SYMMETRY OF MINIMIZERS OF A GAUSSIAN ISOPERIMETRIC PROBLEM

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1 SYMMETRY OF MINIMIZERS OF A GAUSSIAN ISOPERIMETRIC PROBLEM MARCO BARCHIESI AND VESA JULIN Abstract. We study an isoperimetric problem described by a functional that consists of the standard Gaussian perimeter and the norm of the barycenter. This second term has a repulsive effect, and it is in competition with the perimeter. Because of that, in general the solution is not the half-space. We characterize all the minimizers of this functional, whenthe volumeisclosetoone, byprovingthatthe minimizeriseitherthe half-space or the symmetric strip, depending on the strength of the repulsive term. As a corollary, we obtain that the symmetric strip is the solution of the Gaussian isoperimetric problem among symmetric sets when the volume is close to one. 00 Mathematics Subject Class. 49Q0, 60E5. Contents. Introduction. Notation and set-up 4 3. Reduction to the two dimensional case 9 4. Reduction to the one dimensional case 5. The one dimensional case 4 Acknowledgments 6 References 6. Introduction The Gaussian isoperimetric inequality proved by Borell [7] and Sudakov-Tsirelson [9]) states that among all sets with given Gaussian measure the half-space has the smallest Gaussian perimeter. Since the half-space is not symmetric with respect to the origin, a natural question is to restrict the problem among sets which are symmetric, i.e., either central symmetric E = E) or coordinate wise symmetric n-symmetric). This problem turns out to be rather difficult as every known method that has been used to prove the Gaussian isoperimetric inequality, such as symmetrization [4] and the Ornstein-Uhlenbeck semigroup argument [], seems to fail. In fact, at the moment it is not even clear what the solution to this problem should be. The Gaussian isoperimetric problem for symmetric sets or its generalization to Gaussian noise is stated as an open problem in [8, 8]. In the latter it was conjectured that the solution should be the ball or its complement, but this was recently disproved in []. Another natural candidate for the solution is the symmetric strip or its complement. Indeed, in [4] Barthe proved that if one replaces the standard Gaussian perimeter by a certain anisotropic perimeter, the solution of the isoperimetric problem among n-symmetric sets is the symmetric strip or its complement. We mention also a somewhat similar result by Latala and Oleszkiewicz [6, Theorem 3] who proved that the symmetric strip minimizes the Gaussian perimeter weighted with the width of the set among convex and symmetric sets with volume constraint. For the standard perimeter Date: May 8, 08.

2 M. BARCHIESI AND V. JULIN the problem is more difficult as a simple energy comparison shows see []) that when the volume is exactly one half, the two-dimensional disk and the three-dimensional ball have both smaller perimeter than the symmetric strip in dimension two and three, respectively. Similar difficulty appears also in the isoperimetric problem on sphere for symmetric sets, where it is known that the union of two spherical caps does not always have the smallest surface area see [4]). However, it might still be the case that the solution of the problem is a cylinder B k r R n k, or its complement, for some k depending on the volume see [, Conjecture.3]). Here B k r denotes the k-dimensional ball with radius r. At least the results by Heilman [, ] and La Manna [5] seem to indicate this. To the best of the authors knowledge there are no other results directly related to this problem. In [3] Colding and Minicozzi introduced the Gaussian entropy, which is defined for sets as Λ) = sup P γ t x 0 R n 0 E {x 0})),,t 0 >0 where P γ is the Gaussian perimeter defined below. The Gaussian entropy is important as it is decreasing under the mean curvature flow and for this reason in [3] the authors studied sets which are stable for the Gaussian entropy. It was conjectured in [] that the sphere minimizes the entropy among closed hypersurfaces at least in low dimensions). This was proved by Bernstein and Wang in [5] in low dimensions and more recently by Zhu [33] in every dimension. This problem is related to the symmetric Gaussian problem since the Gaussian entropy of a self-shrinker equals its Gaussian perimeter. In this paper we prove that the symmetric strip is the solution of the Gaussian isoperimetric problem for symmetric set when the volume is close to one. Similarly, its complement is the solution when the volume is close to zero). Our proof is direct and thus we could give an explicit estimate on how close to one the volume has to be. In particular, the bound on the volume is independent of the dimension. But as our proof is rather long and the bound on the volume is obtained after numerous inequalities, we prefer to state the result in a more qualitative way in order to avoid heavy computations. In order to describe the main result more precisely, we introduce our setting. Given a Borel set E R n, γe) denotes its Gaussian measure, defined as γe) := π) n E e x dx. If E is an open set with Lipschitz boundary, P γ E) denotes its Gaussian perimeter, defined as P γ E) := e x dh n x), ) π) n where H n is the n )-dimensional Hausdorff measure. We define the non-renormalized) barycenter of a set E as be) := xdγx) and define the function φ : R 0,) as E φs) := π s e t dt. Moreover, given ω S n and s R, H ω,s denotes the half-space of the form while D ω,s denotes the symmetric strip H ω,s := {x R n : x,ω < s}, D ω,s := {x R n : x,ω < as)}, where as) > 0 is chosen such that γh ω,s ) = γd ω,s ).

3 SYMMETRY OF MINIMIZERS 3 We approach the problem by studying the minimizers of the functional FE) := P γ E)+ π/ be) ) under the volume constraint γe) = φs). Note that the isoperimetric inequality implies that for = 0 the half-space is the only minimizer of ), while it is easy to see that the quantity be) is maximized by the half-space. Therefore the two terms in ) are in competition and we call the barycenter term repulsive, as it prefers to balance the volume around the origin. It is proven in [, 7] that when is small, the half-space is still the only minimizer of ). This result implies the quantitative Gaussian isoperimetric inequality see also [, 30, 3, 3]). It is clear that when we keep increasing the value, there is a threshold, say s, such that for > s the half-space H ω,s is no longer the minimizer of ). In this paper we are interested in characterizing the minimizers of ) after this threshold. Our main result reads as follows. Main Theorem. There exists s 0 > 0 such that the following holds: when s s 0 there is a threshold s such that for [0, s) the minimizer of ) under volume constraint γe) = φs) is the half-space H ω,s, while for s, ) the minimizer is the symmetric strip D ω,s. As a corollary this provides the solution for the symmetric Gaussian problem, because symmetric sets have barycenter zero. Corollary. There exists s 0 > 0 such that for s s 0 it holds P γ E) P γ D ω,s ) = + ln ) s +o/s ) e s, for any symmetric set E with volume γe) = φs), and the equality holds if and only if E = D ω,s for some ω S n. Another corollary of the theorem is the optimal constant in the quantitative Gaussian isoperimetric inequality see [, 7]) when the volume is close to one. Let us denote by βe) the strong asymmetry βe) := min ω S n be) bhω,s ), which measures the distance between a set E and the family of half-spaces. Corollary. There exists s 0 > 0 such that for s s 0 it holds P γ E) P γ H ω,s ) c s βe), for every set E with volume γe) = φs). The optimal constant is given by c s = πe s / P γ D ω,s ) P γ H ω,s )) = π ln s +o/s ). It would be interesting to obtain a result analogous to Corollary in the Euclidean setting, where the minimization problem which corresponds to ) was introduced in [6], and on the sphere [6]. The motivation for this is that, by the result of the second author [3], the optimal constant for the quantitative Euclidean isoperimetric inequality implies an estimate on the range of volume where the ball is the minimizer of the Gamov s liquid drop model [9]. This is a classical model used in nuclear physics and has gathered a lot attention in mathematics in recent years [9, 0, 4]. We also refer to the survey paper [5] for the state-of-the-art in the quantitative isoperimetric and other functional inequalities. The main idea of the proof is to study the functional ) when the parameter is within a carefully chosen range s,, s, ), and to prove that within this range the only local minimizers, which satisfy certain perimeter bounds, are the half-space H ω,s and the symmetric strip D ω,s. We have to choose the lower bound s, large enough so that the symmetric strip is a local minimum of ). On the other hand, we have to choose the upper bound s, small enough so

