On the isoperimetric problem with perimeter density r^p

In this paper the author studies the isoperimetric problem in $\re^n$ with perimeter density $|x|^p$ and volume density $1.$ We settle completely the case $n=2,$ completing a previous work by the author: we characterize the case of equality if $0\leq p\leq 1$ and deal with the case $-\infty<p<-1$ (with the additional assumption $0\in\Omega$). In the case $n\geq 3$ we deal mainly with the case $-\infty<p<0,$ showing among others that the results in $2$ dimensions do not generalize for the range $-n+1<p<0.$


Introduction
Let Ω ⊂ R n be a bounded open set with Lipschitz boundary ∂Ω. We study the inequality nL n (Ω) ω n−1 n+p−1 n ≤ 1 ω n−1 ∂Ω |x| p dH n−1 (x). (1) where L n (Ω) denotes the n dimensional Lebesgue measure of Ω, H n−1 is the n − 1 dimensional Hausdorff measure and ω n−1 = H n−1 (S n−1 ) is the surface area of the unit n − 1 sphere. For p = 0 this is the classical isoperimetric inequality. Note that for balls centered at the origin there is always equality. For p < 0, we introduce a new condition: Ω shall contain the origin. The inequality (1) is a particular case among a broader class of problems called isoperimetric problems with densities. Given two positive functions f, g : R n → R one studies the existence of minimizers of I(C) = inf ∂Ω g dH n−1 : Ω ⊂ R n and Ω f dL n = C .
Concerning the inequality (1) the following results are already known: In the case 0 ≤ p < ∞ the inequality (1) always holds. This has been first proved in [3] for p ≥ 1. Another proof of the same result can be found in [15], Section 7. Later (1) was shown by the author [11] for 0 ≤ p ≤ 1 if n = 2, and then by [1] for any n ≥ 3. See also the very recent paper [20], which uses variational methods and the study of the Euler equations satisfied by a minimizer. The method of [1] contains a very original interpolation argument with which (1) is deduced from the classical isoperimetric inequality. For the sake of completeness we repeat this proof in the special case of starshaped domains, see Proposition 15. The general case follows from a more standard, but weighted, symmetrization argument, see [1].
In the case p < 0 the only available result is [11]. This result shows that if n = 2 and −1 ≤ p ≤ 0 then (1) holds true under the additional assumptions Ω is connected and contains the origin. The motivation for studying negative values of p and adding these additional assumptions come from the singular Moser-Trudinger functional [2]. These assumptions arise naturally in the harmonic transplantation method of Flucher [18] (see [12] and [13]) to establish the existence of extremal functions for the singular Moser-Trudinger functional. Without going into the details, the connection is the following: one uses (1) for the level sets of the Greens function G Ω,0 with singularity at 0. Obviously these level sets will always contain the origin and will be connected by the maximum principle.
In the present paper we make the following further contributions. It turns out that there are big differences depending on the dimension n.
The case n = 1. The inequality (1) is elementary, but we have included it for completeness. The inequality hols for all p ∈ R\(0, 1). If p ∈ (0, 1), then minimzers of the weighted perimeter still exist, but they are not intervals centered at the origin.
The case n = 2. We settle the case of equality if 0 < p < 1 showing that balls centered at the origin are the unique sets satisfying equality. Moreover we also deal with p < −1 and prove also the sharp form in that case. Thus we settle completely the 2 dimensional case and the results are summarized in Theorem 3.
The case n ≥ 3. We completely tackle the case p < 0. We prove that (1) remains true if p < −n + 1. This proof is basically the same as that of [3], with a slight difference in the proof of the unicity result. However, the rather surprising result is that the inequality does not hold true for any p between −n + 1 and 0. Particularly interesting is the case −n + 1 < p < −n + 2. In that range the variational method shows that balls centered at the origin are stationary and stable, but nevertheless they are not global minimizers. We actually prove something stronger: there are no minimizers of the corresponding variational problem (2) if −n + 1 < p < 0 and the infimum is zero.

The 1 dimensional case
In this case H 0 is the counting measure, ω 0 = 2 and the inequality (1) becomes We assume that Ω is the union of disjoint open intervals, such that the closed intervals do not intersect. If p = 0, and there is just one interval, then the inequality trivially reads as 1 = 1. We have the following proposition.
Proposition 1 (i) If p ≥ 1, then inequality (3) holds true for any Ω. In case of equality Ω must be single interval and if p > 1 this interval has to be centered at the origin.
(iii) If p < 0, then inequality (3) holds true for all Ω containing the origin. In case of equality Ω has to be a single interval centered at the origin.

