ON POSITIVE
CENTERS, INTEGRAL
GEOMETRY AND THE
ISOPERIMETRIC
DEFICIT
MICHAEL E. GAGE
UNIVERSITY OF ROCHESTER
THE isoperimetric
inequality
2
L
≥ 4π
A
• For a plane object the length of the
boundary squared is at least as big as 4π
times the area. Equality is obtained for the
circle.
• In two dimensions -- many proofs in addition
to symmetrization. In higher dimensions the
analytic proofs are missing.
I.D.
the Isoperimetric Deficit
L2
A
− 4π = I.D.
• By estimating the I.D. you can determine
how far “away” from a circle you are:
A Bonnesen inequality
π2
2
I.D. ≥
(rout − rin )
A
• When the ID is small
the object is trapped
between two circles
of nearly the same
radius.
• Only the circle has zero
ID
Plan
• Some integral geometry (geometric
probability)
• Derive the Santalo proof of the
isoperimetric and Bonnesen inequalities
•
Goal: “best” proof of
for convex curves.
LA
≥
π
!
• Used in “curve shortening” problem
2
h ds
Plan
• Prove goal for symmetric curves
• Symmetrization trick for non-symmetric
curves
• Positive centers
• Minimal annulus
• averaging trick
• extension to Minkowski geometries
Support function h
• h can be considered as
as a function of either
θ or the arclength s
• h depends on the choice of
the origin
h(!)
!
s
Area from support
function
!
2A =
2dA = h ds
h ds
• Integrating the
h(!)
ds
!
support function
against arc length
gives twice the
area.
• The red triangle has base ds and height h
Crofton formula
2L =
!
n drdθ
• Twice length is the
•
expected value of the
number of intersections
with random
lines
!
L=
• Length is
h(θ)dθ
π
times
the average width
r
!
Poincare formula
4L1 L2 =
!
n dxdydϕ
n is number of intersection
points of boundaries
of a random object with a
fixed one
n=4
n=2
n=6
Blaschke formula
!
ν dxdyφ = 2π(A1 + A2 ) + L1 L2
ν is the number of components in
the intersection of interiors
ν=1
ν=1
ν=1
Topology
=>
Geometry
!
!
n
dxdydϕ = 2L1 L2
2
ν dxdydϕ = 2π(A1 + A2 ) + L1 L2
• Assume the two objects are congruent and
subtract
! "
#
n
2
0≤
− ν dxdydϕ = (L1 − 4πA)
2
Bonnesen functional
• Take the second body to be a circle of radius
r. Write the area and length in terms of
the radius and simplify to get
! "
#
n
2
− ν dxdy = rL1 − A1 − πr
2
inradius and outradius
• n − 2ν will be non-
negative for circles with
radius
r ∈ [rin , rout ]
• some parts of the
boundary circle have to be
inside the object and some
outside if the disk and the
object intersect at all
Bonnesen functional
! "
#
n
2
0≤
− ν dxdy = rL1 − A1 − πr
2
When r ∈ [rin , rout ]
• Since the polynomial has a real root the
discriminant must be positive:
L21 − 4πA ≥ 0
Bonnesen functional
B(r) = rL − A − πr
2
r1
r2
rin
• Roots lie outside the interval [r√in , rout ]
(rout − rin ) ≤ (r2 − r1 ) ≤
L2 − 4πA
π
hence
L − 4πA ≥ π (rout − rin )
2
2
2
rout
Topology Picture
• In orange region
elsewhere
n
−ν =1
2
n
−ν =0
2
• If circle is too small nor too large then
possibly
2
−ν <0
Symmetric curve
• For a convex curve symmetric through the
origin the inradius circle and the outradius
circle are both centered at the origin.
• In this case the support function h from the
origin satisfies
rin ≤ h ≤ rout
Goal:
LA
≥
π
!
2
h ds
Which we
can
obtain
by
integrating
!
hL − A − πh ds ≥ 0
2
to get
2AL − AL − π
!
2
h ds
which proves the result for any convex curve
that has concentric inradius and outradius
circles.
Counter-example for
non-convex curves
•
LA
<
π
!
2
h ds
Symmetrization trick
• Choose points L/2
apart on boundary
• Move until the tangents
are parallel
• (Intermediate value theorem)
• We can cut this to create two
symmetric convex objects with
continuous tangents and apply
the previous theorem, then add.
2LA1
≥
π
2LA2
≥
π
!
A1
2
h ds
A1
2C1
!
A1
A2
2
h ds
2C2
L(A1 + A2 )
≥
π
A2
!
