TV denoising of the characteristic function of two balls in
the plane
∗
†
Vicent Caselles, Matteo Novaga, Christiane Pöschl
‡
Abstract
The aim of this paper is to compute the explicit solution of the total variation
denoising problem corresponding to the characteristic function of a set which is the
union of two balls in the plane with disjoint interiors.
1
Introduction
The purpose of this paper is to compute explicit solutions of the total variation denoising
problem
1
|Du| +
2λ
R2
Z
Minu∈BV (R2 )∩L2 (R2 )
where
f = χS
and
S ∈ R2
Z
R2
|u − f |2 dx ,
(1)
is the union of two balls whose interiors are disjoint sets.
The study of explicit solutions of (1) was initiated in [8, 9], where the authors studied
the bounded sets of nite perimeter
S
in
R2
for which the solution of (1) is a multiple of
χS .
Such sets, which were called calibrable, produce solutions of the total variation ow which
evolve at constant speed without distortion of the boundary. They where characterized in
[8] by the existence of a vector eld
−div ξ =
P (S)
|S| χS , where
ξ ∈ L∞ (R2 , R2 )
|ξ| ≤ 1, ξ · DχS = |DχS |,
S and |S| denotes the area of
such that
P (S) denotes the perimeter of
S . For bounded connected sets S ⊂ R2 , calibrable sets are characterized as the convex,
(S)
where κ(x) denotes the curvature of
C 1,1 sets satisfying the bound ess supx∈∂S κ(x) ≤ P|S|
∂S at the point x. The paper [8] gives also a characterization of non connected calibrable
2
2
sets in R . The paper [9] describes the explicit solution of (1) for sets S ⊂ R of the form
S = C0 \ ∪ki=1 Ci where the sets Ci , i = 0, 1, . . . , k , are convex and satisfy some bounds on
the curvature of its boundary. The explicit solution when the set S is a convex subset of
R2 was described in [4] (also the case of a set S which is a union of convex sets which are
and
suciently far from each other). The explicit solution corresponding to a general convex set
DTIC, Universitat PompeuFabra, Roc Boronat 138, 08018 Barcelona, Spain, email: [email protected]
†
Dipartimento di Matematica, University of Padova, via Trieste 63, 35121 Padova, Italy, email: [email protected]
‡
Computational Science Center, Nordbergstraÿe 15, 1090 Wien, Austria, email:
[email protected]
∗
1
in
RN
was described in the papers [3, 11, 2] (covering also the case of the union of convex
sets which are suciently far in a precise sense).
When
S
S1 , S2 ,
is the union of two convex sets
the situation may become more com-
plicated, even when both sets are calibrable. In particular, if the sets are near each other,
more precisely if the perimeter of its convex hull is smaller than the sum of the perimeters
of the two sets, then the sets interact, and the solution outside
of
λ).
This is the case, for instance, when
S
S
is not null (for small values
is the union of two balls in
R2 ,
which is the
object of this paper.
Let us mention in this context the work of Allard [1] who calculated the solution of (1)
S is the union of two balls with the same radius. He also computed the solution of (1)
S is the union of two squares. In this paper we extend the result of Allard [1] to the
2
case where S is the union of any two balls in R . The interesting case is when the perimeter
of the convex hull of S is less than the perimeter of S , since otherwise the solution can be
when
when
described as the sum of the solutions corresponding to each ball [8]. Our approach diers
from the one in [1] even for the case of two balls of the same radius. While the solution in
[1] is obtained by a explicit computation, we describe it in a shorter way by means of more
general geometric arguments.
of (1) if and only if the sets
Our starting point is the observation that
[uλ ≥ s]
Fs,λ (X) := P (X) +
uλ
is a solution
minimize the variational problem [12, 1]
s
(1 − s)
|X \ S| −
|X ∩ S|
λ
λ
s ∈ [0, 1], λ > 0 ,
(2)
P (X) is the perimeter of X (and we understand that P (X) = +∞ if χX 6∈ BV (R2 )).
us point out that, for λ > 0 xed, the solutions of (2) are monotonically decreasing as
where
Let
s
increases, and can be then packed together to build up a function which solves (1) [3, 12].
Thus, to compute the solution of (1) we study the solution of (2) for any value of
and
s ∈ [0, 1],
λ>0
and those solutions can be constructed by means of geometric arguments.
Cs,λ is a minimizer
s
and −
λ outside S . When S is
On one hand, the Euler-Lagrange equation of (2) tells us that, if
1,1 ,
of (2), then ∂Cs,λ is C
∂Cs,λ has curvature 1−s
λ inside
S
the union of two balls with disjoint interiors, we prove that the intersection of
Si or ∅.
∅, S1 , S2 , S, Close λ (S),
with the minimizing sets is either
the following ones:
Si , i = 1, 2,
This reduces the possible minimizers of (2) to
where
s
Closer (S)
denotes the closing of the set
r contained in R2 \ S ) and co(S) its
convex hull (co(S) corresponds to Close λ (S) with s = 0). The essential of the proof is then
s
to compare the energy Fs,λ of those ve sets to decide which is the solution of (2).
S
(the complement of the union of the balls of radius
The computation of explicit examples of TV denoising permits to exhibit qualitative
features of the solution. In particular, the appearance of new level lines is an undesirable
feature for a denoising. Let us mention that better denoising algorithms have been developed
in the last few years [10].
Let us nally describe the plan of the paper.
In Section 2 we review some known
results that permit to set the context of our analysis. In Section 3 we describe the generic
properties of the minimizers
the intersection of
Cs,λ
with
Cs,λ of (2) and we prove that when S is the union of two balls
Si , i = 1, 2, is either Si or the emptyset. This permits to reduce
2
the set of possible minimizers of (3) to the following ve ones:
∅, S1 , S2 , S, Close λ (S).
This
s
is proved in Section 4, which also contains the description of the solution of (1).
2
Preliminaries
2.1
Let
Total variation and perimeter
Ω
be an open subset of
RN .
A function
u ∈ L1 (Ω)
whose gradient
Du
in the sense of
distributions is a (vector valued) Radon measure with nite total variation in
function of bounded variation. The class of such functions will be denoted by
Du
total variation of
on
Ω
Z
sup
u
div z
Ω
Ω is called a
BV (Ω). The
turns out to be
dx : z ∈
C0∞ (Ω; RN ), kzkL∞ (Ω)
:= ess sup |z(x)| ≤ 1 ,
x∈Ω
P
2
by
v = (v1 , . . . , vN ) ∈ RN we set |v|2 := N
i=1 vi ) and will be denoted
R
|Du|(Ω) or by Ω |Du|. BV (Ω) is a Banach space when endowed with the norm Ω |u| dx +
|Du|(Ω). We recall that BV (RN ) ⊆ LN/(N −1) (RN ).
N −1 the (N − 1)-dimensional Haussdorf measure.
Let us denote by H
N is said to be of nite perimeter in RN if χ ∈ BV (RN ) of
A measurable set E ⊆ R
E
E . The perimeter of E is dened as P (E) := |DχE |(RN ). We recall that when E is a
nite-perimeter set with regular boundary (for instance, Lipschitz), its perimeter P (E) also
N −1 (∂E). For sets of nite perimeter E one
coincides with the more standard denition H
∗
can dene the essential boundary ∂ E , which is countably (N − 1) rectiable with nite
HN −1 measure, and compute the outer unit normal ν E (x) at HN −1 almost all points x of
∂ ∗ E , where HN −1 is the (N −1) dimensional Haussdorf measure. Moreover, |DχE | coincides
N −1 to ∂ ∗ E . For more properties and references on functions of
with the restriction of H
(where for a vector
R
bounded variation we refer to [5].
2.2
Review of some basic results
The following result was proved in [1, 12].
Let S ⊂ R2 . Then there is a unique solution uλ of (1). Then for any
s ∈ R, {uλ ≥ s} (resp. {uλ > s}) is the maximal (resp. the minimal) solution of
Proposition 2.1.
min Fs,λ (X) := P (X) +
X⊂R2
s
(1 − s)
|X \ S| −
|X ∩ S|.
λ
λ
(3)
In particular, for all t but a countable set the solution of (3) is unique.
