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domain need not be traces of least gradient functions
Mickael dos Santos
To cite this version:
Mickael dos Santos. Characteristic functions on the boundary of a planar domain need not be traces
of least gradient functions. 2017. �hal-01216155v2�
MICKAËL DOS SANTOS
Abstract. Given a smooth bounded planar domainΩ, we construct a compact set on the boundary such that its characteristic function is not the trace of a least gradient function. This generalizes the construction of Spradlin and Tamasan [ST14] whenΩis a disc.
Contents
1. Introduction 1
2. Strategy of the proof 2
2.1. The model problem 2
2.2. Outline of the proof of Theorem 1 3
3. Notation, definitions 4
4. Construction of the Cantor setK 5
4.1. First step: localization of∂Ω 5
4.2. Step 2: Iterative construction 5
5. Construction of a sequence of functions 7
6. Basic properties ofB∞and ΨN 8
6.1. Basic properties ofB∞ 8
6.2. Basic properties ofΨN 9
7. Proof of Theorem 1 10
7.1. Upper bound 10
7.2. Optimality of the upper bound 10
7.3. Transverse argument 12
Appendices 13
Appendix A. A smoothing result 13
Appendix B. Proofs of Lemma 2, Lemma 3, Lemma 4 and Lemma 16 15
B.1. Proof of Lemma 2 15
B.2. Proof of Lemma 3 16
B.3. Proof of Lemma 4 16
B.4. Proof of Lemma 16 16
Appendix C. Results related to the Cantor setK 17
C.1. Justification of Remark 6.(1) 17
C.2. Two preliminary results 17
C.3. Proof of Proposition 13 19
Appendix D. A fundamental ingredient in the construction of theΨ˜N’s 20
Appendix E. Adaptation of a result of Giusti in [Giu84] 21
References 23
1. Introduction
We let Ω be a boundedC2 domain of R2. For a functionh ∈L1(∂Ω,R), the least gradient problem with boundary datumhconsists in deciding whether
(1) inf
ßZ
Ω|Dw|; w∈BV(Ω) andtr∂Ωw=h
™
2010Mathematics Subject Classification. 26B30, 35J56 . 1
is achieved or not.
In the above minimization problem,BV(Ω)is the space of functions of bounded variation. It is the space of functionsw∈L1(Ω)having a distributional gradientDw which is a bounded Radon measure.
If theinfimumin (1) is achieved, minimal functions are calledfunctions of least gradient.
Sternberg, Williams and Ziemmer proved in [SWZ92] that if h: ∂Ω → R is a continuous map and if ∂Ω satisfies a geometric properties then there exists a (unique) function of least gradient. For further use, we note that the geometric property is satisfied by Euclidean balls.
On the other hand, Spradlin and Tamasan [ST14] proved that, for the disc Ω ={x∈R2;|x|<1}, we may find a functionh0∈L1(∂Ω)which is not continuous s.t. theinfimumin (1) is not achieved. The functionh0 is the characteristic function of a Cantor type setK ⊂S1={x∈R2;|x|= 1}
The goal of this article is to extend the main result of [ST14] to a generalC2 bounded open setΩ⊂R2. We prove the following theorem.
Theorem 1. Let Ω⊂R2 be a bounded C2 open set. Then there exists a measurable setK ⊂∂Ω such that the infimum
(2) inf
ßZ
Ω|Dw|; w∈BV(Ω) andtr∂Ωw=1IK
™
is not achieved.
The calculations in [ST14] are specific to the caseΩ =D. The proof of Theorem1 relies on new arguments for the construction of the Cantor setKand the strategy of the proof.
2. Strategy of the proof
2.1. The model problem. We illustrate the strategy developed to prove Theorem1 on the model caseQ= (0,1)2. Clearly, this model case does not satisfy theC2 assumption.
Nevertheless, the flatness of ∂Q allows to get a more general counterpart of Theorem 1. Namely, the counterpart of Theorem1[see Proposition1below] is no more an existence result of a setK ⊂∂Qs.t. Problem (2) is not achieved. It is a non existence result of a least gradient function for h =1IM for any measurable domainM ⊂[0,1]× {0} ⊂∂Qwith positive Lebesgue measure.
We thus prove the following result whose strategy of the proof is due to Petru Mironescu.
Proposition 1. [P. Mironescu] Let M ⊂˜ [0,1]be a measurable set with positive Lebesgue measure. Then the infimum in
(3) inf
ßZ
Q|Dw|; w∈BV(Q)andtr∂Qw=1IM×{0}˜
™
is not achieved.
This section is devoted to the proof of Proposition 1. We fix a measurable set M ⊂˜ [0,1] with positive measure and we let h=1IM×{0}˜ . We argue by contradiction: we assume that there exists a minimizer u0 of (3). We obtain a contradiction in 3 steps.
Step 1. Upper bound and lower bound
This first step consists in obtaining two estimates. The first estimate is the upper bound (4)
Z
Q|Du0| ≤ k1IM×{0}˜ kL1(∂Q)=H1( ˜M).
Here,H1( ˜M)is the length ofM˜.
Estimate (4) follows from Theorem 2.16 and Remark 2.17 in [Giu84]. Indeed, by combining Theorem 2.16 and Remark 2.17 in [Giu84] we may prove that forh∈L1(∂Ω)and for allε >0there exists a mapuε∈BV(Ω)
s.t. Z
Ω|Duε| ≤(1 +ε)khkL1(∂Ω) andtr∂Ωuε=h.
The proof of this inequality whenΩ is a half space is presented in [Giu84]. It is easy to adapt the argument whenΩ =Q= (0,1)2. The extension for aC2 set Ωis presented in AppendixE.
