľ
[C].
Theorem.
Suppose
Î
n
is a uniformly bounded sequence of closed piecewise
C
1
curves in the plane, enclosing compact regions
T
n
(or the
T
n
are convex) which are
converging to a compact set
T
bounded by a closed piecewise
C
1
curve
Î
(or
T
is
convex) in the sense of Hausdorff convergence. Suppose also that
L(Î
n
)
â¤
L
and
L(Î)
â¤
L
are uniformly bounded by
L <
â
. Then
lim
n
ââ
Area(
T
n
) = Area(
T
)
.
27
Suppose furthermore that
Î
n
is converging to
Î
in the sense that Fr´echet distance
of the parametrizations
ÎŚ(
Îł
n
, Îł
)
â
0
. (
Î
n
=
Îł
n
(
S
1
)
.) Then
lim inf
n
ââ
L(Î
n
)
âĽ
L(Î)
.
Proof.
Let
r
=
Ρ
(
T
n
, T
). First approximate Î
n
and Î by polygons that are close in
Hausdorff distance, length and area ([B]. p. 28.) There is a closed polygonal curve
q
enclosing
Q
with
|
Area(
Q
)
â
Area(
T
)
|
< r
,
|
L(
q
)
â
L(Î)
|
< r
and
Ρ
(
Q, T
)
<
r
. Similarly there is a closed polygonal curve
q
n
enclosing the region
Q
n
with
|
Area(
Q
n
)
â
Area(
T
n
)
|
< r
,
|
L(
q
n
)
â
L(Î
n
)
|
< r
and
Ρ
(
Q
n
, T
n
)
< r
. Note that
T
â
Q
r
so that
T
n
â
T
r
â
Q
2
r
. Similarly,
T
â
(
Q
n
)
2
r
. Then, since Steinerâs
Inequality applies to polyhedra by the Corollary to the First Variation Formula,
Area(
T
n
)
â
Area(
T
)
â¤
Area(
Q
2
r
)
â
Area(
T
)
â¤
Area(
Q
) + 4 L(
q
)
r
+ 4
Ďr
2
â
Area(
T
)
â¤
r
+ 4
Lr
+ (4 + 4
Ď
)
r
2
Area(
T
n
)
â
Area(
T
)
âĽ
Area(
T
n
)
â
Area((
Q
n
)
2
r
)
âĽ
Area(
T
n
)
â
Area(
Q
n
)
â
4 L(
q
n
)
r
â
4
Ďr
2
⼠â
r
â
4
Lr
â
(4 + 4
Ď
)
r
2
Thus if
n
1 so that
Ρ
(
T
n
, T
)
â¤
1,
|
Area(
T
n
)
â
Area(
T
)
| â¤
(5 + 4
L
+ 4
Ď
)
Ρ
(
T
n
, T
)
â
0
as
n
â â
.
The length formula follows by the semicontinuity of Jordan length [C], [R] with
respect to Fr´echet convergence of curves.
If we are willing to restrict ourselves to simple closed curves (Jordan curves
without self intersections), then the statement is much simpler. For piecewise
C
1
simple curves the arclength of the boundary curve of âŚ
â
R
2
is given by the
perimiter
Per(âŚ
,
O
) =
Z
O
|
DĎ
âŚ
|
where the characteristic function
Ď
âŚ
(
x
) = 1 if
x
â
⌠and
Ď
âŚ
(
x
) = 0 otherwise
and where the integral means the variation. Thus for
O â
R
2
an open set and
f
â
L
1
(
O
) we define
Z
O
|
Df
|
= sup
Z
O
f
(
g
x
+
h
y
)
dxdy
:
g, h
â
C
1
0
(
O
)
, g
2
+
h
2
â¤
1
.
If Per(âŚ
,
O
)
<
â
for all bounded open sets
O
then we say ⌠is a Cacciopoli set.
Then Per(âŚ
,
O
) =
H
1
(
O âŠ
â
âŚ) is the Hausdorff measure of the boundary[Gi]. The
subspace
BV
(âŚ) =
{
f
â
L
1
(âŚ) :
R
Ď
|
Df
|
<
â}
with the norm
k
f
k
BV
(âŚ)
=
k
f
k
L
1
(âŚ)
+
Z
âŚ
|
Df
|
is a Banach space. Semiconinuity holds for perimeter [Gi].