C1 ∪C2
2
h ds
A2
r1
r2
rin
• Can you find a smaller r than the inradius
that ensures that B(r) is positive?
rout
inradius symmetry point
• The inradius at c
rin (c) is the radius of
the largest inscribed circle centered at c.
• c is an inradius symmetry point if it is
the midpoint of a secant and the inradius
circle touches the boundary on both sides.
• If an object has an inradius symmetry point
then
B(rin (c)) ≥ 0
inradius symmetry point
• Facts to check:
• After symmetrizing r(c) is the inradius for
each half, hence
B2 (r(c)) + B1 (r(c)) ≥ 0
• One half will not be convex, but this
doesn’t matter.
• A similar result is true for outradius.
• Theorem:
For all convex plane curves
B(r) ≥ 0 for all r ∈ [min rin (c), max rout (d)]
c
d
Positive centers
• Is it possible to choose the origin point for
non-symmetric curves so that
?
B(h) = hL − A − πh ≥ 0
2
• We’ll call such a point a positive center.
• A point which is both inradius symmetry and
outradius symmetry is a positive center
Minimum width annulus
•
• It’s equivalent to choosing
It’s unique.
1
a, b and c to give the
best approximation of the
support function in the formula
4
max |a cos(θ) + b sin(θ) + c − h(θ)|
θ
• One can find 4 boundary contacts
with the
inner and outer circles which alternate.
• Intermediate value theorem implies that the
center is a positive center
2
3
Bonnesen functional
B(r) = rL − A − πr
2
r1
ρin
B(r) ≥ 0 for all r ∈ [ρin , ρout ]
• Roots lie outside the interval
the radii for the minimal annulus of any
convex curve.
√
(ρout − ρin ) ≤ (r2 − r1 ) ≤
L2 − 4πA
π
r2
ρout
B(r) ≥ 0 for all r ∈ [ρin , ρout ]
•
B(h) ≥ 0 for the minimal annulus so we
can prove the goal, BUT
• The proof still requires cutting the original
figure into pieces and symmetrizing.
• Can one see directly that the Bonnesen
functional is always positive for the minimal
annulus?
Averaging trick
n
Green: 2 − ν ≥ 1
n
Red:
− ν ≥ −1
2
The green and red
regions are congruent
so they cancel to give a
a non-negative quantity.
Hence the integral is positive.
! "
#
n
0≤
− ν dxdy = rL1 − A1 − πr2
2
If the circle center is in red area then the circle is
completely inside the hull. If the circle center is in green
area circle intersects hull boundary in at least 4 points.
If the circle center is in red area then the circle is
completely inside the hull. If the circle center is in green
area circle intersects hull boundary in at least 4 points.
If the circle center is in red area then the circle is
completely inside the hull. If the circle center is in green
area circle intersects hull boundary in at least 4 points.
If the circle center is in red area then the circle is
completely inside the hull. If the circle center is in green
area circle intersects hull boundary in at least 4 points.
Outcircle picture
Green area guarantees 4 points of
intersection
Circles with 0 points of intersection can
only have centers which lie within the red
area
Only the inradius circle and the tangency points and lines
are needed to construct the regions.
Minkowski geometry
• The theorem extends to Minkowski
geometries where the unit ball is a convex
set centered at the origin and the “distance
rulers” change with direction.
• e.g. “taxicab geometry”
Example of unit ball and
isoperimetrix
Isoperimetrix
Ball
• The isoperimetrix has least Minkowski
boundary length for a given area
Bonnesen formula
! "
#
n
2
0≤
− ν dxdy = rL − A1 − πr
2
• Where L is the length measured in
Minkowski geometry
• The hexagon is the isoperimetrix, not the
unit ball
Averaging still holds
n/2 − ν ≥ 1
n/2 − ν ≥ −1
n/2 − ν = 0
Pentagon
• What happens if
the isoperimetrix
is not
symmetric?
• Averaging fails
to be positive
for inradius
• open!
LA
≥
π
h2 ds ??
References
• Gage, M. An isoperimetric inequality with applications
to curve shortening., Duke Math. J. v. 50 (1983)
• Gage, M. Positive centers and the Bonnesen inequality,
Proceedings of the AMS, 1990
• Gage, M. Evolving plane curves by curvature in relative
geometries. Duke Math. J. 1993
• Green, M. and Osher, S. Steiner polynomials,Wulff
flows and some new isoperimetric inequalities for
convex plane curves. Asian J. Math, 1999
References
• R. Osserman, 'Bonnesen-style Isoperimetric
Inequalities', Amer. Math. Monthly 86 (1979).
• R. Osserman, The isoperimetric inequality,
Bull. Amer. Math. Soc. 84 (1978).
• L. Santalo, “Integral geometry and Geometric
Probability”
Thank you