Quantitative Isoperimetric Inequalities on the Real Line

In a recent paper A. Cianchi, N. Fusco, F. Maggi, and A. Pratelli have shown that, in the Gauss space, a set of given measure and almost minimal Gauss boundary measure is necessarily close to be a half-space. Using only geometric tools, we extend their result to all symmetric log-concave measures \mu on the real line. We give sharp quantitative isoperimetric inequalities and prove that among sets of given measure and given asymmetry (distance to half line, i.e. distance to sets of minimal perimeter), the intervals or complements of intervals have minimal perimeter.


The isoperimetric inequality on the real line
This section presents the standard isoperimetric inequality for the log-concave measures, and the asymmetry which measures the gap between a given set and the sets of minimal perimeter.
1.1.The standard isoperimetric inequality for the log-concave measures.Let µ be a measure with density function f .Throughout this paper, we assume that (i) the function f is supported and positive over some interval (a f , b f ), where a f and b f can be infinite, (ii) the measure µ is a probability measure, (iii) the measure µ is a log-concave measure, (iv) and the measure µ is symmetric with respect to the origin.
Observe that the point (iv) is not restrictive.As a matter of fact, the measure µ(.+ α), where α ∈ R, shares the same isoperimetric properties as the measure µ.By the same token, the assumption (ii) is obviously not restrictive.
1.1.1.The µ-perimeter.Let Ω be a measurable set.Following [Fed69], define the set Ω d of all points with density exactly d ∈ [0, 1] as where L 1 is the Lebesgue measure over the real line and B ρ (x) the ball with center x and radius ρ.Define the essential boundary ∂ M Ω as the set R \ Ω 0 ∪ Ω 1 , consisting of points with neither empty nor full density.Define the µ-perimeter as (2) where H 0 is the Hausdorff measure of dimension 0 over R and H 0 µ the measure of density f with respect to H 0 .The isoperimetric function I µ of the measure µ is defined by (3) I µ (r) = inf µ(Ω)=r P µ (Ω).
In the log-concave case, we can give an explicit form to the isoperimetric function using the function J µ .
1.1.2.The function J µ .Denote F the cumulative distribution function of the measure µ.Since the function f is supported and positive over some interval (a f , b f ) then the cumulative distribution function is increasing on the interval (a f , b f ).Define (4) where the quantile r ranges strictly from 0 to 1, J µ (0) = J µ (1) = 0, and F −1 denotes the inverse function of F.
Using (5), one can check [Bob94] that the measure µ is log-concave if and only if J µ is concave on (0, 1).Furthermore it is known [Bor75] that the infima of (3) are exactly the intervals (−∞, σ − ) and (σ + , +∞), where σ − = F −1 (r) and where Ω is a Lebesgue measurable set.This shows that, in the log-concave case, the isoperimetric function coincides with the function J µ .
1.2.The asymmetry.We concern with quantifying the difference between any measurable set Ω and an isoperimetric infimum (i.e.any measurable set such that the isoperimetric inequality ( 6) is an equality) with the same µ-measure.
Remark.The name asymmetry [FMP08] is inherited from the case of the Lebesgue measure on R n .In this case, the sets with minimal perimeter are balls, hence very symmetric.
Define the isoperimetric projection of a set Ω as the open half-line achieving the minimum in (7).In the case where this minimum is not unique we can chose whatever infima as an isoperimetric projection.

