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On the Homogeneity of Global Minimizers for the Mumford-Shah Functional when K is a Smooth Cone

ANTOINELEMENANT(*)

ABSTRACT- We show that if (u;K) is a global minimizer for the Mumford-Shah functional inRN, and ifKis a smoothenoughcone, then (modulo constants)uis a homogenous function of degree1

2. We deduce some applications inR3 as for instance that an angular sector cannot be the singular set of a global minimizer, that ifK is a half-plane thenuis the corresponding cracktip function of two variables, or that ifK is a cone that meetsS2withan union ofC1curvilinear convex polygones, then it is aP,YorT.

Introduction.

The functional of D. Mumford and J. Shah [18] was introduced to solve an image segmentation problem. IfVis an open subset ofR2, for example a rectangle, and g2L1(V) is an image, one can get a segmentation by minimizing

J(K;u):ˆ

Z

VnK

jruj2dx‡ Z

VnK

(u g)2dx‡H1(K)

over all the admissible pairs (u;K)2 Adefined by

A:ˆ f(u;K); KV is closed; u2Wloc1;2(VnK)g:

Any solution (u;K) that minimizesJrepresents a ``smoother'' version of the image and the setKrepresents the edges of the image.

Existence of minimizers is a well known result (see for instance [11]) usingSBV theory.

(*) Indirizzo dell'A.: Universite Paris XI, Bureau 15 BaÃtiment 430, Orsay, 91400, France.

E-mail: antoine.lemenant@math.u-psud.fr

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The question of regularity for the singular setKof a minimizer is more difficult. The following conjecture is currently still open.

CONJECTURE 1 (Mumford-Shah). [18] Let (u;K) be a reduced mini- mizer for the functional J. Then K is the finite union of C1arcs.

The term ``reduced'' just means that we cannot find another pair (~u;K)~ suchthatKK~ and~uis an extension ofuinVnK.~

Some partial results are true for the conjecture. For instance it is known thatK isC1almost everywhere (see [7], [4] and [2]). The closest to the conjecture is probably the result of A. Bonnet [4]. He proves that if (u;K) is a minimizer, then every isolated connected component of K is a finite union ofC1-arcs. The approach of A. Bonnet is to use blow up limits.

If (u;K) is a minimizer inVandyis a fixed point, consider the sequences (uk;Kk) defined by

uk(x)ˆ 1

tk

p u(y‡tkx); Kk ˆ1

tk(K y); Vkˆ1

tk(V y):

When ftkg tends to infinity, the sequence (uk;Kk) may tend to a pair (u1;K1), and then (u1;K1) is called a Global Minimizer. Moreover, A.

Bonnet proves that if K1 is connected, then (u1;K1) is one of the list below:

1ST CASE:K1ˆ= andu1is a constant.

2ND CASE:K1is a line andu1 is locally constant.

3RD CASE: ``Propeller'':K1 is the union of 3 half-lines meeting with 120 degrees andu1 is locally constant.

4TH CASE: ``Cracktip'':K1 ˆ f(x;0);x0gandu1(rcos(u);rsin(u))ˆ

ˆ

2 p r

r1=2 sinu

2‡C, forr>0 andjuj5p(Cis a constant), or a similar pair obtained by translation and rotation.

We don't know whether the list is complete without the hypothesis that K1 is connected. This would give a positive answer to the Mumford-Shah conjecture.

The Mumford-Shah functional was initially given in dimension 2 but there is no restriction to define Minimizers for the analogous functional in RN. Then we can also do some blow-up limits and try to think about what should be a global minimizer inRN. Almost nothing is known in this di- rection and this paper can be seen as a very preliminary step. Let state some definitions.

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DEFINITION2. LetV RN,(u;K)2 Aand B be a ball such thatB V.

A competitor for the pair(u;K)in the ball B is a pair(v;L)2 Asuch that uˆv

KˆL

o in VnB

and in addition if x and yare two points inVn(B[K)that are separated by K then theyare also separated byL.

The expression ``be separated byK'' means thatxandylie in different connected components ofVnK.

DEFINITION3. A global minimizer inRNis a pair (u;K)2 A(with V ˆRN) such that for everyball B inRNand everycompetitor(v;L)in B we have

Z

BnK

jruj2dx‡HN 1(K\B) Z

BnL

jrvj2dx‡HN 1(L\B)

where HN 1denotes the Hausdorff measure of dimension N 1.

Proposition9onpage267of[8]ensuresthatanyblowuplimitofaminimizer for the Mumford-Shah functional inRN, is a global minimizer in the sense of Definition 3. As a beginning for the description of global minimizers inRN, we can firstly think about what should be a global minimizer inR3. Ifuis locally constant, thenK is a minimal cone, that is, a set that locally minimizes the Hausdorffmeasureofdimension2inR3.Thenby[9]weknowthatKisaconeof typeP(hyperplane),Y(three half-planes meeting with 120 degrees angles) or of typeT(cone over the edges of a regular tetraedron centered at the origin).

Those cones became famous by the theorem of J. Taylor [20] which says that any minimal surface inR3is locallyC1equivalent to a cone of typeP,YorT.

Fig. Cones of typeYandTinR3.

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To be clearer, this is a more precise definition ofYandT, as in [10].

DEFINITION4. Define PropR2by

Propˆ f(x1;x2);x10;x2ˆ0g [ f(x1;x2);x10;x2ˆ 

p3 x1g [ f(x1;x2);x10;x2ˆ 

p3 x1g:

Then let Y0ˆPropRR3: The spine of Y0 is the line L0ˆ

ˆ fx1ˆx2ˆ0g. A cone of typeYis a set Y ˆR(Y0)where R is the com- position of a translation and a rotation. The spine of Y is then the line R(L0).