Remark 2
In case (ii) one can still ask the question whether there is a minimizer for the weighted perimeter, under the constraint L 1 (Ω) = c. One can verify that the unique minimizers are the intervals (0, c) or (−c, 0).
Proof (i) Let a = min{x| x ∈ ∂Ω} and b = max{x| x ∈ ∂Ω}. Then L 1 (Ω) ≤ (b − a) ≤ |a| + |b| and the inequality follows from the fact that the map s → s p is increasing and convex.
Using that s → s p is decreasing and convex one obtains easily the result.

The 2 dimemsional case
The following theorem summarizes the works by Betta-Brock-Mercaldo-Posteraro [3], Csató [11] and the results proven in the present paper. In this section |Ω| = L 2 (Ω) shall denote the area of a set and σ is the 1-dimensional Hausdorff measure. We will often just say that a set Ω is C k meaning that its boundary ∂Ω is a C k curve.

Theorem 3
Let Ω ⊂ R 2 be a bounded open Lipschitz set. Regarding the inequality the following statements hold true: (i) If p ≥ 0, then (4) holds for all Ω.
(iv) In case (i), if Ω is C 2 and there is equality in (4), then Ω must be a ball; centered at the origin if p = 0. If there is equality in case (ii) and (iii) then also Ω must be a ball. [21] that if one has equality in (4) (i.e. Ω is a minimizer), then ∂Ω\{0} is smooth. On the other hand if one shows the inequality (4) for smooth sets, then it also holds for Lipschitz sets by approximation. So we can work with smooth sets and have to be careful only if 0 ∈ ∂Ω.

Remark 4 It follows from Morgan
Proof (i) and (ii) have been proven in [11] and [3], (iii) will be proven in Theorem 12. The case of equality has been proven for (ii) in [11] and for (iii) it is again a special case of Theorem 12. So it remains to deal with the case of equality in (i). This has also been dealt with in [11] as long as 0 / ∈ ∂Ω or p > 1. The case 0 ∈ ∂Ω is Proposition 5 The rest of this section is devoted to the proof of the following Proposition 5. It is based on variational methods and a careful analysis of the resulting Euler-Lagrange equation.
then 0 cannot lie on the boundary ∂Ω.
The proof of Proposition 5 is based on two lemmas. The first one, Lemma 6, is the variational formula that we need, establishing the Euler-Lagrange equation satisfied by a minimizer. It is the generalization of the statement that minimizers of the classical isoperimetric problem have constant curvature, with the difference that one introduces a generalized curvature. Such formulae are broadly used to deal with isoperimetric problems with densities, see for instance [5], [10], [14] [15], respectively [22] for a summary. Since in the present case the derivation is very short and elementary, we provide the proof to make the presentation self-contained. Afterwards, in Lemma 8, one uses the symmetry properties of the Euler-Lagrange equation to show that minimizers are symmetric with respect to any line through the origin and a point P on ∂Ω with maximal distance from the origin. Such symmetrization arguments are also well known, see for instance Lemma 2.1 in [14]. Then to conclude the proof of Proposition 5 one essentially compares the generalized curvature, which has to be constant, at the point P and at the origin, which will lead to a contradiction.
This can always be achieved by reparametrization and chosing the orientation properly. The prime (·) shall always denote the derivative with respect to the argument of γ. If |γ | = 1 then the curvature κ of ∂Ω calculates as κ(t) = γ (t), ν (t) .

