BEST CONSTANT IN POINCARÉ INEQUALITIES WITH TRACES: A FREE DISCONTINUITY APPROACH

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1 BEST CONSTANT IN POINCARÉ INEQUALITIES WITH TRACES: A FREE DISCONTINUITY APPROACH DORIN BUCUR, ALESSANDRO GIACOMINI, AND PAOLA TREBESCHI Abstract For Ω R N oen bounded and with a Lischitz boundary, and < +, we consider the Poincaré inequality with trace term C (Ω) u L (Ω) u L (Ω;R N ) + u L ( Ω) on the Sobolev sace W, (Ω) We show that among all domains Ω with rescribed volume, the constant is minimal on balls The roof is based on the analysis of a free discontinuity roblem Keywords: Free discontinuity roblems, functions of bounded variation, sets of finite erimeter, Robin boundary conditions 200 Mathematics Subject Classification: 49Q0, 26A45, 35R35, 35J65, 35J9,49K20 Contents Introduction 2 Notation and reliminaries 4 2 A numerical inequality 4 22 Functions of bounded variation and sets of finite erimeter 5 23 Almost quasi-minimizer of the Mumford-Shah functional 6 24 Symmetrization techniques 7 3 The sace SBV (R N ) 7 4 The free discontinuity roblem 0 4 First roerties of minimizers 0 42 Regularity roerties of the minimizers 4 43 Minimizers are suorted on balls 8 5 Proof of the main result 23 References 25 Introduction Let Ω R N be an oen, bounded set with Lischitz boundary Given [, + [, an equivalent norm on the Sobolev sace W, (Ω) is given by u L (Ω;R N ) + u L ( Ω) As a consequence, there exists a maximal constant C (Ω) > 0 such that () C (Ω) u L (Ω) u L (Ω;R N ) + u L ( Ω) for every u W, (Ω) Inequality () can be seen as a Poincaré inequality with trace term The main result of the aer states that balls are the sets which minimize the constant in () among domains with a given volume Theorem (The main result) Let [, + [ Then for every oen, bounded set with Lischitz boundary Ω R N we have C (B) C (Ω), where B R N is a ball such that B = Ω Moreover equality holds if and only if Ω is a ball

2 2 D BUCUR, A GIACOMINI, AND P TREBESCHI Essentially the whole aer is concerned with the range < < + The case = is well known in the literature since C (Ω) is recisely the Cheeger constant of the set Ω In this case, the roof of Theorem comes by symmetrization We shall comment this issue in last section of the aer, Remark 5 For >, no symmetrization argument is known to work In order to describe our aroach, let us comment the articular case = 2 Moreover, we concentrate on a variant of () given by C 2 (Ω) u 2 L 2 (Ω) u 2 L 2 (Ω;R N ) + u 2 L 2 ( Ω) for every u W,2 (Ω) We get easily that the maximal constant is given by Ω C 2 (Ω) = min u 2 dx + Ω u2 dh N, u W,2 (Ω),u 0 Ω u2 dx so that C 2 (Ω) coincides with the first eigenvalue of the Robin-Lalace oerator on Ω with constant β = : more recisely we deduce that C 2 (Ω) = λ R,(Ω), where for β > 0 the quantity λ R,β (Ω) is characterized by the existence of a nontrivial function u W,2 (Ω) such that u = λ R,(Ω)u in Ω u ν + βu = 0 on Ω u 0 in Ω, where ν denotes the outer normal to the boundary The Robin conditions u ν + βu = 0 on Ω are associated to the resence of the boundary term in the Rayleigh quotient: they are somehow intermediate between the Neumann conditions (β = 0) and the Dirichlet conditions (obtained formally for β ) The otimality of the ball for the constant C 2 is a consequence of the Faber-Krahn inequality for the Robin-Lalacian, ie, (2) λ R,β(B) λ R,β(Ω), where B is a ball such that B = Ω (the equality holds only if Ω is itself a ball) This inequality has been roved by Bossel [2] (for two dimensional simly connected smooth domains) and Daners [3] (in the N-dimensional setting, under Lischitz regularity for the boundary) Their roof, by a dearrangement rocedure, involves a direct comarison between Ω and B based not on the Rayleigh quotient reresentation for λ R,β (Ω), but on a different one which involves a different quantity, named the H-functional Coming back to the kind of inequality we are interested in, if we consider C 2 (Ω) u L2 (Ω) u L2 (Ω;R N ) + u L2 ( Ω), we see that the new constant C 2 (Ω) is given now by (3) C 2 (Ω) = min u W,2 (Ω),u 0 u L 2 (Ω;R N ) + u L 2 ( Ω) u L2 (Ω) Even in this case we have a connection with the Robin-Lalacian: if the minimum in (3) is attained on a nonconstant function u W,2 (Ω), which we may assume to be nonnegative, then by exloiting its otimality we deduce that u is the first eigenfunction for the Robin-Lalacian with constant and moreover β u := u L 2 (Ω;R N ), u L 2 ( Ω) u L 2 (Ω) C 2 (Ω) = λ R,β u u (Ω) L 2 (Ω;R N )

3 BEST CONSTANT IN POINCARÉ INEQUALITIES WITH TRACES: A FREE DISCONTINUITY APPROACH 3 This reresentation shows that the link with the Robin-Lalacian is too weak to infer the otimality of the ball in Theorem from the Faber-Krahn inequality (2) Moreover, the aroach by Bossel and Daners cannot be alied to the characteristic value (3), since the Rayleigh quotient involved is now nonlinear (sums of norms are involved), and no analogue of the H-functional is known in this situation In order to rove Theorem in the case >, we follow the strategy roosed in [5] and [6] to deal with the Faber-Krahn inequality for the Robin-Lalacian, and based on the analysis of free discontinuity functionals More recisely, for ], + [, we concentrate on the free discontinuity functional u (R dx ) ( / ) / + N J u [(u + ) + (u ) ] dh N F (u) := defined on the set of functions (R N u dx ) / SBV (R N ) := {u L (R N ) : u SBV (R N ), u 0} Here SBV denotes the sace of secial functions of bounded variation (see [] and Subsection 22) The basic remark, which leads to the study of the functional F, and which was the motivation for [5] and [6], is that if u 0 is an otimal function for (3), then its extension to zero outside Ω is such that u Ω SBV (R N ) with C (Ω) = F (u Ω ) This observation, leads to the following natural inequality Given some constant m > 0, inf{c (Ω) : Ω oen, bounded, Lischitz, Ω = m} inf{f (u) : u SBV (R N ), {u > 0} = m} Now, if we rove that the infimum in the right hand side is attained by a function u which is the extension by 0 of a minimizer for (3) on a ball (of volume m), then we achieve the roof of Theorem and, even more, we rovide a Poincaré inequality in SBV (R N ), with an otimal constant Following the strategy of [6], the roof of Theorem is thus obtained by showing that minimizers of F, under a volume constraint for the suort, are functions suorted on balls Our analysis, shaed after [6], can be summarized as follows (a) We focus on the roblem inf{f (u) : u SBV (R N ), {u > 0} = m}, and rove existence of a solution A regularity argument of toological tye à la De Giorgi, Carriero and Leaci (see Subsection 42) shows that every minimizer u of F (under a volume constraint) is such that H N (J u ) < +, H N (J u \ J u ) = 0, and the associated suort is given by an oen connected set Ω with Ω = J u In articular the boundary of Ω is an hyersurface in the weak sense of geometric measure theory, and Ω turns out to have finite erimeter (see Subsection 22) (b) By means of a reflection technique (Proosition 40), it is shown that F admits minimizers of the form ψ Ω, where Ω is symmetric around the origin and ψ :]0, + [ R is smooth, radial symmetric, ositive and bounded from above and below on Ω This yields that H N ( Ω \ Ω) = 0,

