Bonnesen s inequality for John domains in R n

Transcription

1 Bonnesen s inequality for John domains in R n Kai Rajala Xiao Zhong Abstract We prove sharp quantitative isoperimetric inequalities for John domains in R n. We show that the Bonnesen-style inequalities hold true in R n under the John domain assumption which rules out cusps. Our main tool is a proof of the isoperimetric inequality for symmetric domains which gives an explicit estimate for the isoperimetric deficit. We use the sharp quantitative inequalities proved in [6] and [4] to reduce our problem to symmetric domains. Mathematics Subject Classification ): 3C65, 5A4. Introduction The sharp isoperimetric inequality states that.) nαn /n E n )/n P E) always holds for Borel sets E R n, n, with finite n-measure E. Here P E) is the n )-measure of the boundary the distributional perimeter, see Section ), and α n = B n. Equality holds in.) if and only if E is an n-ball up to a set of measure zero). The quantitative isoperimetric inequalities are estimates which improve the latter statement: if.) is almost an equality for a set E, then E is almost a ball with respect to some geometric quantity. A classical estimate of this type is Bonnesen s inequality.) l Ω) 4π Ω 4πR s), valid for Jordan domains Ω R. Here R and s are the circumradius and inradius of Ω, respectively. There is a large collection of similar inequalities for planar Jordan domains. Such inequalities are called Bonnesen-style inequalities, see [9]. Bonnesen s inequality does not hold for general domains in dimensions higher than two. This can be seen, for example, by gluing long, thin cusps to the unit n-ball. Recently, however, important quantitative isoperimetric Both authors were supported by the Academy of Finland.

2 inequalities for general Borel sets have been obtained in all dimensions. Let E R n be a Borel set and r its volume radius, that is, E = B n r) for a ball B n r). The isoperimetric deficit δe) is and the Fraenkel asymmetry λe) is P E) δe) = nαn /n, E n )/n E \ B n x, r) λe) = min. x R n E The following sharp result, which was conjectured by Hall [8], gives an estimate for the asymmetry of a set in terms of the isoperimetric deficit. Theorem. Fusco, Maggi, and Pratelli [6]). Let E be Borel measurable, < E <. Then λe) CδE) /, where C depends only on n. Related problems around isoperimetric, Sobolev, and other geometric and functional inequalities are currently under active investigation, cf. [6], [5], [3], and the references therein. In this paper, we take a different point of view concerning the extension of.) to higher dimensions. Namely, we want to find the natural extensions of the sharp Bonnesen-style inequalities by restricting the class of domains in a suitable manner. For convex domains, the following extension was found by Fuglede [4]. The metric distortion βω) of a bounded domain Ω is { } R s βω) = inf : there exists x such that Bx, s) Ω Bx, R). s Theorem. Fuglede [4]). Let Ω R n, n 3, be a bounded convex domain. Then /, C δω) log βω) δω)) n = 3 CδΩ) /n+), n 4. The constants depend only on n. Theorem. gives a Bonnesen inequality in all dimensions for the class of convex domains. Fuglede also gives examples to show that the theorem is sharp in all dimensions, except for the constants. On the other hand, the counterexample we mentioned above hints that such inequalities might hold in much greater generality, namely for domains for which cusps do not occur. The main purpose of this paper is to prove sharp estimates which show that this is indeed the case. We will apply the familiar John domain condition, which in particular rules out outward cusps.