DEFINITION 5. Let A1ˆ(1;0;0), A2ˆ 1

3;2 

p2 3 ;0

, A3ˆ

ˆ 1

3;

2 p

3 ;

6 p

3

, and A4ˆ 1

3;

2 p

3 ;

6 p

3

the four vertices of a regular tetrahedron centered at0. Let T0be the cone over the union of the6 edges[Ai;Aj]i6ˆj. The spine of T0is the union of the four half lines[0;Aj[. A cone of typeTis a set TˆR(T0)where R is the composition of a translation and a rotation. The spine of T is the image byR of the spine of T0.

So the pairs (u;Z) whereuis locally constant andZis a minimal cone, are examples of global minimizers inR3. Another global minimizer can be obtained withK1a half-plane, by settingu:ˆCraktipR(see [8] section 76). These examples are the only global minimizers inR3that we know.

Note that if (u;K) is a global minimizer inRN, thenulocally minimizes the Dirichlet integral inRNnK. As a consequence,uis harmonic inRNnK.

Moreover, ifBis a ball suchthatK\Bis regular enough, then the normal derivative ofuvanishes onK\B.

In this paper we wish to study global minimizers (u;K) for whichKis a cone. It seems natural to think that any singular set of a global minimizer is a cone. But even if all known examples are cones, there is no proof of this fact. In addition, we will add some regularity onK. We denote bySN 1the unit sphere inRNand, ifVis a open set,W1;2(V) is the Sobolev space. We will say that a domainV onSN 1 has a piecewise C2 boundary, if the to- pological boundary of V, defined by @VˆVnV, consists of an union of N 2 dimensional hypersurfaces of classC2. This allows some cracks, i.e.

whenVlies in eachsides of its boundary. We will denote byS~ the set of all the singular points of the boundary, that is

S~ :ˆ fx2@V;8r>0;B(x;r)\@V is not aC2 hypersurfaceg:

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DEFINITION6. A smooth cone is a set K of dimension N 1inRNsuch that K is conical, centered at the origin, and such that SN 1nK is a domain with piecewise C2 boundary. Moreover we assume that the embedding W1;2(SN 1nK)!L2(SN 1nK)is compact. Finallywe suppose that we can stronglyintegrate byparts in B(0;1)nK. More precisely, denoting bySthe set of singularities

S:ˆ ftx;(t;x)2R‡Sg;~

we want that Z

B(0;1)nK

hru;rWi ˆ0

for everyharmonic function u in B(0;1)nK with @

@nuˆ0on KnS, and for allW2W1;2(B(0;1)nK)with vanishing trace on SN 1nK.

REMARK7. For instance ifKis the cone over a finite union ofC2-arcs on S2, then we can strongly integrate by parts inB(0;1)nK. Another example in RNisgivenbytheunion of admissibleset of faces (as inDefinition (22.2)of [5]).

Now this is the main result.

THEOREM15. Let(u;K)be a global minimizer inRNand assume that K is a smooth cone. Then there is a1

2-homogenous function u1such that u u1is locallyconstant.

As we shall see, this result implies that if (u;K) is a global minimizer in RN, and ifKis a smooth cone other than a minimal cone, then3 2N

4 is an eigenvalue for the spherical Laplacian inSN 1nKwithNeumann boundary conditions. In section 2 we will give some applications about global mini- mizers inR3, using the estimates on the first eigenvalue that can be found in [6], [5] and [14]. More precisely, we have:

PROPOSITION17. Let(u;K)be a global Mumford-Shah minimizer in R3such that K is a smooth cone. Moreover, assume that S2\K is a union of convex curvilinear polygons with C1 sides. Then u is locallyconstant and K is a cone of typeP,YorT.

Another consequence of the main result is the following.

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PROPOSITION19. Let(u;K)be a global Mumford-Shah minimizer in R3 such that K is a half plane. Then u is equal to a function of type cracktipR, that is, in cylindrical coordinates,

u(r;u;z)ˆ

2 p r

r12sinu 2‡C for05r5‡ 1, p5u5pwhere C is a constant.

Finally, we deduce two other consequences from Theorem 15. Let (r;u;z)2R‡[ p;p]R be the cylindrical coordinates inR3. For all v2[0;p] set

@Gv:ˆ f(r;u;z)2R3;uˆ v oruˆvg:

and

Sv:ˆ f(r;u;z)2R3;zˆ0; r>0; u2[ v;v]g (1)

Observe thatS0is a half line,Sp2,@G0and@Gpare half-planes, and that Sp and@Gp2are planes.

PROPOSITION18. There is no global Mumford-Shah minimizer inR3 such that K is wing of type@Gvwithv62

0;p

2;p

.

PROPOSITION23. There is no global Mumford-Shah minimizer inR3 such that K is an angular sector of type(u;Sv)forv62

p 2;p

.

1. IfKis a cone thenuis homogenous.

In this section we want to prove Theorem 15. Notice that this result is only useful if the dimensionN3. Indeed, in dimension 2, ifKis a cone then it is connected thus it is in the list described in the introduction.

1.1 ±Preliminary.

Let us recall a standard uniqueness result about energy minimizers.

PROPOSITION 8. LetV be an open and connected set of RN and let I@V be a hypersurface of class C1. Suppose that u and v are two

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functions in W1;2(V)such that uˆv a.e. on I (in terms of trace), solving the minimizing problem

minE(w):ˆ

Z

V

jrw(x)j2dx

over all the functions w2W1;2(V)that are equal to u and v on I. Then uˆv:

PROOF. This comes from a simple convexity argument which can be found for instance in [8], but let us write the proof since it is very short. By the parallelogram identity we have

E u‡v 2

ˆ1

2E(u)‡1

2E(v) 1

4E(u v):

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On the other hand, sinceu‡v

2 is equal touandvonI, and by minimality of uandvwe have

E u‡v 2

E(u)ˆE(v):

Now by (2) we deduce thatE(u v)ˆ0 and sinceVis connexe, this implies thatu vis a constant. Butu vis equal to 0 onIthusuˆv. p REMARK9. The existence of a minimizer can also be proved using the convexity ofE(v).