4 4 D BUCUR, A GIACOMINI, AND P TREBESCHI where Ω is the reduced boundary of Ω (see Subsection 22), and (Ω F (ψ Ω ) = ψ ( x ) dx ) / + ( Ω ψ ( x ) dx ) / (Ω ψ ( x ) dx ) / We obtain thus a candidate otimal shae, on which the functional F gains a geometrical flavor, with ψ (and its gradient) acting as volume and surface densities on Ω and its boundary (c) It is shown (Proosition 43) that the radial symmetry of ψ entails that also Ω has a circular symmetry, being either a ball or an annulus A direct comarison shows that the annulus is not otimal, which yields that minimizers are suorted on balls Along with the new form of the Poincaré inequality we deal with, the main technical novelties of the resent aer concerning the revious analysis are the following () The circular symmetry of the otimal domain Ω in oint (c) is obtained by making use of the sherical ca symmetrization technique, taking advantage of the radial symmetry of ψ and of the geometric flavor of the roblem mentioned in oint (b) This aroach yields a simlified roof of the otimality of the ball also for the semilinear variants of the first eigenvalue of the Robin-Lalacian studied in [6]: in the resent context, it roves to be an efficient tool to coe with the nonlinear structure of F, involving sums of norms (2) The structure of F and the resence of a general exonent entail some technical difficulties, esecially when dealing with the regularity analysis of oint (a) (see in articular Theorem 45 and Theorem 47 where technical maniulations are needed to get rid of the norms) The uniqueness issue is settled in Theorem 44, by exloiting some equality cases in a chain of inequalities which are at the core of the reflection argument mentioned in oint (b) We conclude this introduction by remarking that Sobolev inequalities with trace terms (raised at a suitable exonent) have been treated in [7] using mass transortation techniques, and showing a suitable otimality for the ball As an extension, a Poincaré tye inequality has been derived in [6], involving L norms for the functions and its trace, and L norm for the gradient, again roving an otimality roerty for the ball It is worth also to notice the result of [3], where it is roved that the ball may not be otimal, at least for some choices of norms The aer is organized as follows In Section 2 we introduce the notation and recall some basic roerties of functions of bounded variation and sets of finite erimeter emloyed in the subsequent analysis In Section 3 we define the functional sace SBV (R N ), recalling the associated comactness and lower semicontinuity roerties Section 4 is devoted to the analysis of the free discontinuity functional F and of its minimizers, along the lines described above in oints (a), (b) and (c) Finally the roof of Theorem is carried out in Section 5 2 Notation and reliminaries Throughout the aer, B r (x) will denote the oen ball of center x R N and radius r > 0 We say that A B if Ā is comact and contained in B If E RN, we will denote its volume by E, its comlement by E c, and E will stand for its characteristic function, ie, E (x) = if x E and E (x) = 0 if x E We set ω N := B (0) Moreover H N will stand for the (N )-dimensional Hausdorff measure, which coincides with the usual area measure on regular hyersurfaces For A R N oen set and, L (A) will denote the usual Lebesgue sace of -summable functions, while W, (A) will denote the Sobolev sace of functions in L (A) whose gradient in the sense of distributions is -summable Moreover u will stand for the su-norm of u, while su(u) will denote the set {u 0}, well defined u to sets of negligible Lebesgue measure 2 A numerical inequality The following inequality will be fundamental for our analysis

5 BEST CONSTANT IN POINCARÉ INEQUALITIES WITH TRACES: A FREE DISCONTINUITY APPROACH 5 Lemma 2 Let ], + [ Then for every a, a 2, b, b 2 0 and c, c 2 > 0 we have { } (a + a 2 ) / + (b + b 2 ) / a / + b / (2) min, a/ 2 + b / 2 (c + c 2 ) / c / c / 2 Moreover, if equality holds, then (a + a 2 ) / + (b + b 2 ) / = a/ + b / (c + c 2 ) / c / Proof By contradiction, let us assume that In articular we get which gives (a + a 2 ) / + (b + b 2 ) / (c + c 2 ) / < min { a / + b / c / (a + a 2 ) / + (b + b 2 ) / (c + c 2 ) / < a/ (22) c ((a + a 2 ) / + (b + b 2 ) /) < (c + c 2 ) Similarly we get (23) c 2 ((a + a 2 ) / + (b + b 2 ) /) < (c + c 2 ) = a/ + b / c / 2 + b / 2 c / 2 }, a/ 2 + b / 2 c / 2, ( ) a / + b / ( ) a / 2 + b / 2 Summing (22) and (23) we get ((a + a 2 ) / + (b + b 2 ) /) ( ) ( ) < a / + b / + a / 2 + b / 2 Choosing we get a i := A i, b i := B i, i =, 2, (24) (A + A 2 )/ + (B + B 2 )/ < ((A + B ) + (A 2 + B 2 ) ) /, which is against the triangle inequality of the -norm on R 2 Let us assume now that { a / (a + a 2 ) / + (b + b 2 ) / (c + c 2 ) / = min Then we necessarily have { a / + b / min c / }, a/ 2 + b / 2 c / 2 + b / c / = a/ + b / c / }, a/ 2 + b / 2 c / 2 = a/ 2 + b / 2, because otherwise one of the relations (22) and (23) would become an equality, the other remaining a strict inequality, which yields again to (24), a contradiction The roof is thus concluded 22 Functions of bounded variation and sets of finite erimeter Let Ω R N be an oen set We say that u BV (Ω) if u L (Ω) and its derivative in the sense of distributions is a finite Radon measure on Ω, ie, Du M b (Ω; R N ) BV (Ω) is called the sace of functions of bounded variation on Ω BV (Ω) is a Banach sace under the norm u BV (Ω) := u L (Ω)+ Du Mb (Ω;R N ) We refer the reader to [] for an exhaustive treatment of the sace BV Concerning the fine roerties, a function u BV (Ω) (or better every reresentative of u) is ae aroximately differentiable on Ω (see [, Definition 370]), with aroximate gradient u L (Ω; R N ) Moreover, the jum set J u is a H N -countably rectifiable set, ie, J u i N M i c / 2