3 Definition.3. A bounded domain Ω R n with a distinguished point x Ω is called a John domain if there exists a constant b > such that for all x Ω there is a path γ : [, l] Ω, parametrized by arclength, such that γ) = x, γl) = x, and dist γt), R n \ Ω) bt for every t l. We will also use the notation Ω, x ) if we want to emphasize x. John domains naturally occur in connection with several areas of mathematics, such as conformal and quasiconformal geometry, and the theory of Sobolev functions. Our first main theorem gives a sharp estimate for the outer metric distortion of a John domain. Let Ω be a bounded domain with circumradius R and volume radius r. We call αω) = R r)/r the outer metric distortion of Ω. Theorem.4. Let Ω R n, n 3, be a John domain. Then /, C δω) log αω) ϕδω)) := δω)) n = 3.3) CδΩ) /n ), n 4. The constants depend only on n and the John domain constant. In order to achieve a Bonnesen inequality corresponding to.) and Theorem., we also need to exclude inward cusps. Let Ω, x ) be a John domain. We denote Ω c = Bx, diamω)) \ Ω. Theorem.5. Let Ω R n, n 3, be a John domain such that also Ω c is a John domain. Moreover, assume that Ω = Ω c. Then the estimate.3) holds with α replaced by β. Theorems.4 and.5 are sharp, except for the constants. In dimension three this follows from the examples of Fuglede mentioned above. For the higher dimensional case, the sharpness is seen by gluing small truncated cones to the unit n-ball. It is perhaps surprising that in dimension three our estimate takes the same form as in the case of convex domains. A particular class of domains for which Theorem.5 can be applied is provided by quasiconformal mappings. Namely, let f : R n R n be a K- quasiconformal mapping. Then fb n satisfies the assumptions of Theorem.5, with John domain constant depending only on K and n. Results similar to Theorem.5 in connection with quasiconformal analysis have been obtained in [] and []. Our main method is a new proof of the isoperimetric inequality for domains symmetric with respect to a line, see Theorem 3. and Corollary

4 The proof is elementary and gives an explicit estimate for the isoperimetric deficit δ. We believe that this method is of independent interest. Another step is to reduce the case of general domains to the -dimensional case above. This is done by using symmetrization methods. In this step the Theorem. of Fusco, Maggi, and Pratelli, is applied, as well as Fuglede s Theorem.. Acknowledgements. We thank Juha Lehrbäck for inspiring discussions. Preliminaries We denote the k-balls with center x and radius r by B k x, r). Also, B k, r) = B k r), and B k ) = B k. We use the notation α k = B k, and ω k = kα k is the k )-measure of the boundary S k of B k R k. The k-dimensional Hausdorff measure H k is normalized such that H k B k ) = α k. Let E R n be Borel measurable, and Ω R n a domain. The perimeter of E in Ω, P E, Ω), is { } P E, Ω) = sup div ϕ dx : ϕ [Cc Ω)] n, ϕ. E For smooth sets E, P E, Ω) = H n E Ω). As usual, we denote P E, R n ) by P E). See [] for other basic properties. Recall that the Schwarz symmetrization F of a Borel set E with respect to a line, say the n:th coordinate axis, is defined as follows: F {x n = t} = B n r t ), where we identify R n with {x n = t}, and r t is defined by α n r n t = H n E {x n = t}). The Steiner symmetrization G of E with respect to a hyperplane, say {x n = t}, is defined as follows: given s R n, G {x = s, x n )} is the open line segment centered at t and with the same -measure as E {x = s, x n )}. Both Schwarz and Steiner symmetrizations preserve the n-measure of E, and do not increase the perimeter, cf. [6] and the references therein. 3 Isoperimetric deficit for domains symmetric with respect to a line Let f C, ) be a non-increasing and non-negative function such that f) =, f), and 3.) ft) f) t for all t [, ]. 4