1.2 ±Spectral decomposition.

The key ingredient to obtain the main result will be the spectral theory of the Laplacian on the unit sphere. Sinceuis harmonic, we will decompose uas a sum of homogeneous harmonic functions just like we usually use the classical spherical harmonics. The difficulty here comes from the lack of regularity ofRNnK.

It will be convenient to work withconnected sets. So let V be a con- nected component ofSN 1nK, and letA(r) be

A(r):ˆ ftx;(x;t)2V[0;r[g:

We also set

A(1):ˆ ftx;(x;t)2VR‡g:

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All the following results are using that the embeddingW1;2(V) inL2(V) is compact. Recall that this is the case by definition, sinceKis a smoothcone.

Notice that for instance the cone property insures that the embedding is compact (see Theorem 6.2. p. 144 of [1]).

Consider the quadratic form Q(u)ˆ

Z

V

jru(x)j2dx

of domainW1;2(V) dense into the Hilbert spaceL2(V). SinceQis a positive and closed quadratic form (see for instance Proposition 10.61 p. 129 of [16]) there exists a unique selfadjoint operator denoted by Dn of domain D( Dn)W1;2(V) suchthat

8u2D( Dn); 8v2W1;2(V);

Z

V

hru;rvi ˆ Z

V

h Dnu;vi:

PROPOSITION10. The operator Dn has a countablyinfinite discrete set of eigenvalues, whose eigenfunctions span L2(V).

PROOF. The proof is the same as ifVwas a regular domain. Consider the new quadratic form

Q(u)~ :ˆQ(u)‡ kuk22

withthe same domainW1;2(V). The formQ~ has the same properties thanQ and the associated operator is Id Dn. MoreoverQ~is coercive. As a result, the operator Id Dn is bijective and its inverse goes from L2(V) to D( Dn)W1;2(V). By hypothesis the embedding ofW1;2(V) intoL2(V) is compact. Thus the resolvant (Id Dn) 1 is a compact operator, and we conclude using the spectral theory of operators with a compact resolvant

(see [19] Theorem XIII. 64 p. 245). p

REMARK 11. The domain of Dn is not known in general. IfV was smooth, then we could show that the domain is exactly D( Dn

ˆ u2W2;2(V);@u

@nˆ0 on@V

. Here, the boundary ofV has some sin- gularities so this result doesn't apply directly. But knowing exactly the domain of Dn will not be necessary for us.

Now we want to study the link between the abstract operatorDn and the classical spherical Laplacian DS on the unit sphere. Recall that if we compute the Laplacian in spherical coordinates, we obtain the following

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equality

Dˆ@2

@r‡N 1

r

@

@r‡ 1 r2DS: (3)

PROPOSITION 12. For everyfunction f 2D( Dn) such that Dnf ˆlf we have

i) f 2C1(V)

ii) DSf ˆ Dnf ˆlf inV iii) @f

@n exists and is equal to0 on K\VnS

PROOF. Let W be a C1 function withcompact support in V and f 2D( Dn). Then the Green formula in the distributional sense gives

Z

V

rf:rWˆ h DSf;Wi

where the left and right brackets mean the duality in the distributional sense. On the other hand, by definition of Dnand sincef is in the domain D( Dn), we also have

Z

V

rf:rWˆ h Dnf;Wi

where this time the brackets mean the scalar product inL2. Therefore Dnf ˆDSf inD0(V):

In other words, DSf ˆlf in D0(V). But now since f 2W1;2(V), by hy- poellipticity of the Laplacian we know thatf isC1 and that DSf ˆlf in the classical sense. That proves i) and ii). We even know by the elliptic theory that, sinceKnSis regular,f is regular at the boundary onKnS.

Now consider a ballBsuch that the intersection withK\V does not meetS. Assume thatBis cut in two partsB‡andB byK, and thatB‡is one part in V. Possibly by modifying B in a neighborhood of the inter- section withK, we can assume that the boundary ofB‡andB isC2. Th e definition ofDnimplies that for all functionW2C2(V) that vanishes out of B‡we have

Z

B‡

hrf;rWidxˆ Z

B‡

h Dnf;Widxˆl Z

B‡

hf;Widx:

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On the other hand, integrating by parts, Z

B‡

hrf;rWidxˆ Z

B‡

h DSf;Wi ‡ Z

@B‡

@u

@nW

ˆl Z

@B‡

hf;Wi ‡ Z

@B‡

@f

@nW

thus Z

@B‡

@f

@nWˆ0:

In other words the function f is a weak solution of the mixed boundary value problem

DSuˆlf inB‡ uˆf on@B‡nK

@u

@nˆ0 onK\@B‡

Therefore, some results from the elliptic theory imply thatfis smoothinB

and is a strong solution (see [21]). p

Let us recapitulate what we have obtained. For all functionf 2L2(V), there is a sequence of numbersaisuchthat

f ˆX‡1

iˆ0

aifi (4)

where the sum converges inL2. The functionsfiare in C1(V)\W1;2(V), verify DSfiˆlifiand@fi

@nˆ0 onK\VnS. Moreover, we can normalize thefi in order to obtain an orthonormal basis onL2(V), in particular we have the following Parseval formula

kfk22ˆX‡1

iˆ0

jaij2:

Note that if f belongs to the kernel of Dn (i.e. is an eigenfunction with eigenvalue 0), then

hrf;rfi ˆ h Dnf;fi ˆ0

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and sinceVis connected that means thatfis a constant. Thus 0 is the first eigenvalue and the associated eigenspace has dimension 1. Then we can suppose thatl0ˆ0 and that all thelifori>0 are positive.