Step 2. Optimality of (4) [see (5)]
The optimality of (4) is obtained viathe following lemma.
Lemma 2. Foru∈BV(Q)we have Z
Q|D2u| ≥ Z 1
0 |tr∂Qu(·,0)−tr∂Qu(·,1)|. Here, fork∈ {1,2} we denoted
Z
Q|Dku|= sup ßZ
Q
u∂kξ;ξ∈Cc1(Q)and|ξ| ≤1
™
whereCc1(Q)are the set of real valuedC1-functions with compact support included inQ. Lemma2 is proved in AppendixB.1.
From Lemma2 we get Z
Q|D2u0| ≥ Z 1
0 |tr∂Qu0(·,0)−tr∂Qu0(·,1)|= Z 1
0
1IM×{0}˜ =H1( ˜M).
Since we have Z
Q|Du0| := sup ßZ
Q
udiv(ξ) ; ξ= (ξ1, ξ2)∈Cc1(Q,R2)andξ21+ξ22≤1
™
≥ Z
Q|D2u0| ≥H1( ˜M), (5)
we get the optimality of (4).
Step 3. A transverse argument From (4) and (5) we may prove (6)
Z
Q|D1u0|= 0.
Equality (6) is a direct consequence of the following lemma.
Lemma 3. Let Ωbe a planar open set. Ifu∈BV(Ω) is s.t.
Z
Ω|Du|= Z
Ω|D2u|, then
Z
Ω|D1u|= 0.
Lemma3 is proved in AppendixB.2.
In order to conclude we state an easy lemma.
Lemma 4. [Poincaré inequality] Foru∈BV(Q)satisfyingtr∂Qu= 0 in{0} ×[0,1]we have Z
Q|u| ≤ Z
Q|D1u|. Lemma4 is proved in AppendixB.3.
Hence, from (6) and Lemma 4 we have u0 = 0 which is in contradiction with tr∂Qu0 = 1IM×{0}˜ with H1( ˜M)>0.
2.2. Outline of the proof of Theorem 1. The idea is to adapt the above construction and argument to the case of a general C2 domain Ω. If Ω has a flat or concave part Γ of the boundary ∂Ω, then a rather straightforward variant of the above proof shows that 1IM, whereMis a non trivial part of Γ, is not the trace of a least gradient function.
Remark 5. Things are more involved whenΩis convex. For simplicity we illustrate this fact when Ω =D= {x∈R2;|x|<1}. LetM ⊂S1∩ {(x, y)∈R2;x <0} be an arc whose endpoints are symmetric with respect to thex-axis. We let(x0,−y0)and(x0, y0)be the endpoints ofM[herex0≤0andy0>0].
We letC be the chord ofM. On the one hand, ifu∈C1(D)∩W1,1(D)is s.t. trS1u=1IM then, using the Fundamental Theorem of calculus, we have for−y0< y < y0
Z √
1−y2
−√
1−y2|∂xu(x, y)| ≥1.
Thus we easily get Z
D|∇u| ≥ Z
D|∂xu| ≥ Z y0
−y0
dy Z √
1−y2
−√
1−y2|∂xu(x, y)| ≥2y0=H1(C).
Consequently, with the help of a density argument [e.g. Lemma 17in AppendixA] we obtain inf
ßZ
D|Du|; u∈BV(D)andtrS1u=1IM
™
≥H1(C).
On the other hand we letω :={(x, y)∈R2;x < x0}. It is clear that u0 =1Iω ∈ BV(D) and trS1u0 =1IM. Moreover
Z
D|Du0|=H1(C).
Consequentlyu0 is a function of least gradient. We may do the same argument for a domainΩas soon as we have a chord entirely contained in Ω. This example suggest that for a convex setΩ, the construction of a set K ⊂∂Ωs.t. (2) is not achieved has to be "sophisticated".
The strategy to prove Theorem 1 consists of constructing a special set K ⊂ ∂Ω [of Cantor type] and to associate toKa set B∞[the analog ofM ט (0,1)in the model problem] which "projects" ontoKand s.t., ifu0
is a minimizer of (1), then (7)
Z
B∞|X~ ·Du0| ≥H1(K).
Here,X~ is a vector field satisfying|X~| ≤1. It is the curved analog ofX~ =e2 used in the above proof.
By (7) [and Proposition24in AppendixE], ifu0is a minimizer, then (8)
Z
Ω\B∞
|Du0|+ Z
B∞
(|Du0| − |X~ ·Du0|) = 0.
We next establish a Poincaré type inequality implying that anyu0 satisfying (8) andtr∂Ω\Ku= 0is0, which is not possible.
The heart of the proof consists of constructingK, B∞and X~ [see Sections4 and5].
3. Notation, definitions
The ambient space is the Euclidean planR2. We letBcan be the canonical basis ofR2. a) The open ball centered atA∈R2 with radiusr >0 is denotedB(A, r).
b) A vector may be denoted by an arrow when it is defined by its endpoints (e.g. −−→AB). It may be also denoted by a letter in bold font (e.g. u) or more simply by a Greek letter in normal font (e.g. ν).