4 4 M. BARCHIESI AND V. JULIN that no other local minimum than H ω,s and D ω,s exist. Naturally also the threshold value s has to be within the range s,, s, ). Our proof is based on reduction argument where we reduce the dimension of the problem from R n to R. First, we develop further our ideas from [] to reduce the problem from R n to R by a rather short argument. In this step it is crucial that we are not constrained to keep the sets symmetric. The main challenge is thus to prove the theorem in R, since here we cannot apply the previous reduction argument anymore. Instead, we use an ad-hoc argument to reduce the problem from R to R essentially by PDE type estimates from the Euler equation and from the stability condition. We give an independent overview of this argument at the beginning of the proof of Theorem 3 in Section 4. Finally, we solve the problem in R by a direct argument.. Notation and set-up In this section we briefly introduce our notation and discuss about preliminary results and estimates. We remark that throughout the paper the parameter s, associated with the volume, is assumed to be large even if not explicitly mentioned. In particular, our estimates are understood to hold when s is chosen to be large enough. C denotes a numerical constant which may vary from line to line. We denote the n )-dimensional Hausdorff measure with Gaussian weight by Hγ n, i.e., for every Borel set A we define H n γ A) := π) n A e x dh n. We minimize the functional ) among sets with locally finite perimeter and have the existence of a minimizer for every by an argument similar to [, Proposition ]. If E R n is a set of locally finite perimeter we denote its reduced boundary by and define its Gaussian perimeter by P γ E) := Hγ n ). We denote the generalized exterior normal by ν E which is defined on. As introduction to the theory of sets of finite perimeter and perimeter minimizers we refer to [8]. If the reduced boundary is a smooth hypersurface we denote the second fundamental form by B E and the mean curvature by H E, which for us is the sum of the principle curvatures. We adopt the notation from [0] and define the tangential gradient of a function f, defined in a neighborhood of, by τ f := f f ν E )ν E. Similarly, we define the tangential divergence of a vector field by div τ X := divx DXν E,ν E and the Laplace-Beltrami operator as τ f := div τ τ f). The divergence theorem on implies that for every vector field X C0 E;R n ) it holds div τ X dh n = H E X,ν E dh n. If is a smooth hypersurface, we may extend any function f C 0 E) to a neighborhood of by the distance function. For simplicity we will omit to indicate the dependence on the set E when this is clear, by simply writing ν = ν E, H = H E etc... We denote the mean value of a function f : R by f := and its average over a subset Σ by f H n γ, f) Σ := f Hγ n. Σ

5 SYMMETRY OF MINIMIZERS 5 We recall that for every number a R it holds f f) Σ ) dhγ Σ Σ f a) dh γ. Recall that H ω,s denotes the half-space {x R n : x,ω < s} and D ω,s denotes the symmetric strip {x R n : x,ω < as)}, where as) is chosen such that γd ω,s ) = γh ω,s ) = φs). Since we areassumingthat s is large, it is important to know the asymptotic behavior of the quantities φs), as), and PD ω,s ). A simple analysis shows that φs) = π s +o/s ) ) e s. 3) The asymptotic behavior of as) is a slightly more complicated. We will show that as) = s+ ln +o/s). 4) s To this aim we write as) = s+δs), so that 4) is equivalent to lim s sδs) = ln. We argue by contradiction and assume that lim s sδs) > ln. By the volume constraint and therefore as) s+ln/s < lim e t dt s s e t dt e t dt = = lim s s ln e t dt 5) )e s+ln/s) s e s which is a contradiction. We arrive to a similar contradiction if lim s sδs) < ln. Therefore we have 4). For the perimeter of the strip D ω,s we have the following estimate: P γ D ω,s ) = + ln ) s +o/s ) e s. 6) Indeed, since P γ D ω,s ) = e as) /, by differentiating 5) we get + δ )P γ D ω,s ) = e s /. Moreover, by lim s sδs) = ln we get lim s s δ s) = ln and thus lim s s e s P γ D ω,s ) + ln ) ) s e s = lim s δ s +δ + ln ) s = 0. In particular, according to our main theorem the threshold value has the asymptotic behavior π s = ln +o)). 7) s es Thisfollows fromthefact that thresholdvalue s is theuniquevalueof for which thefunctional ) satisfies FH ω,s ) = FD ω,s ), i.e., P γ D ω,s ) = e s + s π e s by taking into account that bh ω,s ) = e s / / π. In order to simplify the upcoming technicalities we replace the volume constraint in the original functional ) with a volume penalization. We redefine F for any set of locally finite perimeter as FE) := P γ E)+ π/ be) +Λ π γe) φs), 8) =,