Sharp quantitative isoperimetric inequalities
This section gives a sharp improvement of (6) involving the asymmetry λ(Ω).In their paper [CFMP10] A. Cianchi, N. Fusco, F. Maggi, and A. Pratelli use a technical lemma (Lemma 4.7, Continuity Lemma) to complete their proof.Their lemma applies in the n-dimensional case and is based on a compactness argument derived from powerful results in geometric measure theory.In the onedimensional case, our approach is purely geometric and does not involve the continuity lemma.
2.1.The shifting lemma.The shifting lemma plays a key role in our proof.This lemma was introduced in [CFMP10] for the Gaussian measure.It naturally extends to even log-concave probability measures.For sake of readability, we begin with the shifting property.
Definition 1 (The shifting property) -We say that a measure ν satisfies the shifting property when for every open interval (a, b), the following is true: - In other words, if an interval is more to the right of 0, shifting it to the right with fixed measure, does not increase the perimeter.
In other words, if an interval is more to the left of 0, shifting it to the left with fixed measure, does not increase the perimeter.
The following remark states that the shifting property can be equivalently formulated with the complement sets.
Remark.As the perimeter is complement-invariant, we may also shift "holes".The shifting property is equivalent to the following property.
-If a + b ≥ 0 then for every (a Roughly, the next lemma shows that, for all measures such that the assumptions (i), (ii), and (iv) hold, the assumption (iii) is equivalent to the shifting property.
Lemma 2.1 (The shifting lemma) -Every log-concave probability measure symmetric with respect to the origin has the shifting property.
-Conversely, let f be a continuous function, positive on an open interval and null outside.If the probability measure with density function f is symmetric with respect to the origin and enjoys the shifting property then it is log-concave.
The log-concavity is equivalent to the shifting property 2.2.Lower bounds on the perimeter.In the following, we show that among sets of given measure and given asymmetry, the intervals or complements of intervals have minimal perimeter.
2.2.1.Structure of the sets with finite perimeter.Let Ω be a set of finite µ-perimeter.
where E is such that µ(E) = 0. Let k be an integer.On the compact K k the function f is bounded from below by a positive real.Thus if Ω ∩ K k has finite µ-perimeter, so it has finite Lebesgue perimeter.As mentioned in [AFP00,Fed69], one knows that every set with finite Lebesgue perimeter can be written as at most countable union of open intervals and a set of measure equal to zero.It holds where for all n in I k and d the euclidean distance over the real line.Denote 1 1 Ω the indicator function of Ω and 1 1 ′ Ω its distribution derivative.The property ( 9) is a consequence of the fact that 1 1 ′ Ω is locally finite (see [Fed69] for instance).Since K k is compact, the set I k is finite.One can check that the decomposition (8) becomes where I is at most countable, E such that µ(E ) = 0, and for all n in I. Notice that µ(E ) = 0. Without loss of generality, assume that Ω is an at most countable union of open intervals such that 1 1 ′ Ω is locally finite.2.2.2.Preliminaries.Let Ω be a set of finite µ-perimeter.As mentioned previously, assume that where I is an at most countable set and (10) holds.Suppose that • an isoperimetric projection of Ω is (−∞, σ − ) (using a symmetry with respect to the origin if necessary), • and that the measure of Ω is at most 1/2 (and we will see at the end of this section how to extend our result to larger measures).Then the real number

where
• Λ − and Λ + are at most countable sets; • and B h is either empty or of the form From Ω we build Ω 0 with same measure, same asymmetry, same isoperimetric projection, and lower or equal perimeter.Denote L = h∈Λ − A h and Using the isoperimetric inequality (6) with L, it follows that P µ (A 0 ) ≤ P µ (L).The same reason gives that there exist a real number α B 0 ≥ σ and a set B 0 = (α B 0 , +∞) with lower or equal perimeter than ∪ h∈Λ + B h (if non-empty).Shift to the left the intervals A ′ h until they reach I or −σ.Shift to the right the intervals B ′ h until they reach J or σ.The above operation did not change the amount of mass on left of −σ and on the right of σ.We build a set Ω 0 with same asymmetry and same isoperimetric projection as Ω and lower or equal perimeter, • I 0 is either empty or of the form I 0 = (α I 0 , β I 0 ) with α I 0 ≤ −σ < β I 0 ; • J 0 is either empty or of the form J 0 = (α J 0 , • and B 0 is either empty or of the form B 0 = (α B 0 , +∞) with α B 0 > σ.A case analysis on the non-emptiness of sets I 0 and J 0 is required to obtain the claimed result.Every step described below lowers the perimeter (thanks to the shifting lemma, Lemma 2.1) and preserves the asymmetry.Before exposing this, we recall that the set Ω 0 is supposed to have (−∞, −σ) as an isoperimetric projection.Thus we pay attention to the fact that it is totally equivalent to ask either the asymmetry to be preserved or the quantity λ(Ω 0 )/2 = µ(Ω 0 ∩ (−∞, −σ)) to be preserved trough all steps described below.
If I 0 and J 0 are both nonempty: Applying a symmetry with respect to the origin if necessary, assume that the center of mass of the hole between I 0 and J 0 is not less than 0. We can shift this hole to the right until it touches σ.Using the isoperimetric inequality (6), assume that there exist only one interval of the form (α ′ B 0 , +∞) on the right of σ.We get the case where I 0 is nonempty and J 0 is empty.If I 0 is nonempty and J 0 is empty: Shift the hole between A 0 and I 0 to the left until −∞ (there exists one and only one hole between A 0 and I 0 since Ω 0 is not a full measure set of (−∞, −σ)).We shift the hole between I 0 and B 0 to the right until +∞ (one readily checks that its center of mass is greater than 0).We get the only interval with same asymmetry and same isoperimetric projection as the set Ω 0 .This interval is of the form (the letter c stands for connected), ( 11) If J 0 is nonempty and I 0 is empty: Shift to the right the hole between J 0 and B 0 to +∞ (there exists one hole between J 0 and B 0 since Ω 0 is not a full measure set of (σ, +∞)).We obtain a set A 0 ∪ J ′ where J ′ is a neighborhood of σ.
• If µ(J ′ ) > µ(A 0 ), then shift J ′ to the right (which has center of mass greater than 0) till J ′ ∩ (σ, +∞) has weight equal to µ(A 0 ) (in order to preserve asymmetry).Using a reflection in respect to the origin, we find ourselves in the case where I 0 is nonempty and J 0 is empty.
which has center of mass greater than 0) to the right until +∞ and get the case where 0 and J 0 are both empty.If I 0 and J 0 are both empty: Then the set Ω 0 is of the form (d stands for disconnected), ( 12) Thus we proved the following lemma.
Lemma 2.2 -Let Ω be a measurable set with µ-measure at most 1/2 and λ(Ω) be the asymmetry of Ω.Then, it holds