Conversely, for any s ∈ R, let Qs be a solution of (3). If s > s0 , then Qs ⊆ Qs0 . The
function
is the solution of (1).
u(x) = sup{s : x ∈ Qs }
3
Thus, in order to built up the solution of (1) it suces to compute the solutions of the
family of problems (3). This will be the strategy we follow to compute the explicit solution
when
S
is the union of two balls.
Let us rst recall a result that permits to compute the value of
uλ = 0.
g ∈ L2 (RN )
the
kgk∗ =
kgk∗ ≤ 1
BV ∗
norm is given by
Z
gu dx.
sup
u∈BV (RN ),|Du|(RN )≤1 RN
if and only if
Z
RN
for any
for which the solution
The result was proved in [16, 8].
Recall that if
Then
λ
gu dx ≤
Z
RN
|Du|
u ∈ BV (RN ). This is equivalent to say that
Z
g ≤ P (F ),
for any set F of nite
perimeter in
RN .
F
Proposition 2.2.
(see [8] ) Let g ∈ L2 (RN ). Let us consider the problem:
Z
Minu∈BV (RN )∩L2 (RN )
R2
|Du| +
1
2λ
Z
RN
|u − g|2 dx.
(4)
The following conditions are equivalent
(i)
(ii)
(iii)
u=0
is a solution of
.
(4)
kgk∗ ≤ λ
There is a vector eld ξ ∈ L∞ (RN , RN ), kξk∞ ≤ 1 such that −div ξ = g.
The following result has been proved in [15, 8, 4, 14].
Let C ⊂ R2 be a bounded set of nite perimeter, and assume that C is
connected. Let γ > 0. The following conditions are equivalent:
Theorem 1.
(i) C decreases at speed γ , i.e, for any λ > 0 uλ := (1 − λγ)+ χC (x) is the solution of (1)
corresponding to χC (x).
(ii) C is convex, γ = γC :=
P (C)
|C|
and minimizes the functional
GγC (X) := P (X) − γC |X|,
X ⊆ C, X of f inite perimeter.
(iii) C is convex, ∂C is of class C 1,1 , γ = γC :=
P (C)
|C|
, and the following inequality holds:
ess sup κ∂C (p) ≤ γC ,
p∈∂C
where κ∂C (p) denotes the curvature of ∂C at the point p.
4
We call a bounded connected set
C ⊆ R2
of nite perimeter
calibrable if it satises any of
the conditions of Theorem 1. The denition of calibrable sets is usually given in a dierent
and more general way (see for instance [8] and the Introduction), but we restrict to the
above formulation since we only need the result of Theorem 1. If
C
(resp.
∅)
is a minimum of
C ⊂ R2
is calibrable, then
min P (X) − µ |X|
(5)
X⊆C
for any
For
(C)
P (C)
P (C)
µ > P|C|
(resp. µ <
|C| ). If µ = |C| , both C and ∅ are minima of
+ := max {0, r}. The following result has been
all r ∈ R, we set r
(5).
proved in [8,
Theorem 7 and Proposition 8].
Lemma 2.3.
Let S1 , S2 ⊂ R2 be two calibrable sets, let S = S1 ∪ S2 and f = χS . Then
P (S1 ) +
P (S2 ) +
χS1 + 1 −
χS2
uλ = 1 −
λ
λ
|S1 |
|S2 |
is a solution of (1) for any λ > 0 if and only if
P (S) ≤ P (co(S)) .
(6)
In other words, the solution of (1) is the sum of the two solutions corresponding to χS1 and
χS2 if and only if (6) holds.
3
The two balls case: reduction of the possible set of minimizers of (3)
As explained in Section 2.2 our purpose is to compute the minimizers of
Fs,λ
when
S
is
the union of two balls with disjoint interiors. In this Section we will reduce the number of
possible minimizers. In Section 4 we will identify them and construct the explicit solution
of (1).
Let S1 , S2 ⊂ R2 be two bounded, convex, disjoint (modulo a null set), and
calibrable sets, and S = S1 ∪ S2 . Let λ > 0, s ∈ [0, 1]. Let Cs,λ be a minimizer of Fs,λ . Then
if the set Cs,λ is non-empty, the boundary ∂Cs,λ is of class C 1,1 and ∂Cs,λ \ S is formed by
> P|S(Si |i ) , Cs,λ ∩ Si = Si or
arcs of circle of curvature − λs . Moreover, Cs,λ ∩ Si = Si if (1−s)
λ
= P|S(Si |i ) , i = 1, 2.
∅ if (1−s)
λ
Assume that S1 and S2 are two open disjoint balls. Then Cs,λ ∩ Si = Si or ∅ for any
λ > 0, s ∈ [0, 1], i = 1, 2. In particular, when they exist, the arcs of ∂Cs,λ \ S have radius
λ
s and are tangent to ∂S .
Proposition 3.1.
The previous result is not true when
Proof.
Cs,λ is
∂Cs,λ \ S are
The regularity of
that, if non-empty,
S1 , S2
are general calibrable convex sets [13].
a classical result [6].
The Euler-Lagrange equations say
arcs of circle of curvature
5
− λs .
Step 1.
Let
i ∈ {1, 2}.
Let us prove that
P (Cs,λ ∩ Si ) −
for any set
F
(1 − s)
(1 − s)
|Cs,λ ∩ Si | ≤ P (F ) −
|F |,
λ
λ
of nite perimeter in
Si
such that
Cs,λ ∩ ∂Si ⊆ ∂ ∗ F ∩ ∂Si
(7)
(modulo an
H1
null
set). For simplicity, when considering inclusions of boundaries or essential boundaries we
H1
mean always modulo an
Let
F
null set.
be a set of nite perimeter in
F ∪ (Cs,λ \ Si ).
P (Cs,λ ) +
Clearly
F 0 \ Si = Cs,λ \ Si
j
is index dierent from
P (Cs,λ ) −
since
Cs,λ ∩ ∂Si ⊆ ∂ ∗ F ∩ ∂Si .
F 0 ∩ Si = F ∩ Si . From
such that
and
Let
F0 =
s
(1 − s)
s
(1 − s) 0
|Cs,λ \ S| −
|Cs,λ ∩ S| ≤ P (F 0 ) + |F 0 \ S| −
|F ∩ S|
λ
λ
λ
λ
≤ P (F 0 ) +
where
Si
Cs,λ \ Si = F 0 \ Si
s
(1 − s)
|Cs,λ \ S| −
|F ∪ (Cs,λ ∩ Sj )|,
λ
λ
i,
we get
(1 − s)
(1 − s)
|Cs,λ ∩ Si | ≤ P (F 0 ) −
|F |,
λ
λ
(modulo a null set) and
F 0 ∩ Si = F .
(8)
Now, let us prove
P (Cs,λ ) − P (F 0 ) ≥ P (Cs,λ ∩ Si ) − P (F )
(9)
which, together with (8), implies (7).
Step 2. The proof of (9).
F0
= F ∪ B.
A = Cs,λ ∩ Si , B = Cs,λ \ Si .
that by our assumption on F
Let
Observe also
Cs,λ = A ∪ B
Observe that
and
∂ ∗ A ∩ ∂ ∗ B ⊆ Cs,λ ∩ ∂Si ∩ ∂ ∗ B ⊆ ∂ ∗ F ∩ ∂ ∗ B.
Then
P (Cs,λ ) − P (Cs,λ ∩ Si ) = P (A ∪ B) − P (A) = P (B) − 2H1 (∂ ∗ A ∩ ∂ ∗ B)
≥ P (B) − 2H1 (∂ ∗ F ∩ ∂ ∗ B) = P (F 0 ) − P (F ).
Step 3.
Now, taking
F = Si
in (7) we have
P (Cs,λ ∩ Si ) −
Since
if
Si
(1−s)
λ
Step 4.
(1 − s)
(1 − s)
|Cs,λ ∩ Si | ≤ P (Si ) −
|Si |.
λ
λ
is calibrable, by (5) we have
=
P (Si )
|Si | .