We let also|u|be the Euclidean norm of the vectoru.
c) For a vectoruwe letu⊥ be the direct orthogonal vector tou,i.e., ifu= (x1, x2)thenu⊥= (−x2, x1).
d) ForA, B∈R2, the segment of endpointsAandB is denoted[AB] ={A+t−−→AB;t∈[0,1]}anddist(A, B) =
|−−→AB|is the Euclidean distance.
e) For a setU ⊂R2, the topological interior ofU is denoted byU◦ and its topological closure isU.
f) For k ≥ 1, a Ck-curve is the range of a Ck injective map from (0,1) to R2. Note that, in this article, Ck-curves are not closed sets ofR2.
g) ForΓaC1-curve,H1(Γ)is the1-dimensional Hausdorff measure ofΓ.
h) For k≥1, a Ck-Jordan curve is the range of aCk injective map from the unit circleS1 toR2. i) For ΓaC1-curve or aC1-Jordan curve,C = [AB]is a chord ofΓwhenA, B∈ΓwithA6=B.
j) If Γ is aC1-Jordan curve then, for A, B ∈ Γ &A 6=B, the set Γ\ {A, B} admits exactly two connected components: Γ1&Γ2. These connected components areC1-curves.
By smoothness ofΓ, it is clear that there exists ηΓ >0 s.t. for 0 <dist(A, B) < ηΓ there exists THE smallest connected components: we haveH1(Γ1)<H1(Γ2)orH1(Γ2)<H1(Γ1).
If0<dist(A, B)< ηΓ we may define AB˜by:
(9) AB˜is the closure of the smallest curve between Γ1 andΓ2. k) In this articleΩ⊂R2 is aC2bounded open set.
4. Construction of the Cantor set K
It is clear that, in order to prove Theorem1, we may assume thatΩis a connected set.
We fix Ω⊂R2 a bounded C2 open connected set. The setK ⊂ ∂Ωis a Cantor type set we will construct below.
4.1. First step: localization of ∂Ω. From the regularity ofΩ, there existℓ+ 1C2-open sets, ω0, ..., ωℓ, s.t.
Ω =ω0\ω1∪ · · · ∪ωℓ and
• ωi is simply connected fori= 0, ..., ℓ,
• ωi ⊂ω0fori= 1, ..., ℓ,
• ωi∩ωj =∅ for1≤i < j≤ℓ.
We letΓ =∂ω0. The Cantor type setK we construct "lives" onΓ. Note thatΓis a Jordan-curve.
LetM0∈Γ be s.t. the inner curvature ofΓ at M0 is positive [the existence of M0 follows from the Gauss- Bonnet formula]. Then there existsr0∈(0,1)s.t. [AB]⊂Ωand[AB]∩∂Ω ={A, B},∀A, B∈B(M0, r0)∩Γ.
Note that we may assume2r0< ηΓ [ηΓ is defined in Section3-j].
We fixA, B∈B(M0, r0)∩Γs.t. A6=B. We have:
• By the definition ofM0 andr0, the chordC0:= [AB] is included inΩ.
• We let AB˜ be the closure of the smallest part of Γ which is delimited by A, B (see (9)). We may assume that AB˜ is the graph of f ∈ C2([0, η],R+) in the orthonormal frame R0 = (A,e1,e2) where e1=−−→AB/|−−→AB|.
• The functionf satisfiesf(x)>0 forx∈(0, η)andf′′(x)<0forx∈[0, η].
For further use we note that the length of the chord[AB]isη and that for intervalsI, J⊂[0, η], ifI⊂J then (10)
(kf|I′ kL∞(I)≤ kf|J′ kL∞(J) kf|I′′kL∞(I)≤ kf|J′′kL∞(J)
wheref|I is the restriction off toI.
Replacing the chordC0= [AB]with a smaller chord ofAB˜ parallel toC0, we may assume that (11) 0< η <min
®1
2; 1
16kf′′k2L∞([0,η])
; 1
2kf′kL∞([0,η])kf′′kL∞([0,η])
´ . We may also assume that
• Letting D0+ be the bounded open set s.t. ∂D0+ = [AB]∪AB˜ we haveΠ∂Ω, the orthogonal projection on∂Ω, is well defined and of classC1 inD+0.
• We have
(12) 1 + 4kf′′k2L∞diam(D+0)< 16 9
wherediam(D+0) = sup{dist(M, N) ; M, N ∈D+0}. [Here we used (10)]
4.2. Step 2: Iterative construction. We are now in position to construct the Cantor type setKas a subset ofAB. The construction is iterative.˜
The goal of the construction is to get at step N ≥0 a collection of2N pairwise disjoint curves included in AB˜[denoted by {K1N, ..., K2NN}] and their chords [denoted by {CN
1 , ...,CN
2N}].
The idea is standard: at the stepN ≥0 we replace a curveΓ0 included inAB˜by two curves included inΓ0
(see Figure1).
Initialization. We initialize the procedure by letting K10:=AB˜ andC0
1 =C0= [AB].
At stepN ≥0we have:
• A set of 2N curves included in AB˜, {K1N, ..., K2NN}. The curves KkN’s are mutually disjoint. We let KN =∪2k=1N KkN.
• A set of2N chords, {CN
1 , ...,CN
2N}s.t. fork= 1, ...,2N,CN
k is the chord ofKkN.
Remark 6. (1) Note that since theCN
k ’s are chords ofAB˜ and since in the frameR0= (A,e1,e2),AB˜ is the graph of a function, none of the chordsCN
k is vertical,i.e., directed bye2. Since the chords CN
k are not vertical, for k ∈ {1, ...,2N}, we may define νCN
k as the unit vector orthogonal toCN
k s.t. νCN
k =αe1+βe2withβ >0.
(2) Forηsatisfying (11), if we consider a chordCN
k and a straight lineDorthogonal toCN
k and intersecting CN
k , then the straight lineD intersectKkN at exactly one points. This fact is proved in AppendixC.1.
Induction rules. From stepN≥0to stepN+ 1 we follow the following rules:
(1) For eachk∈ {1, ...,2N}, we letηNk be the length of CN
k . Inside the chordCN
k we center a segmentIkN of length(ηNk )2.
(2) With the help of Remark6.2, we may define two distinct points ofKkN as the intersection of KkN with straight lines orthogonal toCN
k which pass to the endpoints of IkN.