6 6 D BUCUR, A GIACOMINI, AND P TREBESCHI u to a H N -negligible set, with M i a C -hyersurface in R N The measure Du admits the following reresentation for every Borel set B Ω: Du(B) = u dx + (u + u )ν u dh N + D c u(b), B J u B where ν u (x) is the normal to J u at x, and D c u is singular with resect to the Lebesgue measure and concentrated outside J u D c u is usually referred to as the Cantor art of Du u ± are the uer and lower aroximate limits of u at x The normal ν u coincides H N -ae on J u with the normal to the hyersurfaces M i The direction of ν u (x) is chosen in such a way that u ± (x) is the aroximate limit of u at x on the sets {y R N : ν u (x) (y x) 0} Moreover, u ± coincide H N -almost everywhere on J u with the traces γ ± (u) of u on J u which are defined by the following Lebesgue-tye limit quotient relation lim r 0 r N B ± r (x) u(x) γ ± (u)(x) dx = 0 where B ± r (x) := {y B r (x) : ν u (x) (y x) 0} (see [, Remark 379]) The sace of secial functions of bounded variation on Ω is defined as SBV (Ω) := {u BV (Ω) : D c u = 0} Such a sace roved to be very useful in the study of free discontinuity roblems arising in different contexts, like for examle image segmentation or fracture mechanics Given E R N measurable, we say that E has finite erimeter if { } P er(e; R N ) := su div(ϕ) dx : ϕ Cc (R N ; R N ), ϕ < + E If E < +, then E has finite erimeter if and only if E BV (R N ) It turns out that D E = ν E H N E, P er(e; R N ) = H N ( E), where E is called the reduced boundary of E, and ν E is the associated inner aroximate normal (see [, Section 35]) We have that E is H N -countably rectifiable and it is contained in the toological boundary E Moreover, the oints in E have density /2 with resect to E, with H N ( e E \ E) = 0, where e E (the essential boundary) is the set of oints whose density with resect to E is neither zero nor one 23 Almost quasi-minimizer of the Mumford-Shah functional In Section 4, we will use the notion of almost quasi minimality for SBV functions with resect to Mumford-Shah tye functionals Definition 22 (Almost quasi-mimimality) Let Ω R N be oen and < < + We say that u SBV (Ω) is an almost quasi-minimizer for the Mumford-Shah functional with exonent if there exist r 0, c, c 2, c 3 > 0 such that for every r < r 0, x 0 Ω and v SBV loc (Ω) with {v u} B r (x 0 ) Ω we have u dx + c H N (J u B r (x 0 )) v dx + c 2 H N (J v B r (x 0 )) + c 3 r N The revious notion is a variant of the minimality roerty emloyed by De Giorgi, Carriero and Leaci [4] to study regularity roerties of minimizers of the Mumford-Shah functional, the main difference lying in the fact that different constants aear in front of the surface terms The analysis of [4] can be extended to cover also this (slightly) more general setting (see [20]), yielding the following result for which we refer to [7, Theorem 23] Theorem 23 Let Ω R N be oen and let u SBV loc (Ω) satisfy the minimality roerty of Definition 22 Then the jum set of u is essentially closed in Ω, ie, H N ( (J u \ J u ) Ω ) = 0

7 BEST CONSTANT IN POINCARÉ INEQUALITIES WITH TRACES: A FREE DISCONTINUITY APPROACH 7 24 Symmetrization techniques In Section 43 we will emloy some basic roerties of the radial symmetric decreasing rearrangement for functions and of the sherical ca symmetrization of sets We recall here their definitions and the basic roerties we will emloy: we refer the reader to eg [2] and [8, Section 92] for further details (a) Radial symmetric decreasing rearrangement of a function Let Ω R N be an oen set and let u be a measurable nonnegative function defined on Ω If B is a ball centered at the origin with B = Ω, we define the radial symmetric decreasing rearrangement of u as the radial function u defined on B such that for every c > 0 (25) (26) It turns out that {u c} = {u c} B (u ) dx = Ω u dx for every [, + [ Moreover, if in addition u W, 0 (Ω), then u W, 0 (Ω) with u dx u dx B (b) Sherical ca symmetrization of a set Let E R N be a measurable set For every shere B r (0), let C r be the sherical ca centered at (0, 0,, r) such that H N (C r ) = H N (E B r (0)) Ω The sherical ca symmetrization of E is given by Ẽ RN such that B r (0) Ẽ = C r for every r > 0 It turns out that (see eg [8, Section 92], [8, Remark 4] or [9, Section 6]) if E has finite erimeter, also Ẽ is of finite erimeter, and for every radial ositive measurable function g(r) g(r) dh N g(r) dh N Ẽ E The revious inequality states that any erimeter with radial density decreases by sherical ca symmetrization: this roerty is reminiscent of the more usual one regarding the Schwartz symmetrization across an hyerlane Remark 24 Following [2, Lemma ], if u is a Lischitz continuous function on Ω oen and bounded, it turns out that u is also Lischitz continuous on B Moreover, the inequality (25), based on the use of the coarea formula and of the isoerimetric inequality, still holds true rovided that for almost every c > 0 H N ({u = c} Ω) H N ({u = c}), ie, an isoerimetric control is available for the inner boundaries of uer levels We will use this roerty in the final ste of the roof of Proosition 43 for a smooth function defined on an annulus 3 The sace SBV (R N ) In this section we introduce a suitable sace of functions of bounded variation tye which will be fundamental for our analysis Given ], + [ we set SBV (R N ) := {u L (R N ) : u SBV (R N ), u 0} In the case = 2, the sace has been introduced in [5] to study the Faber-Krahn inequality for the first eigenvalue of the Robin-Lalacian, and it has used in [6] to address some related semilinear variants including the case of the torsional rigidity In the following lemma we detail some basic roerties of elements in SBV (R N ): the roof follows closely [5, Lemma ] and will not be given Lemma 3 Let u SBV (R N ) Then the following items hold true

8 8 D BUCUR, A GIACOMINI, AND P TREBESCHI (a) u is ae aroximately differentiable (see [, Definition 370]) with aroximate gradient u such that (u ) = u u ae in R N (b) The jum set J u is H N -countably rectifiable and a normal ν u can be chosen in such a way that the jum art of the derivative is given by D j (u ) = [(u + ) (u ) ] ν u H N J u (c) For every ε > 0 and Ω R N oen and bounded we have (u ε) + SBV (Ω) The following comactness and lower semicontinuity roerties are a straightforward variant of [5, Theorem 2] Theorem 32 Let (u n ) n N be a sequence in SBV (R N ) and let C > 0 be such that for every n N u n dx + R N [(u + n ) + (u n ) ] dh N + J un u n dx C R N Then there exist u SBV (R N ) and a subsequence (u nk ) k N such that the following items hold true (a) Comactness: u nk u strongly in L loc (RN ) (b) Lower semicontinuity: for every oen set A R N we have u dx lim inf u nk dx, k and J u A A [(u + ) + (u ) ] dh N lim inf k A J unk A [(u + n k ) + (u n k ) ] dh N In the subsequent sections, we will make use of the following inequality Proosition 33 Given m > 0, there exists λ (m) > 0 such that for every u SBV (R N ) with su(u) m u dx + R N [(u + ) + (u ) ] dh N λ (m) J u u dx R N Moreover, for m we have (3) λ (m) m /N λ () Proof Let us set λ (m) := inf u SBV (R N ),u 0 su(u) m u dx + R N J u [(u + ) + (u ) ] dh N, u R dx N and let us check that λ (m) > 0 By contradiction assume that there exists u n SBV (R N ), u n 0, with su(u n ) m, (32) u ndx =, R N and such that u n dx + [(u + n ) + (u n ) ] dh N 0 R N J un