2
, +∞ , are sets with same measure and same asymmetry as Ω.
Let us emphasize that the sets Ω c and Ω d have fixed isoperimetric projection (i.e.(−∞, −σ)), asymmetry, and measure.Observe that these properties are satisfied only for particular values of µ(Ω) and λ(Ω).

Domains of sets with minimal perimeter given measure and given asymmetry.
We are concerned here with the domain D = {(µ(Π), λ(Π)), Π measurable set}.The next lemma shows that asymmetry and perimeter are complement invariant.

Lemma 2.3 -For every sets A and B with finite µ-perimeter the following is true:
• • the perimeter is complement-invariant: P µ (A) = P µ (A c ), • and it holds m(A) = m(A c ) where m(A) = min {µ(A), 1 − µ(A)}.
Proof.One can check that the symmetric difference is complement-invariant (remark that 1 The essential boundary is also complement- invariant, thus Definition 2 shows that the µ-perimeter is complement-invariant.
Considering the symmetry of the isoperimetric function J µ , we pretend that the isoperimetric projections are complements of the isoperimetric projections of the complement.This latter property and the fact that the symmetric difference is complement-invariant give that the asymmetry is complement-invariant.The last equality is easy to check since µ is a probability measure.
Since asymmetry is complement-invariant, the domain D is symmetric with respect to the axis x = 1/2.Furthermore, we have the next lemma.
By construction, the sets Ω c and Ω d verify three properties: (1) their measure is µ(Ω), (2) their asymmetry is λ(Ω), (3) their isoperimetric projection is (−∞, −σ).We recall that µ(Ω) ≤ 1/2.Using Lemma 2.4, it is easy to check that Ω c satisfies these properties if and only if Using the definition of the isoperimetric projection, one can check that Ω d satisfies these properties if and only if Notice that on domain 0 ≤ λ(Ω) ≤ µ(Ω) both sets exist.On this domain, /2, we deduce from the concavity and the symmetry of the isoperimetric function that P µ (Ω d ) ≤ P µ (Ω c ) with equality if and only ifµ(Ω) = 1/2.Using (13) and ( 14), we have the following result.
Theorem 2.6 -Let Ω be a measurable set and λ(Ω) be the asymmetry of and these inequalities are sharp.
Proof.Let Ω be a measurable set.If Ω has infinite µ-perimeter the result is true, hence assume that Ω has finite µ-perimeter.We distinguish four cases as illustrated in Figure 3.
Lemma 2.2 says that Ω has greater or equal µ-perimeter than then from Lemma 2.5 we know that Ω d does not exist for such range of asymmetry.Necessary, it follows that P µ (Ω) ≥ P µ (Ω c ).Using (17), we complete (16).
If Ω has measure greater than 1/2, then 1 − µ(Ω) = m(Ω).The Lemma 2.3 shows how to deal with sets of large measure and allows us to consider either Ω or its complement.
where Ω c has same asymmetry and measure equal to m(Ω).Using (17), we get (16).This case analysis ends the proof.
The equalities (17) and the case analysis of the proof of Theorem 2.6 give the explicit lower bounds on µ-perimeter.
Proposition 2.7 -The sets (see Figure 3) • have the lowest perimeter given measure µ and given asymmetry λ.
Remark.The proof of Proposition 2.7 shows that, up to a negligible set, Ω c , Ω d , Ω c d and Ω c c are optimal given measure and given asymmetry.Moreover, it shows that the bounds in Theorem 2.6 are sharp.