We define the scalar product inW1;2(V) by hu;viW1;2:ˆ hu;viL2‡ hru;rviL2:

PROPOSITION13. The familyffigis orthogonal in W1;2(V). Moreover if f 2W1;2(V)and if its decomposition in L2(V) is f ˆ‡1P

iˆ0aifi, then the sum‡1P

iˆ0jaij2krfik22 converges and X‡1

iˆ0

jaij2krfik22ˆ krfk22: (5)

PROOF. We know thatffigis an orthogonal family inL2(V). In addition ifi6ˆjthen

Z

V

rfirfjˆ Z

V

Dnfifj

ˆli Z

V

fifj

ˆ0 thusffigis also orthogonal inW1;2(V) and

kfik2W1;2:ˆ kfik22‡ krfik22ˆ1‡li:

Consider now the orthogonal projection (for the scalar product ofL2) Pk:f 7!Xk

iˆ0

aifi:

The operator Pk is the orthogonal projection on the closed subspace Ak generated by ff0;. . .;fkg. More precisely, we are interested in the restriction of Pk to the subspace W1;2(V)L2(V). Also denote by P~k:W1;2!Akthe orthogonal projection on the same subspace but for the scalar product of W1;2. We want to show thatPkˆP~k. To prove this, it suffice to show that for all sets of coefficientsfaigiˆ1:::kandfbigiˆ1:::k,

f Xk

iˆ0

aifi;Xk

iˆ0

bifi

* +

W1;2

ˆ0:

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Since we already have

f Xk

iˆ0

aifi;Xk

iˆ0

bifi

* +

L2

ˆ0;

all we have to show is that Z

V

rf Xk

iˆ0

airfi;Xk

iˆ0

birfi

* +

dxˆ0:

Now Z

V

rf Xk

iˆ0

airfi;Xk

iˆ0

birfi

* +

ˆ Z

V

rf;Xk

iˆ0

birfi

* +

Xk

iˆ0

aibikrfik22

ˆXk

iˆ0

bih Dnfi;fiL2

Xk

iˆ0

aibili

ˆXk

iˆ0

aibili Xk

iˆ0

aibili

ˆ0 thusPkˆP~kand therefore, by Pythagoras

kPk(f)k2W1;2 kfk2W1;2: By lettingktend to infinity we obtain

X‡1

iˆ0

a2ikrfik22 krfk22: (6)

From this inequality we deduce that the sum is absolutely converging in W1;2(V). Therefore, the sequence of partial sum PK

iˆ0aifi is a Cauchy se- quence for the normW1;2(V). Thus, since the sumP

aifialready converges tofinL2(V), by uniqueness of the limit the sum converges tofinW1;2(V), so we deduce that (6) is an equality and the proof is over. p Once we have a basisffigonVSN 1, we consider for a certainr0>0, the functions

hi(x)ˆra0ifi x r0

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defined onr0V. The exponentaiis defined by aiˆ (N 2)‡ 

(N 2)2‡4li q

2 :

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The functionshi form a basis ofW1;2(r0V). Indeed, iff 2W1;2(r0V), then f(r0x)2W1;2(V) thus applying the decomposition onV we obtain

f(r0x)ˆX‡1

iˆ0

bifi(x) thus

f(x)ˆX‡1

iˆ0

aihi(x) with

aiˆbir0ai: (8)

Notice that sincekhik22ˆr2a0i‡N 1we also have X1

iˆ0

a2ikhik22ˆX1

iˆ0

a2ir2a0i‡N 1ˆ kfk2L2(r0V)5‡ 1:

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Moreover, applying Proposition 13 we have that X1

iˆ0

b2ikrfik22ˆ krf(r0x)k225‡ 1:

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We are now able to state our decomposition inA(r0).

PROPOSITION14. Let K be a smooth cone inRN, centered at the origin and let V be a connected component of SN 1nK. Then there exist some harmonic homogeneous functions gi, orthogonal in W1;2(A(1)), such that for everyfunction u2W1;2(A(1)) harmonic in A(1) with @u

@nˆ0 on

K\A(1)nS, and for everyr02]0;1[, we have that uˆX‡1

iˆ0

aigi in A(r0)

where the aido not depend on radius r0and are unique. The sum converges in W1;2(A(r0))and uniformlyon all compact sets of A(1). Moreover

kuk2W1;2(A(r0))ˆX‡1

iˆ0

a2ikgik2W1;2(A(r0)): (11)

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PROOF. Sinceu2W1;2(A(1)) then for almost everyr0in ]0;1] we have that

ujr0V2W1;2(r0V):

Thus we can apply the decomposition onr0Vand say that uˆX‡1

iˆ0

aihi onr0V:

Definegiby

gi(x):ˆ kxkaifi x kxk

whereaiis defined by (7). Since thefiare eigenfunctions for DS, we deduce from (3) that

Dgiˆ@2

@rgi‡N 1 r

@

@rgi‡ 1 r2DSgi

ˆai(ai 1)rai 2fi‡N 1

r airai 1fi rai 2lifi

ˆ(a2i‡(N 2)ai li)rai 2fi

ˆ0

by definition ofai, thus thegiare harmonic inA(‡ 1). Notice that thegi are orthogonal inL2(A(1)) because they are homogeneous and orthogonal in L2(V). Note also thathiis equal togionr0V. Moreover for all 05r1 we have

kgik2L2(A(r))ˆ Z

A(r)

jgij2ˆ Zr

0

Z

@B(t)\A(1)

jgi(w)j2dwdt (12)

ˆ Zr

0

Z

V

tN 1jgi(ty)j2dydtˆ Zr

0

t2ai‡N 1 Z

V

jgi(y)j2dydt

ˆ r2ai‡N

2ai‡Nkfik2L2(V)ˆ r2ai‡N 2ai‡N1:

On the other hand, since thefiand their tangential gradients are ortho- gonal inL2(V), we deduce that the gradients ofgiare orthogonal inA(1).