9 BEST CONSTANT IN POINCARÉ INEQUALITIES WITH TRACES: A FREE DISCONTINUITY APPROACH 9 By assumtion we have that u n SBV (R N ): by emloying the embedding BV (R N ) L N N (R N ), Hölder s and Young s inequalities, for every ε > 0 we can find c ε > 0 such that C N ( u N N n R N ( ( ε ) N N dx Du n (R N ) un u n dx + R N [(u + n ) + (u n ) ] dh N J un ) ( ) + { ε u n dx u n dx R N R N } u n dx + c ε u n dx R N R N u N N n R N ) N N su (un ) N + cε [(u + n ) + (u n ) ] dh N J un + [(u + n ) + (u n ) ] dh N Ju n R N u n dx + Ju n [(u + n ) + (u n ) ] dh N Letting ε be sufficiently small we can absorb the first integral of the right-hand side in the left-hand side to get for some C ε > 0 But then ( u N N n dx R N ) N ( ) N Cε u n dx + [(u + n ) + (u n ) ] dh N 0 R N Ju n ( u n dx u N N n dx R N R N ) N N su(un ) N 0 which is against (32) We conclude thus that λ (m) > 0 Let us rove inequality (3) Let u SBV (R N ) with su(u) m Setting v(x) := u(tx) we obtain that su(v) = su(u) /t N If we choose t := m /N, v is an admissible function to comute λ () so that If m we get so that inequality (3) easily follows λ () t u dx + t R N J u [(u + ) + (u ) ]dh N u R dx N u dx + λ () m /N R N J u [(u + ) + (u ) ]dh N, u R dx N Remark 34 It turns out that λ (m) is equal to the first eigenvalue of the -Lalace oerator under Robin boundary conditions with constant β = on a ball B with B = m For details, we refer the reader to Remark 52 at the end of the aer We conclude the section recalling the following result (see [7, Theorem 35 and Remark 37]), which is crucial for our analysis, which will be used in the roof of Theorem 45 Proosition 35 Given u SBV (R N ), assume that there exist ε 0, c, c 2 > 0 such that for ae 0 < δ < ε < ε 0 we have u dx + c δ H N ( {δ < u < ε} J u ) c 2 ε H N ( {u > ε} \ J u ) {u ε} Then there exists α > 0 such that u α ae on su(u)

10 0 D BUCUR, A GIACOMINI, AND P TREBESCHI 4 The free discontinuity roblem Given ], + [ and m > 0, we concentrate in this section on the free discontinuity roblem (4) min u SBV (R N ),u 0 su(u) m F (u) where SBV (R N ) is the sace introduced in Section 3, while F is the free discontinuity functional F (u) := u + E s (u) / u with E s (u) := [(u + ) + (u ) ] dh N J u Existence of minimizers can be roved by erforming a concentration comactness alternative as in [5, Theorem 4] Theorem 4 (Existence of minimizers) The minimum roblem (4) admits a solution Proof It is sufficient to adat the roof of [5, Theorem 4] using the comactness and lower semicontinuity roerties given in Theorem 32 and emloying the numerical inequality of Lemma 2 (to exclude the dichotomy case) Remark 42 Notice that if u is a minimizer of roblem (4), then su(u) = m This is a consequence of the following simle rescaling roerty: for every t min v SBV (R N ),v 0 su(v) tm F (v) t N min u SBV (R N ),u 0 su(u) m F (u) 4 First roerties of minimizers This subsection is devoted to the roof of some ivotal roerties of minimizers In articular we are interested in bounds from above and below (on the suort) Let us start with the bound from above Theorem 43 (L -bound) Let u be a minimizer of (4) Then u L (R N ) Proof It is not restrictive to assume u L (R N ) = Let us assume by contradiction that u / L (R N ) We divide the roof in several stes Ste Assume that u 0 and J u By exloiting the Euler-Lagrange equation satisfied by u, for every ϕ SBV (R N ) such that J ϕ J u we have u 2 u ϕ dx R (42) ( N u R dx ) J + u [(u + ) γ (ϕ) + (u ) γ 2 (ϕ)] dh N ( ) N [(u+ ) + (u ) ] dh N Ju = F (u) u ϕ dx, R N where γ (ϕ) and γ 2 (ϕ) are the traces of ϕ on the rectifiable set J u oriented by the normal ν u By multilying equation (42) with ( ) ( ) u dx + [(u + ) + (u ) ] dh N, R N J u we obtain the inequality (43) u 2 u ϕ dx+ [(u + ) γ (ϕ)+(u ) γ 2 (ϕ)] dh N 2F (u) u ϕ dx R N J u R N

11 BEST CONSTANT IN POINCARÉ INEQUALITIES WITH TRACES: A FREE DISCONTINUITY APPROACH Notice that inequality (43) still holds true if either J u =, u 0 or J u, u 0, in the last case requiring that also ϕ 0 Ste 2 Let us consider for every M > 0 the function u M = (u M) + SBV (R N ) Recalling that u M = u {u M} and J um J u, u M turns out to be an admissible test function for inequality (43), so that we infer (44) u M dx + R N [(u + ) u + M + (u ) u M ] dhn 2F (u) J u u u M dx R N Notice that [(u + ) u + M + (u ) u M ] dhn J u [(u + M ) + (u M ) ]dh N J um and u u M dx = (u M + M) u M dx C (u R N R N {u M} R M + M u M ) dx N for some constant C > 0 From (44) we conclude (45) u M dx + [(u + M ) + (u M ) ]dh N C 2 (u M + M u M )dx R N J um R N for some C 2 > 0 Ste 3 Let Since we are assuming u L (R N ), we have α(m) := {u M} M > 0 : α(m) > 0 and lim α(m) = 0 M + In view of Proosition 33, inequality (45) entails λ (α(m)) u M dx C ( 2 u M + M ) u M dx R N R N By (3), for M sufficiently large we infer (46) u α(m) M dx C 3 M u M dx N R N R N for some C 3 > 0 By Holder inequality and (46) we deduce ( ) u M dx u M dx α(m) R N R N so that (47) ( ) C 3 M α(m) N u M dx α(m) R N R N u M dx C 4 Mα(M) + N( ) for some C 4 > 0 Setting now g(m) = u M dx, R N and recalling that g (M) = α(m), we can rewrite inequality (47) as g(m) C 4 M( g (M)) + N( ) It follows that there exists M 0 > 0 such that for ae M M 0 M γ C g (M) 5 g(m) γ