6 6 M. BARCHIESI AND V. JULIN where we choose Λ = s+. 9) As with the original functional the existence of a minimizer of 8) follows from [, Proposition ]. It turns out that the minimizers of 8) are the same as the minimizers of ) under the volume constraint γe) = φs), as proved in the last section. The advantage of a volume penalization is that it helps us to bound the Lagrange multiplier in a simple way. The constants π/ and π in front of the last two terms are chosen to simplify the formulas of the Euler equation and the second variation. As we explained in the introduction, the idea is to restrict the parameter in 8) within a range, which contains the threshold value 7) and such that the only local minimizers of 8), which satisfy certain perimeter bounds, are the half-space and the symmetric strip. To this aim we assume from now on that is in the range 6 π 7 π 5 s es 5 s es. 0) Note that the threshold value 7) is within this interval. If we are able to show that when satisfies 0) the only local minimizers of 8) are H ω,s and D ω,s, we obtain the main result. Indeed, when takes the lower value in 0) it holds FH ω,s ) < FD ω,s ) and the minimizer is H ω,s. It is then not difficult to see that for every value less than this, the minimizer is still H ω,s. Similarly, when takes the larger value in 0) it holds FD ω,s ) < FH ω,s ) and the minimizer is D ω,s. Hence, for every value larger than this, D ω,s is still the minimizer of 8), since it has barycenter zero. Next we deduce a priori perimeter bounds for the minimizer. First, we may bound the perimeter from above by the minimality as P γ E) FE) FD ω,s ) = P γ D ω,s ) + s ) e s. ) To bound the perimeter from below is slightly more difficult. Let E be a minimizer of 8) with volume γe) = φ s). First, it is clear that s 0. Let us show that s s + s, which by the Gaussian isoperimetric inequality implies the following perimeter lower bound P γ E) 4 e s. ) We argue by contradiction and assume s > s+ s. By the Gaussian isoperimetric inequality we deduce FE) P γ E)+Λ s πφs) φ s)) e s +s+) e t dt. Define the function f : [s, ) R, ft) := e t +s+) t f t) = t+s+)e t. s e l s dl. By differentiating we get The function is clearly increasing up to t = s+ and then decreases to the value lim t ft) = s+) dt + s) e s by 3). We also deduce that f t) 4 e s for s t s+/s s e t and therefore fs+/s) + 4s) e s. Hence, if s s+ s FE) f s) min{fs+/s), lim t ft)} we have that + 4s ) e s. But this contradicts FE) FD ω,s ) = PD ω,s ) by 6). Thus we have ). For reader s convenience we summarize the results concerning the regularity of minimizers and the first and the second variation of 8) contained in [, Section 4] in the following theorem.

7 SYMMETRY OF MINIMIZERS 7 Theorem. Let E be a minimizer of 8). Then the reduced boundary is a relatively open, smooth hypersurface and satisfies the Euler equation H x,ν + b,x = λ on. 3) The Lagrange multiplier λ can be estimated by λ Λ. The singular part of the boundary \ is empty when n < 8, while for n 8 its Hausdorff dimension can be estimated by dim H \) n 8. Moreover, the quadratic form associated with the second variation is non-negative F[ϕ] := τ ϕ B E ϕ + b,ν ϕ ϕ ) dhγ n + π ϕxdh n γ 0 for every ϕ C 0 E) which satisfies ϕdhn γ = 0. The Euler equation 3) yields important geometric equations for the position vector x and for the Gauss map ν. For arbitrary ω S n we write x ω = x,ω and ν ω = ν,ω. If {e ),...,e n) } is a canonical basis of R n we simply write x i = x,e i and ν i = ν,e i. From 3) and from the fact τ x ω = H ν ω [7, Proposition ] we have 4) τ x ω τ x ω,x = x ω λν ω + b,x ν ω. 5) Moreover, from 3) and from the fact τ ν ω = B E ν ω + τ H,ω [0, Lemma 0.7] we get τ ν ω τ ν ω,x = B E ν ω + b,ν ν ω b,ω. 6) By the divergence theorem on we have that for any function ϕ C0 E) and for any function ψ C ), ) div τ e x ψ τ ϕ dh n = H e x ψ τ ϕ,ν E dh n = 0. The previous equality gives us an integration by parts formula ψ τ ϕ τ ϕ,x )dhγ n = τ ψ, τ ϕ dhγ n. We will use along the paper the above formula with ϕ = x ω or ϕ = ν ω. Also if they do not belong to C 0 E), we are allowed to do so by an approximation argument see [, 3]). Remark. We associate the following second order operator L with the first four terms in the quadratic form 4), L[ϕ] := τ ϕ+ τ ϕ,x B E ϕ+ b,ν ϕ ϕ, 7) where ϕ C0 E). By integration by parts the inequality 4) can be written as L[ϕ]ϕdHγ n + ϕxdhγ n 0. π Note that when the vector ω is orthogonal to the barycenter, i.e., ω,b = 0, then by 6) the function ν ω is an eigenfunction of L and satisfies L[ν ω ] = ν ω.

8 8 M. BARCHIESI AND V. JULIN For every ω S n it holds by the divergence theorem in R n that ν ω dhγ n x) = divωe x )dx π) n = π E E x,ω dγx) = π b,ω. In particular, when ω,b = 0 the function ϕ = ν ω has zero average. Therefore by Remark it is natural to use ν ω with ω,b = 0 as a test function in the second variation condition 4). The equality ν ωdhγ n = π b,ω for every ω S n also implies In particular, we have by 0)-) νp γ E) = πb. 8) 4s ν b s ν. 9) We conclude this preliminary section by providing further regularity properties from5) for the minimizers of 8). We call the estimates in the following lemma Caccioppoli inequalities since they follow from 5) by an argument which is similar to the classical proof of Caccioppoli inequality known in elliptic PDEs. This result is an improved version of [, Proposition ]. Lemma Caccioppoli inequalities). Let E R n be a minimizer of 8). Then for any ω S n it holds s+) and x ω dhn γ x ω x ω ) dh n γ s+) Proof. Let us first prove 0). To simplify the notation we define { x, b b if b 0, x b := 0 if b = 0. We multiply 5) by x ω and integrate by parts over to get = λ ν ω x ω dhγ n + τ x ω dhγ n x ω dhn γ ν ω dhn γ +8P γ E) 0) ν ω ν ω ) dh n γ +8P γ E). ) + b x b ν ω x ω dhγ n ) We estimate the right-hand-side of ) in the following way. We estimate the first term by Young s inequality λ ν ω x ω dhγ n x ωdhγ n + λ ν ωdh γ n x ωdhγ n + s+) ν ωdhγ n, where the last inequality follows from the bound on the Lagrange multiplier λ s+ 3) given by Theorem and by our choice of Λ in 9). Since τ x ω = νω, we may bound the second term simply by τ x ω dhγ n P γ E).