5 Moreover, we assume that 3.) f) ft) Mt for every t [, ], for some M >. We denote 3.3) Ω f = {x : x n < f x ), x < }, where x = x, x n ) R n and x = x,..., x n ). Notice that 3.) implies that f) is the circumradius of Ω f. In this section, we prove that Theorem.4 holds for such a domain Ω f. Theorem 3.. Let f satisfy the above assumptions. Then the estimate.3) holds for Ω f, with constants depending only on M and n. We first notice that Assumption 3.) implies that Ω f contains a ball of radius ɛf), where ɛ depends only on M. Therefore, αω f ) is bounded from above by a constant depending only on M. This implies that in the proof of Theorem 3. we may assume that δω f ) <. We use the following auxiliary functions 3.4) φt) = for t [, ]. Then t + f s) ) s n ds, ψt) = fs)s n ds Ω f = ω n ψ), P Ω f ) = ω n φ). In Lemma 3. below, as well as in the proof, all of the inequalities become equalities when Ω f is the unit ball, that is, when ft) = f t) := t. In this case, we denote 3.5) φ t) = Thus, we have Moreover, we also have t s n s ds, ψ t) = t t s s n ds. α n = B n = ω n ψ ), P B n ) = ω n φ ). 3.6) nψ ) = φ ). We denote by φ the inverse function of φ. We use also the following auxiliary functions in the proof of Theorem 3.. We define h : [, φ)] [, φ)/φ )] as 3.7) ht) = φ) φ ) [ φ 5 ) φ ) φ) t)],

6 and define g : [, φ)) [, ) as [ gt) = φ ] φ ) 3 n φ) t). In the case n = 3, the function g is the identity function and ht) = t. It is easy to check that the function h is the prime function of g, that is, 3.8) h t) = gt), t, φ)). Now, we go back to Theorem 3.. The isoperimetric deficit of Ω f is 3.9) δω f ) = P Ω f ) nα n Ω f n n = ψ ) n n φ ) φ) ψ) n n. The circumradius of Ω f is f), and the volume radius is ω n ψ)/α n ) n = ψ)/ψ )) n. Therefore, 3.) αω f ) = ψ ) n f). ψ) n Theorem 3. claims that 3.) αω f ) C δω f ) log δω f ) CδΩ f ) ), n = 3 n, n 4, where C = Cn, M) >. We will prove 3.) in this section. The crux of the proof is the inequality in the following lemma. In the lemma, the functions ψ, φ and ψ, φ are defined as in 3.4) and 3.5), respectively. Lemma 3.. Let f be as in Theorem 3.. We have 3.) ψ) ψ )a n cn)af + G), where cn) >, a = φ)/φ )) n, 3.3) F = G = when n = 3; and G = when n 4. f t) a + f t) ) ) t n t + f t) ) a n gφt)) t n gφt)) n 3 ) + f t) ) t n dt dt, 6

7 Proof. To prove 3.), we start with the left hand of the equality. Integration by parts gives us ψ) = Then we use the trivial equality ft)t n dt = n f t)t n dt. xy = x + y x y), x, y R to write f t)t n = + f t) ) t n + a f t) t n a + f t) ) a f t) a + f t) ) ) t n t + f t) ). Here and in the following, the constant a = φ)/φ )) n in the lemma. Thus we arrive at 3.4) n )ψ) = a + a + f t) ) t n dt f t) + f t) ) t n dt af, is the same as where F is as in 3.3). The remaining part of the proof is to estimate the first integral in the right hand side of 3.4). We claim that 3.5) + f t) ) t n dt a t n + f t) ) dt + n )ψ )a n+ cn)a G, where G is defined as in the lemma. Combining 3.4) and 3.5), we obtain that n )ψ) aφ) + n )ψ )a n af cn)ag, from which, together with the fact 3.6), follows 3.). Therefore, it only remains to prove the claimed inequality 3.5). We need the following consequence of the Cauchy-Schwarz inequality. 3.6) t n = n ) where 3.7) Jt) = t t s n ds n )Jt) Ht), s n hφs)) + f s) ) 7 ds,