Proof
Step 1. Without loss of generality we assume the second case 0 ∈ ∂Ω and assume that γ(0) = γ(L) = 0. We can also assume that (5) holds, since (7) is independent of the parametrization. Let h ∈ C ∞ c (0, L) be arbitrary and define the curve g s (t) = γ(t) + sh(t)ν(t). For all s small enough this is also a simple closed curve bounding a domain Ω s . Because Ω is a minimizer, we claim that Let us show (8). If this is not the case, this means that the function h is such that the first equality in (8) holds true but Then define Ω s = λ s Ω s where λ s = |Ω|/ |Ω s |. Note that | Ω s | = C for all s and that, by the first iquality in (8), d/ds (λ s ) = 0 at s = 0. Hence we obtain that Thus for sufficiently small s (negative or positive depending on the sign of the last inequality) the weighted perimeter of Ω s is strictly smaller than that of Ω. This contradicts the fact that Ω is a minimizer.
Step 2. We now calculate the derivatives in (8) explcitly. First one gets We therefore obtain which leads to, using partial integration to get rid of derivatives of h, On the other hand we have We therefor obtain that d ds ∂Ωs |x| p dσ Note that Thus we get Let us now calculate B. As above This leads to d ds Setting this into B and adding A + B we finally obtain that (8) This implies the claim of the lemma.
Lemma 8 Let p ∈ R, C > 0 and suppose that Ω is a C 2 minimizer of (6). Assume Then γ is symmetric with respect to the line through γ(t 0 ) and the origin.
The functions F 1 and F 2 are C 1 and hence Lipschitz as long as γ = 0. They have the properties Using these properties, it can be easily verified, by evaluating the differential equation at −t, that also the curve ω(t) = (γ 1 (−t), −γ 2 (−t)) satisfies the initial value problem. By uniqueness we obtain that ω(t) = γ(t). Since (F 1 , F 2 ) is locally Lipschitz, by the theorey of ordinary differential equations (see for instance [24] page 68) the solution exists either for all times t, it blows up or goes out of the region of definition of (F 1 , F 2 ). In the present case this means that the solution γ exists and is unique for all times t, unless |γ| goes to 0 or to ∞. But it cannot go to infinity, because then the weighted perimiter would go to infinity and then Ω cannot be a minimizer. So we conclude, using that γ is continuous by assumption, that ω(t) = γ(t) for all t and this shows the claim of the lemma. Proof of Proposition 5 Step 1. Let us show first that Ω has to be a simply connected set. If Ω is not connected, let us say Ω = Ω 1 ∪ Ω 2 and Ω 1 ∩ Ω 2 = ∅ for two nonempty sets, Using the hypothesis that there is equality in Proposition 4 and Theorem 3 part (i) for Ω i we get In the last inequality we have assumed that 0 < p < 1 and used that a s + b s > (a + b) s , if 0 < s < 1 and a, b > 0. If p = 1 there is still strict inequality because for one of the i = 1, 2 we must have ∂Ωi |x| p > 2πR If not, assume that we have shown Theorem 3 (iv) first for connected sets. Then both Ω 1 and Ω 2 would have to be balls centered at the origin, a contradiction to Ω 1 ∩ Ω 2 = ∅.
Assume now that Ω is not simply connected and is of the form Ω = Ω 0 \Ω 1 for some Ω 1 ⊂ Ω. Then we obtain using Theorem 3 (i) that which is again a contradiction.
Step 2. We now assume that 0 ∈ ∂Ω and show that this leads to a contradiction. Let γ : [0, L] → ∂Ω be as in (5) and γ(0) = γ(L) = 0. Without loss of generality, by rotating the domain, we know by Lemma 8 that the maximum of |γ| has to be achieved at t = L/2, that for some d > 0 and that γ is symmetic with respect to the x 1 axis {(x 1 , x 2 ) ∈ R 2 | x 2 = 0}. Using this symmetry and the chosen orientation of γ, one obtains that Note also that ν(L/2) = (1, 0). Hence we obtain for the generalized curvature at L/2 that Step 3. Using again Lemma 8 we obtain that γ 1 (−t) = γ 1 (t) and therefore γ 1 (0) = 0 and γ 2 (0) = −1. Therefore γ 2 is invertible near zero and . This new parametrization allows us to reduce the problem to a 1-dimensional one, analyzing the function f. Since α = (f , 1), α = (f , 0) and keeping the reversed orientation in mind, we have the following formulas for the outer normal ν and curvature κ where m(t) is just an abbreviation for the right hand side.
This will be a contradiction to (10) and the fact that k has to be constant, by Lemma 6.
Using the facts that p > 0, lim t→0 |α(t)| = 0, lim t→0 1 + f 2 = 1, and f ∈ C 2 , to prove (12), it suffices to show that It follows from de l' Hopital rule that which proves that (13) holds and the claim of Step 4.