(3) These intersection points are the endpoints of a curveK˜kN included inKkN. We letK2k−1N+1 and K2kN+1 be the connected components ofKkN \K˜kN. We let also
• CN+1
2k−1 and CN+1
2k be the corresponding chords;
• KN+1 =∪2k=1N+1KkN+1.
Notation 7. A natural terminology consists in defining the father and the sons of a chord or a curve:
• F(CN+1
2k−1) =F(CN+1
2k ) =CN
k is thefatherof the chordsCN+1
2k−1 andCN+1
2k . F(K2k−1N+1) =F(K2kN+1) =KkN is thefatherof the curvesK2k−1N+1 andK2kN+1.
• S(CN
k ) ={CN+1
2k−1,CN+1
2k } is the set of sons of the chordCN
k ,i.e.,F(CN+1
2k−1) =F(CN+1
2k ) =CN
k . S(KkN) ={K2k−1N+1, K2kN+1} is the set of sons of the curveKkN,i.e.,F(K2k−1N+1) =F(K2kN+1) =KkN. The inductive procedure is represented in Figure1.
(ηkN)2 ηNk Figure 1. Induction step
In Figure2&3the two first iterations of the process are represented.
Figure 2. First iteration of the process Figure 3. Second iteration of the process We now define the Cantor type set
(13) K= \
N≥0
KN. The Cantor type setK is fat:
Proposition 8. We haveH1(K)>0.
This proposition is proved in AppendixC.3.
5. Construction of a sequence of functions
A key argument in the proof of Theorem 1is the use of the coarea formula to calculate a lower bound for (2). The coarea formula is applied to a function adapted to the setK.
ForN = 0we let
• D0+ be the compact set delimited byK0=AB˜ andC10:= [AB] the chord ofK0.
• We recall that we fixed a frameR0= (A,e1,e2)wheree1=−−→AB/|−−→AB|. Forσ= (σ1,0)∈C10, we define:
(14) Iσ is the connected component of{(σ1, t)∈Ω ; t≤0}which containsσ.
[Iσ is a vertical segment included inΩ].
• D0−=∪σ∈C0
1Iσ.
• We now define the maps
Ψ˜0: D−0 → C0
1
x 7→ ΠC10(x)
and Ψ0: D0−∪D+0 → C10
x 7→
®Π∂Ω(x) ifx∈D0+ Π∂Ω[ ˜Ψ0(x)] ifx∈D0− whereΠ∂Ωis the orthogonal projection on∂ΩandΠC0
1 is the orthogonal projection onC10. Note that, in the frameR0, forx= (x1, x2)∈D0−, we haveΠC10(x) = (x1,0).
ForN = 1andk∈ {1,2}we let:
• Dk1 be the compact set delimited byKk1 andC1
k ;
• Tk1be the compact right-angled triangle (with its interior) havingC1
k as side adjacent to the right angle and whose hypothenuse is included inC10;
• Hk1be the hypothenuse ofTk1.
We now defineD1−= ˜Ψ−10 (H11∪H21), T1=T11∪T21 andD+1 =D11∪D12. We first consider the map
Ψ˜1: T1∪D1− → C1
1 ∪C1
2
x 7→
®ΠC1
k(x) ifx∈Tk1 ΠC1
k[ ˜Ψ0(x)] ifx∈D1−.
In Appendix D[Lemma 22and Remark 23], it is proved that the trianglesT11 and T21 are disjoint. Thus the mapΨ˜1is well defined
By projectingC1
1 ∪C1
2 on∂Ωwe get
Ψ1: T1∪D1−∪D+1 → K1
x 7→
®Π∂Ω(x) ifx∈D1+ Π∂Ω[ ˜Ψ1(x)] ifx∈T1∪D1−
.
σ
D12 T21
ψ˜0−1(H21) D11∪T11∪Ψ˜−10 (H11)
Figure 4. The sets defined at StepN = 1and the dashed level line ofΨ1associated to σ∈ K1
ForN ≥1, we first constructΨ˜N+1and then ΨN+1 is obtained fromΨ˜N+1and Π∂Ω. Fork∈ {1, ...,2N+1}, we let
• DkN+1 be the compact set delimited byKkN+1 andCN+1
k [recall thatCN+1
k is the chord associated to KkN+1] ;
• TkN+1 be the right-angled triangle (with its interior) having CN+1
k as side adjacent to the right angle and whose hypothenuse is included in F(CN+1
k ). HereF(CN+1
k ) is the father ofCN+1
k (see Notation 7);
• HkN+1⊂ F(CN+1
k )be the hypothenuse ofTkN+1. We denoteTN+1=
2N+1
[
k=1
TkN+1,D−N+1= ˜Ψ−1N Ñ2N+1
[
k=1
HkN+1 é
andD+N+1=
2N+1
[
k=1
DkN+1.
K2kN+1−1
DN+12k−1 T2kN+1−1 H2kN+1−1 DkN K2kN+1
Figure 5. Induction. The bold lines correspond to the new iteration Remark 9. It is easy to check that forN ≥0:
(1) TN+1⊂DN+,
(2) ifx∈T◦N thenx /∈TN′ forN′ ≥N+ 1 [hereT0=∅].
We now define
Ψ˜N+1: TN+1∪D−N+1 → ∪2k=1N+1CN+1
k
x 7→
(ΠCN+1
k (x) ifx∈TkN+1 ΠCN+1
k [ ˜ΨN(x)] ifx∈Ψ˜−1N (∪2k=1N+1HkN+1) .
In AppendixD [Lemma22 and Remark23], it is proved that forN ≥1, the trianglesTkN fork= 1, ...,2N are mutually disjoint. recursively, we find that all theΨ˜N’s are well-defined.
