On Hadwiger’s containment formula
Alina Stancu
Polytechnic University of New York
&
Université de Montréal
1
2
Notations
Here and thereafter K1 and K2 are two convex bodies in R2 with non-empty
interiors.
We denote by Ai, i = 1, 2 their respective areas, and by Li, i = 1, 2 the
lengths of their boundaries. Let ∆i = L2i − 4πAi denote the isoperimetric
deficit of Ki.
Theorem 1 (Hadwiger, 1942). If
(1)
2π(A1 + A2) − L1 · L2 > 0,
then there exists a Euclidean motion g such that either K1 ⊆ int gK2 or
K2 ⊆ int gK1.
Remark 1. (1) is equivalent to |L2 − L1| >
p
∆1 + ∆ 2 .
Remark 2. Denoting by H(K1, K2) := 2π(A1 + A2) − L1 · L2, then
∆1 + ∆ 2
H≥−
.
2
3
Questions
(1) Is there a functional
∆
+
∆
1
2
f : K2 × K2 → −
, 0
2
such that
2π(A1 + A2) − L1 · L2 > f (K1, K2)
implies the containment of Ki in the interior of Kj , up to a rigid motion?
(2) Is there hope to find such a condition which is not only sufficient, but
also necessary?
(3) Can such a containment formula give additional information about the
two bodies, for example some estimates on the relative inradius, respective, the relative outradius of Ki with respect to Kj ?
Remark 3. One immediate observation is that an optimal f must satisfy
a homogeneity condition of the form
f (tK1, tK2) = t2f (K1, K2),
∀t > 0.
Remark 4. In fact, if such an optimal f exists then
p
f (K1, K2) ≥ 16π 2A1A2 − L1L2.
4
Theorem 2 (S., 2005). If
(2)
p
2π(A1 + A2) − L1 · L2 > − ∆1 · ∆2,
then there exists a rigid motion g such that either K1 ⊆ int g(K2) or
K2 ⊆ int g(K1).
Comments:
• It reduces to Hadwiger’s inequality if one of the bodies is a circle.
• It is definitely not a necessary condition as well. See the case of a circle of
radius 1 with a circumscribed square.
p
p
• (2) is equivalent to |L2 − L1| > | ∆1 − ∆2 |.
Theorem 3 (E. Grinberg, Delin Ren, Jiazu Zhou, ≥ 1993).
Let Di, i = 1, 2 be domains in the plane X k (R2, RP 2, H 2). Then either
of the following conditions implies that there exists an isometry g of X k
so that Di ⊃ gDj . (Here ∆ denotes L2 − 4πA + kA2.)
2
σ(Di, Dj ) := ∆i + ∆j − k(Ai − Aj ) ≥ 0
and
σ(Di, Dj ) < 0.
Li − Lj >
q
σ(Di, Dj ).
5
p
Heuristic argument which led to f (K1, K2) = − ∆1 · ∆2
Let K1 be the parallel body of a segment L (whose length is l). Thus
K1 = L + T · B, where B is the unit ball in R2 and T > 0 is some positive number.
Let K2 ⊂ R2 be such that, for any t, 0 ≤ t ≤ T , K2 − t · B is a convex body
of nonempty interior with ∂(K2 − t · B) ∈ C 2 or better.
Then hi(θ, t) = hi(θ) − t, i = 1, 2 are support functions of convex bodies
with C 2 or better boundary, up to T , and they are also solutions to the flow
dhi(θ, t)
= −1, 0 ≤ t ≤ T, i = 1, 2.
dt
p
Under this flow, Ht ≡ 0, thus H constant, and so is − ∆1 · ∆2. In fact,
each ∆i remains constant for 0 ≤ t ≤ T .
We note that
∃g rigid motion s.t. K1 ⊆ int gK2
iff
∃g rigid motion s.t. K1(T ) ⊆ int gK2(T )
(∃g rigid motion s.t. L ⊆ int gK2(T )).
But, at t = T , the inequality (2) is 2πA2(T ) − 2lL2(T ) > −2l
or
√
L2(T ) + ∆2
2l <
< L2(T ). 2
p
∆2
6
However this approach does not work for general convex bodies.
Recall then
p
(2) 2π(A1 + A2) − L1 · L2 > − ∆1 · ∆2
and note first that, for K1, K2 rectangles of dimensions li, Li, i = 1, 2,
respectively,
q
2π(A1 + A2) − L1 · L2 > − (L21 − 4πA1) · (L22 − 4πA2)
↓
↓
↓
q
2 · 4(A1 + A2) − L1 · L2 > − (L21 − 4 · 4A1) · (L22 − 4 · 4A2)
the modified isoperimetric deficit: ∆ri = 4(Li −li)2 and, overall, the inequality
(2) becomes
8(L1 − L2)(l1 − l2) > 0.
The latter is a necessary and sufficient condition for two rectangles to be such
that one contains the other, up to a rigid motion.
7
This led to the following argument for proving Theorem 2.
Let Ki, i = 1, 2, convex bodies with nonempty interiors for which we denote
by hi : [−π, π] → R their support functions.
p
Note that if we show that equation (2): 2π(A1 + A2) − L1 · L2 > − ∆1 · ∆2,
implies (3) below, for i 6= j ∈ {1, 2} fixed, then the theorem is proved:
(3)
hi(θ) + hi(θ + π) < hj (θ) + hj (θ + π),
for all θ ∈ [−π, π].