9 SYMMETRY OF MINIMIZERS 9 Finally we bound the last term again by Young s inequality and by b proved in 9)) s b x b ν ω x ω dhγ n s x ω dhn γ + s x b dhn γ. By using these three estimates in ) we obtain ) s x ω dhn γ s+) ν ω dhn γ +P γ E)+ s x b dhn γ. 4) If the barycenter is zero the claim follows immediately from 4). If b 0, we first use 4) with ω = b b and obtain ) s x b dhn γ s+) νb dhn γ +P γ E) ) s+) + P γ E). This implies Therefore we have by 4) ) s This yields the claim. x ω dhn γ x b dhn γ s P γ E). 5) s+) ν ω dhn γ +3P γ E). The proof of the second inequality is similar. We multiply the equation 5) by x ω x ω ) and integrate by parts over to get x ω x ω ) dhγ n = λ x ω x ω )ν ω ν ω )dhγ n + τ x ω dhγ n + b x b ν ω x ω x ω )dhγ n. By estimating the three terms on the right-hand-side precisely as before, we deduce ) s x ω x ω ) dhγ n s+) ν ω ν ω ) dhγ n +P γ E)+ s s+) ν ω ν ω ) dh n γ where the last inequality follows from 5). This implies ). 3. Reduction to the two dimensional case +3P γ E), x b dhn γ In this section we prove that it is enough to obtain the result in the two dimensional case. More precisely, we prove the following result. Theorem. Let E be a minimizer of 8). Then, up to a rotation, E = F R n for some set F R. Proof. Let {e ),...,e n) } be an orthonormal basis of R n. We begin with a simple observation: if i j then by the divergence theorem x i ν j dhγ n = π x i x j dγ. E

10 0 M. BARCHIESI AND V. JULIN In particular, the matrix A ij = x iν j dhγ n is symmetric. We may therefore assume that A ij is diagonal, by changing the basis of R n if necessary. In particular, it holds x i ν j dhγ n = 0 for i j. 6) By reordering the elements of the basis we may also assume that x j dhγ n x j+dhγ n 7) for j {,...,n }. Since we assume n 3, we may choose a direction ω S n which is orthogonal both to the barycenter b and to e ). To be more precise, we choose ω such that ω,b = 0 and ω span{e ),e 3) }. Since ω,b = 0, 8) yields ν ω = 0. In other words, the function ν ω has zero average. We use ϕ = ν ω as a test function in the second variation condition 4). According to Remark we may write the inequality 4) as L[ν ω ]ν ω dh n γ + π ν ω xdh n 0, where the operator L is defined in 7). Since ω is orthogonal to b we deduce by Remark that ν ω is an eigenfunction of L and satisfies L[ν ω ] = ν ω. Therefore we get νωdh γ n + ν ω xdhγ n 0. 8) π The crucial step in the proof is to estimate the second term in 8), by showing that it is small enough. This is possible, because ω is orthogonal to e ). Indeed, by using 6) and the fact that ω span{e ),e 3) }, and then Cauchy-Schwarz inequality, we get ν ω xdhγ n = x ν ω dh n γ x +x 3 dhn γ ) + ) γ x 3 ν ω dhγ n ) ν ω dhn γ We estimate the first term on the right-hand-side first by 7), then by the Caccioppoli estimate 0) and finally by ) x +x 3 )Hn γ x 3 +x +x 3 )Hn γ [s+) ] ν +ν +ν 3 3)H γ n +4P γ E) 9) [ s+) +4 ] P γ E) s e s. Since we assume 7 π e s 5s see 0)), the previous two inequalities yield ν ω xdhγ n 63 π 65 Then, by collecting 8) and 30) we obtain. ) ν ωdh n γ. 30) ν ωdh n γ 0. 3) This implies ν ω = 0. We have thus reduced the problem from n to n. By repeating the previous argument we reduce the problem to the planar case.

11 SYMMETRY OF MINIMIZERS Remark. We have to be careful in our choice of direction ω, and in general we may not simply choose any direction orthogonal to the barycenter b. Indeed, if ω,v S n are vectors such that b,ω = 0 and ν ω xdhγ n = ν ω xdhγ n,υ = ν ω x,υ dhγ n. 3) Then, by using Cauchy-Schwarz inequality, we may estimate the second term in 8) by ν ω xdh n ) ) x υdhγ n νωdh γ n. π π γ We may estimate the term π x υdhγ n by the Caccioppoli estimate 0), and by ) and 0) π ν ω xdhγ n 8 5 ν ω dhn γ instead of 30). Unfortunately this estimate is not good enough. Note that we cannot shrink, since we have the constrain given by 7). Remark 3. We may further reduce the problem to the one dimensional case if b = 0, since we may use ω = e ) in the previous argument ν ω has zero average and x is small enough). However, this is a special case and a priori nothing guaranties that b = 0. Because of that we have to handle the reduction to the one dimensional case in a different way. 4. Reduction to the one dimensional case In this section we will prove a further reduction of the problem, by showing that it is enough to obtain the result in the one dimensional case. This is technically more involved than Theorem and requires more a priori information on the minimizers. Theorem 3. Let E be a minimizer of 8). Then, up to a rotation, E = F R n for some set F R. Thanks to Theorem we may assume from now on that n =. In particular, by Theorem the boundary is regular and =. Moreover the Euler equation and 6) simply read as k = λ+ x,ν b,x, 33) τ ν ω τ ν ω,x = k ν ω + b,ν ν ω b,ω, 34) where k is the curvature of. The idea is to proceed by using the second variation argument once more, but this time in a direction that it is not necessarily orthogonal to the barycenter. This argument does not reduce the problem to R, but gives us the following information on the minimizers. Lemma. Let E R be a minimizer of 8). Then Moreover, there exists a direction v S such that k dhγ s. 35) ν v ν v ) dhγ 0 s ν v. 36) Observe that the above estimate implies that ν v is close to a constant. In particular, this excludes the minimizers to be close to the disk.

12 M. BARCHIESI AND V. JULIN Proof. We begin by showing that for any ω S it holds + ν ω k dhγ ν ω ν ω dhγ s ν ωp γ E)+ ν ω ν ω )xdhγ. π To this aim we choose ϕ = ν ω ν ω as a test function in the second variation condition 4). We remark that because ω is not in general orthogonal to the barycenter b, neither ν ω or ν ω ν ω is an eigenfunction of the operator L associated with the second variation defined in Remark. We multiply the equation 34) by ν ω and integrate by parts to obtain τ ν ω k νω + b,ν νω ) dh γ = b,ω ν ω P γ E), 38) and simply integrate 6) over to get k ) ν ω b,ν ν ω dh γ = b,ω P γ E). 39) Hence, by also using νp γ E) = πb see 8)), we may write τ ν ω k ν ω ν ω ) + b,ν ν ω ν ω ) ) dhγ 37) = ν ω = ν ω ν ω k dh γ + b, ν ν ωp γ E) b,ω ν ω P γ E) k dh γ + π ν ) ν ω P γ E) k dh γ + s ν ωp γ E), where in the last inequality we have used the estimates 0) and ). The above inequality and the second variation condition 4) with ϕ = ν ω ν ω imply 37). Let us consider an orthonormal basis {e ),e ) } of R and assume x dh γ x dh γ. As in 9), we use the Caccioppoli estimate 0) and ) to get x dhγ x +x )Hγ [ s+) +6 ] P γ E) [ s+4) ] 40) e s. We choose a direction v S which is orthogonal to the vector x ν ν)dh γ. Since ffl x ν v ν v )dhγ = ffl x ν ν)dh γ,v = 0, we have xν v ν v )dhγ = x ν v ν v )dhγ. Then, by the above equality, by Cauchy-Schwarz inequality and by 40) we have ) xν v ν v )dhγ = x ν v ν v )dhγ ) ) x dhγ ν v ν v ) dhγ ) s+4) e s ν v ν v ) dhγ.