And, as in the Initialization Step, we getΨN+1 fromΨ˜N+1by projecting∪2k=1N+1CN+1
k on∂Ω:
ΨN+1: TN+1∪DN−+1∪D+N+1 → KN+1
x 7→
®Π∂Ω[ ˜ΨN+1(x)] ifx∈TN+1∪D−N+1 Π∂Ω(x) ifx∈D+N+1
.
It is easy to see thatΨN+1(TN+1∪D−N+1∪D+N+1) =KN+1.
6. Basic properties ofB∞ andΨN
6.1. Basic properties of B∞. We set BN = TN∪D+N∪DN−. It is easy to check that for N ≥0 we have BN+1⊂BN andK ⊂∂BN. Therefore we may define
B∞=∩N≥0BN
which is compact and satisfiesK ⊂∂B∞. We are going to prove:
Lemma 10. The interior ofB∞ is empty.
Proof of Lemma 10. From Lemma22[and Remark23] in Appendix Dcombined with Hypothesis (11), we get two fundamental facts:
(1) The trianglesT1N, ... ,T2NN+1 are mutually disjoint.
(2) We have:
(15) H1(HkN+1)< H1(F(CN
k ))
2 .
For a non empty setA⊂R2 we let
rad(A) = sup{r≥0 ;∃x∈As.t. B(x, r)⊂A}. Note that the topological interior ofAis empty if and only ifrad(A) = 0.
On the one hand, it is not difficult to check that for sufficiently largeN
(16) rad(BN) = rad(BN ∩DN−).
On the other hand, using (15) we obtain forN ≥1:
(17) rad(BN+1∩DN−+1)≤rad(BN ∩D−N)
2 .
Consequently, by combining (16) and (17) we get the existence ofC0 s.t.
(18) rad(BN)≤ C0
2N. SinceB∞=∩N≥0BN, from (18) we get that rad(B∞) = 0.
6.2. Basic properties of ΨN. We now prove the key estimate forΨN:
Lemma 11. There exists bN =oN(1) s.t. for N ≥1 andU a connected component of BN, the restriction of ΨN toU is(1 +bN)-Lipschitz.
Proof. LetN ≥1andU be a connected component ofBN. The restriction ofΨ˜N toU∩(TN∪D−N)is obtained as composition of orthogonal projections on straight lines and thus is1-Lipschitz.
There existsbN =oN(1)s.t. the projectionPN := Π∂Ω defined inD+N is (1 +bN)-Lipschitz. The functions ΨN are either the composition of Ψ˜N with PN or ΨN = PN. Consequently the restriction of ΨN to U is
(1 +bN)-Lipschitz.
In the following we will not use ΨN but "its projection" on R. For N ≥ 1 and k ∈ {1, ...,2N}, we let BkN := Ψ−1N (KkN)and we define
Πk,N: BkN → R x 7→ H1(AΨ¸N(x))
whereAΨ¸N(x) ⊂ AB˜is defined by (9) as the smallest connected component of∂Ω\ {A,ΨN(x)}ifΨN(x)6=A andAΨ¸N(x) ={A}otherwise.
Lemma 12. For N ≥ 1 there exists cN ∈ (0,1) with cN = oN(1) s.t. for k ∈ {1, ...,2N} the function Πk,N :BkN →Ris(1 +cN)-Lipschitz.
Proof. LetN≥1,k∈ {1, ...,2N}and letx, y∈BNk be s.t. ΨN(x)6= ΨN(y). It is clear that we have
|Πk,N(x)−Πk,N(y)|=H1(ΨˇN(y)ΨN(x))
whereΨˇN(y)ΨN(x)⊂KkN is defined by (9) as the smallest connected component of∂Ω\ {ΨN(y),ΨN(x)}. Moreover, from Lemma20in AppendixC.2, we have the existence ofC≥1independent ofN andks.t. for x, y∈BNk s.t. ΨN(x)6= ΨN(y)we have [denotingX := ΨN(x), Y := ΨN(y)]
dist (X, Y)≤H1Ä
¯XYä
≤dist (X, Y) [1 +Cdist (X, Y)]
and
H1(KkN)≤H1(CN
k )
1 +CH1(CN
k ) . From Step 1 in the proof of Proposition13[Appendix C.3] we have
k=1,...,2maxN
H1(CN
k )≤ Å2
3 ãN
. Thus lettingaN :=
Å2 3
ãNñ 1 +C
Å2 3
ãNô
we haveaN →0 and since¯XY ⊂KkN we get:
dist (X, Y)≤H1Ä XY¯ä
≤H1(KkN)≤H1(CN
k )
1 +CH1(CN
k )
≤aN(1 +CaN).
Thus, lettinga˜N = max{aN(1 +CaN),|bN|}wherebN is defined in Lemma11, we get H1Ä
¯XYä
=|Πk,N(x)−Πk,N(y)| ≤ H1([ΨN(y)ΨN(x)]) (1 +C˜aN)
≤ (1 + ˜aN) (1 +C˜aN)|x−y|.
Therefore, letting cN be s.t. 1 +cN = (1 + ˜aN) (1 +C˜aN) we have cN = oN(1), cN is independent of k ∈
{1, ...,2N}andΠk,N is(1 +cN)-Lipschitz.
7. Proof of Theorem1
We are now in position to prove Theorem1. This is done by contradiction. We assume that there exists a mapu0∈BV(Ω)which minimizes (2).
7.1. Upper bound. The first step in the proof is the estimate (19)
Z
Ω|Du0| ≤ k1IKkL1(∂Ω)=H1(K).