8 and 3.8) Ht) = t hφs)) + f s) ) s n ds = t hφs)) d φs)), where the function h is defined as in 3.7) and φ is as in 3.4). Now we estimate the first integral in the right hand side of 3.4) as follows. We write 3.9) + f t) ) t n dt = t d φt)). Then we use the inequality u γ v γ γu + γ)v c u v), u, v, for all γ, ), where c = cγ) >. We let γ = /n ) and We obtain u = ut) = a 3 n gφt))t n, v = vt) = a gφt)) n 3. t = ut) n vt) n 3 n n ut) + n 3 n vt) cn) ut) vt)), where cn) > is a constant when n 4. We comment on the above inequality in the special case n = 3. In this case we have γ =. We let cn) =. So, the last two terms in the above inequality vanish in this case. Recall that gs) = when n = 3. So the above inequality is trivial. Thus we have from 3.9) that 3.) + f t) ) t n dt n + n 3 n ut) d φt)) vt) d φt)) cn)a G, where G is defined as in the lemma. We remark here that when n = 3, this term vanishes. In this case, we let G =. To finish the proof of 3.5), we only need to estimate the two integrals on the right side of 3.). For the second one, we have 3.) vt) d φt)) = a n+ n )ψ ), where ψ is defined as in 3.5). Indeed, by changing of variables, and integrating by parts, vt) d φt)) =a φ) [ =a =a φ) φ ) gφt)) n 3 d φt)) φ ] φ ) φ) s) ds r n r dr = an+ n )ψ ). 8

9 This proves 3.). Then we apply 3.6) to estimate the first integral on the right side of 3.). We have 3.) ut) d φt)) = a 3 n gφt))t n d φt)) =a 3 n t n d hφt))) a 3 n n ) a 3 n n ) Jt) Ht) d hφt))) ) Jt) d hφt))) Ht) d hφt)))), where the second equality follows from the fact 3.8) that the function h is the prime function of g, and the last inequality from the Cauchy-Schwarz inequality. Recall that the functions J and H are defined in 3.7) and 3.8), respectively. We observe that by integration by parts, 3.3) Jt) d hφt))) = t n + f t) ) dt. We claim that 3.4) Ht) d hφt))) = a 3n 3 ψ ). Indeed, by changing of variables and integration by parts, we have Ht) d hφt))) = = = t φ) φ) φ) hφs)) d φs)) d hφt))) w hw) dw hr) dr dhw)) ) φ) φ) [ ] = φ φ ) φ ) φ) w) dw ) φ) 3 = r φ ) r n dr = a 3n 3 ψ ), which proves 3.4). Now we plug the equalities 3.3) and 3.4) into 3.) and obtain 3.5) ut) d φt)) n )ψ ) n+3 a t n + f t) ) dt. 9

10 To conclude the proof of 3.5), we go back to 3.). Plugging the estimates 3.5) and 3.) into it, we obtain + f t) ) t n dt ψ ) a n+3 t n + f t) ) + n 3)a n+ ψ ) cn)a G, dt from which 3.5) follows by the Cauchy-Schwarz inequality. This proves the claimed inequality 3.5), and hence the lemma. Now we can rewrite Lemma 3. to get an explicit estimate for the isoperimetric deficit δω f ) from below. Corollary 3.3. Let f be as in Theorem 3.. Suppose that δω f ). Then we have 3.6) F + G cn)a n δω f ), where F, G and a are as in Lemma 3.. Proof. We start with the isoperimetric deficit δω f ) and rewrite 3.9) as 3.7) + δω f ) = where a = φ)/φ )) n 3.) as follows. F + G cn) ψ) a ) n ψ ) n ψ) an, is the same as in Lemma 3.. Then we rewrite ) ψ ) ψ) an cn)a n + δω f )) n n ) cn)a n δω f ), since we assume that δω f ). This proves the corollary. In the remaining part of this section, we prove 3.), and hence Theorem 3.. We will show that the outer metric distortion αω f ) of the domain Ω f can be controlled by the integrals F and G in Lemma 3., and therefore by the isoperimetric deficit δω f ). Proof of Theorem 3.. We will estimate the integrals F and G in Lemma 3. from below. First, we deal with F. Let, ) be chosen later. By Hölder s inequality, f t) + f t) ) t ) dt a F + f t) ) t n dt.