28
Lemma. Semicontinuity for BV.
If
âŚ
â
R
2
is open and
{
f
j
}
a sequence in
BV
(âŚ)
which converges in
L
1
loc
(âŚ)
to a function
f
then
Z
âŚ
|
Df
| â¤
lim inf
j
ââ
Z
âŚ
|
Df
j
|
.
This implies that if
T
n
are uniformly bounded Jordan domains such that
T
j
â
T
in Hausdorff distance (so
Ď
T
j
â
Ď
T
in
L
1
loc
(âŚ)) then Per(
T,
O
)
â¤
lim inf Per(
T
j
,
O
).
Equivalently, this is a statement about the semicontinuity of the mass of a 1-current
with respect to the flat norm [M].
We deduce some other facts that are being used in the discussion above. We
would like to find the rate of change of arclength for a one parameter family of
curves.
First Variation Formula for closed curves in the plane.
Suppose
f
(
s, t
)
is a
one parameter family of closed continuous, piecewise
C
2
curves in the plane defined
on
C
2
pieces
f
i
:
{
(
s, t
) :
â
Îľ < s < Îľ, l
i
(
s
)
â¤
t
â¤
u
i
(
s
)
} â
R
2
where
l
i
(
s
)
, u
i
(
s
)
are differentiable functions on
(
â
Îľ, Îľ
)
such that
f
i
â
1
(
s, u
i
â
1
(
s
)) =
f
i
(
s, l
i
(
s
))
and
f
1
(
s, l
1
(
s
)) =
f
k
(
s, u
k
(
s
))
for all
i
= 1
, . . . , k
. The length is
L
(
s
) =
P
i
L(
f
i
(
s,
¡
))
.
Let
T
= (
f
i
)
t
/
|
(
f
i
)
t
|
be the tangent vector,
N
be the outward normal,
Ď
the arclength
along
f
(0
, t
)
, and
Îş
the curvature. Then
L
0
(0) =
k
X
i
=1
Z
u
i
(
s
)
l
i
(
s
)
h
ÎşN, V
i
dĎ
+
+
k
X
i
=1
|
(
f
i
â
1
)
t
(
s, u
i
â
1
)
|
d u
i
â
1
ds
â |
(
f
i
)
t
(
s, l
i
)
|
d l
i
ds
+
h
T, V
i
u
i
l
i
.
Proof.
The length of the piecewise
C
2
curve is given by
L
(
s
) =
k
X
i
=1
Z
u
i
(
s
)
l
i
(
s
)
|
(
f
i
)
t
(
s, t
)
|
dt.
Set
f
0
=
f
k
and
u
0
=
u
k
. Differentiating,
L
0
(
s
) =
k
X
i
=1
|
(
f
i
)
t
(
s, u
i
)
|
u
0
i
â |
(
f
i
)
t
(
s, l
i
)
|
l
0
i
+
Z
u
i
(
s
)
l
i
(
s
)
h
f
t
, f
st
i
|
f
t
(
s, t
)
|
dt
!
=
k
X
i
=1
|
(
f
i
)
t
(
s, u
i
)
|
u
0
i
â |
(
f
i
)
t
(
s, l
i
)
|
l
0
i
â
Z
u
i
(
s
)
l
i
(
s
)
d
dt
f
t
|
f
t
|
, f
s
dt
+
h
f
t
, f
s
i
|
f
t
|
u
i
l
i
!
.
29
Thus at
s
= 0, writing the variation vector field as
V
(
t
) =
f
s
(0
, t
),
L
0
(0) =
k
X
i
=1
Z
u
i
(
s
)
l
i
(
s
)
h
ÎşN, V
i|
(
f
i
)
t
|
dt
+
+
k
X
i
=1
|
(
f
i
â
1
)
t
(
s, u
i
â
1
)
|
u
0
i
â
1
â |
(
f
i
)
t
(
s,
(
l
i
))
|
l
0
i
+
h
T, V
i
u
i
l
i
.