13 SYMMETRY OF MINIMIZERS 3 With the bound 7 π e s 5s π see 0)), the previous inequality yields ν v ν v )xdhγ 4 ν v ν v ) dh 5 γ. Hence, the inequality 37) implies ν v k dhγ + ν v ν v ) dhγ 5 s ν vp γ E). From this inequality we have immediately 36), and also 35), if ν v is not zero. If instead ν v = 0, then also ν v = 0 by 36). Thus is flat, k = 0 and 35) holds again. We will also need the following auxiliary result. Lemma 3. Let E R be a minimizer of 8). Then, for every x it holds x s. 4) Proof. We argue by contradiction and assume that there exists x such that x < s. We claim that then it holds H B / x)) s. 4) We remark that H is the standard Hausdorff measure, i.e., H B / x)) denotes the length of the curve. We divide the proof of 4) in two cases. Assume first that there is a component of, say Γ, which is contained in the disk B / x). By regularity, Γ is a smooth Jordan curve which encloses a bounded set Ẽ, i.e., Γ = Ẽ. Note that then it holds Ẽ B R for R = s /. We integrate the Euler equation 33) over Ẽ with respect to the standard Hausdorff measure and obtain by the Gauss-Bonnet formula and by the divergence theorem that π = kdh = x,ν +λ b,x ) dh Γ Γ Ẽ + λ + ) 43) H Γ), s where in the last inequality we have used b s proved in 9)) and the fact that for all x Ẽ it holds x s /. The isoperimetric inequality in R implies Ẽ 4π H Γ). Therefore since λ s+ we obtain from 43) that π π H Γ) +s+)h Γ). This implies H Γ) s and the claim 4) follows. Letusthenassumethatnocomponentof iscontainedinb / x). Inthiscasetheboundary curve passes x and exists the disk B x, ). In particular, it holds H B / x)) / which implies 4). Since for all x B / x) it holds x s /, the estimate 4) implies P γ E) π B / x) e x dh π e s /) H B / x)) e s. This contradicts ).

14 4 M. BARCHIESI AND V. JULIN For the remaining part of this section we choose a basis {e ),e ) } for R such that e ) = v, wherev is thedirection inlemmaande ) is anorthogonal direction tothat. Thedisadvantage of Lemma is that the argument does not seem to give us any information on ν = ν,e ). However, by studying closely the proof of Lemma we may reduce to the case when it holds ν ν ) dhγ ) Indeed, we conclude below that if 44) does not hold then the argument of the proof of Lemma yields that the minimizer is one-dimensional. In fact, by the one dimensional analysis in Section 5 we deduce that if 44) does not hold then the minimizer is the half-space. To show 44), we argue by contradiction, in which case it holds ffl ν ν ) dh γ < 4 7. Then the Caccioppoli estimate ) yields x x ) dhγ 4 7 s+) +8 while again by ) and by 36) from Lemma we have x ) x dhγ s+) ν ν ) dhγ +8 C. Let e S be orthogonal to the barycenter b. We now apply the argument in the proof of Lemma for the test function ϕ = ν e. By the two above inequalities and by ) we have xν e dhγ ) ) = x x )ν e dhγ + x x )ν e dhγ ) x x ) +x x ) dhγ νe dh γ 9 3 s e s νe dhγ. e s 5s In other words, since 7 π we conclude that the crucial estimate 30) in the proof of Theorem holds for a direction orthogonal to the barycenter and thus by Remark 3 we conclude that ν e = 0. Hence, we may assume from now on that 44) holds. Let us define = {x : x > 0} and Σ = {x : x < 0}. In the next lemma we use 36) from Lemma and 44) to conclude first that and Σ are flat in shape. The second estimate in the next lemma states roughly speaking that the Gaussian measure of {x : x s 3 } is small. The latter estimate implies that, from measure point of view, and Σ are almost disconnected. This enables us to variate and Σ separately, which will be crucial in the proof of Theorem 3. Recall that, given a function f : R, we denote f) Σ+ = ffl f dhγ and f) Σ := ffl Σ f dhγ. Lemma 4. Let E R be a minimizer of 8) and assume 44) holds. Then we have the following: x i x i ) Σ± ) CP γ E), for i =, 45) Σ ± x dh { x s 3 } γ C ν dh γ 46)

15 SYMMETRY OF MINIMIZERS 5 Proof. Inequality 45). We first observe that the claim 45) is almost trivial for i =. Indeed, by thecaccioppoli estimate ) andby 36) fromlemma recall that we have chosen e ) = v) we have x x ) Σ+ ) dhγ x x ) dhγ Σ + x x ) dhγ s+) ν ν ) dhγ +8P γe) Thus we need to prove 45) for i =. We first show that CP γ E). ν ν ) Σ+ ) dh γ C s P γe). 47) Note that 44) implies ffl ν dh γ 4 7. By Jensen s inequality we then have ν Therefore we deduce by 48) and by 36) ) ν ν dhγ = 8 Since ) ν ν ) Σ+ dh γ ν dh γ ) C s P γe). ν ν )) ν + ) ν dhγ ν ν ) dh γ ν ν ) dh γ ) ν ν dhγ we have 47). Toprove theinequality 45) for i = we multiply theequation 5), withω = e, byx +λν ) and integrate by parts x +λν ) dhγ τ x +λν ), τ x b,x ν x +λν )dhγ. We estimate the first term on the right-hand-side by Young s inequality and by λ s+ and the second as τ x +λν ), τ x τ x +λ τ ν +s+) k b,x ν x +λν ) b x +s+) ν ). Hence, we have by b proved in 9)), ) and 35) that s x +λν ) dhγ +s+) k + 4 ) s x +s+) ν) CP γ E). dh γ