11 We denote ρ) = + f t) ) t dt. The integral on the left side of the above inequality is f) ρ)/a. Thus we obtain 3.8) ρ) f) a F + f t) ) t n dt We continue to estimate the integral on the right side of the above inequality. It is easy to prove that 3.9) t ϕt) := t + f s) ) ds M + )t, t [, ]. Indeed, the first inequality is trivial, and the second follows from the assumption 3.) on the non-increasing function f, which we have not used before, ϕt) t f s)) ds = t + f) ft)) M + )t. This proves 3.9). Here, we also estimate φ). We have n φ) = + f t) ) t n dt ϕ) M +. Therefore, we have the following estimate for a = φ)/φ )) n 3.3) a. Here and in the following, we denote A B if two quantities A and B are comparable, that is, if there are constants c >, c > depending only on n and M, such that c A B c A. Now we apply 3.9) to estimate the integral on the right side of 3.8):. + f t) ) t n from which we deduce dt M + ) n =M + ) n + f t) ) ϕt) n dt dϕt)), ϕt) n + f t) ) t dt cm) log ϕ) ϕ) M + M + ) log,

12 and 3.3) + f t) ) t n dt cn, M) 3 n, when n 4. Now it follows from 3.8) that 3.3) ρ) f) a cf Φ) cδωf ) Φ), where c = cn, M) > and Φ) = { log M+, n = 3; 3 n, n 4. Here we used the estimate 3.6) on F and the estimate 3.3) on a. Second, we deal with the integral G. We will show that 3.33) ρ) a cg Φ) + c, where c = cn, M) > and Φ is as above. We divide the proof of 3.33) into two cases: n = 3 and n 4. In the case n = 3, we have ρ) = φ) and a = φ)/φ ) = φ). Thus we have by 3.9) that ρ) a = φ) φ) = + f t) ) t dt ϕ) c, which proves 3.33) in the case n = 3. To prove 3.33) in the case n 4, we start with the following estimate. It follows from 3.9) that n tn φ) φt) = t The function φ has the following property. +f s) ) s n ds t n ϕt) cn, M)t n, φ φ ) s) s n, s [, φ )]. Thus, from the above two estimates, we deduce that and therefore φ φ ) φ) φt)) t, t [, ], 3.34) gφt)) t 3 n.

13 Then Now we go back to the integral G. We denote ũt) = a n gφt)) t n, ṽt) = gφt)) n 3. G = We note that it follows from 3.34) that ũt) ṽt) dt. 3.35) ũt) + ṽt) t, t [, ]. Let It) = a 3 n gφt)) t 3 n. We will show that 3.36) It) ct n ũt) ṽt), t [, ]. Indeed, we apply the elementary inequality x γ y γ cγ) x y x + y) γ, x, y, γ >, cγ) > with x = ũt), y = ṽt) and γ = n 3)/n ) to obtain a 3 n gφt)) n 3 n t n 3 gφt)) n c ũt) ṽt) ũt) + ṽt)) n 5 n ct n 5 n ũt) ṽt), where the last step follows from 3.35). The left side of the above inequality is t n 3 gφt)) n It). Thus It) ct 3 n t n 5 n gφt)) n ũt) ṽt), from which, together with 3.34), the claimed inequality 3.36) follows. Now by 3.36) and the Cauchy-Schwarz inequality, we have 3.37) It) d φt)) G + f t) ) t n dt cg Φ), where c = cn, M) >. The last inequality follows from 3.3). We estimate the integral of I as follows. We note that It) d φt)) = a 3 n gφt)) d φt)) ρ). We have by changing variables that gφt)) d φt)) = φ) s φ) ds = φ ) s φ ) = an. 3