This estimate is obtained by proving that for allε >0there exists uε∈W1,1(Ω)s.t. tr∂Ωuε=1IK and (20) k∇uεkL1(Ω)≤(1 +ε)ktr∂ΩuεkL1(Ω)= (1 +ε)H1(K).
Proposition24in AppendixEgives the existence of suchuε’s.
Clearly (20) implies (19).
7.2. Optimality of the upper bound. In order to have a contradiction we follow the strategy of Spradlin and Tamasan in [ST14]. We fix a sequence (un)n⊂C1(Ω) s.t.
(21) un∈W1,1(Ω) ;un →uin L1(Ω) ; Z
Ω|∇un| → Z
Ω|Du0|; tr∂Ωun= tr∂Ωu0. Note that (21) implies
(22)
Z
F|∇un| → Z
F|Du|for allF ⊂Ωrelatively closed set.
Such a sequence can be obtainedviapartition of unity and smoothing ; see the proof of Theorem 1.17 in [Giu84].
For the convenience of the reader a proof is presented in AppendixA[see Lemma17].
For further use, let us note that the sequence(un)n constructed in AppendixAsatisfies the following addi- tional property:
Ifu0= 0outside a compact setL⊂Ωand ifω is an open set s.t. dist(ω, L)>0 then, for largen,un= 0in ω . Forx∈B0 we let
(23) V0(x) =
®νΠ∂Ω(x) ifx∈D+0 (0,1) ifx∈D−0 , and forN ≥0,x∈BN+1 we let
(24) VN+1(x) =
VN(x) ifx∈BN\T◦N+1 νCN+1
k ifx∈T◦N+1k
, where, forσ∈∂Ω,νσ is the normal outward ofΩinσand νCN+1
k is defined in Remark6.1.
We now prove the following lemma.
Lemma 13. When N→ ∞ we may defineV∞(x) a.e. x∈B∞ by
(25) V∞: B∞ → R2
x 7→ limN→∞VN(x) . Moreover, from dominated convergence, we have:
VN1IBN →V∞1IB∞ in L1(Ω).
Proof. Ifx∈B∞\ ∪N≥1TN, then we haveVN(x) =V0(x)for allN ≥1. ThuslimN→∞VN(x) =V0(x).
For a.e. x∈ B∞∩ ∪N≥1TN there exists N0 ≥1 s.t. x∈T◦N0. Therefore for allN > N0 we have VN(x) =
VN0(x). ConsequentlylimN→∞VN(x) =VN0(x).
This section is devoted to the proof of the following lemma:
Lemma 14. For all w∈C∞∩W1,1(Ω) s.t. tr∂Ωw=1IK we have Z
B∞∩Ω|∇w·V∞| ≥H1(K) whereV∞ is the vector field defined in (25).
Remark 15. Since|V∞(x)|= 1for a.e. x∈B∞, it is clear that Lemma14implies that for allnwe have Z
B∞∩Ω|∇un| ≥H1(K).
From (22) we have:
Z
B∞∩Ω|Du0| ≥H1(K).
Section7.3is devoted to a sharper argument than above to get Z
B∞∩Ω|∇un| ≥ Z
B∞∩Ω|∇un·V∞|+δ with δ > 0 is independent of n. The last estimate will imply R
B∞∩Ω|Du0| ≥ H1(K) +δ which will be the contradiction we are looking for.
Proof of Lemma 14. We will first prove that forw∈C∞∩W1,1(Ω)s.t. tr∂Ωw=1IK we have (26)
Z
BN∩Ω|∇w·VN| ≥ H1(K) 1 +oN(1). whereVN is the vector field defined in (23) and (24).
Granted (26), we conclude as follows: if w∈C∞∩W1,1(Ω) s.t. tr∂Ωw=1IK, then Z
B∞∩Ω|∇w·V∞| = lim
N→∞
Z
BN∩Ω|∇w·VN|
≥ lim
N→∞
H1(K)
1 +oN(1) =H1(K), by dominated convergence.
It remains to prove (26). We fixw∈C∞∩W1,1(Ω)s.t. tr∂Ωw=1IK. Using the Coarea Formula we have for N ≥1andk∈ {1, ...,2N}, with the help of Lemma12, we have
(1 +cN) Z
BN(k)∩Ω|∇w·VN| ≥ Z
B(k)N ∩Ω|∇Πk,N||∇w·VN|
= Z
R
dt Z
Π−k,N1 ({t})∩Ω|∇w·VN|. Here, ifΠ−1k,N({t})is non trivial, thenΠ−1k,N({t})is a polygonal line:
Π−1k,N({t}) =Iσ(t,k,N)∪Ik,N,t1 ∪ · · · ∪Ik,N,tN+1 where
• σ(t, k, N)∈[AB]is s.t. [AB]∩Π−1k,N({t}) ={σ(t, k, N)},
• Iσ(t,k,N) is defined in (14),
• forl= 1, ..., N we haveIk,N,tl = Π−1k,N({t})∩TN+1−l,
• Ik,N,tN+1 = Π−1k,N({t})∩D+N.
From the Fundamental Theorem of calculus and from the definition ofVN, denoting
• Iσ(t,k,N)= [M0, M1][whereM0∈∂Ω\˜AB andM1=σ(t, k, N)],
• Ik,N,tl = [Ml, Ml+1],l= 1, ..., N+ 1andMN+2∈KkN, we have for a.e. t∈Πk,N(KkN)and using the previous notation,
Z
[Ml,Ml+1]|∇w·VN| ≥ |w(Ml+1)−w(Ml)|. Here we used the conventionw(Ml) = tr∂Ωw(Ml)forl= 0&N+ 2.