Some results in general dimensions n ≥ 3
In this section we shall use the following notations and abbreviations: let n ∈ N Recall that α n = ω n−1 /n. If the dimension is obvious we omit n in the superscript e.g. B δ = B n δ and the same for S n δ . We also set |Ω| := L n (Ω). The following is the main theorem, summarizing the results of [1], [3] and the present paper.
the following statements hold true: (i) If p ≥ 0 then (15) holds true for all Ω.
(ii) If −n + 1 < p < 0, then we have for any C > 0 inf ∂Ω |x| p dH n−1 : 0 ∈ Ω ⊂ R n connected and |Ω| = C = 0, where the infimum is taken over all bounded open smooth sets. In particular (15) cannot hold for all Ω containing the origin.
(iv) If there is equality in case (i) or (iii), then Ω must be a ball, centered at the origin if p = 0.
Remark 10 In (ii), if one omits the condition that Ω has to contain the origin, then it is trivial that the infimum is zero, by taking any domain with constant volume and shifting it to infinity. This is true in any dimension, in particular also if n = 2, compare with Theorem 3 (ii).
Proof Part (i) and its sharp form (iv) have been proven in [1] and [3], respectively the classical isoperimetric inequality. See also Proposition 15 for the case 0 < p < 1 and for starshaped domains. Part (ii) will be proven at the end of this section. Part (iii) and its sharp form (iv) will follow from Theorem 12, by setting a(t) = t p .
Although our results will not need the variational method, it should be mentioned, since it gives immediately some results on the nonvalidity of (1) for the range −n + 2 < p < 0. It can be easily seen with this method that if one takes a ball centered at the origin and moves it slightly away in any direction, then the weighted perimeter decreases. It actually continues to decrease even more as one moves it further and further away. Moreover, for the range −n + 1 < p < −n + 2, Part (ii) of Theorem 9 gives an interesting example for a case when balls centered at the origin are local but not global minimizers of the weighted perimeter. We shall illustrate first the variational method with an example.
Example 11 Fix r > 0 and let Ω s = B r (0) + (s, 0), where we will understand from now on (s, 0) = (s, 0, . . . , 0) ∈ R n and s ∈ R. Obviously |Ω s | = α n r n for all s and Ω s contains the origin for all s < r. We shall denote the weighted perimeter as P (Ω s ) = ∂Ωs |x| p dH n−1 .
Assume that F = (F 1 , F 2 , . . . , F n ) = (F 1 ,F ) : U ⊂ R n−1 → ∂B r (0) is a parametrization of ∂B r (0) up to a set of H n−1 measure 0. Let g(u)du denote the surface element of ∂B r (0) in this parametrization, with u ∈ U. Then F + (s, 0) = (F 1 + s,F ) is a parametrization of ∂Ω s and one obtains that This gives d ds P (Ω s ) s=0 = p ∂Br |x| p−2 x 1 dH n−1 = 0 for any p ∈ R.
One can now easily verify that if n ≥ 3 In view of (17) and the third inequality in (18), one obtains that P (Ω s ) decreases for increasing s near the origin. This shows that (15) cannot hold true if −n + 2 < p < 0.
The equations (18) remain true for much more general variations. We will not use them but nevertheless summarize the result in the present case. For the methods and proofs we refer to [5], [16] and [22]. Let ϕ s : R n → R n be a diffeomorphism for small s.
Define Ω s = ϕ s (B r (0)) and assume that the variaton is normal: ∂ϕ s /∂s = u is normal to ∂Ω at s = 0. Moreover we assume that the variaton is such that it is volume preserving: |Ω s | = |Ω| for all s. This implies, calculating the first variation of the volume ∂Br udH n−1 = 0.
Assume we are given a radial perimeter density G : (0, ∞) → (0, ∞), smooth in a neighborhood of r, and P G is defined by Then the second variation for P G is given by where ∇ u is the covariant derivative along ∂B r . By the standard isoperimetric inequality one has that M 1 ≥ 0 for any u satisfying (19). This follows again by variational methods (taking G = 1), or equivalently, it is the Poincaré inequality on the sphere with optimal constant, see [16] Example (2.13). Moreover one can show by the first variation of P G that balls centered at the origin are always critical points of P G under the volume constraint. So this implies that B r can be a minimizer of P G if and only if M 2 ≥ 0, i.e.
(n − 1) In the present case G(r) = r p , so the last inequality holds true if and only if p(n+p−2) ≥ 0. This is precisely again (18). Note that if −n + 1 < p < −n + 2, then p(n + p − 2) > 0, but nevertheless Theorem 9 (ii) shows that balls centered at the origin are not global minimizers.
We shall now prove Part (iii) of Theorem 9. However this can be done for much more general densities and we state the result in that form. It follows the same idea as in the proof of the corresponding result when p ≥ 1 in [3]. Then any bounded open Lipschitz set Ω ⊂ R n containing the origin satisfies where B R is the ball of radius R centered at the origin and with the same volume as Ω.
If there is equality in (21) and a(t) > 0 for all t ∈ (0, ∞), then Ω must be a ball centered at the origin.