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Then, by a computation similar to (12) we obtain for all 05r1 krgik2L2(A(r))ˆ

Zr

0

Z

@B(t)\A(1)

@gi

@r

2‡jrtgij2dwdt (13)

ˆ Zr

0

Z

@B(t)\A(1)

aitai 1fi w t

2‡tairtfi w

t

1 t

2dwdt

ˆa2i Zr

0

t2(ai 1) Z

@B(t)\A(1)

fi w

t

2dwdt

‡ Zr

0

t2(ai 1) Z

@B(t)\A(1)

rtfi w

t

2dwdt

ˆa2i Zr

0

t2(ai 1) Z

V

jfi(w)j2tN 1dwdt

‡ Zr

0

t2(ai 1) Z

V

jrtfi(w)j2tN 1dwdt

ˆa2i r2(ai 1)‡N

2(ai 1)‡Nkfik2L2(V)‡ r2(ai 1)‡N

2(ai 1)‡Nkrtfik2L2(V)

ˆ r2(ai 1)‡N

2(ai 1)‡N(a2i ‡li)kfik2L2(V)

Cr2ai(a2i‡li)

becausekrtfik22ˆlikfik22,r1 andai0. Moreover the constantCde- pends on the dimensionNbut does not depend oni.

We denote bygthe function defined inA(1) by g:ˆX‡1

iˆ0

aigi: Thenglies inL2(A(r0)) because using (12) and (9)

kgk2L2(A(r0))ˆX‡1

iˆ0

jaij2kgik2L2(A(r0))X‡1

iˆ0

jaij2r2a0 i‡N5‡ 1:

We want now to show thatgˆu.

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First step:We claim thatgis harmonic inA(r0). Indeed, since thegi

are all harmonic in A(r0), the sequence of partial sumssk :ˆPk

iˆ0aigi is a sequence of harmonic functions, uniformly bounded for the L2 norm in eachcompact set ofA(r0). By the Harnack inequality we deduce that the sequence of partial sums is uniformly bounded for the uniform norm in each compact set. Thus there is a subsequence that converges uniformly to a harmonic function, which in fact is equal togby uniqueness of the limit.

Therefore,gis harmonic inA(r0).

Second step: We claim that g belongs to W1;2(A(r0)). Firstly, since u2W1;2(r0V), by (8) and (10) we have that

X‡1

iˆ0

a2ir2a0 ikrtfik2L2(@B(0;1)nK)5‡ 1:

(14)

In addition, sincekrtfik22ˆlikfik22 andkfik2ˆ1, we deduce X‡1

iˆ0

a2ir2a0 ili5‡ 1 (15)

and sinceaiandliare linked by the formula (7) we also have that X‡1

iˆ0

a2ir02aia2i5‡ 1:

(16)

Now, sinceP

aigiconverges absolutely on every compact set, we can say that

rgˆX‡1

iˆ0

airgi

thus using (13), (15), (16), and orthogonality, krgk2L2(A(r0))ˆX‡1

iˆ0

a2ikrgik2L2

CX‡1

iˆ0

a2ir2a0 i(a2i‡li)5‡ 1: Therefore,g2W1;2(A(r0)).

Third step :We claim that@g

@nˆ0 onK\A(r0)nS. We already know that@gi

@nˆ0 onKnS(because thefihave this property). We want to show

(17)

thatgis so regular that we can exchange the order of @

@nandP

. So letx0 be a point ofK\A(r0)nS and letBbe a neighborhood of x0 in RN that doesn't meet S and suchthat K separates B in two parts B‡ and B . Assume that B‡ is a part in A(r0). The sequence of partial sums sk:ˆPk

iˆ0aigiis a sequence of harmonic functions inB‡. Since@B‡\Kis C2we can do a reflection to extendskinB . For allk, this new functionskis the solution of a certain elliptic equation whose operator become from the composition of the Laplacian with the application that makes@B‡\Kflat.

Thus sinceP

aigiconverges absolutely for theL2 norm, by the Harnack inequalityP

aigiconverges absolutely for the uniform norm in a smaller neighborhood B0B that still contains x0. Th us sk converges to a C1 function denoted bys, which is equal togonB‡. And since@sk

@n(x0)ˆ0, by the absolute convergence of the sum we can exchange the order of the derivative and the symbolP

so we deduce that@s

@n(x0)ˆ0. Finally, sinces is equal togonB‡we deduce thatgisC1at the boundary and@g

@nˆ0 atx0. Fourth step:we claim thatgis equal touonr0V. Letrbe a radius such that r5r0. Then the function x7!gr(x):ˆg rx

r0

is well defined for x2r0V, and since thegiare homogeneous we have

g

rx r0

ˆX‡1

iˆ0

aigi

rx r0

ˆX‡1

iˆ0

r r0

ai

aigi(x)ˆX‡1

iˆ0

r r0

ai

aihi(x):

We deduce that the functionx7!g r r0x

is inL2(r0V) and its coefficients in the basisfhigare r

r0

ai

ai

. We want to show thatkgr ukL2(r0V)tend to 0. Indeed, writinguin the basisfhig

uˆX‡1

iˆ0

aihi; we obtain

kgr uk22ˆX‡1

iˆ0

r r0

ai

1

2

a2ikhik22

(18)

which tends to zero when r tends to r0 by the dominated convergence theorem because r

r0 ai

1

2

1. Therefore, there is a subsequence for whichgrtends toualmost everywhere. On the other hand, sincegis har- monic, the limit ofgrexists and is equal tog. That means thatgtends tou radially at almost every point ofr0V.