12 2 D BUCUR, A GIACOMINI, AND P TREBESCHI with γ < and C 5 > 0 Integrating between M 0 and M we get g(m) γ g(m 0 ) γ C 5 ( Being g(m) γ > 0, the above inequality entails in articular g(m 0 ) γ C 5 ( M γ M γ M γ M γ which yields a contradiction letting M + We thus conclude that u L (R N ), and the roof is concluded Let us now come to the bound from below on the suort of minimizers We need the following erturbation lemma Lemma 44 Let u be a minimizer of roblem (4) Then the following items hold true (a) There exist ε > 0 and k > 0 such that for every v SBV (R N ) with (48) su(u) < su(v) < su(u) + ε, then (49) F (u) + k su(u) F (v) + k su(v) (b) There exist ε > 0 and ˆk > 0 such that for every v SBV (R N ) with (40) su(u) ε < su(v) < su(u), then (4) F (u) + ˆk su(u) F (v) + ˆk su(v) Proof The roof is very similar to that of [6, Lemma 62] Let us start with oint (a) By contradiction, let us assume that for every ε > 0 and k > 0 there exists v SBV (R N ) satisfying (48) but for which (49) is violated Let us consider ε n 0 and k n + and let us denote by v n the associate function such that and 0 0 ), su(u) < su(v n ) < su(u) + ε n, (42) F (u) + k n su(u) > F (v n ) + k n su(v n ) Let us set Then t n > and t n If we set we obtain t n := ( ) su(vn N ) su(u) w n (x) := v n (t n x) su(w n ) = su(v n) = su(u), which imlies (being w n admissible for roblem (4)) (43) F (u) F (w n ) = t n v n + t / n E s (v n ) / v n t n F (v n ) Since su(v n ) = t N n su(u), by using (42) and (43), we get so that t N n F (u) + k n su(u) > F (v n ) + k n su(v n ) t n F (u) + k n t N n su(u), k n t n F (u) t N n su(u) But the right hand side is bounded as n +, against k n + The roof of oint (a) is thus concluded )

13 BEST CONSTANT IN POINCARÉ INEQUALITIES WITH TRACES: A FREE DISCONTINUITY APPROACH3 Let us ass to oint (b) We roceed again by contradiction by considering ε n 0, ˆk n 0 and the associated v n SBV (R N ) satisfying (40) but violating (4) Reasoning as above we get F (u) t / n F (v n ) for t n, so that (44) ˆkn t / n F (u) t N n su(u), against ˆk n 0 Point (b) thus follows, and the roof is now comlete We are now in a osition to rove the following bound from below which is essential for our analysis Theorem 45 (Bound from below) Let u be a minimizer for (4) Then there exists α > 0 such that u α ae on su(u) Proof We can assume u L (R N ) = Moreover, thanks to Proosition 43, we have u L (R N ) Assume by contradiction that for every ε small enough (45) {u < ε} > 0 Notice that for ae ε > 0 we have u {u ε} SBV (R N ) In view of Lemma 44, there exists ˆk > 0 such that comaring u and u {u ε} (with ε small enough) we get (46) u + E s (u) + ˆk {u < ε} u {u ε} + E s (u {u ε} ) u {u ε} Moreover we may write u {u ε} = ( {u ε} u dx ) ( = {u<ε} u dx ) < + Cε {u < ε} for some C > 0 As a consequence, for ε small enough inequality (46) entails (47) u + E s (u) u{u ε} + E s (u {u ε} ) Assume that u 0 Recalling that for every a > b > 0 (48) a (a b) a / b / b (a b), we get u dx R u u {u ε} N {u ε} u dx {u<ε} u = u dx u, so that from (47) we deduce {u<ε} u dx u + E s (u) Es (u {u ε} ) Notice that in articular E s (u {u ε} ) E s (u), so that using again (48) we obtain {u<ε} u dx By setting u E s(u {u ε} ) E s (u) E s (u) β u := u E s (u)

14 4 D BUCUR, A GIACOMINI, AND P TREBESCHI we obtain, writing exlicitly E s (u) and E s (u {u ε} ), (49) u dx + β u [(u + ) + (u ) ] dh N {u<ε} J u β u (u + ) dh N + β u J u {u <ε u + } which yields u dx + β u {u<ε} J u {u <u + ε} J u {ε u <u + } [(u + ) + (u ) ] dh N + β u ε H N ( {u > ε} \ J u ), [(u ) + (u + ) ] dh N β u ε H N ( {u > ε} \ J u ) We deduce in articular that for ae 0 < δ < ε (420) u dx + β u δ H N ( {δ < u < ε} J u ) β u ε H N ( {u > ε} \ J u ) {u ε} From Proosition 35 we obtain that (45) cannot hold In the case u 0, inequality (420) still holds rovided that we choose β u := : again we reach a contradiction, so that the roof is concluded Remark 46 The bound from below can be established also by adating the arguments roosed in [9, Theorem 32], where a free discontinuity aroach similar to the resent one has been roosed to deal with free boundary roblems arising in thermal insulation 42 Regularity roerties of the minimizers In this subsection we show that a minimizer of roblem (4) satisfies a local minimality roerty for a Mumford-Shah functional with exonent : this yields some regularity roerties for the jum set and consequently for the suort Theorem 47 (Essential closedness of the jum set) Let u be a minimizer of roblem (4) Then u SBV (R N ) L (R N ), H N (J u ) < + and J u is essentially closed, ie, H N (J u \ J u ) = 0 Proof By Proosition 43 we know that u L (R N ), while Theorem 45 entails (42) u α > 0 ae on su(u) Since u SBV (R N ), the chain rule in BV (see [, Theorem 396] entails u SBV (R N ) Finally we have α H N (J u ) [(u + ) + (u ) ) dh N < +, J u so that H N (J u ) < + Notice moreover that H N (J u ) > 0 (since otherwise u W, (R N ) and (42) cannot hold) In order to conclude the roof, we need to show that J u is essentially closed We will show that u is an almost quasi-minimizer of the Mumford-Shah functional with exonent according to Definition 22, and then we conclude using Theorem 23 We divide the roof in two stes Ste Given r 0 > 0, let us consider v SBV loc (R N ) such that {v u} B r (x 0 ), where r < r 0 Let us assume that (422) v dx + 2 u H N (J v B r (x 0 )) u dx + H N (J u B r (x 0 )), and let us set Since ṽ := min{ v, u } su(ṽ) su(u) + ω N r N,

15 BEST CONSTANT IN POINCARÉ INEQUALITIES WITH TRACES: A FREE DISCONTINUITY APPROACH5 either by the minimality of u or by the erturbed minimality given by Lemma 44, we deduce that if r 0 is small enough, there exists k > 0 such that (423) u + E s (u) u In view of (422) and the definition of ṽ we have ṽ + E s (ṽ) + kω N r N ṽ (424) ṽ + E s (ṽ) C, where C deends only on u Assuming as usual u =, and since ( ṽ u ) dx 2 u ω N r N, u to reducing r 0 we have ṽ = Then, from (423), we infer ( + B ( ṽ r(x 0) u ) dx ) < + 4 u ω N r N (425) u + E s (u) ṽ + E s (ṽ) + C2 r N, where C 2 deends only on u We claim that there exist k, k 2, k 3 > 0 deending only on u such that (426) u dx + k [(u + ) + (u ) ] dh N J u ṽ dx + k 2 J v Thanks to (42) and in view of the very definition of ṽ we obtain (427) u dx + k α H N (J u B r (x 0 )) [(ṽ + ) + (ṽ ) ] dh N + k 3 r N v dx + 2k 2 u H N (J v B r (x 0 )) + k 3 r N U to reducing the constant k and increasing the constants k 2, k 3 (if necessary), we see that inequality (427) still holds even if v does not satisfy assumtion (422) We conclude that u is an almost quasi-minimizer of the Mumford-Shah functional with exonent according to Definition 22 In view of Theorem 23, the essential closedness of J u follows Ste 2 In order to conclude the roof, we need to show that claim (426) holds true U to reducing r 0, we can assume that for every x 0 R N [(u + ) + (u ) ] dh N 2 E s(u) > 0, (428) J u B c r 0 (x 0) B c r 0 (x 0) u dx 2 R N u dx Assume u 0 We may write [ ] u ṽ = c (u, ṽ) u dx ṽ dx R N R (429) [ N ] = c (u, ṽ) u dx ṽ dx,