Cs,λ ∩ Si = Si
if
(1−s)
λ
>
P (Si )
|Si | , and
Cs,λ ∩ Si ∈ {Si , ∅}
S2 are two balls. Let λ > 0 and s ∈ [0, 1].
Let us observe that either Cs,λ ⊂ Si for some i = 1 or 2 or Cs,λ ∩ Si 6= ∅ for both i = 1, 2.
λ
1,1 and ∂C
Indeed this happens because Cs,λ is C
s,λ \ S are arcs of circle of radius s .
Assume that
S1
and
6
L1
K
Cs,λ
Γ0
Γ
G
L2
Figure 1: Illustration of case (i) in Step 4 of the proof of Lemma 3.1.
If
Cs,λ ⊂ Si
i=1
for some
or 2, then
Cs,λ
is a minimizer of
(1 − s)
|X|.
λ
min P (X) −
X⊆Si
Cs,λ ∩ Si = ∅ or Cs,λ ∩ Si = Si .
Cs,λ ∩ Si 6= ∅ for both i = 1, 2. For each i = 1, 2, let Bi
λ
be the ball of radius
1−s such that Bi ∩ Si = Cs,λ ∩ Si . Since both sets are balls, either
∂Bi ∩ ∂Si consists of at most two points or Bi = Si . If for some i ∈ {1, 2} ∂Bi ∩ ∂Si consists
of a single point, then Bi ⊆ Si and Cs,λ ∩ Si = Bi . In that case, Cs,λ ∈ {∅, Si } because
Si is calibrable. Thus we may assume that either Bi = Si for i = 1, 2 and we are done,
or ∂Bi ∩ ∂Si consists of two points for i = 1, 2, or (without loss of generality) ∂B2 ∩ ∂S2
consists of two points and B1 = S1 . In the second case we consider the two subcases:
Since
Si
is calibrable we have that either
Thus, we may assume that
(i) the angular span of
∂Cs,λ ∩ int(Si )
is
≥π
for some
(ii) the angular span of
∂Cs,λ ∩ int(Si )
is
<π
for
i ∈ {1, 2},
i = 1, 2.
(i) holds we modify Cs,λ ∩Si to construct a new set with improved energy. Let us consider
Γ contained in Cs,λ ∩ Si and whose circular part of the boundary (a circular arc
going from p to q ) is contained in K := ∂Cs,λ ∩ int(Si ). In the rst step, we observe that we
can translate Γ in the direction of the tangent at p so that we get a bigger set G such that
Fs,λ (G) = Fs,λ (Cs,λ ). The boundary has now two straight line segments Li (let us denote
0
0
0
them [p , p] and [q , q], respectively) joining the translated half-disc Γ to the rest of the set.
λ
0
Now, we dilate Γ and we get a new half-disc with radius
1−s + η for some η > 0 small. Then
If
a half-disc
we complete the set by adding triangles whose hypotenuse joins the vertex of the enlarged
half-disc to the points
p
and
q
points where the segment ends and
Cs,λ ).
this new set. Then for η > 0 small
< Fs,λ (Cs,λ ). This contradiction proves that the angular span
starts again the original boundary of
(G0 )
Li (the
0
Let G be
of the segments
Fs,λ
of ∂Cs,λ ∩ int(Si ) is < π .
Assume that (ii) holds. Notice that the curvature κs,λ
∂Cs,λ ∩ int(Si ) and ≤ 0 in ∂Cs,λ \ Si . Since
Z
κs,λ dH1 = 2π,
enough we have that
of
Cs,λ
∂Cs,λ
then
Z
κs,λ >0
1
κs,λ dH = 2π −
7
Z
κs,λ ≤0
κs,λ dH1 ≥ 2π,
is positive in the arcs
∂Cs,λ ∩ int(Si ), i = 1, 2, is < π .
∂B2 ∩ ∂S2 consists of two points and B1 = S1 . Since
which is impossible since the angular span of both arcs
We are left with the case where
Fs,λ (Cs,λ ∪ S2 ) ≥ Fs,λ (Cs,λ )
we have
s
1−s
s
1−s
P (Cs,λ ∪ S2 ) + |Cs,λ \ S| −
|(Cs,λ ∪ S2 ) ∩ S| ≥ P (Cs,λ ) + |Cs,λ \ S| −
|Cs,λ ∩ S|.
λ
λ
λ
λ
By canceling common terms we have
P (Cs,λ ∪ S2 ) −
Since
1−s
1−s
|S2 | ≥ P (Cs,λ ) −
|Cs,λ ∩ S2 |.
λ
λ
P (Cs,λ ∪ S2 ) − P (Cs,λ ) = P (S2 ) − P (Cs,λ ∩ S2 )
P (S2 ) −
This implies that
For any set
1−s
1−s
|S2 | ≥ P (Cs,λ ∩ S2 ) −
|Cs,λ ∩ S2 |.
λ
λ
Cs,λ ∩ S2 ∈ {∅, S2 }.
X ⊂ R2 ,
This concludes the proof.
its closing with balls of radius
Closer (X) := R2 \
C 1,1
Now, notice that the only
S
Close λ (S) = co(S),
choices for minimizers of
Fs,λ
is dened as the set
Br (x) .
x:Br (x)⊂R2 \X
connected set that contains
the convex hull of
s
r≥0
[
λ
consists of arcs of circles of radius
s is
of
we have
S.
Close λ (S).
S
whose boundary outside
Moreover for
s
s = 0,
we set
As a consequence of Proposition 3.1, the only
are
∅, S1 , S2 , S
and
Close λ (S) .
s
Hence it remains to compare the energies of these sets in order to decide which one has the
smallest energy.
Before that, let us look for the values of
Lemma 3.2.
then
that lead to the trivial solution
u = 0 is a solution
BV -norm of χS .
According to Proposition 2.2,
interested in estimating dual
λ
of (1) if
kχS k∗ ≤ µ.
u ≡ 0.
Hence we are
Let S = S1 ∪ S2 , S1 , S2 being convex, calibrable and disjoint. If kχS k∗ ≤ µ
µ ≥ max
|S1 |
|S2 |
|S|
,
,
P (S1 ) P (S2 ) P (co(S))
.
(10)
If both S1 , S2 are balls and (10) holds, then kχS k∗ ≤ µ.
Proof.
g ∈ L2 (R2 ) the following equivalence holds:
Z
2
kgk∗ ≤ µ ⇐⇒
g ≤ µP (F )
for any set F ⊆ R
Since for
with nite perimeter,
(11)
F
the fact that
kχS k∗ ≤ µ implies that |F ∩ S| ≤ µP (F ) for any F ⊆ R2
8
with nite perimeter.
F = co(S), F = S1 , and F = S2 we get
|S1 |
|S2 |
|S|
µ ≥ max
.
,
,
P (S1 ) P (S2 ) P (co(S))
In particular, taking
respectively.
Let us assume now that
to prove that
only if
kχS k∗ ≤ µ
S1 , S2
are circles and
µ ≥ max
n
|S1 |
|S2 |
|S|
P (S1 ) , P (S2 ) , P (co(S))
we use the characterization given by (11). Hence
0 ≤ min
F ⊂R2
o
. In order
kχS k∗ ≤ µ if and
1
P (F ) − |F ∩ S| =: min F0,µ (F ),
µ
F ⊂R2
F ⊂ R2 with nite perimeter.
By Lemma 3.1, we have that C0,µ ∩ Si ∈ {∅, Si } for i = 1, 2. This implies that C0,µ ∈
{∅, S1 , S2 , S1 ∪ S2 , co(S)}. By our choice of µ we have
where the minimum can be restricted to the sets
F0,µ (C0,µ ) ≥ 0.
Hence
4
kχS k∗ ≤ µ.
The two balls case: explicit solution of (1)
The aim of this section is to nd an explicit solution of (1), in the case where
of two balls
S1 , S2 .
S
is the union
Recall that, by Lemma 2.3, we may assume that
P (S) > P (co(S)),
otherwise the solution corresponding to
sponding to
S1
and
S2 ,
S
(12)
is described by the sum of the solutions corre-
that is, the two sets do not interact.
By Lemma 3.2 and Proposition 2.2 it is enough to consider
λ ≤ max
|S1 |
|S2 |
|S|
,
,
P (S1 ) P (S2 ) P (co(S))
,
otherwise the solution of (1) is equal to zero.