Fifth step:The functionsuandgare harmonic functions inA(r0), with finite energy, witha normal derivative equal to zero onK\A(r0)nS and that coõÈncide on@A(r0)nK. To show thatuˆginA(r0) we shall prove thatg is an energy minimizer. Proposition 8 will then give the uniqueness.

LetW2W1;2(A(r0))nK) have a vanishing trace on@B(0;r0). Then, set- tingJ(v):ˆ s

A(r0)jrvj2 forv2W1;2(A(r0)) we have J(g‡W)ˆJ(g)‡

Z

A(r0)

rgrW‡J(W):

Now since g is harmonic with Neumann condition on KnS and since W vanishes onr0V, integrating by parts we obtain

J(g‡W)ˆJ(g)‡J(W):

SinceJis non negative andg‡Wdescribes all the functions inW1;2(A(r0)) withtrace equal touonr0V, we deduce thatgminimizesJ. We can do the same withuthusuandgare two energy minimizers withsame boundary conditions. Therefore, by Proposition 8 we know thatgˆu.

Sixth step:The decomposition do not depends onr0. Indeed, letr1be a second choice of radius. Then we can do the same work as before to obtain a decomposition

u(x):ˆX‡1

iˆ0

bigi(x) inB(0;r1)nK:

Now by uniqueness of the decomposition inB(0;min(r0;r1)) we deduce that biˆaifor alli.

In addition,r0was initially chosen almost everywhere in ]0;1[. But since the decomposition does not depend on the choice of radius,r0can be chosen anywhere in ]0;1[, by choosing a radius almost everywhere in ]r0;1[. p THEOREM15. Let(u;K)be a global minimizer inRNsuch that K is a smooth cone. Then for each connected component of RNnK there is a constant uksuch that u ukis1

2-homogenous.

(19)

PROOF. Let V be a connected component of RNnK. We apply the preceding proposition tou. Thus

u(x)ˆX‡1

iˆ0

aigi(x) in A(r0):

for a certain radiusr0chosen in ]0;1[. Let us prove that the same decom- position is true in A(1). Applying Proposition 14 to the function uR(x)ˆu(Rx) we know that there are some coefficientsai(R) suchthat

uR(x)ˆX‡1

iˆ0

ai(R)gi(x) inA(r0):

Now sinceuR x

R ˆu(x) we can use the homogeneity of thegito identify the terms inB(0;r0) thusai(R)ˆaiRai. Now we fixyˆRxand we obtain that

u(y)ˆX‡1

iˆ0

aigi(y) inA(Rr0):

SinceRis arbitrary the decomposition is true inA(1).

In addition for every radiusRwe know that kruk2L2(A(R))ˆX‡1

iˆ0

a2ikrgik2L2(A(R))

(17)

and sincegiisai-homogenous,

krgik2L2(A(R))ˆR2(ai 1)‡Nkrgik2L2(A(1)):

Now, since uis a global minimizer, a classical estimate on the gradient obtained by comparing (u;K) with(v;L) where vˆ1B(0;R)cu and Lˆ@B(0;R)[(KnB(0;R)) gives that there is a constantCsuchthat for all radiusR

kruk2L2(B(0;R)nK)CRN 1: We deduce

X‡1

iˆ0

a2iR2(ai 1)‡Nkrgik2L2(A(1)) CRN 1: Thus

X‡1

iˆ0

a2iR2ai 1krgik2L2(A(1))C:

(20)

This last quantity is bounded whenRgoes to infinity if and only ifaiˆ0 wheneverai>1=2. On the other hand, this quantity is bounded whenR goes to 0, if and only ifaiˆ0 whenever 05ai51=2. Therefore,u a0is a finite sum of terms of degree1

2. p

REMARK16. In Chapter 65 of [8], we can find a variational argument that leads to a formula in dimension 2 that links the radial and tangential derivatives of u. For all j2K\@B(0;r), we call uj2 0;p

2 h i

the non or- iented angle between the tangent toKat pointjand the radius [0;j]. Then we have the following formula

Z

@B(0;r)nK

@u

@r

2dH1ˆ Z

@B(0;r)nK

@u

@t

2dH1‡ X

j2K\@B(0;r)

cosuj 1

rH1(K\B(0;r)):

Notice that for a global minimizer inR2withK a centered cone we find Z

@B(0;r)nK

@u

@r

2dH1ˆ Z

@B(0;r)nK

@u

@t

2dH1: (18)

Now suppose that (u;K) is a global minimizer inRN with K a smooth cone centered at 0. Then by Theorem 15 we know thatuis harmonic and 1

2-homogenous. Its restriction to the unit sphere is an eigenfunction for the spherical Laplacian with Neumann boundary condition and asso- ciated to the eigenvalue 2N 3

4 . We deduce th at krtuk2L2(@B(0;1))ˆ2N 3

4 kuk2L2(@B(0;1)): On the other hand

@u

@r(x)ˆ1 2kxk 12u

x kxk

thus

k@u

@rk2L2(@B(0;1))ˆ1

4kuk2L2(@B(0;1)): So krtuk2L2(@B(0;1))ˆ(2N 3)k@u

@rk2L2(@B(0;1)): In particular, forNˆ2 we have the same formula as (18).