Consider the Fourier expansions of hi(θ). We indicate them by writing
∞
X
h1(θ) ∼
(ak cos kθ + bk sin kθ) (with b0 = 0)
h2(θ) ∼
k=0
∞
X
(ck cos kθ + dk sin kθ) (with d0 = 0).
k=0
It is well known [Groemer, “Handbook of convex geometry”, Vol. B] that
L1 = 2πa0 and L2 = 2πc0.
∞
Also A1 =
πa20
πX 2
−
(k − 1)(a2k + b2k ), therefore
2
k=2
∆1 = L21 − 4πA1 = 2π 2
∞
X
k=2
and similarly for A2, ∆2.
(k 2 − 1)(a2k + b2k ),
8
Furthermore
p
Φ(K1, K2) := 2π(A1 + A2) − L1 · L2 + ∆1 · ∆2 =
v
2
v
u∞
u∞
uX
uX
2
2
2 t
2
2
2
2π (a0 − c0) − π
(k − 1)(ak + bk ) − t (k 2 − 1)(c2k + d2k )  .
k=2
k=2
Meanwhile, consider, for each θ ∈ [−π, 0], and for some θ0 ∈ [−π, π],
the circumscribed rectangles to Ki of dimensions
[h1(θ) + h1(θ + π)], [h1(θ + π/2) + h1(θ + 3π/2)]
and [h2(θ + θ0) + h2(θ + θ0 + π)], [h2(θ + θ0 + π/2) + h2(θ + θ0 + 3π/2)].
Calculate
Φr = 8(h1(θ) + h1(θ + π) − h2(θ + θ0) − h2(θ + θ0 + π))(h1(θ + π/2)
+h1(θ + 3π/2) − h2(θ + θ0 + π/2) − h2(θ + θ0 + 3π/2)).
The essential part is that there exists θ0 such that
Φr ≥ Φ(K1, K2)> 0,
for all θ ∈ [−π, 0] which implies that, for a given direction θ, either both
h1(θ) + h1(θ + π) < h2(θ + θ0) + h2(θ + π + θ0)
h1(θ + π/2) + h1(θ + 3π/2) < h2(θ + π/2 + θ0) + h2(θ + 3π/2 + θ0)
hold, or both
h1(θ) + h1(θ + π) > h2(θ + θ0) + h2(θ + π + θ0)
h1(θ + π/2) + h1(θ + 3π/2) > h2(θ + π/2 + θ0) + h2(θ + 3π/2 + θ0)
hold.
9
The strict inequalities, combined with a continuity argument, implies that
the width of one body is, in all directions, smaller than the corresponding
width of the other.
Without any loss of generality, one can assume that a0 > c0 in the Fourier
expansions of h1, h2, thus L1 > L2 and K2 is contained in K1, up to a rigid
motion. Proposition 1. Let K1, K2 be planar convex bodies.
p
p i
1 h
For any t such that 0 < t <
(L1 − L2) − | ∆1 − ∆2| , if such t
2π
exists, then K2 + tB ⊆ K1, up to a rigid motion.
p
p i
1 h
For any t such that t >
(L1 − L2) + | ∆1 − ∆2| , then K2 +tB ⊇
2π
K1, up to a rigid motion.
The proof is immediate by looking at the roots of the quadratic equation:
Φ(K1, K2 + tB) = 0.
Remark 5. Similarly one can look at Φ(K1, tK2) to get estimates on the
Inradius, respectively Outradius, of K2 relative to K1, where
Rin := sup{r > 0 | (rK2) ⊆ g(K1), for some rigid motion g}
Rout := inf{r > 0 | (rK2) ⊇ g(K1), for some rigid motion g}.
10
∆
+
∆
1
2
Conjecture: If f : K2 × K2 → −
, 0 such that
2
• f is symmetric in its arguments,
• f homogenous of degree 1 in one (thus in both) factor(s),
then 2π(A1 + A2) − L1 · L2 > f (K1, K2) cannot be a sufficient AND necessary condition for K1 to contain gK2 for some g rigid motion.
Why? If such a necessary and sufficient condition would exist, i.e.
Ψ(K1, K2) := 2π(A1 + A2) − L1 · L2 − f (K1, K2) > 0,
then the roots of the equation
Ψ(K1, tK2) := 2π(A1 + t2A2) − L1 · tL2 − tf (K1, K2) = 0,
will be precisely
Rin(K2) := sup{r > 0 | (rK2) ⊆ g(K1), for some rigid motion g}
Rout(K2) := inf{r > 0 | (rK2) ⊇ g(K1), for some rigid motion g}.
Then the two roots (radii) would be at equal distance from the vertex of the
L1L2 + f (K1, K2)
parabola y = Ψ(t), thus from
.
2πA2
Similarly, and simultaneously, for R∗(K1), where the distance is from
L1L2 + f (K1, K2)
.
2πA1
Hard to believe that such a precise expression of Rin , Rout can be given for arbitrary convex bodies in terms
of A and L.