aa r X i v : . [ m a t h . M G ] F e b Reverse isoperimetric inequalities for parallel sets
Piotr Nayar ∗ February 9, 2021
Abstract
We consider the family of r -parallel sets in R d , that is sets of the form A r = A + rB n ,where B n is the unit Euclidean ball and A is arbitrary Borel set. We show that theratio between the upper surface area measure of an r -parallel set and its volume isupper bounded by d/r . Equality is achieved for A being a single point.As a consequence of our main result we show that the Gaussian upper surface areameasure of an r -parallel set is upper bounded by 18 d max( √ d, r − ). Moreover, weobserve that there exists a 1-parallel set with Gaussian surface area measure at least0 . · d / .2010 Mathematics Subject Classification : Primary 52A40; Secondary 60G15.
For sets
A, B is R d we define their Minkowski sum A + B = { a + b : a ∈ A, b ∈ B } .Suppose K is some compact convex set and let r >
0. A set of the form A r,K = A + rK is called ( r, K )-parallel. If K = B d is a unit Euclidean ball then A r,K will simply becalled r -parallel and will be denoted by A r . For a Borel set A we shall write | ∂A | + = lim sup ε → + | A ε | − | A | ε , | ∂ K A | + = lim sup ε → + | A ε,K | − | A | ε , where | · | stands for the Lebesgue measure. The quantity | ∂A | + is called the uppersurface area measure of A .In [2] Jog considered reverse isoperimetric inequalities for parallel sets. He provedthat for any compact set A in R d one has | ∂A r | ≤ d d − | A r | r . In this note we prove thefollowing sharp result.
Theorem 1.
Let A be a Borel set and let K be compact and convex. Then | ∂ K A r,K | + ≤ dr · | A r,K | , which is tight for A = { } . In particular | ∂A r | + ≤ dr · | A r | , . ∗ The author was supported by the National Science Centre, Poland, grant 2018/31/D/ST1/01355. n [2] also the Gaussian case was treated. The upper Gaussian surface area of ameasurable set is defined as γ + d ( ∂A ) = lim sup ε → + γ d ( A ε ) − γ d ( A ) ε , where γ d stands for the standard Gaussian measure, that is measure with density(2 π ) d/ e −| x | / . Jog proved the inequality γ + d ( ∂A r ) ≤ d − d d max(1 , r − ). We shallprove this bound with a better dependence on the dimension. Theorem 2.
For any Borel set A we have γ + d ( ∂A r ) ≤ d max( √ d, r − ) . We mention that it is not possible to remove the dimension dependence in the aboveestimate: there exists a 1-parallel set whose Gaussian surface area is of order d / . Thisfollows from a simple observation: every set of the form K c (complement of K ), where K is open and convex, is r -parallel for every r >
0. It is easy to verify that any closedhalfspace H is r -parallel for every r > H = A r for an appropriate halhspace A ). Since every open convex set K is of the form K = T i ∈ I H i for some family of openhalfspaces ( H i ) i ∈ I , we have K c = [ i ∈ I H ci = [ i ∈ I ( A i ) r = (cid:16) [ i ∈ I A i (cid:17) r , where sets A i satisfy ( A i ) r = H ci (note that H ci are closed halfspaces). According to theresult of Nazarov from [3], there exists a convex set K such that γ + d ( ∂K ) ≥ . · d / . We first prove Theorem 1.
Proof of Theorem 1.
According to the result of Fradelizi and Marsiglietti from [1] (Propo-sition 2.1), for any compact set A in R d and any compact convex set K in R d thefunction ( s, t ) sA + tK | is non-decreasing on R + × R + in each coordinate. This is a consequence of the resultof Stach´o from [4]. Since for r > | A + rK | r d = | r − A + K | , the left hand side isnon-increasing. Thus, for any ε > ≥ ε (cid:18) | A + ( r + ε ) K | ( r + ε ) d − | A + rK | r d (cid:19) = 1( r + ε ) d · | A + ( r + ε ) K | − | A + rK | ε + | A + rK | · r + ε ) d − r d ε . Taking ε → + we arrive at 0 ≥ | ∂ K A r,K | + r d − dr d +1 | A r,K | . The proof is completed. heorem 1 implies Theorem 2. For the proof we follow the strategy developed in[2]. Proof of Theorem 2.