(21)

2. Some applications.

As it was claimed in the introduction, here is some few applications of Theorem 15.

PROPOSITION17. Let(u;K)be a global minimizer inR3such that K is a smooth cone. Moreover, assume that S2\K is a union of convex curvilinear polygons with C1 sides. Then u is locallyconstant and K is a cone of typeP,YorT.

PROOF. In each polygon we know by Proposition 4.5. of [6] that the smallest positive eigenvalue for the operator minus Laplacian with Neu- mann boundary conditions is greater than or equal to 1. Thus it cannot be3 anduis locally constant. ThenKis a minimal cone inR3and we know from4

[9] that it is a cone of typeP,YorT. p

Let (r;u;z)2R‡[ p;p]Rbe the cylindrical coordinates in R3. For everyv2[0;p] set

Gv:ˆ f(r;u;z)2R3; v5u5vg of boundary

@Gv:ˆ f(r;u;z)2R3;uˆ voruˆvg:

ConsiderVvˆGv\S2and letl1be the smallest positive eigenvalue of DS inVvwithNeumann conditions on@Vv. Then by Lemma 4.1. of [6] we have that

l1ˆmin (2;lv) where

lvˆ p 2v‡1

2

2

1 4: In particular for the cone of typeY,vˆp

3thusl1ˆ2.

Observe that forv6ˆp,lv6ˆ3

4. So we get this following proposition.

PROPOSITION18. There is no global Mumford-Shah minimizer inR3 such that K is wing of type@Gv withv62

0;p

2;p

.

(22)

Another consequence of Theorem 15 is the following. LetPbe the half plane

P:ˆ f(r;u;z)2R3;uˆpg:

PROPOSITION19. Let(u;K)be a global Mumford-Shah minimizer in R3such that KˆP. Then u is equal to cracktipR, that is in cylindrical coordinates

u(r;u;z)ˆ

2 p r

r12sinu 2‡C for05r5‡ 1and p5u5p.

REMARK20. In Section 3 we will give a second proof of Proposition 19.

REMARK 21. We already know that uˆcracktipR is a global minimizer inR3(see [8]).

To prove Proposition 19 we will use the following well known result.

PROPOSITION22 ([5], [13]) . The smallest positive eigenvalue for Dn

in S2nP is3

4, the corresponding eigenspace is of dimension1generated by the restriction on S2 of the following function in cylindrical coordinates

u(r;u;z)ˆr12sinu 2 for05r5‡ 1and p5u5p.

Now the proof of Proposition 19 can be easily deduce from Proposition 22 and Theorem 15.

PROOF OFPROPOSITION19. If (u;P) is a global minimizer, we know that after removing a constant the restriction ofuto the unit sphere is an ei- genfunction for Dn in S2nP associated to the eigenvalue 3

4. Therefore, from Proposition 22 we know that

u(r;u;z)ˆCr12sinu 2

so we just have to determinate the constant C. But by a well known ar- gument about Mumford-Shah minimizers we prove thatCmust be equal to

2 p r

(see [8] Section 61 for more details). p

(23)

Now set

Sv:ˆ f(r;u;0);r>0;u2[ v;v]g

PROPOSITION23. There is no global Mumford-Shah minimizer inR3 such that K is an angular sector of type(u;Sv)for05v5p

2orp

25v5p.

PROOF. According to Theorem 15, if (u;Sv) is a global minimizer, then u u0is a homogenous harmonic function of degree1

2, thus its restriction toS2nSvis an eigenfunction for Dnassociated to the eigenvalue3

4. Now if l(v) denotes the smallest eigenvalue on@B(0;1)nSv, we know by Theorem 2.3.2. p. 47 of [14] that l(v) is non decreasing withrespect to v. Since l p

2 ˆ3

4, we deduce that forv5p

2, we have l(v)3

4: (19)

In [14] page 53 we can find the following asymptotic formula nearvˆp 2 l(v)ˆ3

4‡2

pcosv‡O( cos2v):

(20)

this proves that the case when (19) is a equality only arises whenvˆp 2. Thus such eigenfunctionudoesn't exist.

Consider now the case v>p

2. For vˆp there are tow connected components. Thus 0 is an eigenvalue of multiplicity 2. The second eigen- value is equal to 2. Therefore, forvˆpthe spectrum is

002l3. . . vˆp

By monotonicity, when v decreases, the eigenvalues increase. Since the domain becomes connexe, 0 become of multiplicity 1 thus the second ei- genvalue become positive. The spectrum is now

0l1l2. . . v5p

withli2 fori2. Thus the only eigenvalue that could be equal to3 4isl2 which is increasing from from 0 to3

4, reached forvˆp

2. Now (20) says that the increasing is strict nearvˆp

2. Therefore there is no eigenvalue equal to 3=4 forv>p

2and there is no possible global minimizer. p

(24)

3. Second proof of Propositions 19 and 22.

Here we want to give a second proof of Proposition 19, without using Theorem 15, and which do not use Proposition 22. In a remark at the end of this section, we will briefly explain how to use this proof of Proposition 19 in order to obtain a new proof of Proposition 22 as well.

Let assume thatK is a half plane inR3. We can suppose for instance that

KˆP:ˆ fx2ˆ0g \ fx10g (21)

We begin by studying the harmonic measure inR3nP.

LetB:ˆB(0;R) be a ball of radiusRand letgbe the trace operator on

@B(0;R)nP. We denote byTthe image ofW1;2(BnK) byg. We also denote byC0b(@BnK) the set of continuous and bounded functions on@B(0;1)nP.