We now provide the estimate for the growth of the boundary curve needed for
Steinerâs inequality. Let
P
â
R
2
be a compact bounded by a polygonal curve
with finitely many vertices. Then the boundary of the parallel set
P
r
=
P
rB
consists of finitely many line segments and arcs of circles, thus a piecewise
C
2
curve. Suppose that going around the polygon consists of vertices
X
i
and closed
line segments corresponding to the edges
e
i
=
X
i
X
i
+1
where
i
= 1
, . . . , k
and
X
k
+1
=
X
1
. For every point in the plane
X
â
R
2
â
P
has a certain distance to
P
given by
d
= dist(
X, P
). The closest point on
P
to
X
may be one point that is
interior to an edge or it may be a vertex of
P
, or it may be several closest points,
each of which is in some edge or some vertex. Label each
X
by the set of closest
edges and vertices. The exterior region is then subdivided into zones with the same
labelling. These may be single points, curves, or open regions. They may be infinite
or finite, but there are only finitely many zones in the plane. The zones with more
than two closest elemets are points. The zones with two labels occur on the are
curves, for example the curve equidistant between two points (a straight line) or
the locus of points equidistant from a vertex and a line (a straight line if the vertex
is an endpoit of the line, or a parabola, in case the point is not an endpint of the
edge.) In both cases, the boundary of the parallel set is transverse to the locus of
two closest elements. It follows that for only finitely many
r
â {
r
1
< r
2
<
¡ ¡ ¡
< r
q
}
such that the boundary of the parallel set contains points that are closest to more
than two elements. Thus if
r
j
< r < r
j
+1
(which we call a
generic
distance) then
âP
r
consists of
k <
â
segments which are line segments or arcs of circles, and such
that the endpoints of the segments depend in a
C
1
fashion on
r
since they are given
by the intersection of
âP
r
with parabolas and lines (the two element zones.) For
convenience let
t
be the arclength parameter,
N
the outward normal and
f
(
r, t
) the
parameterization of
âP
r
. Then
f
r
=
N
and
Îş
= 0 on line segments in
âP
r
which
correspond to âone edge zonesâ and circles whose curvatures are always
Îş
= 1
/r
,
that correspond to âone vertex zones.â Note that
âP
r
contains no concave circular
zones. Thus we have the corollary.
Corollary. Steinerâs Inequality for Polygons.
Suppose
P
is the compact re-
gion of area
A
bounded by a closed polygonal curve
p
of length
L
. Let
r >
0
be a
generic distance and
L
(
r
) = L(
âP
r
)
be the length of the boundary of the equidistant
domain. Then
L
0
(
r
)
â¤
2
Ď
. Hence, for all
r
âĽ
0
,
L(
âP
r
)
â¤
L
+ 2
Ďr
and
Area(
P
r
)
â¤
A
+
Lr
+
Ďr
2
,
30
with equality for convex sets.
Proof.
Let each circular or linear piece of the boudary be parameterized by arc-
length
l
i
(
r
)
â¤
t
â¤
u
i
(
r
) so
T
=
f
t
,
V
=
N
. By the first variation formula,
(19)
L
0
(0) =
X
i
=circle
Z
u
i
(
s
)
l
i
(
s
)
1
r
dt
+
k
X
i
=1
d u
i
â
1
dr
â
d l
i
dr
.
The segments of
âP
r
donât necessarily correspond to all vertices and edges of
P
. It may happen that at some edpoint of a circular arc, two vertices
X
i
and
X
j
of
P
are the closest points. This may happen if there is a âconcave sectionâ of
âP
between
X
i
and
X
j
. Let
θ
i
denote the âtotal curvatureâ corresponding to the
arc
f
i
of
âP
. If
f
i
is a line, then
θ
i
= 0. If
f
i
is a circular arc corresponding
to a âconvexâ vertex, say
X
j
, then
θ
i
=
(
e
j
â
1
, e
j
) is the angle deficit form the
edge
e
j
â
1
to
e
j
. If one takes the outward normal
N
j
â
1
of the
E
j
â
1
edge and
N
j
corresponding to the
e
j
edge, then the zone labelled
X
j
is the sector in the plane
whose vertec is
X
j
and whose sides are rays in the directions
N
j
â
1
and
N
j
and
whose angle
(
N
j
â
1
, N
j
) =
θ
i
. The points of the arc
f
i
are a distance
r
from
X
j
in this sector.