16 6 D BUCUR, A GIACOMINI, AND P TREBESCHI where for a suitable 0 < θ < c (u, ṽ) := (θ u dx + ( θ ) ṽ dx R N R N Analogously (430) E s (ṽ) Es (u) = c2 (u, ṽ) (E s (ṽ) E s (u)), with c 2 (u, ṽ) := (θ 2E s (u) + ( θ 2 )E s (ṽ)) and 0 < θ 2 < Substituting (429) and (430) into (425) we get (43) u dx + c 2(u, ṽ) c (u, ṽ) E s(u) ) ṽ dx + c 2(u, ṽ) c (u, ṽ) E s(ṽ) + crn c (u, ṽ) In view of (428), (422) and (424), and recalling the very definition of ṽ, we may write ϑ u R dx + ( ϑ ) ṽ dx u dx u dx > 0 N R N Br c(x0) 2 R N and ϑ u R dx + ( ϑ ) ṽ dx ϑ u dx + ( ϑ ) v dx N R N R N R N = u dx + v dx C 3, R N where C 3 deends only on u Similarly we have θ 2 E s (u)+( θ 2 )E s (ṽ) [(u + ) +(u ) ] dh N [(u + ) +(u ) ] dh N > 0, J u Br c(x0) 2 J u and θ 2 E s (u) + ( θ 2 )E s (ṽ) [(u + ) + (u ) ] dh N + J u where C 4 deends only on u Combining the revious inequalities, we conclude that k c 2(u, ṽ) c (u, ṽ) k 2 Jṽ [(ṽ + ) + (ṽ ) ] dh N 2 u (H N (J u ) + H N (J v B r (x 0 ))) C 4, and c c (u, ṽ) k 3 for suitable constants k i deending only on u, so that inequality (43) entails claim (426) Assume now u 0 Coming back to (425), we can raise the inequality to the ower getting for some constants ˆk, ˆk 2 deending only on E s (u) ˆk ṽ dx + ˆk E s (ṽ) + ˆk 2 r N, from which we deduce again claim (426) (u to reducing r 0 if necessary) We are now in a osition to draw the main regularity roerties we need for minimizers of our free discontinuity roblem Theorem 48 (Regularity roerties of minimizers) Let u be a minimizer of roblem (4) The following items hold true (a) u SBV (R N ) L (R N ) with H N (J u ) < + and H N (J u \ J u ) = 0

17 BEST CONSTANT IN POINCARÉ INEQUALITIES WITH TRACES: A FREE DISCONTINUITY APPROACH7 (b) The suort of u is an oen and connected set Ω such that Ω = J u In articular Ω has finite erimeter in R N (c) The restriction of u to Ω is an element of W, (Ω) such that, either it is constant or (432) u = λu in Ω, for a suitable λ > 0 In articular u C,γ (Ω) for some 0 < γ <, and there exists α > 0 such that (433) u α on Ω Proof Point (a) follows by Theorem 47 Let us come to oint (b) Notice that u W, (R N \ J u ) Assume u 0 By exloiting the Euler-Lagrange equation for the functional F on R N \ J u, we get for every ϕ Cc (R N \ J u ) u 2 u ϕ dx = λ u ϕ dx, R N \J u R N \J u where By regularity (see eg [5]) we infer λ := u u F (u) u C,γ (R N \ J u ) for some 0 < γ < Let us decomose the oen set R N \ J u into its connected comonents, and select those on which u is not identically zero If we denote by Ω their union, it turns out Ω = J u so that H N ( Ω) = H N (J u ) = H N (J u ) < + In articular Ω has finite erimeter and equation (432) holds true Moreover, in view of the regularity of u and of the bound from below given by Theorem 45, we get easily that Ω is the suort of u, and that (433) is satisfied (indeed Ω cannot contain strictly the suort of u, otherwise the function should aroach continuously zero) Let us show that Ω is connected By contradiction, let us assume that Ω = Ω Ω 2 with Ω, Ω 2 oen sets such that Ω, Ω 2 and Ω Ω 2 = Note that Ω i has finite erimeter for i =, 2 since Ω i Ω Let us set u i := u Ωi i =, 2 Since u L (R N ), by [, Theorem 384] we get u i SBV (R N ) with Du i = Du Ω () i u Ω i ν Ωi H N Ω i, where Ω () i denotes the oint of density of Ω i, ν Ωi stands for the exterior normal to Ω i, while u Ω i denotes the trace on Ω i of u coming from Ω i Notice that su(u i ) = Ω i Moreover by construction the following additivity relation concerning the surface energy holds true: (434) [(u + ) + (u ) ] dh N = J u [(u + ) + (u ) ] dh N + J u [(u + 2 ) + (u 2 ) ] dh N J u2

18 8 D BUCUR, A GIACOMINI, AND P TREBESCHI By emloying the numerical inequality of Lemma 2, we may assume = (R N u dx ) / + ( J u [(u + ) + (u ) ] dh N ) / ( u R dx ) N u (R N dx + u R N 2 dx ) ( / + J u [(u + ) + (u ) ] dh N + ) / J u2 [(u + 2 ) + (u 2 ) ] dh N ( u R N dx + u R N 2 dx) u (R N dx ) ( / ) / + J u [(u + ) + (u ) ] dh N ( R N u dx) so that u is a minimizer of the free discontinuity roblem (4) with su(u ) < su(u) This is in contradiction with Remark 42 If u 0, we can follow the revious arguments and conclude that u is constant on Ω, so that the roof is concluded Remark 49 The roof of the connectedness of Ω deends heavily on the additivity relation (434): the roof for = 2, carried out in detail in [6, Theorem 65], extends readily to this case 43 Minimizers are suorted on balls In this subsection we want to show that minimizers of (4) are suorted on balls, and more recisely that they are of the form ψ B, where B is a ball of volume m, and ψ is a radial function with resect to the center of B Let us start with the following result Proosition 40 Let v be a minimizer of roblem (4) We can associate to v, by means of successive reflections across N orthogonal hyerlanes, a new minimizer u such that the following items hold true (a) Proerties of the suort The suort Ω of u is oen, connected, and such that (435) H N ( Ω \ Ω) = 0 U to a translation, we may assume that Ω is symmetric with resect to the origin (b) Radiality of the function There exists a function ψ : I ]0, + [ of class C, where I = [0, + [ or I =]0, + [, such that either ψ is constant or (436) ( r N ψ 2 ψ ) = λ ψ 2 ψr N, and, u to a translation, for every x Ω (437) u(x) = ψ( x ) Here λ > 0 is the constant aearing in equation (432) satisfied by v on its suort (c) We have (Ω (438) F (u) = ψ ( x ) dx ) / + ( Ω ψ ( x ) dx ) / (Ω ψ ( x ) dx ) / Proof We divide the roof in several stes Ste : existence of a symmetric minimizer Let us consider an hyerlane π arallel to x = 0 which slits the suort of v in two arts of equal measure, and let us set v := v π + and where π ± are the two half saces determined by π v 2 := v π,,