Therefore for a.et∈Πk,N(KkN)we have Z
Π−k,N1({t})∩Ω|∇w·VN| ≥ |tr∂Ωw(MN+2)−tr∂Ωw(M0)|=1IK(MN+2).
SinceK ⊂ KN =∪2k=1N KkN, we may thus deduce that (1 +cN)
Z
BN∩Ω|∇w·VN|= (1 +cN)
2N
X
k=1
Z
BN(k)∩Ω|∇w·VN| ≥ Z
ˆ
AB
1IK=H1(K).
The last estimate clearly implies (26) and completes the proof of Lemma14.
7.3. Transverse argument. We assumed that there exists a mapu0 which solves Problem (2).
We investigate the following dichotomy:
• u06≡0 inΩ\B∞;
• u0≡0 inΩ\B∞.
We are going to prove that both cases lead to a contradiction.
7.3.1. The case u06≡0 inΩ\B∞. We thus have Z
Ω\B∞
|u0|>0. In this case, since(tr∂Ωu0)|∂Ω\∂B∞ ≡0, we have
(27) δ:=
Z
Ω\B∞
|Du0|>0.
Estimate (27) is a direct consequence of the following lemma applied on each connected components ofΩ\B∞. Lemma 16. [Weak Poincaré lemma] Let ω ⊂R2 be an open connected set. Assume that there exist x0 ∈∂ω andr >0 s.t. ω∩B(x0, r) is Lipschitz.
If u∈BV(ω)satisfies tr∂ω∩B(x0,r)= 0 andR
ω|Du|= 0thenu= 0.
Lemma16is proved in Appendix B.4.
Recall that we fixed a sequence(un)n ⊂C1∩W1,1(Ω)satisfying (21).
In particular, for sufficiently largen, we have Z
Ω\B∞
|∇un|>δ 2. Thus, from Lemma14and the fact that|V∞(x)|= 1 for a.e. x∈B∞,
Z
Ω|∇un| ≥ Z
B∞
|∇un·V∞|+ Z
Ω\B∞
|∇un| ≥H1(K) +δ 2. This implies
Z
Ω|Du0|= lim
n
Z
Ω|∇un| ≥H1(K) +δ 2 which is in contradiction with (19).
7.3.2. The case u0 ≡ 0 in Ω\B∞. We first note that, since tr∂D+
0u0 6≡ 0, there exists a triangle TkN0 s.t.
R
TkN0|u0|>0. We fix such a triangleTkN0 and we letαbe the vertex corresponding to the right angle.
We let R˜ = (α,e˜1,˜e2) be the direct orthonormal frame centered in α where ˜e2 = νCN0 k
[νCN0 k
is defined Remark6.1],i.e., the directions of the new frame are given by the side of the right-angle ofTkN0.
It is clear that forN ≥N0we haveVN ≡˜e2in
◦
TkN0.
By construction of B∞, TkN0∩B∞ is a union of segments parallel toe˜2, i.e. 1IB∞|TkN0(s, t)depends only on the first variable "s" in the frameR˜.
SinceR
TkN0|u0|>0, in the frameR˜, we may finda, b, c, d∈Rs.t., considering the rectangle (whose sides are parallel to the direction ofR˜)
P :={α+s˜e1+t˜e2; (s, t)∈[a, b]×[c, d]} ⊂TkN0 we have
Z
P|u0|>0.
Since from Lemma 10 the set B∞ has an empty interior [and that 1IB∞|TkN0(s, t) depends only on the first variable in the frameR˜], we may finda′ < b′ s.t.
• [a′, b′]×[c, d]⊂[a, b]×[c, d],
• S ∩B∞=∅withS :={α+s˜e1+t˜e2; (s, t)∈ {a′, b′} ×[c, d]}
• δ:=
Z
P′|u0|>0withP′:={α+s˜e1+t˜e2; (s, t)∈[a′, b′]×[c, d]}.
Moreover, sinceS andB∞are compact sets with empty intersection, we may findV, an open neighborhood of S s.t. dist(V, B∞)>0.
Noting thatu0≡0inΩ\B∞, from Lemma17[in AppendixA] it follows that for sufficiently largenwe have
• un≡0in S,
• Z
P′|un|> δ 2.
Consequently, from a standard Poincaré inequality Z
P′|∂˜e1un| ≥ 2 b′−a′
Z
P′|un|> δ
b′−a′ =:δ′. ThereforeR
P′|∂e˜1un|> δ′, R
P′|∂˜e2un| ≤2H1(K)and then by Lemma 3.3 in [ST14] we obtain:
Z
P′|∇un| ≥ Z
P′|∂˜e2un|+ δ′2 4H1(K) +δ′. Thus, from Lemma14, for sufficiently largen:
Z
Ω|∇un| ≥H1(K) + δ′2
4H1(K) +δ′ −on(1).
From the convergence inBV-norm ofun tou0we have Z
Ω|Du0| ≥H1(K) + δ′2 4H1(K) +δ′. Clearly this last assertion contradicts (19) and ends the proof of Theorem1.
Appendices
Appendix A. A smoothing result We first state a standard approximation lemma forBV-functions.
Lemma 17. Let Ω ⊂ R2 be a bounded Lipschitz open set and let u ∈ BV(Ω). There exists a sequence (un)n ⊂C1(Ω) s.t.
(1) un strictly
−→ uin the sense that un→uin L1(Ω)and Z
Ω|∇un| → Z
Ω|Du|, (2) tr∂Ωun= tr∂Ωu for alln,
(3) fork∈ {1,2}, Z
Ω|∂kun| → Z
Ω|Dku|:= sup ßZ
Ω
u∂kξ;ξ∈Cc1(Ω,R)and|ξ| ≤1
™
(4) If u= 0 outside a compact set L ⊂Ω and if ω is an open set s.t. dist(ω, L)> 0 then, for large n, un= 0in ω.