16 6 M. BARCHIESI AND V. JULIN Therefore it holds recall that x > 0 on ) CP γ E) x +λν ) dhγ x +λν ) Σ dhγ + x λ ν ) dhγ = x λ ν ) dhγ x λ ν ) Σ+ ) dhγ λ ν ν ) Σ+ ) dh γ. Hence, by 47) and λ s+ we deduce x λ ν ) Σ+ ) dh γ CP γ E). The claim then follows from x x ) Σ+ ) dhγ x λ ν ) Σ+ ) dhγ. Inequality 46). We choose a smooth cut-off function ζ : R [0,] such that { for t s ζt) = 3 0 for t s and ζ t) 8 for t R. s We multiply the equation 5), with ω = e, by x ζ x ) and integrate by parts x ζ x )dhγ = λx ν ζ x )+ τ x, τ x ζ x )) + b,x ν x ζ x ) ) dhγ. 49) We estimate the first term on right-hand-side by Young s inequality and by λ s+ λx ν ζ x ) x ζ + s+) ν ζ, where we have written ζ = ζx ) for short. We estimate the second term by using τ ζx ) = ζ x ) x 8 s ν as follows τ x, τ x ζ x )) τ x ζ + 6 s ζ x τ x ν We estimate the third term simply by using b s ζ + 0 x ζ + C s ν. b,x ν x ζ x ) s x ζ. Hence, we deduce from 49) by the three above inequalities that x 0 x 4 s x )ζ dh γ s+) ν ζ + C ) s ν We recall that ζ = 0 when x s and that by 4) we have that x s ) on. In particular, for every x {x : x s } it holds dh γ x = x x s ) s s ) 50)

17 SYMMETRY OF MINIMIZERS 7 and x x. Therefore we deduce 4 x 5 ζ dhγ s+) ν ζ dh γ + C s We write the first term on the right-hand-side of 5) as s+) νζ dh {ν /} γ +s+) νζ dh {ν >/} γ s+) ζ dhγ +s+) {ν ν >/} dhγ 3 x 4 ζ dhγ +s+) {ν ν >/} dh γ, where the last inequality follows from 50). Therefore 5) implies x 0 ζ dhγ s+) {ν ν >/} dhγ + C s Now since x x and ζx ) = for x s 3 we have { x s 3 } x dh γ Cs Hence, we need yet to show that {ν >/} ν dh γ C s {ν >/} ν dh γ + C s to finish the proof of 46). We obtain by 48) and 36) that 3 ν dh γ ν ν dh {ν >/} 7 {ν >/} γ ν ν dhγ C s Thus we have {ν > /} ) C H γ {ν >/} This proves 5) and concludes the proof of 46). ν dh γ. 5) ν dh γ. ν dh γ. ν dh γ 5) 3 ν dh γ C 7 s ν dh γ. ν dh γ. We are now ready to prove the reduction to the one dimensional case. Proof of Theorem 3. We recall that = {x : x > 0} and Σ = {x : x < 0}. As we mentioned in Remark, using ϕ = ν e with e S orthogonal to the barycenter as a test function in the second variation inequality 4), does not provide any information on the minimizersincetheterm ν exdhγ can betoolarge andthus8) becomestrivial inequality. We overcome this problem by essentially variating only while keeping Σ unchanged, and vice-versa see Figure ). To be more precise, we restrict the class of test function by assuming ϕ C ) to have zero average and to satisfy ϕx) = 0 for every x {x s 3 } or

18 8 M. BARCHIESI AND V. JULIN x s/3 s/3 x Σ Figure. The sets and Σ. ϕx) = 0 for every x {x s 3 }). The point is that for these test function an estimate similar to 30) holds, ϕxdhγ ϕ dhγ π. 53) Indeed, by writing ϕxdhγ = = x ϕdhγ) + ) x ϕdhγ x x ) Σ+ )ϕdhγ) + x x ) Σ+ )ϕdhγ and estimating both the terms by 45) and 46) we have ) x i x i ) Σ+ )ϕdhγ) = x { x s i x i ) Σ+ )ϕdh 3 } γ ) 8 x i x i ) Σ+ ) + x dh { x s 3 } γ ) CP γ E) ϕ dhγ, for i =,. e s 5s ) ϕ dh γ Hence, we get 53) thanks to ) and 7 π from 0). In order to explain the idea of the proof, we assume first that and Σ are different components of. This is of course a major simplification but it will hopefully help the reader to follow the actual proof below. In this case we may use the following test functions in the second variation condition, ϕ i := { ν i ν i ) Σ+ on 0 on Σ 54) )

19 SYMMETRY OF MINIMIZERS 9 for i =,, where ν i ) Σ+ is the average of ν i on. We use ϕ i as a test functions in the second variation condition 4) and use 53) to obtain τ ϕ i k ϕ i + b,ν ϕ i ) ϕ i dhγx) 0. By using equalities 38) and 39), rewritten on, we get after straightforward calculations ν i ) k b,ν )dhγ + ν i ν i ) Σ+ ) dhγ b,e i ν i dhγ, i =,. 55) By summing up the previous inequality for i =, we get [ν ) +ν ) ] k b,ν )dhγ + [ ν ) Σ + ν ) ]dhγ b,ν dhγ. This can be rewritten as [ ν ) ν ) ] b,ν + ) dhγ +[ν ) Σ + +ν ) ] k dhγ 0. By Jensen inequality ν ) ν ) 0, while b,ν s which follows from 9). Therefore k = 0 and is a line. It is clear that a similar conclusion holds also for in Σ. When and Σ are connected the argument is more involved, since we need a cut-off argument in order to separate and Σ. This is possible due to 46), which implies that the perimeter of the minimizer in the strip { x s/3} is small. Therefore the cut-off argument produces an error term, which by 46) is small enough so that we may apply the previous argument. However, the presence of the cut-off function makes the equations more tangled and the estimates more complicated. Since the argument is technically involved we split the rest of the proof in two steps. Step. In the first step we prove ) ν ) +ν ) k dhγ + where the remeinder term satisfies ) ν ) ν ) dhγ R, 56) R C Pγ E) ) s 4 HγΣ ν + ). 57) We do this by proving the counterpart of 55), which now reads as ) k ν i ) b,ν dhγ + ν i ν i ) Σ+ ) dhγ b,e i ν i dhγ +R, 58) for i =,, where the reminder R satisfies 57). Let us show first how 56) follows from 58). Indeed, by b given by 9) we have s ) ν i ) Σ b,ν dhγ νi +Σ dh γ b,ν dhγ + s ν i ν i ) Σ+ ) dhγ. Therefore we have ν i ) b,ν dhγ + ) ν i ν i ) Σ+ ) dhγ νi dhγ b,ν dh 4 γ.