14 We also have by 3.34) and 3.9) that gφt)) d φt)) c + f t) ) t dt cϕ) c. It then follows from the above three inequalities that It) d φt)) ρ) a c, from which, together with 3.37), follows the claimed inequality 3.33). Finally, we combine 3.3) and 3.33) to obtain f) a cδω f ) Φ) + c. We assume 3.) on f, that is, f) f) M, [, ]. Thus we have 3.38) f) a cδω f ) Φ) + c. Now we estimate the outer metric distortion αω f ). By 3.) and 3.7), αω f ) = + δω f ) ) n f) a c f) a + cδω f ). Thus, by 3.38), we obtain the following estimate for αω f ) αω f ) cδω f ) Φ) + c + cδωf ), which holds for all, ). Then we choose = δω f ) n and we obtain the desired estimate 3.). The proof of Theorem 3. is finished. 4 Proof of Theorem.4 We now prove Theorem.4. We first assume that δω) δ, where δ > is a small number to be determined later. The John domain condition now implies that there exists a ball Bx, ɛ diamω)) Ω, where ɛ > depends only on the John domain constant. Since the circumradius R is bounded by diamω), we have αω) diam Ω r r ɛ Cɛ, δ )ϕδω)). Thus.3) holds when δω) δ. The case δω) δ is proved by combining Proposition 4. below, where the choice of δ is made, with Theorem 3.. 4

15 Proposition 4.. Let Ω R n be a John domain. Then there exists δ >, such that if δω) δ, then the following holds: there exists f : [, ] [, f)] such that f satisfies the assumptions of Theorem 3., and such that Ω f satisfies δω f ) δω) and αω f ) αω). Here δ and constant M in 3.) depend only on n and the John domain constant. Proof. Let B n x, r) be a ball which realizes the asymmetry λω). By translating Ω, if necessary, we may assume that x =. Let u be the smallest radius such that Ω B n u). Then the circumradius of Ω is not larger than u. Fix a point a Ω S n u). By rotating Ω about the origin, if necessary, we may assume that a = ue n. We now perform Schwarz symmetrization with respect to the x n -axis. We thus obtain Ω such that Ω {x n = t} is an n )-ball with center,,..., t). Moreover, Ω = Ω, δω ) δω), and ue n belongs to the boundary of Ω. Next, let ν R satisfy Ω + ν = Ω \ Ω + ν, where Then, if u r < ν, we have Ω + ν = Ω {x n ν}. α n r n λω) B n r) \ Ω B n r) {x n < ν} \ Ω B n r) / + Cνr n B n r) / Cνr n because Ω \ Ω + ν = B n r) /. Using Theorem., we conclude that u r r CλΩ) CδΩ) / CϕδΩ)) when δ is small enough. Therefore, we can assume that 4.) ν u r We define when δω) is small enough. Ω = Ω + ν T Ω + ν ), where T is the reflection with respect to {x n = ν}. Then Ω = Ω = B n r), and the closure of Ω contains both a = ue n and T a) = ν u)e n. We next estimate the deficit δω ). By the relative isoperimetric inequality, nαn /n Ω \ Ω + ν n )/n ω n r n /. We have P Ω ) ω n r n / + P Ω + ν, {x n > ν}) nαn /n Ω n )/n / + P Ω )/. 5