Through the proof c is a universal constant independent of the di-mension, whose value may change from one line to the next. Note that for a measurableset A we have γ d ( A ) = (2 π ) − d Z A e −| x | / d x = (2 π ) − d Z A Z ∞| x | te − t / d t d x = (2 π ) − d Z ∞ Z R d te − t / {| x |≤ t } A d x d t = (2 π ) − d Z ∞ te − t / | A ∩ tB d | d t. Let us fix ε > < ε < ε . Let A t = A ∩ ( t + r + ε ) B d . We have γ d ( A r + ε ) − γ d ( A r ) = (2 π ) − d Z ∞ te − t / | ( A r + ε \ A r ) ∩ tB d | d t = (2 π ) − d Z ∞ te − t / | (( A t ) r + ε \ ( A t ) r ) ∩ tB d | d t ≤ (2 π ) − d Z ∞ te − t / | ( A t ) r + ε \ ( A t ) r | d t. Dividing by ε , taking the limit ε → + and applying Theorem 1 gives γ + d ( ∂A r ) ≤ (2 π ) − d Z ∞ te − t / | ∂ ( A t ) r | + d t ≤ (2 π ) − d · dr Z ∞ te − t / | ( A t ) r | d t ≤ (2 π ) − d · dr Z ∞ te − t / | ( t + 2 r + ε ) B d | d t = | B d | (2 π ) d · dr Z ∞ te − t / ( t + 2 r + ε ) d d t. Taking the limit ε → + yields γ + d ( ∂A r ) ≤ | B d | (2 π ) d · dr Z ∞ te − t / ( t + 2 r ) d d t. For p > − m p := Z ∞ e − t / t p d t = 2 p − Γ (cid:18) p + 12 (cid:19) . We also have | B d | = π d/ Γ( d +1) . As a consequence m d +1 = 2 d Γ (cid:0) d + 1 (cid:1) = (2 π ) d / | B d | andthus γ + d ( ∂A r ) ≤ dr · m d +1 · Z ∞ te − t / ( t + 2 r ) d d t = dr · m d +1 · d X i =0 (cid:18) di (cid:19) m i +1 (2 r ) d − i = dr · d X i =0 (cid:18) di (cid:19) Γ( i + 1)Γ( d + 1) 2 i − d (2 r ) d − i . sing the standard bounds √ πxx x e − x ≤ Γ( x + 1) ≤ √ πxx x e − x , x ∈ [1 , ∞ ) ∪ { / } . and (cid:0) di (cid:1) ≤ d d − i ( d − i )! we get d X i =0 (cid:18) di (cid:19) Γ( i + 1)Γ( d + 1) 2 i − d (2 r ) d − i ≤ d X i =0 d d − i ( d − i )! · ( i/ i/ e − i/ ( d/ d/ e − d/ ( √ r ) d − i = 2 d X i =0 d d − i ( d − i )! · i i/ d d/ (2 √ er ) d − i . Let us now assume that r ≤ r ∗ := d − / √ e . Then2 d X i =0 d d − i ( d − i )! · i i/ d d/ (2 √ er ) d − i ≤ d X i =0 d d − i ( d − i )! · i i/ d d/ · d d − i = 2 d X i =0 d − i )! · (cid:18) id (cid:19) i/ ≤ d X i =0 d − i )! ≤ e. Therefore, for r ≤ r ∗ we get γ + d ( ∂A r ) ≤ der .If r > r ∗ one can use the bound for r = r ∗ , since every r -parallel set is r ′ parallelfor every r ′ < r . Thus in this case we get γ + d ( ∂A r ) ≤ e / d / < d / . We provedthat always γ + d ( ∂A r ) ≤ max(18 d / , der ) ≤ d max( √ d, r ). References [1] M. Fradelizi, A. Marsiglietti, On the analogue of the concavity of entropy power inthe Brunn-Minkowski theory, Advances in Applied Mathematics 57, 2014, 1–20.[2] V. Jog, Reverse Euclidean and Gaussian isoperimetric inequalities for parallel setswith applications, arXiv:2006.09568[3] F. L. Nazarov, On the maximal perimeter of a convex set in R n with respect toGaussian measure, Geometric Aspects of Func. Anal. 1807, 2003, 169–187.[4] L. Stach´o, On the volume function of parallel sets, Acta Scientiarum Mathemati-carum 38, 1976,365–374.Institute of MathematicsUniversity of WarsawBanacha 2, 02-097, Warsaw, Polandemail: [email protected] respect toGaussian measure, Geometric Aspects of Func. Anal. 1807, 2003, 169–187.[4] L. Stach´o, On the volume function of parallel sets, Acta Scientiarum Mathemati-carum 38, 1976,365–374.Institute of MathematicsUniversity of WarsawBanacha 2, 02-097, Warsaw, Polandemail: [email protected]