19 BEST CONSTANT IN POINCARÉ INEQUALITIES WITH TRACES: A FREE DISCONTINUITY APPROACH9 Note that the term F (v) [(v + ) + (v ) ] dh N J v π which (eventually) aears in the surface art of the free discontinuity functional F can be reinterreted as [γ(v ) + γ(v 2 ) ] dh N, J v π where γ(v ), γ(v 2 ) are the traces of v and v 2 on π By using the numerical inequality of Lemma 2 we may assume ( / ( v dx) 2 π + + J v π [(v + ) + (v ) ] dh N + ) + J v π γ(v ) dh N ( π + ) /, v dx so that by reflecting v across π we obtain a new minimizer of roblem (4) symmetric with resect to π, still denoted v We oerate in the same way on v by emloying an hyerlane π 2 arallel to x 2 = 0, and obtaining a new minimizer v 2 symmetric with resect to both π and π 2 Proceeding in this by considering hyerlanes arallel to x i = 0 for i = 3,, N, we end u with a minimizer v N =: u which turns out to be symmetric with resect to N orthogonal hyerlanes, whose intersection we may take as the new origin of R N Ste 2: radiality of the minimizer Let Ω R N oen and connected be the suort of u according to Theorem 48: we have that u is C,γ on Ω for some γ > 0 If x Ω, and π is a hyerlane through x and the origin, we can reflect again u across π obtaining a new minimizer û: indeed the arguments of Ste can be alied since by the symmetry roerties of u, the hyerlane π slits Ω in two arts of equal measure If ˆΩ is the associated suort, we have that x ˆΩ, and by the symmetry and the regularity of û we conclude that D ν u(x) = 0, where ν is orthogonal to π This means that u is locally radial Assume that u is not constant on Ω, and let x 0 Ω with r 0 := x 0 Let ψ : I R be the solution of ( ψ 2 ψ r N ) = λ ψ 2 ψr N with ψ(r 0 ) = u(x 0 ) and ψ (r 0 ) = r u(x 0 ), where λ is the constant aearing in the equation (432) satisfied by the nonconstant function u on Ω In view of [22, Section 3], there exists a unique solution for this roblem, for which either I = [0, + [ or I =]0, + [ Clearly u(x) = ψ( x ) locally near x 0 : but since Ω is connected, the equality extends to the entire Ω Point (b) is thus roved Ste 3: conclusion In order to comlete the roof of oint (a), we need to show that (435) holds true Let x Ω be a oint of density zero or one for Ω: since the function ψ is of class C, we deduce that x J u Since Ω = J u (thanks to Theorem 48), we deduce (recall that e Ω is the essential boundary of Ω, see Subsection 22) so that Ω \ Ω ( J u \ J u ) ( e Ω \ Ω), H N ( Ω \ Ω) H N ( J u \ J u ) + H N ( e Ω \ Ω) = 0 Point (a) is thus comletely roved Finally, oint (c) is an immediate consequence of the fact that u = ψ Ω, Ω = J u and H N ( Ω \ Ω) = 0, so that the roof is concluded

20 20 D BUCUR, A GIACOMINI, AND P TREBESCHI Remark 4 The revious roof shows that every minimizer v of roblem (4) generates new minimizers v, v 2,, v N := u, such that v i+ is obtained from v i through a reflection across the hyerlane x i = k which slits the suort of v i in two arts with the same volume: since the inequalities of tye (2) on which the argument is based are indeed equalities, both v i {xi<k} and v i {xi>k} generate an admissible v i+ Remark 42 The roof of Proosition 40 shows that given any minimizer v of (4) and any hyerlane π which slits the associated suort in two arts with the same volume, both the restrictions v π ± generate by reflection across π a new minimizer of the functional We now show that the suort of the symmetric minimizer constructed in Proosition 40 is indeed a ball Proosition 43 Let u be the minimizer of (4) given by Proosition 40 Then the associated suort Ω is a ball Proof If u = ψ Ω with ψ constant, then by (438) we deduce F (u) = HN ( Ω) / Ω / If B is a ball such that B = Ω, by considering the admissible function B we obtain H N ( Ω) / Ω / = F (u) F ( B ) = HN ( B) / B /, which yields, in view of the isoerimetric roerty of the ball, that Ω coincides u to negligible sets with a ball Assume that ψ is not constant on Ω We divide the roof in several stes Ste Let Ω be the sherical ca symmetrization of Ω (see Subsection 24 for the definition): such a set is oen and with finite erimeter By assing to olar coordinates, and using the fact that H N ( Ω B r (0)) = H N (Ω B r (0)), we deduce that ψ ( x ) dx = Ω Ω ψ ( x ) dx and ψ ( x ) dx = Ω Ω ψ ( x ) dx Thanks to inequality (26) on erimeters with radial densities alied to g = ψ, we get in view of (438) (439) ( Ω ψ ( x ) dx ) / + ( Ω ψ ( x ) dx ) / ( Ω ψ ( x ) dx ) / (Ω ψ ( x ) dx ) / + ( Ω ψ ( x ) dx ) / (Ω ψ ( x ) dx ) / = F (u) Notice that ψ W, ( Ω) L ( Ω) with (recall (433) and (437)) ψ α > 0 on Ω, for some α > 0 Thanks to [, Theorem 384] we get that the function ψ Ω belongs to SBV (R N ) L (R N ) and in articular it is thus an element of SBV / (R N ) Since ( Ω F (ψ Ω) = ψ ( x ) dx ) / + ( Ω ψ ( x ) dx ) / ( Ω ψ ( x ) dx ), / we deduce from (439) that also ψ Ω is a minimizer of (4) In articular Ω (which is the associated suort) is also connected