20 0 M. BARCHIESI AND V. JULIN Thus we obtain from 58) ν i ) k dh γ ν i dh γ ) b,ν dhγ+ ν i ν i ) Σ+ ) dhγ 4 b,e i ν i dhγ +R. Note that i= b,e i ν i dhγ = b,ν dhγ. Therefore, by adding the above estimate with i =, we obtain ν ) +ν ) )Σ+ k dh γ b,ν dhγ + ) ν ν ) Σ 4 + +ν ν ) b,ν dhγ +R, which implies 56). Hence, we need to prove 58). We prove 58) by using the second variation condition 4) with test function ϕ i := ν i α i )ζx ) for i =,. Here ζ : R [0,] is a smooth cut-off function such that { for t 0, ζt) = and ζ t) 4 for all t R 0 for t s/3, s and α i is chosen so that ϕ i has zero average. This choice is the counterpart of 54) in the case when is connected. In particular, the cut-off function ζ guarantees that ϕ i x) = 0, for x {x s 3 }. Therefore the estimate 53) holds and the second variation condition 4) yields τ ϕ i k ϕ i + b,ν ϕ i ) ϕ i dhγ x) 0. 59) Let us simplify the above expression. Recall that the test function is ϕ = ν i α i )ζ, where ζ = ζx ). By straightforward calculation ) τ ϕ i dhγ = ϕ i τ ϕ i + ϕ i,x dhγ = ϕ i ζ τ ν i + τ ν i,x )+ν i α i ) τ ζ ) dhγ. Therefore we have by the above equality and by multiplying the equation 34) with ϕ i and integrating by parts τ ϕ i dhγ = k b,ν )ζ ν i ν i α i )dhγ +R, 60) where the remainder term is R = dh γ b,ei ϕ i ζ +ν i α i ) τ ζ ) dh γ. 6) On the other hand, multiplying 34) with ζ and integrating by parts yields α i k b,ν )ν i ζ ) dhγ = α i τ ν i + τ ν i,x )ζ b,e i ζ ) dhγ = α i b,e i ζ dhγ +R, 6)

21 SYMMETRY OF MINIMIZERS where the remainder term is R = α i Collecting 59), 60), 6) yields α i k b,ν )+ ) ν i α i ζ dhγ α i ζ τ ν i, τ ζ dh γ. 63) b,e i ζ dh γ +R +R, 64) where the remainder terms R and R are given by 6) and 63) respectively. Let us next estimate the remainder terms in 64). We note that 36) recall that ν v = ν ) implies ffl ν dh γ + 0 s ) ν. Therefore we deduce from 4) and 46) that H γ {x : x s/3} ) C s P γe) ν. 65) Therefore since τ ζx) 4/s, for x s/3, and τ ζx) = 0 otherwise, 65) yields ζ dhγ C s 4P γe) ν. 66) We may therefore estimate R given by 63)) by Young s inequality and by 66) as + R α i τ ν i ζ dhγ τ ζ dhγ α i k ζ dhγ + C s 4P γe) ν. Similarly we may estimate 6) as R b,e i ϕ i ζdh γ + C s 4P γe) ν. To estimate the first term in R we recall that ϕ idhγ = 0 and therefore ϕ iζdhγ = ϕ iζ )dhγ. Since ϕ i ζ ) = 0 on { x > s/3}, we deduce by b s and by 65) that b,e i ϕ i ζdhγ C s 4P γe) ν. Hence, we may write 64) as k α i )+ b,ν ν i α i )ζ dh γ α i b,e i ζ dhγ + R, 67) where the remainder term R satisfies R C s 4P γe) ν. 68) Byasimilarargumentwemayalsoget ridofthecut-off functionin67). Indeedby b /s and 65) we have b,ν ζ dhγ b,ν dhγ R, where R satisfies 68). Similarly we get α i b,e i ζ dhγ α i b,e i dhγ + R. Therefore we obtain from 67) k α i )+ b,ν ) ν i α i dhγ α b,e i dhγ + R, 69) i where the remainder term R satisfies 68).

22 M. BARCHIESI AND V. JULIN We need yet to replace α i by ν i ) Σ+ in order to obtain 58). We do this by showing that α i is close the average ν i ) Σ+. To be more precise we show that α i ν i ) Σ+ C ) Pγ E) s Hγ Σ ν. 70) +) Indeed, since ζ = on we may write Hγ )α i ν i ) Σ+ ) = α i ν i ) Σ+ )ζdhγ. Since ζ = 0 when x s/3 we may estimate HγΣ + ) αi ν i ) Σ+ α i ν i ) Σ+ )ζdhγ +Hγ {x : x s/3} ). The inequality 70) then follows from α i ν i ) Σ+ )ζdhγ = ϕ idhγ = 0 and from 65). We use 35) and b to conclude that Σ s + k + b,ν dhγ C P s γ E). Therefore we may estimate 69) by 70) and get k ν i ) )+ Σ+ b,ν ) ν i α i dhγ ν i) b,e i dhγ Σ+R, + where the remainder term R satisfies 57). Finally the inequality 58) follows from ν i ν i ) Σ+ dhγ ν i α i dhγ. Step. Precisely similar argument as in the previous step, gives the estimate 68) also for Σ, i.e., ) ) ν ) Σ +ν ) Σ k dhγ + ν ) Σ ν ) dhγ R, Σ Σ 7) where the remainder satisfies R C Pγ E) ) s 4 HγΣ ν ). 7) Let us next prove that H γ ) 0 P γe) and H γσ ) 0 P γe). 73) Without loss of generality we may assume that H γ Σ ) H γ ). In particular, we have H γ Σ ) P γe) and therefore 7) and 7) imply Σ ν ν ) Σ ) dh γ C s 4P γe). 74) We need to show the first inequality in 73). We use 44) and 74) to deduce 4 7 P γe) ν ν ) dhγ ν ν ) Σ ) dhγ = ν ν ) Σ ) dhγ + ν ν ) Σ ) dhγ Σ C s 4P γe)+4h γ ). Hence we obtain H γ ) 0 P γe). Thus we have 73).

23 SYMMETRY OF MINIMIZERS 3 We conclude from 73) and from 35) that ffl k dh γ C s. Therefore we have by 56) and 73) that k dh γ C s 4 P γe) ν. Similarly we get and therefore Σ k dh γ C s 4 P γe) ν k dh γ C s 4 ν. 75) We are now close to finish the proof. We proceed by recalling the equation 34) for ν, i.e., We integrate this over, use 75) and get ν ν,x = k ν + b,ν ν b,e ). b,e ) + b,ν ν dhγ k dh γ C s 4 ν. Note that by νp γ E) = πb proved in 8)) we have ν b,e ) = b ν. Thus we deduce from the above inequality that b ν b ν We proceed by concluding from 44) that This implies 4 7 ν ν ) dhγ ν dhγ + C s 4 ν ν. 76) ν ν ) +ν ν ) ) dh γ = ν ν. ν = ν + ν 3 7. Using this and the inequality ffl ν dh γ + 0 ) ν s given by 36)) we estimate ) 3 / b ν ν dhγ ν 7 b dhγ 4 b ν 3. Therefore we deduce from 76) We use b 4s ν from 9)) to conclude This yields ν = 0. But then 36) implies 4 b ν C s 4 ν ν. s ν ν C s 4 ν ν. ν = 0 and we have reduced the problem to the one dimensional case.