Without loss of generality, we may assume that the radius of
of
S2 ,
S1
is bigger than the radius
that is
|S2 |
|S1 |
≤
.
P (S2 )
P (S1 )
(13)
We start with the simple case where we compare the energies of
Lemma 4.1.
S1 , S2 , S
and
∅.
Let λ > 0. We have
min {0, Fs,λ (S) , Fs,λ (S1 ) , Fs,λ (S2 )} =
9



Fs,λ (S)
Fs,λ (S1 )


0
P (S2 )
1−s
λ ≥ |S2 | ,
P (S1 )
P (S2 )
1−s
|S1 | ≤ λ ≤ |S2 | ,
P (S1 )
1−s
λ ≤ |S1 | .
s
1 − λ P|S(S22| )
1 − λ P|S(S11| )
∅
S1
S
|S2 |
P (S2 )
|S1 |
P (S1 )
λ
(λ, s) this diagram shows which one of the
P (S1 )
P (S2 )
For (λ, s) = (λ, 1 − λ
|S1 | ), (λ, 1 − λ |S2 | ),
Figure 2: Illustration of Lemma 4.1. For any point
four sets
S1 , S2 , S, ∅ has the minimal Fs,λ
value.
the minima are not unique.
Proof.
Observe that, for
i = 1, 2,
Fs,λ (Si ) = P (Si ) −
1−s
|Si | ≤ 0
λ
if and only if
1−s
P (Si )
≥
.
λ
|Si |
P (S2 )
1−s
λ ≥ |S2 | , then Fs,λ (S) = Fs,λ (S1 ) + Fs,λ (S2 ) ≤ min {0, Fs,λ (S1 ) , Fs,λ (S2 )}. If
P (S2 )
P (S1 )
1−s
|S1 | ≤ λ < |S2 | , then Fs,λ (S2 ) > 0 and Fs,λ (S) = Fs,λ (S1 ) + Fs,λ (S2 ) > Fs,λ (S1 ). In
P (S1 )
1−s
case
λ < |S1 | , Fs,λ (S1 ) > 0, Fs,λ (S2 ) > 0 and min {0, Fs,λ (S) , Fs,λ (S1 ) , Fs,λ (S2 )} =
If
0.
We dene special values of
way. The parameters
λ
in order to classify the minimizers of
Fs,λ
in a compact
λ1 , λ2 , λ3 introduced in the following proposition are well dened (they
exist and are unique). This is shown in the proof of Proposition 4.6.
Let S1 , S2 be two balls. Assume that (12) and (13) hold. Let Rc (S) be
the minimal radius r such that Closer (S) is connected.
Proposition 4.2.
(i) There is a unique value R1 ∈ [Rc (S), ∞) such that
P (CloseR1 (S)) +
1
|CloseR1 (S) \ S| = P (S).
R1
Let s2 (λ) = 1 − λ P|S(S22| ) and λ1 be given by
[0, P|S(S22| ) ] and
Fs2 (λ),λ Close
λ1
s2 (λ1 )
= R1 .
Then λ1 =
(14)
R1 |S2 |
R1 P (S2 )+|S2 |
∈
λ
s2 (λ)
(S)
> Fs2 (λ),λ (S)
(resp.
=, <)
(15)
for any λ < λ1 (resp =, >). The value λ1 = 0 if and only if R1 = 0 and this happens
if and only if Rc (S) = 0, in other words if both balls touch. If Rc (S) > 0, then
R1 > Rc (S).
10
(ii) (a) If
|S1 |
P (S1 )
<
|S|
P (co(S))
, there is a unique value R2 ∈ [Rc (S), ∞) such that
P (CloseR2 (S)) +
|S|
1
|CloseR2 (S) \ S| = P (S1 )
.
R2
|S1 |
Let s1 (λ) = 1 − λ P|S(S11| ) and λ2 be given by
i
λ1 , P|S(S11| ) and
λ2
s1 (λ2 )
= R2 .
Then λ2 =
(16)
R2 |S1 |
R2 P (S1 )+|S1 |
h
∈
Fs1 (λ),λ Close
λ
s1 (λ)
(S)
> Fs1 (λ),λ (S1 )
(resp.
=, <)
(17)
for any λ < λ2 (resp =, >). We have that R1 = R2 if and only if P|S(S11| ) = P|S(S22| )
if and only if λ1 = λ2 . And R2 = 0 (in that case also R1 = 0 and λ1 = λ2 = 0)
if and only if Rc (S) = 0.
|S|
|S2 |
(b) If P (co(S))
≤ P|S(S11| ) set λ2 := P (co(S))−P
(S1 ) .
(iii) Set λ3 := max
n
|S|
|S1 |
P (S1 ) , P (co(S))
Remark 4.3. In the case
|S1 |
P (S1 )
=
.
o
|S2 |
P (S2 ) ,
λ1 = λ2 .
Sn = S1n ∪ S2n , where S1n , S2n are balls, is a sequence of
|Sn |
|S1 |
|S|
|S1n |
sets such that
P (S1n ) < P (co(Sn )) and Sn → S = S1 ∪ S2 with P (S1 ) = P (co(S)) , the value of
|S2 |
2n |S1n |
λ2n = R2n PR(S
given in (ii)(a) converges to the value of λ2 =
P (co(S))−P (S1 ) given in
1n )+|S1n |
|S|
(ii)(b). Indeed, if P|S(S1n1n| ) → P (co(S))
, then with the same notation as in Proposition 4.2, we
Remark 4.4. Observe that if
have
P (CloseR2n (Sn )) +
1
|Sn |
|CloseR2n (Sn ) \ Sn | = P (S1n )
→ P (co(S)).
R2n
|S1n |
R2n is bounded, modulo a subsequence, we may assume that R2n converges
R2 < ∞. Then CloseR2n (Sn ) → CloseR2 (S) and passing to the limit in the last
If we assume that
to some
equation we get
P (CloseR2 (S)) +
1
|S|
|CloseR2 (S) \ S| = P (S1 )
= P (co(S))
R2
|S1 |
P (co(S)) < P (CloseR2 (S)), hence we can conclude
2n |S1n |
λ2n = R2n PR(S
→ P|S(S11| ) .
1n )+|S1n |
On the other hand if we look at the denition of λ2 in (ii) (b) and consider the
|S1 |
|S|
|S2 |
|S1 |
P (S1 ) = P (co(S)) , we also obtain that λ2 = P (co(S))−P (S1 ) = P (S1 ) .
But this is not possible since
R2n → ∞.
Remark 4.5. Observe that when
|S1 |
P (S1 )
=
|S|
P (co(S)) ,
Now we are ready to list the minimizers of
by
Cs,λ
that
This implies that
the largest minimizer of
Fs,λ
Fs,λ .
λ2 =
|S1 |
P (S1 )
case
= λ3 .
For simplicity, from now on we denote
(see Proposition 2.1).
Assume that S1 , S2 are circles such that P (co(S)) < P (S) and (13) holds.
Let λ1 , λ2 , λ3 be as in Proposition 4.2. Then Cs,λ , minimizer of Fs,λ , is given by:
Proposition 4.6.
11
(a) λ ∈ [0, λ1 ]. There is a value 0 < sa (λ) ≤ 1 − λ P|S(S22| ) such that
Cs,λ


Close λ (S)


s


S
=

S1




∅
0 ≤ s ≤ sa (λ)
sa (λ) < s ≤ 1 − λ P|S(S22| )
1 − λ P|S(S22| ) < s ≤ 1 − λ P|S(S11| )
1 − λ P|S(S11| ) < s
(the third interval is empty in the case
|S1 |
P (S1 )
=
|S2 |
P (S2 )
).
(b) λ ∈ (λ1 , λ2 ]. There is a value 0 < sb (λ) ≤ 1 − λ P|S(S11| ) such that
Cs,λ =


Close λ (S)


s
0 ≤ s ≤ sb (λ)
sb (λ) < s ≤ 1 − λ P|S(S11| )
S1



∅
1 − λ P|S(S11| ) < s .