16 Therefore, since Ω = Ω = Ω, P Ω ) nαn /n Ω n )/n P Ω ) nαn /n Ω n )/n). We conclude that δω ) δω ) δω). We next estimate αω ). By 4.), the circumradius of Ω is bounded from below by a T a) / u u r)/ = u r)/ + r. But the circumradius of Ω is at most u, so αω ) αω) when δ is small enough. Using the John domain condition, and Theorem., we see that for every ν µ < a = u, there exists a point yµ) such that yµ)) n = µ, and such that Byµ), Cu yµ) n )) Ω. Here C depends only on the John domain constant. In particular, 4.) H n Ω {x n = t}) Cu t) n when t ν. By symmetry, similar estimate holds when t < ν. Finally, we perform another symmetrization, this time a Steiner symmetrization with respect to the hyperplane {x n = ν}. We translate such that ν =. Then the resulting domain is of the form Ω f as in 3.3), where 4.3) f x ) = H {gt) > x }). By 4.), and 4.3), f x ) H {u t > C x }) = H {t < u C x }) = u C x, and so f) f x ) = u f x ) C x. Also, f is clearly non-increasing and satisfies f) = and f x ) f) x after rescaling. Moreover, we can approximate f by smooth functions f j satisfying the assumptions of Theorem 3., such that both the outer metric distortions and the deficits converge, cf. [6, Theorem 6.]. The proof is complete. 5 Proof of Theorem.5 In this section we assume that Ω satisfies the assumptions of Theorem.5. Also, the constants appearing will depend only on n and the John domain constants of Ω and Ω c. We first notice that the argument in the beginning 6

17 of Section 4 shows that we can assume δω) δ, where δ can be chosen to be as small as desired. Let r be the volume radius of Ω. By translating Ω if necessary, we may assume that B n r) realizes the Fraenkel asymmetry of Ω. Let s and t be the largest and smallest radii, respectively, such that B n s) Ω B n t). Then βω) t s, r so it suffices to estimate t s)/r. We have the following nonsharp version of Theorem.5. Lemma 5.. If δ is small enough, then t s r CλΩ) /n CδΩ) /n. Proof. Using the John domain property of Ω, and the fact S n t) Ω, we find a ball B n x, Ct r)) Ω \ B n r). We conclude that 5.) Ct r) n λω)r n. Similarly, since S n s) \ Ω Ω c if s = ), the John domain property of Ω c gives a ball We conclude that B n y, Cr s)) Ω c B n r) = B n r) \ Ω. 5.) Cr s) n λω)r n. The first inequality follows by combining 5.) and 5.), and the second by Theorem.. Lemma 5. implies that if δω) is assumed to be small enough, then s is close to r, and strictly positive in particular. We can estimate t using Theorem.4. Lemma 5.. We have t s r ϕδω)) + r s. r 7

18 Proof. Let R be the circumradius of Ω, and Ω B n x, R). Since B n s) Ω B n x, R), we have x < R s. We conclude that In particular, Ω B n x, R) B n R + R s)). t s R + R s) s = R r) + r s). But R r can be estimated by rϕδω)) using Theorem.4. The lemma follows. In view of Lemma 5., Theorem.5 follows if we can estimate r s properly. We start by defining functions g and h as follows: gu) = H n S n u) \ Ω) and hu) = H n S n u) Ω) = ω n u n gu). Notice that gu) = when u < s, and gu) = ω n u n when u > t. The John domain properties of Ω c and Ω, respectively, show that there exists C > such that 5.3) gu) Cu s) n, and hu) Ct u) n for every s u t. We now perform spherical symmetrization on Ω; we obtain Ω such that Ω S n u) is a relatively open spherical cap in S n u), with center ue n, and n )-measure hu). Then Ω = Ω. Moreover, by [3, Theorem.4.], we may assume that Ω is polyhedral. Then P Ω ) P Ω), cf. [7]. We split the rest of the proof into two cases. We first assume that 5.4) Ω { t < x n < s + 3r s)/4} =: V. We consider the following problem: find a Borel set E V with E = Ω = Ω, such that P E) P F ) for every F V with the same properties. Now a solution to this problem exists, and by slightly modifying V, we can apply [] to conclude that E is convex. Then we are in the setting of Theorem., so, in particular, 5.5) βe) ϕδe)) ϕδω )) ϕδω)). On the other hand, the volume radius of E is r, and, since E V, the inradius is at most t + s)/ + 3r s)/8. Thus, as in the proof of Lemma 5., 5.6) βe) r t s 3r s)/4 r 8 r s 8r CϕδΩ)).