24 4 M. BARCHIESI AND V. JULIN 5. The one dimensional case In this short section we finish the proof of the main theorem which states that the minimizer of 8) is either the half-space H ω,s or the symmetric strip D ω,s. By the previous results it is enough to solve the problem in the one-dimensional case. Theorem 4. When s is large enough the minimizer E R of 8) is either,s), s, ) or as),as)). Proof. As we explained in Section, we have to prove that, when is in the interval 0), the only local minimizers of 8) are,s), s, ) and as),as)). Let us first show that the minimizer E is an interval. Recall that since E R is a set of locally finite perimeter it has locally finite number of boundary points. Moreover, since there is no curvature in dimension one the Euler equation 3) reads as xνx)+ bx = λ. 77) By 4) we have that s+,s ) E. It is therefore enough to prove that the boundary has at most one positive and one negative point. Assume by contradiction that has at least two positive points the case of two negative points is similar). If x is a positive point which is closest to the origin on then νx) =. On the other hand, if y is the next boundary point, then νy) =. Then the Euler equation yields x+ bx = y + by. By b s proved in 9)) we conclude that s ) y s ) x, which is a contradiction when since x,y > 0. The minimizer of 8) is thus an interval of the form E = x,y), where s x,y. Without loss of generality we may assume that x y. Therefore we have e x Pγ E) e x. Using the bounds on the perimeter ) and ) we conclude that s /s x s+3/s. The Euler equation 77) yields x+ bx = y by. Hence, we conclude from b s that s s x y s+ 8 s. 78) Let us next prove that the minimizer has the volume γe) = φs). Indeed, it is not possible that γe) < φs), because by enlarging E we can decrease its perimeter, barycenter and the volume penalization term in 8). Also γe) > φs) is not possible. Indeed, in this case we can perturb the set E by E t = x+t,y), t > 0. Then φs) γe t ) < γe) and d dt FE t) t=0 = xe x + be)xe x s+)e x + s ) xe x s+)e x,

25 SYMMETRY OF MINIMIZERS 5 taking again into account that be). But since x s+3/s the above inequality yields s d dt FE t) t=0 < 0, which contradicts the minimality of E. Let usfinallyshowthat ifalocal minimizerisafiniteinterval E = x,y) forx y <, then necessarily x = y = as). We study the value of the functional 8) for intervals E t = αt),t), which have the volume γe t ) = φs). By the inequality 78) we need to only study the case when as) t s+ 8 s. This leads us to study the function f : [as),s+ 8 s ] R, The volume constraint reads as t ft) := FE t ) = e t +e α t) + π αt) e l e α t) e t ). dl = πφs). By differentiating this we obtain α t)e α t) = e t. 79) From 79) we conclude that for t αt) it holds 0 > α t) >. By differentiating f once and by using 79) we get f t) = t+αt)+ )) t+αt)) e α t) e t e t. π Therefore at a critical point it holds ) t+αt)) e α t) e t = t αt). 80) π We are interested in the sign of f t) at critical points on the interval t [as),s+ 8 s ]. Let us denote the barycenter of E t by b t := be t ) = π e α t) e t By differentiating f twice and by using 79) and 80) we obtain f t) = b t )+α t)+ b t )+ ) t+αt)) e t e t π at a critical point t. Let us write = 0 π e s s, where In order to analyze the sign of f t) at critical points we define g : [as),s+ 8 s ] R as ). gt) := b t )+α t)+ b t )+ 0 s t+αt)) e t s e. By recalling that by 78) αt) s /s, we have b t π e α t) 4. s Note that the end point t = αt) = as) is of course a critical point of f. Let us check that it is a local minimum. We have for the barycenter b as) = 0, α as)) = by 79), as) = s+ ln s +o s ) by 4) and e as) = + ln s +o/s ) gas)) + 0 C s > 0 ) e s by 6). Therefore it holds when s is large. In particular, we deduce that t = as) is a local minimum of f. Let us next show that g is strictly decreasing. Let us first fix a small number δ > 0, which value will be clear later. We obtain by differentiating 79) that α = α αα t).

26 6 M. BARCHIESI AND V. JULIN Byrecalling that α t) andthat by 78)αt) s+8/s, weget that α t) s α t) +6/s for t [as),s+ 8 s ]. Moreover, we estimate b t C/s, where b t = d dt b t. We may then estimate the derivative of g as g t) α t)+ b t )++α t)) b t 0 s tt+αt)) e t e s + 0 s α t) 4 0se t e s +δ s t+αt))+α t))e t when t [as),s+ 8 s ] and α [s s,as)]. To study 8) it is convenient to write t = s+ lnz s where δ z e 8. We obtain from the volume condition t similarly as in 4) we obtain αt) = s+ ) z s ln + εz) z s and from 79) that αt) e l e s 8) dl = πφs) arguing α t) = z +εz), where εz) is a function which converges uniformly to zero as s. Keeping these in mind we may estimate 8) as g t) s z 4 0s +δs s 7z) +δs. z 5zz ) Since δ z e 8, the above inequality shows that g t) < 0 when δ is chosen small enough. Hence, we conclude that g is strictly decreasing. Recall that gas)) > 0. Since g is strictly decreasing, there is t 0 as),s + 8 s ) such that gt) > 0 for t [as),t 0 ) and gt) < 0 for t t 0,s + 8 s ]. Therefore the function f has no other local minimum on [as),s+ 8 s ] than the end point t = as). Indeed, if there were another local minimum on as),t 0 ] there would be at least one local maximum on as),t 0 ). This is impossible as the previous argument shows that f t) > 0 at every critical point on as),t 0 ). Moreover, from gt) < 0 for t t 0,s+ 8 s ] we conclude that there are no local minimum points on t 0,s+ 8 s ]. This completes the proof. Acknowledgments The first author was supported by INdAM and by the project VATEXMATE. The second author was supported by the Academy of Finland grant 347. References [] D. Bakry & M. Ledoux. Lévy-Gromov isoperimetric inequality for an infinite dimensional diffusion generator. Invent. Math ), [] M. Barchiesi, A. Brancolini & V. Julin. Sharp dimension free quantitative estimates for the Gaussian isoperimetric inequality. Ann. Probab., 45 07), [3] M. Barchiesi & V. Julin. Robustness of the Gaussian concentration inequality and the Brunn-Minkowski inequality. Calc. Var. Partial Differential Equations, 56 07), art. n. 80. [4] F. Barthe, An isoperimetric result for the Gaussian measure and unconditional sets. Bull. London Math. Soc ),

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