The remaining circular arcs
f
i
of
âP
r
correspond to points
X
j
that are not
convex, in the sense that at an endpoint of
f
i
is not closest to an adjacent edge of
X
j
. Say the endpoint
u
i
is closest to the vertex
X
j
and the vertex
X
m
of the edge
e
m
not adjacent to
X
j
. The total angle of such an arc is not the angle between
adjacent edges at
X
j
but rather some part of it. The remaining angle of the vertex
X
j
has to be counted in the âtotal curvatuereâ of the âconcave part of
âP
.â Thus,
the angle corresponding to the endpoint
u
i
(same as
l
i
+1
) is the remainder of the
angle at
X
j
plus the total curvature (the sum of the angle deficits of intervening
vertices and whatever angle there is left between
N
m
â
1
and
X
m
X
. Call the total
ν
i
which is negative. The total curvature around
P
is thus
X
θ
i
>
0
θ
i
+
X
ν
i
<
0
ν
i
=
Z
âP
r
Îş dt
+
X
i
ν
i
= 2
Ď.
which is just the GauĂ Bonnet Theorem for
âP
r
. It remains to compute
u
0
i
and
l
0
i
.
At a convex vertex, the arc
f
i
is either a sector or a line segment whose normal at the
endpoint
u
i
coincides with
N
j
of
N
j
â
1
which is the normal of one of the adjacent
edges of
X
j
, one of the closest points to
f
i
(
u
i
). It follows that
u
0
i
=
l
0
i
+1
= 0.
Suppose that the endpoint
f
i
â
1
(
u
i
â
1
) is closest to a pair of vertices
X
j
and
X
m
of
P
. In that case, the curve
r
â
Y
(
r
) =
f
i
â
1
(
r, u
i
â
1
(
r
)) =
f
i
(
r, l
i
(
r
)) is on
the perpendicular bisector of
X
j
X
m
, and the arcs on both sedes of this point are
decreasing by the same amount. The angle between the perpendicular bisector and
the vector
X
j
Y
(
r
) is
â
ν
i
/
2. It follows that
(20)
u
0
i
â
1
(
r
) = tan(
ν
i
/
2)
and
l
0
i
(
r
) =
â
tan(
ν
i
/
2)
.
31
The final case is if
Y
(
r
) is closest to a vertex, say
X
j
and a line segment
e
m
. Let
Z
m
â
e
m
be the interior point where dist(
Y
(
r
)
, X
j
) = dist(
Y
(
r
)
, Z
m
). Now the lo-
cus os points equidistant from
X
j
and
e
m
is a parabola. By the geometry of parabo-
las, the tangent line to the parabola at
Y
(
r
) bisects the angle
(
Y
(
r
)
X
j
, Y
(
r
)
Z
m
).
It follows that (20) holds in this case as well. Finally, it may happen that
Y
(
r
) is
equidistant to two line segments
e
j
and
e
m
. The locus of points equidistant to two
lines is the angle bisector between the lines. Thus if
X
j
â
e
j
Z
m
â
e
m
are inte-
rior points where dist(
Y
(
r
)
, e
j
) = dist(
Y
(
r
)
, e
m
) = dist(
Y
(
r
)
, X
j
) = dist(
Y
(
r
)
, Z
m
)
then
â
ν
i
/
2 =
(
Y
(
r
)
X
j
, Y
(
r
)
Z
m
) and therefore (20) holds in this case as well.
Thus we can complete the estimate using
dt
=
r dθ
on circular arcs in (19) so
L
0
(
r
) =
X
θ
i
>
0
θ
i
+
X
ν
i
<
0
2 tan
ν
i
2
<
X
θ
i
>
0
θ
i
+
X
ν
i
<
0
ν
i
= 2
Ď.