(c) λ ∈ (λ2 , λ3 ]
•
|S1 |
P (S1 )
<
|S|
P (co(S))
. There is a value sc (λ) > 1 − λ P|S(S11| ) such that
Cs,λ
•
|S|
P (co(S))
≤

Close λ (S)
s
=
∅
else.
|S1 |
P (S1 )
Cs,λ =
0 ≤ s ≤ 1 − λ P|S(S11| )
(
S1
else.
∅
Figure 4 shows solutions of (1) for dierent
radii
0 ≤ s ≤ sc (λ)
r1 = 1.2, r2 = 1
and distance
d = 0.05
(case
λ,
when
|S1 |
P (S1 )
<
S
is the union of two balls with
|S|
P (co(S)) ).
In order to prove Propositions 4.2, 4.6 we need the following Lemmas:
Lemma 4.7.
Let 0 < R < r and dene
Z
R
Gr (h) :=
p
1+
h0 (x)2
−R
1
− h(x)
r
dx.
(18)
The function that represents an arc of a circle with radius r (angular span smaller than π)
from −R to R, minimizes Gr (h) under all functions h with h(−R) = 0, h(R) = 0.
Proof.
See [7, Lemma 4.29].
Let Rc (S) be the minimal radius r such that Closer (S) is connected. Then
for s ∈ [0, 1], 0 < µ < λ, such that Rc (S) ≤ µs ,
Lemma 4.8.
Fs,λ Close λ (S) ≤ Fs,λ Close µ (S) .
s
s
12
h
P
Q
g̃
f
g
S2
a
a0
b0
S1
f˜
b
b0
a0
Figure 3: The geometric conguration of the proof of Lemma 4.8.
Proof.
Let us take the
x-axis
as the axis joining the centers of the two
circles.
is symmetric with respect to the
∂ Close µ (S)
s
x-axis.
Then the upper parts of
Then
∂ Close λ (S)
S
and
s
are representable as functions
f, g : [a, b] → R,
such that
!
Z
Z
p
1
s
1
Fs,λ Close λ (S) =
1 + (f 0 )2 +
f − |S|
s
2
λ
2
[a,b]
[a,b]
!
Z
Z
p
1
s
1
Fs,λ Close µ (S) =
1 + (g 0 )2 +
g − |S|
s
2
λ
2
[a,b]
[a,b]
P, Q be the points in the positive y -plane where ∂Close λ (S) intersects with S1 and
s
S2 . Dene h := [a0 , b0 ] → R as the ane function from P = (a0 , f (a0 )) to Q = (b0 , f (b0 ))
0 0
0 0
(see Figure 3). Set f˜ := [a , b ] → R, f˜ = h − f and g̃ := [a , b ] → R, g̃ = h − g , then
2
2
0
0
0
0
0
0
2
0
0
g̃(a ) = f˜(a ) = 0, g̃(b ) = f˜(b ) = 0, g = (g̃ ) and f = (f˜ )2 . Note that f˜ is an arc of
Let
circle with radius
λ
s.
Using functional
Gr
as in (18) with
r=
λ
s , replacing the domain of integration
0 0
by [a , b ], and Lemma 4.7 (see also [7, Lemma 4.29]) we have
G λ (f˜) ≤ G λ (g̃).
s
[−R, R]
Hence
s
1
1
Fs,λ Close λ (S) − Fs,λ Close µ (S) = G λ f˜ − G λ (g̃) ≤ 0.
s
s
s
s
2
2
Next we show some properties of the functions
s, λ → Fs,λ Close λ (S) .
s
Lemma 4.9.
Let S := S1 ∪ S2 , S1 , S2 being two bounded, convex, disjoint sets.
(i) Let κ > 0 and sκ (λ) = 1 − , λ ≥ 0. Then the mapping
λ
κ
is strictly decreasing and continuous as long as
Close λ (S) is connected.
sκ (λ)
13
λ
sκ (λ)
λ → Fsκ (λ),λ Close
≥ Rc (S),
λ
sκ (λ)
(S)
i.e., as long as the set
(ii) For λ > 0, the mapping s → Fs,λ Close λ (S) is continuous and strictly increasing
s
for s ∈ [0, Rcλ(S) ], i.e. as long as the set Close λ (S) is connected.
s
(iii) For r ∈ [Rc (S), ∞] the functions
1
|Closer (S)|
r
1
r → P (Closer (S)) + |Closer (S) \ S|
r
r → P (Closer (S)) +
are continuous and strictly decreasing in r.
Remark 4.10. Although the statement of the Lemma is redundant, it will be convenient
for later application.
Proof. (i)
that
λ1
s1
λ1 < λ2 and set si := 1 − λκi , i = 1, 2.
≥ Rc (S). By Lemma 4.8 we have that
Let
Then
s2 < s 1
and
λ1
s1
<
λ2
s2 . Assume
Fs2 ,λ2 Close λ2 (S) ≤ Fs2 ,λ2 Close λ1 (S)
s2
s1
s2 1 = P Close λ1 (S) − Close λ1 (S) ∩ S +
Close
λ1 (S) \ S
κ
λ2 s1
s1
s1
< Fs1 ,λ1 Close λ1 (S) .
s1
Hence the mapping
Rc (S).
(ii)
λ
sκ (λ)
(S)
is strictly decreasing as long as
λ
sκ (λ)
≥
The continuity follows from the continuity of the involved functions.
and
s ∈ [0, 1] → Fs,λ Close λ (S)
s
the continuity of the involved functions. Assume 0 ≤ s1 < s2 ≤ 1 and λ such
(S) , Close λ (S) are connected. Since Rc (S) ≤ sλ2 < sλ1 , Lemma 4.8 gives
(iii)
follows from
that
λ → Fsκ (λ),λ Close
Close λ
s1
Let
λ > 0.
The continuity of the function
s2
Fs1 ,λ Close λ (S) ≤ Fs1 ,λ Close λ (S) ,
s1
s2
hence
s1
Fs1 ,λ Close λ (S)
= P Close λ (S) +
s1
s1
λ
s1
≤ P Close λ (S) +
s2
λ
s2
< P Close λ (S) +
s2
λ
= Fs2 ,λ Close λ (S) .
s2
14
Close λ (S) − 1 |S|
λ
s1
Close λ (S) − 1 |S|
λ
s2
Close λ (S) − 1 |S|
λ
s2
(19)
This proves that the mapping
s → Fs,λ Close λ (S)
is strictly increasing on the interval
s
h
n
oi
0, min 1, Rcλ(S) .
λ
s2 we have that
Adding
0 < r2 < r1
P (Closer1 (S)) +
1
λ
|S|
to both sides of the inequality and setting
r1 :=
λ
s1 , r2
:=
and
1
1
|Closer1 (S)| < P (Closer2 (S)) + |Closer2 (S)| .
r1
r2
P (Closer (S)) + 1r |Closer (S)| is strictly decreasing in r for r ∈ [Rc (S), ∞] if Rc (S) >
0, or in (Rc (S), ∞] if Rc (S) = 0. We understand that the function is +∞ when r = Rc (S) =
0.
Hence
Writing again (19) as
1 − s1
s1 −
Close
(S)
\
S
|S|
Close λ (S) +
λ
s1
s1
λ
λ
1 − s1
s1 ≤ P Close λ (S) +
Close λ (S) \ S −
|S|
s2
s2
λ
λ
1 − s1
s2 Close λ (S) \ S −
< P Close λ (S) +
|S| ,
s2
s2
λ
λ
P
1
0), and adding 1−s
λ |S| to
1
both sides of the inequality above, we get that P (Closer (S)) + |Closer (S) \ S| is strictly
r
decreasing in r for r ∈ [Rc (S), ∞] if Rc (S) > 0, or in (Rc (S), ∞] if Rc (S) = 0. Notice that
setting
r1 :=
λ
s1 , r2
:=
λ
s2 (notice that these values are not
the function is also continuous in that range.
It remains to consider the case
Rc (S) = 0
P (Closer (S)) +
is continuous at
r = 0.
and to prove that
1
|Closer (S) \ S|
r
This follows if we prove that
1
|Closer (S) \ S| → 0 +
r
To prove (20), let us estimate the area of
Hence
r → 0+.