Proof. The first assertion is quite standard. It is for example proved in [AG78] [Theorem 1]. We present below the classical example of sequence for such approximation result [we follow the presentation of [Giu84], Theorem 1.17].
LetΩ⊂R2 be a bounded Lipschitz open set and letu∈BV(Ω).
Forn≥1, we letε= 1/n. We may fixm∈N∗ sufficiently large s.t. letting fork∈N Ωk =
ß
x∈Ω ; dist(x, ∂Ω)> 1 m+k
™
we have
Z
Ω\Ω0
|Du|< ε.
We fix nowA1:= Ω2 and fori∈N\ {0,1} we letAi = Ωi+1\Ωi−1. It is clear that(Ai)i≥1 is a covering ofΩ and that each point inΩbelongs to at most three of the sets(Ai)i≥1.
We let (ϕi)i≥1 be a partition of unity subordinate to the covering (Ai)i≥1, i.e., ϕi ∈ Cc∞(Ai),0 ≤ϕi ≤1 andP
i≥1ϕi = 1inΩ.
We letη ∈Cc∞(R2)be s.t. supp(η)⊂B(0,1),η≥0,R
η= 1and for x∈R2 η(x) =η(|x|). Fort >0we let ηt=t−2η(·/t).
As explained in [Giu84], fori≥1, we may chooseεi∈(0, ε)sufficiently small s.t.
supp(ηεi∗(uϕi))⊂Ai
Z
Ω|ηεi∗(uϕi)−uϕi|< ε 2i Z
Ω|ηεi∗(u∇ϕi)−u∇ϕi|< ε 2i
.
Here∗ is the convolution operator.
Define
un:=X
i≥1
ηεi∗(uϕi).
In some neighborhood of each pointx∈Ωthere are only finitely many nonzero terms in the sum defining un. Thusun is well defined and smooth inΩ.
Moreover, we may easily check that kun−ukL1(Ω)+
Z
Ω|Du| − Z
Ω|∇un|
< ε[hereε= 1/n].
Thus the previous estimate proves that(un)satisfies the first assertion,i.e,un strictly
−→ u.
As claimed in [Giu84] [Remark 2.12] we havetr∂Ωun= tr∂Ωufor alln. Thus the second assertion is satisfied.
We now prove the third assertion. Since un → u in L1(Ω), by inferior semi continuity we easily get for k∈ {1,2}
Z
Ω|Dku| ≤lim inf
n→∞
Z
Ω|∂kun|. We now prove
Z
Ω|Dku| ≥lim sup
n→∞
Z
Ω|∂kun|.
Letξ∈Cc1(Ω,R)with|ξ| ≤1. Sinceη is a symmetric mollifier andPϕi= 1we have
Z
Ω
un∂kξ = X
i≥1
Z
Ω
ηεi∗(uϕi)∂kξ
= X
i≥1
Z
Ω
uϕi∂k(ηεi∗ξ)
= X
i≥1
Z
Ω
u∂k[ϕi(ηεi∗ξ)]−X
i≥1
Z
Ω
u∂kϕi(ηεi∗ξ)
= X
i≥1
Z
Ω
u∂k[ϕi(ηεi∗ξ)]−X
i≥1
Z
Ω
ξ[ηεi∗(u∂kϕi)−u∂kϕi].
On the one hand we have [note thatϕi(ηεi∗ξ)∈Cc1(Ai)and|ϕi(ηεi∗ξ)| ≤1]
X
i≥1
Z
Ω
u∂k[ϕi(ηεi∗ξ)]
= Z
A1
u∂k[ϕi(ηεi∗ξ)] +X
i≥2
Z
Ai
u∂k[ϕi(ηεi∗ξ)]
≤ Z
Ω|Dku|+X
i≥2
Z
Ai
|Dku|
≤ Z
Ω|Dku|+ 3 Z
Ω\Ω0
|Dku|
≤ Z
Ω|Dku|+ 3ε.
Here we used that each point inΩbelongs to at most three of the sets(Ai)i≥1, fori≥2we haveAi⊂Ω\Ω0
and
Z
Ω\Ω0
|Dku| ≤ Z
Ω\Ω0
|Du|< ε.
On the other hand, since fori≥1 Z
Ω|ηεi∗(u∇ϕi)−u∇ϕi|< ε
2i, we get
X
i≥1
Z
Ω
ξ[ηεi∗(u∂kϕi)−u∂kϕi]
≤X
i≥1
Z
Ω|ηεi∗(u∂kϕi)−u∂kϕi|< ε.
Consequently
sup ßZ
Ω
un∂kξ;ξ∈Cc1(Ω,R)and|ξ| ≤1
™
= Z
Ω|∂kun| ≤ Z
Ω|Dku|+ 4ε and thuslim sup
n
Z
Ω|∂kun| ≤ Z
Ω|Dku|.This inequality in conjunction withlim inf
n
Z
Ω|∂kun| ≥ Z
Ω|Dku| proves the third assertion of Lemma17.
The last assertion of Lemma17is a direct consequence of the definition of theun’s.
Appendix B. Proofs of Lemma 2, Lemma 3, Lemma 4 and Lemma16 B.1. Proof of Lemma 2. Letu∈BV(Q). We prove that
Z
Q|D2u| ≥ Z 1
0 |tr∂Qu(·,0)−tr∂Qu(·,1)|. From Lemma17, there exists(un)n ⊂C1(Q)s.t. tr∂Qun = tr∂Qu,un
strictly
−→ uand Z
Q|∂2un| → Z
Q|D2u|.