If
P
was convex, then there are no zones corresponding to the concave parts of
P
so no
ν
1
terms and this inequality is equality. Suppose 0 =
r
0
â¤
r
1
< . . . < r
j
â¤
r
,
then
L
(
r
) =
L
(0) +
j
X
i
=1
Z
r
j
r
i
â
1
L
0
(
r
)
dr
+
Z
r
r
j
L
0
(
r
)
dr
â¤
L
+
j
X
i
=1
Z
r
j
r
i
â
1
2
Ď dr
+
Z
r
r
j
2
Ď dr
=
L
+ 2
Ďr.
Finally, since
A
0
(
r
) =
Z
âP
r
h
V, N
i
dt
it follows that since
V
=
f
r
=
N
that
A
(
r
) =
A
(0) +
Z
r
0
L
(
r
)
dr
â¤
A
+
Lr
+
Ďr
2
and the proof is complete.
Remarks.
These notes were inspired by my talk âMy Favorite Proofs of the Isoperimetric In-
equalityâ given in the Undergraduate Colloquium at the University of Utah on Nov. 20, 2001.
They are offered as an instructional module, which may be useful for the course on Curves and
Surfaces, geometric methods in education, beginning analysis, advanced calculus or for an intro-
duction to proofs. The notes represent a self contained discussion of the isoperimetric inequality.
One of the objectives of the author was to show by example that that many of the arguments
are interconnected. Methods that prove one inequality also can be used to prove the others. We
mention as exercises specific instances of these interconnections. For example, in one problem the
student is asked to show that the Wirtinger inequality implies the Brunn-Minkowski inequality.
A second objective of the author is to show the importance of inequalities. However, we deduce
our inequalities by producing a sequence of equalities and then discarding the obviously negative
terms to obtain the desired inequality. This gives an easy way to treat the case of equalty. It
sometimes makes for a slightly circuitous route to a result for the sake of self-containedness. Thus
we donât invoke the Schwarz inequality when it is just as easy to complete the square and discard
a negative term.
32
We have made a selection of proofs which are usually not presented together in the same
reference. For example, in Chernâs beautiful book
Global Differential Geometry
, [C], the proofs of
Schmidt and Hurwitz are presented. We select one of these proofs. do Carmoâs undergraduate text
[dC] on Curves and Surfaces presents Schmidtâs, Hurwitzâs and Santaloâs proof of the isoperimetric
inequality. Blashkeâs text [B] on the Circle and Sphere presents the Steiner symmetrization proof
and the hinge proof, of which we borrow the hinge proof, but we donât deduce the Brunn-Minkowski
inequality using symmetrization as Blaschke does. Blaschke & Reichardtâs text [BR] on differential
geometry presents Crone & Frobeniusâs proof, Knotheâs proof, Hurwitzâs proofs, Sz. Nagyâs and
Bolâs proof and Santaloâs proofs, from which we borrow Knotheâs (which is attributed to Blaschke
in [S].) Crone and Frobeniusâs method is used instead to prove the Brunn inequality in Ljusternikâs
text [L] on convex figures, from which the isoperimetric inequality is deduced in turn. In Blaschke
and LeichtweiĂâs differential geometry text [BL], the calculus of variations argument is given, as
well as Schmidtâs and Hurwitzâs proofs of the isoperimetric inequality and GroĂâs proof of the
convergence of Steiner symmetrization. The quadrilateral and Ptolemyâs inequalities are one of
hundreds of inequalities given in [BD]. Fourrey gives delightful historical commentary [Fo] as well
as several proofs of the Pythagorean Theorem and Heronâs Formula.
We have included references for most of the facts that are quoted. Space prohibits a really
careful discussion, particularly about the meaning of area, length and the regularity needed to
make these notions rigorous. Although we hope to have provided ample references to where these
issues are discussed. Many of the sources are accessible to undergraduate students.
There are several places where the isoperimetric inequality is discussed in detail. We mention
specifically Ossermanâs article [O] and the book of Burago-Zalgaller [BZ].
References
[B]
W. Blaschke,
Kreis und Kugel
, Chelsea, New York, 1949, Originally published by Teubner,
Leipzig, 1916.
[BL]
W. Blaschke & K. LeichtweiĂ,
Elementare Differentialgeometrie 5th. ed.
, Springer, Berlin,
1973.