Closer (S) \ S .
≤ 2r and whose height is less than
|Closer (S) \ S| = O(r3/2 ) and (20) holds.
whose basis has length
√
O( r).
as
(20)
This set is contained in triangle
√
√
√
r max( 2R1 − r, 2R2 − r) =
Proof of Proposition 4.2.
(i) Assume rst that Rc (S) > 0.
P (CloseRc (S) (S)), and we have
In that case, because of the convexity of
P (S) < P CloseRc (S) (S) +
S1 , S2 , P (S) <
1 CloseRc (S) (S) \ S .
Rc (S)
On the other hand
lim P (Closer (S)) +
r→∞
1
|Closer (S) \ S| = P (co(S)) < P (S) .
r
Hence, by Lemma 4.9.(iii), (14) has a unique solution in
15
R1 ∈ (Rc (S), ∞).
(21)
Rc (S) = 0. Then we have CloseRc (S) (S) = S , moreover, since f (r) :=
P (Closer (S)) + |Closer (S) \ S| is a continuous and decreasing function in [Rc (S), ∞) by
Assume now that
1
r
Lemma 4.9.(iii), we have
lim P (Closer (S)) +
r→0+
1
|Closer (S) \ S| ≥ lim P (Closer (S)) = P (S).
r→0+
r
R1 = 0 ∈ [0, ∞) satises (14).
r → P (Closer (S)) + 1r |Closer (S) \ S| is strictly decreasing.
On the other hand, we also have (21). Thus, the
is unique since
This value
To prove (15), let us observe that
s
Fs,λ Close λ (S) − Fs,λ (S) = P Close λ (S) + Close λ (S) \ S − P (S).
s
s
s
λ
s = s2 (λ) and r = s2λ(λ)
function of λ we have that
Setting
in the above equality and observing that
Fs2 (λ),λ Close
= 0, < 0)
Notice that λ1 = 0
only if Rc (S) = 0.
is
>0
(resp.
(ii)(a)
S1
is
(S) − Fs2 (λ),λ (S)
λ < λ1 (resp. λ = λ1 , λ > λ1 ).
if R1 = 0 and we have proved that
if and only if
if and only
this happens if and
P (S1 )
|S1 | |S|. Let us rst assume that the radius of
P (S)
P (S1 )
than the radius of S2 , hence
|S| > |S1 | . If Rc (S) > 0, then
We are assuming that
>
λ
s2 (λ)
λ
s2 (λ) is an increasing
P (co(S)) <
P (Closer (S)) + 1r |Closer (S) \ S| → P CloseRc (S) (S) +
> P (S) >
and
P (Closer (S)) +
in case that
Rc (S) = 0.
P (Closer (S)) +
1
Rc (S)
P (S1 )
|S1 | |S|
as
CloseR (S) (S) \ S c
r → Rc (S)
P (S1 )
1
|Closer (S) \ S| → P (S) >
|S|
r
|S1 |
as
r → 0+
On the other hand
1
P (S1 )
|Closer (S) \ S| → P (co(S)) <
|S|.
r
|S1 |
as
r→∞.
R2 ∈ (Rc (S), ∞) satisfying (16).
P (S1 )
, then both equations (14) and (16) are the same and we can take
|S1 |
(S)
R1 . Hence λ2 = λ1 . Clearly, if R2 = R1 , then P|S|
= P|S(S11| ) . Notice that if λ1 = λ2 ,
Thus there is a unique value
Now, if
R2 =
P (S)
|S|
=
then
0 ≥ (R1 − R2 )|S1 ||S2 | = R1 R2 (|S1 |P (S2 ) − |S2 |P (S1 )) ≥ 0.
R1 = R2 .
λ2 = λ1 = 0.
Thus
Note that if
P (S)
|S|
=
P (S1 )
|S1 | and
To prove (17) we proceed as in the proof of
P (S1 )
|S1 |
≤
P (S)
|S| . From the explicit formula for
λ2 ,
16
Rc (S) = 0,
(i).
by
(i), R2 = R1 = 0
The fact that
it follows that
λ2 ≤
λ2 ≥
P (S1 )
|S1 | .
λ1
and
follows since
Proof of Proposition 4.6.
them we compute
Cs,λ
For each of
sa (λ), λ ∈ [0, λ1 ], such that
(respectively =, >)
Close λ (S) < Fs,λ (S)
Let us prove that there is a function
Fs,λ
if and only if
Fs,λ (S)
λ.
s ∈ [0, 1].
for any
(a) λ ∈ [0, λ1 ].
Step a1.
We consider the three dierent intervals for
(22)
s
s ∈ [0, sa (λ)),
resp.
s = sa (λ), s > sa (λ).
Fs,λ Close λ (S) ≤
Notice that
s
if and only if
s
P Close λ (S) + Close λ (S) \ S ≤ P (S) .
s
s
λ
(23)
Clearly, by Proposition 4.2(i), if we dene
sa (λ) =
1
λ λ ∈ [0, λ1 ] ,
R1
then the equality in (23) holds identically. Now, by Lemma 4.9, the left hand side of (23)
is an increasing function of
s,
and the identity in (23) only holds at
s = sa (λ).
Thus (22)
holds.
Remark that (22) holds for any value of
Step a2. Identication of Cs,λ .
sa (λ) ≤ 1 −
λ.
h λ ∈ [0, λ1 ] iis equivalent
s ∈ 0, 1 − λ P|S(S22| ) we have
One can easily check that
λ P|S(S22| ) . Recall that, by Lemma 4.1, for any
to
Fs,λ (S) = min {Fs,λ (S) , Fs,λ (S1 ) , Fs,λ (S2 ) , 0} .
Thus
when
Cs,λ = Close λ (S) if s ∈ [0, sa (λ)], Cs,λ = S
s
s = sa (λ), S is also a minimizer of Fs,λ .
Using (22) and Lemma 4.1 we clearly have that
and
(b)
Cs,λ = ∅
Let
if
s > 1 − λ P|S(S11| ) .
λ ∈ (λ1 , λ2 ].
if
s ∈ (sa (λ), 1 − λ P|S(S22| ) ].
i
Cs,λ = S1 if s ∈ 1 − λ P|S(S22| ) , 1 − λ P|S(S11| )
In this case, let us prove that there is a function
Fsb (λ),λ Close
λ
sb (λ)
Notice that
(S) = Fsb (λ),λ (S1 )
sb (λ)
λ ∈ [λ1 , λ2 ] .
such that
(24)
Let us consider two cases.
Step b1. Proof of
(24)
assuming that
|S1 |
P (S1 )
have
sa (λ) =
<
|S|
P (co(S))
.
In this case
λ
P (S1 )
≤1−λ
.
R1
|S1 |
if and only if
λ≤
R1 |S1 |
:= λ̄1 .
R1 P (S1 ) + |S1 |
17
λ2 =
R2 |S1 |
R2 P (S1 )+|S1 | . We
Observe that
λ̄1 ≤ λ2
R1 ≤ R2 . h
i
λ
interval s ∈ 0,
R1
since
(*) Let us work in the
for all
λ ∈ [λ1 , λ2 ].
Let us prove that
F0,λ (co(S)) ≤ F0,λ (S1 ) .
Indeed, since
|S1 |
P (S1 )
≤
|S|
P (co(S)) , after some simple computations we deduce that
R2 |S1 |
|S2 |
≤
.
R2 P (S1 ) + |S1 |
P (co(S)) − P (S1 )
λ2 =
|S2 |
P (co(S))−P (S1 ) and this is equivalent to (25).
|S1 |
|S|
Moreover, by Proposition 4.2.(ii) (assuming that
P (S1 ) < P (co(S)) ), for
= s1 (λ) = 1 − λ P|S(S11| ) we have
Thus, if
s
(25)
λ ≤ λ2 ,
then
λ≤
Fs1 (λ),λ (S1 ) < Fs1 (λ),λ Close
with equality if
λ = λ2 ,
and for
F
λ
,λ
R1
λ ∈ (λ1 , λ2 ]
(S1 ) < F
λ
,λ
R1
and
s=
(S) = F
λ
s1 (λ)
λ < λ2
and
(S) ,
λ
R1 , we have
λ
,λ
R1
(CloseR1 (S))
(26)
λ > λ1 ). Since both functions s → Fs,λ (S1 )
are continuous in s, they have to intersect for some s ∈
and s → Fs,λ Close λ (S)
soi
h
n
0, min Rλ1 , 1 − λ P|S(S11| ) . Hence there is at least one value s that satises (24). Let
sb (λ) be the smallest value of s satisfying (24).