19 Combining 5.5) and 5.6) yields r s)/r CϕδΩ)), which, in view of Lemma 5., yields the claim when we assume 5.4). Next we assume that 5.4) does not hold. Notice that Ω { t < x n }. Let µ > s + 3r s)/4 be the smallest number such that Ω {x n < µ}. Then Ω {x n = µ} consists of one or more possibly infinitely many) n )- spheres S n u) {x n = µ}. We denote the smallest such u by v. Then B n v) {x n < µ} \ Ω has a unique component U containing {,..., x n ) : s < x n < µ}. Let U be the reflection of U with respect to {x n = µ}, and define Ω as the interior of Ω U U. Then P Ω ) P Ω ), and the circumradius R satisfies and the volume is R = t + µ s t + s + 3r s)/ Ω + U Ω + U B n r) + B n t) \ B n r) r + r s)/4, α n r n + α n λω)r n + α n t n r n ) α n r n + CϕδΩ))) by Lemma 5. and Theorem.. So the volume radius r of Ω is at most r + CϕδΩ))), and αω ) = R r r C r + r s)/4 r CrϕδΩ)). r Therefore, r s αω ) + CϕδΩ)). 4r Now we can use Theorem 3. to estimate αω ). Namely, using 5.3), and applying Schwarz and Steiner symmetrizations as in the proof of Theorem.4, we obtain a domain Ω f satisfying the assumptions of Theorem 3., such that αω f ) αω ), and δω f ) δω ). Combining Theorem 3. with the above estimates yields r s r as desired. The proof is complete. References CϕδΩ)), [] Ambrosio, L., Fusco, N., Pallara, D.: Functions of bounded variation and free discontinuity problems, Oxford Mathematical Monographs, Oxford University Press,. 9

20 [] Cianchi, A., Fusco, N., Maggi, F., Pratelli, A.: The sharp Sobolev inequality in quantitative form, J. Eur. Math. Soc. JEMS), 9), no. 5, [3] Figalli, A., Maggi, F., Pratelli, A.: A mass transportation approach to quantitative isoperimetric inequalities, Invent. Math., 8 ), no., 67. [4] Fuglede, B.: Stability in the isoperimetric problem for convex or nearly spherical domains in R n, Trans. Amer. Math. Soc., ), no., [5] Fusco, N., Maggi, F., Pratelli, A.: The sharp quantitative Sobolev inequality for functions of bounded variation, J. Funct. Anal., 44 7), no., [6] Fusco, N., Maggi, F., Pratelli, A.: The sharp quantitative isoperimetric inequality, Ann. of Math. ), 68 8), no. 3, [7] Gehring, F. W.: Symmetrization of rings in space, Trans. Amer. Math. Soc., 96), [8] Hall, R. R.: A quantitative isoperimetric inequality in n-dimensional space, J. Reine Angew. Math., 48 99), [9] Osserman, R.: Bonnesen-style isoperimetric inequalities, Amer. Math. Monthly, ), no., 9. [] Rajala, K.: The local homeomorphism property of spatial quasiregular mappings with distortion close to one, Geom. Funct. Anal., 5 5), no. 5, 7. [] Rajala, K.: Quantitative isoperimetric inequalities and homeomorphisms with finite distortion, preprint 376, University of Jyväskylä, 9. [] Rosales, C.: Isoperimetric regions in rotationally symmetric convex bodies, Indiana Univ. Math. J., 5 3), no. 5, 4. [3] Ziemer, W. P.: Extremal length and conformal capacity, Trans. Amer. Math. Soc., 6 967), University of Jyväskylä Department of Mathematics and Statistics P.O. Box 35 MaD) FI-44 University of Jyväskylä Finland kai.i.rajala@jyu.fi, xiao.x.zhong@jyu.fi