Finally setA:ˆT\Cb0. ObviouslyAis not empty. To every functionf 2A, Proposition 15.6. of [8] associates a unique energy minimizing function u2W1;2(BnK) suchthatg(u)ˆf on@BnP. Sinceuis harmonic we know that it isC1inBnK. Lety2BnKbe a fixed point and consider the linear formmydefined by

my:A!R f7!u(y):

(22)

By the maximum principle for energy minimizers, we know that for all f 2Awe have

jmy(f)j kfk1

thusmyis a continuous linear form onAfor the normk k1. We identifymy withits representant in the dual space of A and we call it harmonic measure.

Moreover, the harmonic measure is positive. That is, iff 2Ais a non negative function, then (by the maximum principle)my(f) is non negative.

By positivity ofmy, iff 2Ais a non negative function andg2Ais suchthat fg2A, then since (kgk1‡g)f and (kgk1 g)f are two non negative functions ofAwe deduce that

jhfg;myij kgk1hf;myi:

(23)

Now here is an estimate on the measuremRy.

(25)

LEMMA24. There is a dimensional constant CNsuch that the following holds. Let R be a positive radius. For05l5R

2 consider the spherical do- main

Cl:ˆ fx2R3; jxj ˆR and d(x;P)lg:

LetWl2C1(@B(0;R))be a function between0and1, that is equal to1onCl and0on@B(0;R)nC2land that is symmetrical with respect to P. Then for everyy2B 0;R

2

nP we have

mRy(Wl)Cl R:

PROOF. SinceWlis continuous and symmetrical withrespect toP, by the reflection principle, its harmonic extensionW inB(0;R) has a normal derivative equal to zero onPin the interior ofB(0;R). MoreoverWlis clearly in the spaceA. Thus by definition ofmy,

W(y)ˆ hWl;mRyi:

On the other hand, sinceWlis continuous on the entire sphere, we also have the formula with the classical Poisson kernel

W(y)ˆR2 jyj2 NvNR

Z

@BR

Wl(x) jx yj3ds(x)

withvNequal to the measure of the unit sphere. In other words mRy(Wl)ˆR2 jyj2

NvNR Z

@BR

Wl(x) jx yj3ds(x):

Forx2@BRwe have 1

2R jxj jyj jx yj jxj ‡ jyj 3 2R:

We deduce that

mRy(Wl)CN 1 R2

Z

C2l

ds:

(26)

Now integrating by parts, Z

Cl

dsˆ2 Zl

0

2p 

R2 w2

p dw

ˆ4pl 2



R2 l2

p ‡R2arcsin l

R

CRl because arcsin(x)p

2x. The proposition follows. p Now we can prove the uniqueness ofcracktipR.

SECOND PROOF OF PROPOSITION19. Let us show that u is vertically constant. Let t be a positive real. For xˆ(x1;x2;x3)2R3 set xt:ˆ(x1;x2;x3‡t). We also set

ut(x):ˆu(x) u(xt):

Sinceuis a function associated to a global minimizer, and sinceK is reg- ular, we know that for all R>0, the restriction of u to the sphere

@B(0;R)nK is continuous and bounded on@B(0;R)nKwithfinite limits on eachsides ofK. It is the same for ut. Thus for allx2R3nPand for all R>2kxkwe can write

ut(x):ˆ hutj@B(0;R)nP;mRxi

wheremxis the harmonic measure defined in (22). We want to prove that forx2R3nP,hutj@B(0;R)nP;mRxitends to 0 whenRgoes to infinity. This will prove thatut ˆ0.

So letx2R3nPbe fixed. We can suppose thatR>100(kxk ‡t). LetCl

andWlbe as in Lemma 24. Then write

ut(x)ˆ hutj@B(0;R)nPWl;mRxi ‡ hutj@B(0;R)nP(1 Wl);mRxi:

Now by a standard estimate on Mumford-Shah minimizers (that comes from Campanato's Theorem, see [3] p. 371) we have for allx2RNnP,

jut(x)j C 

pt : Then, using Lemma 24 we obtain

jhutj@B(0;R)nPWl;mRxij C 

pt l R:

(27)

On the other hand, for the points y suchthat d(y;P)l, since u~:u(:) u(y) is harmonic inB(y;d(y;P)) we have, by a classical estimation on harmonic functions (see the introduction of [12])

jr~u(y)j C 1

d(y;P)k~ukL1(@B(y;12d(y;P))): Now using Campanato's Theorem again we know that

k~ukL1(@B(y;12d(y;P)))Cd(y;P)12 thus

jru(y)j C 1 d(y;P)12

and finally by the mean value theorem we deduce that for all the pointsy suchthatd(y;P)l,

jut(y)j C sup

z2[y;yt]jru(z)j:jy ytj t 1 l12: Therefore,

jhutj@B(0;R)nP(1 Wl);mRxij Ct 1 l12: So

jut(x)j C 

pt l R‡Ct 1

l12

thus by settinglˆR12and by lettingRgo to‡1we deduce thatut(x)ˆ0 thusz7!u(x;y;z) is constant.

Now we fix z0ˆ0 and we introduce P0:ˆP\ fzˆ0g. We want to show that (u(x;y;0);P0) is a global minimizer inR2. Let (v(x;y);G) be a competitor foru(x;y;0) in the 2-dimensional ballBof radiusr. LetCbe the cylinderC:ˆB[ R;R]. Define~vandG~ inR3 by

~v(x;y;z)ˆ v(x;y) if (x;y;z)2 C u(x;y;z) if (x;y;z)62 C

G~ :ˆ(C \[G[ R;R]])[(PnC)[(B fRg):

It is a topological competitor because R3nP is connected (thusP doesn't separate any points). Now finally letB~ be a ball that containsC. Then (~v;G)~

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