[BR] W. Blaschke & H. Reichardt,
Einf¨
uhrung in die Differentialgeometrie, 2nd. ed.
, Springer,
Berlin, 1960.
[BF] T. Bonnesen & W. Fenchel,
Theorie der Konvexen K¨
orper
, Springer, Berlin, 1934.
[BD] O. Bottema, R. Djordjevi´
c, R. Jani´
c, D. Mitrinovi´
c, P. Vasi´
c,
Geometric Inequalities
,
Wolters Noordhoff Publishing, Groningen, 1968.
[BZ]
J. Burago & V. Zalgaller,
Geometric Inequalities
, Springer, Berlin, 1988., Originally pub-
lished Nauka, Leningrad, 1980.
[C]
L. Cesari,
Rectifiable Curves and the Weierstrass Integral
, Amer. Math. Monthly
65
(1958),
485-500.
[dC]
M. do Carmo,
Differential Geometry of Curves and Surfaces
, Prentice Hall, Englewood
Cliffs, 1976.
[F]
H. Federer,
Geometric Measure Theory
, Springer, Berlin, 1969, p. 183.
[Fo]
E. Fourrey,
Curiosit´
es G´
eom´
etriques
, Librairie Vuibert, Paris, 1938.
[GT] D. Gilbarg & N. Trudinger,
Elliptic Partial Differential Equations of the Second Order,
2nd ed.
, Springer, Berlin, 1977, p. 157.
[Gi]
E. Giusti,
Minimal Surfaces and Functions of Bounded Variation
, Birkh¨
aser, Boston, 1984.
[G]
H. Guggenheimer,
Differential Geometry
, Dover, Mineola, 1977, Originally published by
McGraw-Hill, New York 1963.
[G2]
H. Guggenheimer,
Applicable Geometry: Global and Local Convexity
, Krieger, Huntington,
1977.
[H]
H. Hadwiger,
Vorlesungen ¨
Uber Inhalt, Oberfl¨
ache und Isoperimetrie
, Springer, Berlin,
1957, p. 153.
[H2]
H. Hadwiger,
Altes und Neues ¨
uber Konvexe K¨
orper
, Birkh¨
auser, Basel, 1955.
[LT]
P. Li & A. Treibergs,
Applications of Eigenvalue Techniques to Geometry
, Contemporary
Geometry: J.-Q, Zhong Memorial Volume, H. H. Wu, ed., Plenum Press, New York, 1991,
pp. 22â95.
33
[L]
Ljusternik,
Convex Figures and Polyhedra
, Heath & Co., Boston, 1966.
[M]
F. Morgan,
Geometric Measure Theory: a Beginners Guide
, Academic Press, San Diego,
1988, p. 58.
[O]
R. Osserman,
The isoperimetric inequality
, Bull. American Math. Soc,
84
(1978), 1128â
1238.
[P]
G. P´
olya,
Mathematics of Plausible Reasoning
, Volume I: Induction and Analogy in Math-
ematics, Princeton U. Press, Princeton, 1954.
[PS]
G. P´
olya & G. Szeg¨
o,
Isoperimetric Inequalities in Mathematical Physics
, Annals of Math.
Studies, vol. 27, Princeton U. Press, Princeton, 1951.
[R]
T. Rad´
o,
Length and Area
, Colloquium Publications, vol. 30, American Mathematical So-
ciety, New York, 1948.
[S]
L. Santal´
o,
Integral Geometry and Geometric Probability
, Encyclopedia of Mathematics
and its Applications, Vol. 1, Addison Wesley Publ. Co., Reading, 1976.
[St]
D. Struik,
A Concise History of Mathematics, 3rd. ed.
, Dover, New York, 1967.
[Wa] W. Wade,
An Introduction to Analysis, 2nd. ed.
, Prentice Hall, Upper Saddle River, N. J.,
2000, p. 263.
[W]
H. Weinberger,
A First Course in Partial Differential Equations
, Dover, Mineola, 1997,
Originally published by Xerox College Publishing, Lexington, 1965.
University of Utah,
Department of Mathematics
155 South 1400 East, Rm 233
Salt Lake City, UTAH 84112-0090
E-mail address:
treiberg@math.utah.edu