P (S1 )
λ
Notice that we have that sb (λ) <
R1 for any λ ∈ (λ1 , λ2 ] and sb (λ) < 1 − λ |S1 | for any
λ < λ2 (with equality if λ = λ2 ).
oi
h
n
Step b2. sb (λ) is the unique value of s ∈ 0, min Rλ1 , 1 − λ P|S(S11| ) satisfying (24) if λ ∈
(λ1 , λ2 ].
n
o
P (S1 )
P (S1 )
λ
Clearly, if λ = λ2 , then sb (λ) = 1 − λ
and (24) holds. Our
=
min
,
1
−
λ
R1
|S1 |
|S1 |
(the rst inequality being true because
assertion is true in this case.
Assume that
we have
λ ∈ (λ1 , λ2 ).
Suppose that we nd
Let us prove that for any
n
oi
s ∈ sb (λ), min Rλ1 , 1 − λ P|S(S11| )
Fs,λ (S1 ) < Fs,λ (Close λ (S)) .
s
n
o
λ
sb (λ) < t1 < t2 < min R1 , 1 − λ P|S(S11| )
(27)
where
Ft1 ,λ (S1 ) < Ft1 ,λ (Close λ (S)) ,
(28)
Ft2 ,λ (S1 ) > Ft2 ,λ (Close λ (S)) .
(29)
t1
t2
Let us compute
(by (28)), and
Ct1 ,λ .
Observe that
Close λ (S)
t1
S1
is better than
is better (i.e. it has less energy) than
S
because
18
t1 <
λ
R1 . Also
Close λ (S)
t1
Ft1 ,λ (S1 ) < 0
because
t1 < 1 − λ P|S(S11| ) .
S1 .
Let us compute
Close λ (S)
Ft1 ,λ (S) = Ft1 ,λ (S1 ) + Ft1 ,λ (S2 ) ≤ Ft1 ,λ (S2 ).
Thus
Ct2 ,λ .
is better than
t2
S
Close λ (S)
Observe that
is better than
t2
because
t2 <
λ
R1 . Also
Ft2 ,λ (S) = Ft2 ,λ (S1 ) + Ft2 ,λ (S2 ) ≤ Ft2 ,λ (S2 ).
Thus
Thus the optimum is
Ft2 ,λ (S1 ) ≤ 0
S1
(by (29)).
because
Thus the optimum is
And
t2 < 1 − λ P|S(S11| ) .
Close λ (S).
t2
t1 < t2 and the optimum att2 containsnthe optimum atot1 . We
Fs,λ (S1 ) ≤ Fs,λ (Close λ (S)) for all s ∈ sb (λ), min Rλ1 , 1 − λ P|S(S11| ) . The
It is not possible that
conclude that
s
t < t0 such that the minima
of Ft,λ are both S1 and Close λ (S) and the minima of Ft0 ,λ are both S1 and Close λ (S). Then
inequality has to be strict. Otherwise there would be two points
t0
t
S1
would contain
Close λ (S),
t0
a contradiction. Thus (27) is proved.
Step b3. Computation of Cs,λ for λ ∈ (λ1 , λ2 ] and any s.
s ∈ [0, sb (λ)) we argue as for t2 and we deduce that the optimum is Close λ (S). Letting
s
s → sb (λ) we deduce that Csb (λ),λ = Close λ (S).
n
oi sb (λ)
P (S1 )
λ
If s ∈ sb (λ), min
,
1
−
λ
(notice that this implies that λ ∈ (λ1 , λ2 )) we argue
R1
|S1 |
as for t1 and
o the optimum is S1 .
n we deduce that
P (S1 )
λ
= Rλ1 , i.e., if λ ∈ (λ1 , λ̄1 ], let us compute the optimum for
If min
R1 , 1 − λ |S1 |
If
s ∈ ( Rλ1 , 1 − λ P|S(S11| ) ]. On this interval Fs,λ (S1 ) ≤ 0. Thus Fs,λ (S) = Fs,λ (S1 ) + Fs,λ (S2 ) ≤
Fs,λ (S2 ). Also (by the denition of R1 ) on this interval Fs,λ (Close λ (S)) > Fs,λ (S). Thus
s
the optimum is either S1 or S . By monotonicity of the optimum with respect to s and the
P (S1 )
λ
λ
fact that the optimum in (sb (λ),
R1 ] is S1 , we deduce that it is also S1 in ( R1 , 1 − λ |S1 | ].
o
n
P (S1 )
λ
= Rλ1 and s > 1 − λ P|S(S11| ) , we have that Fs,λ (S1 ) > 0 and the
If min
R1 , 1 − λ |S1 |
minimum n
is ∅.
o
P (S1 )
λ
= 1 − λ P|S(S11| ) , i.e., if λ ∈ (λ̄1 , λ2 ], as in the previous paragraph
If min
,
1
−
λ
R1
|S1 |
the minimum for
s > 1 − λ P|S(S11| )
Let us point out that for
is
∅.
λ = λ2
and for
s > sb (λ2 ) = 1 − λ2 P|S(S11| ) ,
Fs,λ2 (Close λ2 (S)) > Fsb (λ2 ),λ2 (Close
s
Since also
Fs,λ2 (S1 ) > 0,
Step b4. Assume that
then
|S1 |
P (S1 )
|S2 |
P (co(S)−P (S1 ) .
λ2
sb (λ2 )
we have that
(S)) = Fsb (λ2 ),λ2 (S1 ) = 0.
Cs,λ2 = ∅.
≥
|S|
P (co(S))
. Dene sb (λ) for λ ∈ (λ1 , λ2 ].
In this case
λ2 =
|S2 |
(26) also holds for any
P (co(S)−P (S1 ) , then (25) holds. On the other hand
h
i
λ
λ ∈ (λ1 , λ2 ]. As in paragraph (*) we identify a solution sb (λ) ∈ 0, R1 of (24). Let us take
h
i
P (S1 )
P (S1 )
the smallest one. Let us prove that sb (λ) ∈ 0, 1 − λ
|S1 | . Let s = s1 (λ) = 1 − λ |S1 | .
Since
Then
λ ≤ λ2 =
Fs1 (λ),λ (S1 ) = 0
Fs1 (λ),λ (Close
and
λ
s1 (λ)
(S)) = P (Close
λ
s1 (λ)
(S)) +
19
s1 (λ)
P (S1 )
|Close λ (S) \ S| −
|S|
s
(λ)
λ
|S1 |
1
≥ P (co(S)) −
P (S1 )
|S| ≥ 0 .
|S1 |
|S1 |
P (S1 )
Notice that the second inequality is strict if
|S1 |
P (S1 )
=
we have
If
|S|
P (co(S)) and
s1 (λ) > 0.
sb (λ) ∈ [0, 1 −
|S1 |
P (S1 )
=
and we take
|S|
P (co(S)) , while the rst is strict if
In both cases, since
sb (λ)
is the smallest solution of (24),
|S1 |
P (S1 )
= λ2
and
λ P|S(S11| ) ].
|S|
P (co(S)) , and
s1 (λ) = 0,
then
Fs1 (λ2 ),λ2 (Close
λ2
s1 (λ2 )
λ=
(S)) = Fs1 (λ2 ),λ2 (S1 ) = 0,
sb (λ2 ) = 0.
Step b5. Assume that
If
>
λ ∈ (λ1 , λ̄1 ],
|S1 |
P (S1 )
then
. Computation of Cs,λ for λ ∈ (λ1 , λ2 ] and any s.
|S|
P (co(S))
≤ Rλ1 ≤
≥
sb (λ)
same case in Step b3.
1 − λ P|S(S11| ) .