Sharp estimate on the asymmetry
In this section we use Theorem 2.6 to get a sharp estimate on the asymmetry.As a matter of fact, we show that a set of given measure and almost minimal boundary measure is necessarily close to be half-line.
3.1.The isoperimetric deficit function.We concern with an upper bound on the asymmetry of sets of given measure and given perimeter.Let Ω be a set with finite perimeter.Define the isoperimetric deficit of Ω as From the inequalities (15) and (16) of Theorem 2.6, we can compute a lower bound on the isoperimetric deficit as the asymmetry ranges from 0 to its upper bound min(2 m(Ω), 1 − m(Ω)) (see Lemma 2.4).Define the isoperimetric deficit function K µ as follows.
The isoperimetric deficit function K µ (x, y) is defined on the domain of all the possible values of (m(Ω), λ(Ω)).The isoperimetric deficit function gives the lower bounds found in Theorem 2.6.The next lemma focuses on the variations of K µ .
Proof.The proof is essentially based on the concavity of J µ .
The variation at x is given by Using the symmetry with respect to 1/2 and the concavity of J µ , one can check that This discussion shows that y → K µ (x, y) is non-decreasing and lower semicontinuous on the whole domain.This ends the proof.
3.1.2.Lower bound on the isoperimetric deficit.In this section we give a convenient lower bound on the isoperimetric deficit.Define the function L µ as follows.
This ends the proof.
The lower bound given in Lemma 3.3 is the key tool of the proof of the continuity theorem.
3.2.The continuity theorem.In the following, we improve the lower bound (20) on the asymmetry.We begin with a remark.Consider the exponential case where One gets K exp = 0 on 0 ≤ y ≤ x ≤ 1/2.This means that there exists sets with a positive asymmetry and an isoperimetric deficit null.In the case of the exponential-like distributions (defined later on), the intervals (−∞, F −1 (r)) and (F −1 (1 − r), +∞) are not the only sets with minimal perimeter (up to a set of measure equals to 0) given measure r.
We specify this thought defining a natural hypothesis (H).Furthermore, we prove that the asymmetry goes to zero as the isoperimetric deficit goes to zero.

3.2.1.
The hypothesis H.We can get a better estimate on the asymmetry making another hypothesis.From now, suppose that the measure µ is such that This hypothesis means that J µ is non-linear in a neighborhood of the origin.We can be more specific introducing the property: Since t → J µ (t)/t is non-increasing, it is not difficult to check that (H) is the alternative hypothesis of (H).

The exponential-like case.
The exponential tail measures can be defined by the following property: Proposition 3.4 -The property (H) is equivalent to the property (E x p).
Proof.The proof is essentially derived [Bob94] from the equality (F −1 ) ′ (t) = 1/J µ (t), for all t ∈ (0, 1).Suppose that the measure satisfies (H).Using the above equality for sufficiently small values of r, one can check that F −1 (r) = 1 c log(r) + c ′′ , where c ′′ is a constant.Hence F(x) = exp(c(x − c ′′ )) = c ′ c exp(cx), which gives the property (E x p).Conversely, suppose that the measure satisfies (E x p).A simple computation gives the property (H).

Continuity of the asymmetry for non-exponential distributions.
The hypothesis (H) ensures that the distribution is non-exponential.It is the right framework dealing with continuity as shown in the next theorem.Theorem 3.5 (Continuity) -Assume that the measure µ satisfies the assumption H, then the asymmetry goes to zero as the isoperimetric deficit goes to zero.
Finally, it is easy to check that if K µ > 0 then there exists a neighborhood of 0 such that K µ is increasing.Taking a sufficiently small neighborhood if necessary, one can suppose that K µ is continuous (the only point of discontinuity of K µ is y = x).On this neighborhood, K −1 µ, x is a continuous increasing function.Using (20), this gives the expected result.
Roughly, it uncovers that a set of given measure and almost minimal boundary measure is necessarily close to be a half-line.Moreover we recover the following well-known result.
Corollary -Assume that the measure µ satisfies the assumption H, then the halflines are the only sets of given measure and minimal perimeter (up to a set of µ-measure null).
This last results ensure that the asymmetry is the relevant notion speaking of the isoperimetric deficit.