21 BEST CONSTANT IN POINCARÉ INEQUALITIES WITH TRACES: A FREE DISCONTINUITY APPROACH2 Ste 2 We claim that Ω has a sherical symmetry, that is, Ω is either a ball or an annulus In order to see this, in view of the structure of Ω (which was constructed through a sherical ca symmetrization), it suffices to check that Ω {x N > 0} = Ω {x N < 0} Indeed, if this holds true, we get that for ae r > 0 either B r (0) Ω or B r (0) Ω = In view of its connectedness, we deduce then that Ω is either a ball or an annulus Assume by contradiction that Ω {x N > 0} > Ω {x N < 0} Then we can find an hyerlane x N = k with k > 0 which slits Ω in two arts with the same volume According to Remark 42, one of the two arts generates by reflection a new minimizer ˆψ ˆΩ for the free discontinuity roblem (4) By construction, the suort ˆΩ is already symmetric with resect to the coordinate hyerlanes x i = 0 for i =,, N If we aly Proosition 40 to ˆψ ˆΩ, we deduce that ˆψ is radial with resect to the oint (0, 0,, 0, k) But this is imossible, since ˆψ coincides with a reflection of ψ which is radial about the origin and it is not constant thanks to equation (432) Ste 3 We claim that Ω is a ball In order to see this, we need to exclude that Ω = A r,r 2 := {x R N : r < x < r 2 } Assume by contradiction that u = ψ Ar,r 2 is a minimizer for roblem (4) Then, by exloiting the associated Euler-Lagrange equation, it turns out that ψ satisfies on A r,r 2 the boundary conditions of Robin tye ψ 2 (r )ψ (r ) + βψ (r ) = 0 and ψ 2 (r 2 )ψ (r 2 ) + βψ (r 2 ) = 0 for a suitable constant β > 0 This entails that ψ (r ) > 0 and ψ (r 2 ) < 0 Since from the differential equation (436) we get that r r N ψ (r) 2 ψ (r) is decreasing on [r, r 2 ], we deduce that there exists r 3 ]r, r 2 [ such that ψ is increasing on [r, r 3 ] and decreasing on [r 3, r 2 ] Two situations may haen, leading both to a contradiction (a) Assume ψ(r ) ψ(r 2 ) Let r 4 ]r, r 3 [ be such that ψ(r 4 ) = ψ(r 2 ) Let us consider the radial symmetric decreasing rearrangement ψ of ψ restricted to A r4,r 2 suorted on the ball B r (0) (see Subsection 24 for the definition), and let us extend it with the value ψ(r 2 ) on the larger ball Bˆr (0), where B r (0) = A r4,r 2 and Bˆr (0) = A r,r 2 Let us denote this function with ϕ We have ϕ dx ψ dx, ϕ dx > ψ dx, Bˆr (0) A r,r 2 Bˆr (0) A r,r 2 and ϕ dh N < Bˆr (0) ψ dh N A r,r 2 The inequality on the gradient comes from the general roerties of the radial symmetric decreasing rearrangement, together with the fact that ϕ is constant on Bˆr (0) \ B r (0) The second inequality comes instead from the equality ϕ dx = B r(0) ψ dx A r4,r 2

22 22 D BUCUR, A GIACOMINI, AND P TREBESCHI together with the fact that ψ ψ(r 2 ) on [r, r 4 ] Finally the inequality for the surface integral is a consequence of the fact that ψ dh N > ψ dh N > ψ (r 2 )H N ( Bˆr (0)) = ϕ dh N A r,r 2 B r2 (0) Bˆr (0) We conclude that ( ) / ( ) / ψ dx Ar,r2 + ψ dh N Ar,r2 F (ψ Ar,r 2 ) = ( ) / ψ dx Ar,r2 ( / ( ) / dx) Bˆr(0) ϕ + Bˆr(0) ϕ dh N > ( Bˆr(0) ϕ dx) / = F (ϕ Bˆr (0)), so that ψ Ar,r 2 is not a minimizer for roblem (4) (b) Let us assume that ψ(r ) > ψ(r 2 ) In this case we roceed directly with a radial symmetric decreasing rearrangement of ψ Ar,r 2 We obtain a new function ψ suorted on a ball B r (0) with B r (0) = A r,r 2 In view of the regularity of ψ on A r,r 2, by Remark 24 we get that ψ is Lischitz continuous on B r (0) Moreover, the geometry of ψ entails that for every a > ψ(r 2 ) H N ({ψ = a} A r,r 2 ) > H N ({ψ = a}) (notice that {ψ > a} is an annulus and {ψ > a} is a ball with equal volume, whose radius is therefore strictly less than the external radius of the former) Again by Remark 24, we deduce ψ dx B r(0) ψ dx A r,r 2 Finally, since (ψ ) dx = ψ dx and (ψ ) dh N < B r(0) A r,r 2 B r(0) ψ dh N, A r,r 2 we obtain as above F (ψ Ar,r 2 ) > F (ψ Br (0)), so that ψ Ar,r 2 is not a minimizer for roblem (4) We conclude the section with the following uniqueness result Theorem 44 Any minimizer of roblem (4) is of the form where B is a ball of volume m, and ψ W, (B) is radial with resect to the center of B ψ B Proof Let v be a minimizer of roblem (4) with associated suort Ω Let us aly Proosition 40 and Remark 4 to this minimizer, generating the family of minimizers v, v 2,, v N with associated suorts Ω,, Ω N By Proosition 43, we know that v N is suorted on a ball B, and radially symmetric with resect to the center inside We may assume that B is centered at the origin The function v N has been obtained from v N by means of a reflection across the hyerlane x N = 0 which slits Ω N in two arts Ω ± N of the same volume Moreover we know that both the restrictions of v N to Ω ± N generate an admissible v N We thus conclude that Ω N = B, and the associated function which is radial

23 BEST CONSTANT IN POINCARÉ INEQUALITIES WITH TRACES: A FREE DISCONTINUITY APPROACH23 In turn v N is obtained from v N 2 by means of a reflection across the hyerlane x N = 0 which slits Ω N 2 in two arts Ω ± N 2 of the same volume Again, both the restrictions of v N 2 to Ω ± N 2 generate an admissible v N Recalling that Ω N 2 is connected and symmetric with resect to the hyerlanes we conclude for examle that x = 0, x 2 = 0, x N 2 = 0, Ω + N 2 = B {x N < 0} and Ω N 2 = B {x N > 0}, where B is a ball of volume m with center on the line x = x 2 = = x N = 0 But in the rocedure of Proosition 40 we could exchange the role of the coordinates x N and x N If we symmetrize firstly with resect to a hyerlane arallel to x N = 0, we should get a minimizer suorted on a ball, and this is ossible only if B = B We conclude that Ω N 2 = B with associated function which is radial Proceeding in this way, we come back to Ω which is itself equal to B with associated function which is radial The roof is thus concluded 5 Proof of the main result We are now in a osition to rove our main theorem We consider the case < < + which relies on the free discontinuity analysis of Section 4 The case = is discussed searately and will be dealt directly through rearrangements of BV functions The case > Let Ω R N be oen, bounded and with a Lischitz boundary Let u W, (Ω) on which the Poincaré constant in () is attained, that is C (Ω) = u L (Ω;R N ) + u L ( Ω) u L (Ω) It is not restrictive to assume that u 0 on Ω Notice that thanks to [, Theorem 384] We deduce v := u Ω SBV (R N ) Notice that u Ω SBV (R N ) v = u L (Ω;R N ) and v = u L (Ω) Moreover we have J v Ω with [(v + ) + (v ) ] dh N = J v u dh N = J v since u = 0 on Ω \ J v We conclude that F (v) = u L (Ω;R N ) + u L ( Ω) u L (Ω) Ω = C (Ω) u dx Thanks to Theorem 44, we have that minimizers of the free discontinuity roblem (4) for m = Ω are of the form ψ B, where B is a ball of volume m, and ψ W, (B) is radial with resect to its center In articular we get (5) C (Ω) = F (v) F (ψ B ) = ψ L (B;R N ) + ψ L ( B) v L (B) C (B), which shows the otimality of the ball for the constant in inequality () Let us come to the uniqueness issue If C (Ω) = C (B), then inequalities in (5) become equalities, which yields that v = u Ω is a minimizer for the free discontinuity roblem (4) By Theorem 44, we get that Ω is a ball, and the roof is concluded