Then the argument is identical to the
λ ∈ (λ̄1 , λ2 ], then 1 − λ P|S(S11| ) < Rλ1 . We have that Fs,λ (Close λ (S)) ≤ Fs,λ (S) for
s
h
i
i
h
P (S1 )
λ
all s ∈ 0,
we have Fs,λ (S1 ) ≤ 0. Then Fs,λ (S) = Fs,λ (S1 ) +
R1 . If s ∈ 0, 1 − λ |S1 |
h
i
Fs,λ (S2 ) ≤ Fs,λ (S2 ). Thus the optimum in s ∈ 0, 1 − λ P|S(S11| ) can be only Close λ (S) or
s
S1 .
h
i
λ
As in Step b2 we prove that sb (λ) is the unique value of s ∈ 0,
R1 satisfying (24) if
λ ∈ (λ1 , λ2 ]. Then we proceed as in Step b3 to prove that Cs,λ = Close λ (S) if s ∈ [0, sb (λ)],
s
i
P (S1 )
P (S1 )
. For s > 1 − λ
and Cs,λ = S1 if s ∈ sb (λ), 1 − λ
, Cs,λ = ∅ since Fs,λ (S1 ) > 0.
|S1 |
|S1 |
If
(c) Let
λ ∈ (λ2 , λ3 ].
Step c1. Assume that
and
λ3 =
|S1 |
P (S1 )
<
. Denition of sc (λ).
|S|
P (co(S))
|S|
P (co(S)) . We look for a value
sc (λ), λ ∈ (λ2 , λ3 ],
Fsc (λ),λ Close
Let
λ ∈ (λ2 , λ3 ).
If
s = 0,
In this case
λ2 =
R2 |S1 |
R2 P (S1 )+|S1 |
such that
λ
sc (λ)
(S)
= 0.
(30)
we have
F0,λ (co(S)) < 0.
Let
s3 (λ) = 1 − λ P (co(S))
.
|S|
Since by Lemma 4.9
λ → Fs3 (λ),λ Close
λ
s3 (λ)
(S)
is strictly
decreasing, then
0 = F0,
Then for any
λ ∈ (λ2 , λ3 )
|S|
P (co(S))
(co(S)) < Fs3 (λ),λ Close
there exists
sc (λ) < s3 (λ)
λ
s3 (λ)
(S) .
such that (30) holds.
Let λ = λ3 . Then F0,λ3 (co(S)) = 0 and s3 (λ3 ) = 0. Hence Fs3 (λ),λ Close λ (S)
= 0.
s3 (λ)
Since Fs,λ3 Close λ3 (S) is strictly increasing in s we have that Fs,λ3 Close λ3 (S) > 0
s i
s
λ3
for any s ∈ 0,
Rc (S) . Then sc (λ3 ) = 0 = s3 (λ3 ).
20
Step c2. Assume that
if
λ ≥ λ2 =
Indeed, if
|S1 |
P (S1 )
<
R2 |S1 |
R2 P (S1 )+|S1 | and
λ ≥ λ2
|S|
P (co(S))
.
Let us prove that
h
i
s ∈ 0, 1 − λ P|S(S11| ) ,
h
i
s ∈ 0, 1 − λ P|S(S11| ) ,
and
then
Then, by Lemma 4.9.(iii), we have
then
Fs,λ Close λ (S) ≤ Fs,λ (S1 ).
(31)
s
s ≤ 1 − λ P|S(S11| ) ≤
λ
s
R2 . That is, λ
≤
1
P (S1 )
s
P (Close λ (S)) + |Close λ (S) \ S| ≤ P (CloseR2 (S)) +
|CloseR2 (S) \ S| =
|S|.
s
s
λ
R2
|S1 |
Now, Fs,λ Close λ (S) ≤ Fs,λ (S1 ) if and only if
1
R2 .
(32)
s
P (Close λ (S)) +
s
s
1−s
1−s
1−s
|Close λ (S) \ S| ≤ P (S1 ) −
|S1 | +
|S| = P (S1 ) +
|S2 |.
s
λ
λ
λ
λ
Thus, by (32), it is sucient to prove that
1−s
P (S1 )
|S| ≤ P (S1 ) +
|S2 |.
|S1 |
λ
But this is true since
Since
if
P (S1 )
|S1 |
Fs,λ (S1 ) < 0
λ ≥ λ2 =
for
≤
1−s
. Hence (31) holds.
hλ
s ∈ 0, 1 − λ P|S(S11| ) ,
R2 |S1 |
R2 P (S1 )+|S1 | and
In particular, we have that
then (31) implies
h
s ∈ 0, 1 − λ P|S(S11| ) ,
sc (λ) > 1 − λ P|S(S11| ) .
Observe also that the inequality in (31) is strict if
Step c3. Assume that
optimum in
[0, 1 −
|S1 |
P (S1 )
λ P|S(S11| ) ] is
|S|
P (co(S))
<
Close λ (S).
then
Fs,λ Close λ (S) < 0.
s
h
s ∈ 0, 1 − λ P|S(S11| ) .
. Computation of Cs,λ .
By (31) it cannot be
s
S1 .
Let us prove that the
On the other hand, by
λ ∈ (λ2 , λ3 ], we have
that λ >
λ2 ≥ λ̄1 and, therefore,
P (S1 )
> 1 − λ |S1 | . By the last remark in Step a1, Fs,λ Close λ (S) ≤ Fs,λ (S) in the interval
s
[0, 1 − λ P|S(S11| ) ]. On the other hand, on that interval, Fs,λ (S) ≤ Fs,λ (S2 ). Thus, the optimum
the rst paragraph of Step b1, if
λ
R1
is
Close λ (S)
(its energy being negative).
s
Again, the optimum in
(1 − λ P|S(S11| ) , sc (λ)]
because its energy is positive. By (13), the
The optimum is
Close λ (S). The optimum cannot be S1
s
energy of S2 is also positive. Thus, also is for S .
is
Close λ (S).
s
sc (λ) the optimum
∅. Thus, either there is a maximal interval of values of s,
s
say (sc (λ), sd (λ)], not reduced to sc (λ), where (30) holds, in which case Close λ (S) is the
s
optimum up to sd (λ), or the energy of Close λ (S) becomes positive immediately after sc (λ)
s
and the optimum is ∅. Thus, we may take sc (λ) as the maximal solution of (30).
Notice that the argument in the previous paragraph shows that after
can only be either
Close λ (S)
Step c4. Assume that
|S|
P (co(S))
or
≤
|S1 |
P (S1 )
. Computation of Cs,λ .
21
f = χS
λ = 0.01 < λ1
λ1 < λ = 0.04 < λ2
Figure 4: Levelsets of the solutions
uλ
for dierent values of
r1 = 1.2, r2 = 1, the distance between
λi , i = 1, 2, 3 as in Proposition 4.2 .
In this case,
λ2 =
λ2 < λ = 0.4 < λ3
|S2 |
P (co(S))−P (S1 ) . Since
them is
d = 0.05.
P (co(S)) < P (S),
λ. S1 , S2
In this case
λ > λ2 ,
we have
|S1 |
P (S1 )
<
|S|
P (co(S)) .
F0,λ (co(S)) < F0,λ (S).
i
h
P (S1 )
small
for s ∈ 0, 1 − λ
|S1 |
sets of uλ , S1 is the optimum
we have
F0,λ (S1 ) < F0,λ (co(S)). Thus,
enough the
h the optimumi is S1 . By monotonicity of the level
P (S1 )
P (S1 )
for all s ∈ 0, 1 − λ
|S1 | . For s > 1−λ |S1 | is the emptyset.
If we take
are balls with radius
This concludes the proof.
Acknowledgements. V. Caselles acknowledges partial support by MICINN project, ref-
erence MTM2009-08171, by GRC reference 2009 SGR 773 and by ICREA Acadèmia prize
for excellence in research funded both by the Generalitat de Catalunya.
V. Caselles and
M. Novaga acknowledge partial support by Acción Integrada Hispano Italiana HI2008-0074.
The work of C. Pöschl is supported by the Austrian Science Fund (FWF): Project J-2970
(Schrödinger).
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