Remark 13
The hypothesis of the theorem implies that a itself has to be non-increasing.

Proof
Step 1. Let Q = (0, π) n−2 × (0, 2π) and H : Q → S n−1 denote the hypershperical coordinates and g(ϕ)dϕ the surface element of S n−1 in these coordinates, see Appendix, Section 5. By an approximation argument, using that 0 / ∈ ∂Ω and a continuous, we can assume that Ω is of the following form (this is exactly the same as in [3] proof of Theorem i=1 Ω i , and each Ω i is given in the following way (see Figure 1 giving an idea how to construct Ω i ). We assume that there exists disjoint open subsets T i ⊂ Q, such that and for each i = 1, . . . , l there exists K i ∈ N and functions r i,s for s = 1, . . . , 2K i r i,s ∈ C 1 T i , 0 = r i,1 < r i,2 < . . . < r i,2Ki , such that We can assume that r i,1 = 0 because 0 ∈ Ω. Let G denote the polar coordinates G(t, ϕ) = tH(ϕ) (see Appendix) whose Jacobian determinant is given by t n−1 g(ϕ). Therefore we obtain that Let us define a function r : Q → R, H n−1 almost eveywhere in Q, by r : For p ∈ S n−1 we setr(p) = r(H −1 (p)), and thusr(H(ϕ)) = r(ϕ) almost everywhere in Q. In this way we obtain that Step 2. We now estimate the weighted perimeter. For i = 1, . . . , l us define Γ i by In this way l i=1 Γ i ⊂ ∂Ω and we obtain that with the parametrizations F i,s (ϕ) = r i,s (ϕ)H(ϕ). Using Lemma 17 and the fact that a ≥ 0, we obtain that ag i,s ≥ a r n−1 i,s g for all i, s.
Step 4. We now deal with the case of equality in (21), first with the additional assumption that Ω is a C 1 set starshaped with respect to the origin: i.e. ∂Ω can be parametrized (up to a set of H n−1 measure zero) as ∂Ω = {r(ϕ)H(ϕ)| ϕ ∈ Q}. In other words l = 1, K 1 = 1 and r = r 1,2 ∈ C 1 (Q). By hypothesis we must have equality in (22). Let us show that r is constant. If r is not constant, then by Lemma 17 there exists ϕ 0 and a neighborhood of U in Q containing ϕ 0 such that, recalling (23), Thus we obtain a strict inequality in (25) and we cannot have equality in (21). (Note that if we additionally assume that h is strictly convex, then equality in Jensen inequality also implies thatr must be constant.) We thus obtain that Ω is a ball centered at the origin with radius r.
Step 5. Let us treat now the general case of equality in (21). Since Ω is bounded there exsist M > 0 such that |x| ≤ M for all x ∈ ∂Ω and all x ∈ ∂B R .
Let us defineã byã Note thatã is continuous andã ≥ 0, because a is non-increasing (Remark 13). Moreover h defined byh (t) =ã t Using now the assumption that we have equality in (21) and that a(M ) > 0, we obtain that H n−1 (∂Ω) ≤ H n−1 (∂B R ). Which implies, in view of the classical isoperimetric inequality, that Ω must be a ball. Since 0 ∈ Ω we are in the case of the assumptions of Step 4 and the result follows.
We will now prove Part (ii) of the main theorem. We will use the notation: if x ∈ R n write x = (x 1 , x ), where x ∈ R n−1 .
Proof of Theorem 9 Part (ii). Step 1. It is sufficient to find a sequence of sets Ω such that each Ω is Lipschitz, bounded, connected, 0 ∈ Ω and lim →0 |Ω | → C > 0 but lim Then, by approximation, (28) holds also for a sequence of smooth sets. Then this sequence can be rescaled, defining a new sequence Ω = λ Ω with λ = C 1/n |Ω | −1/n so that | Ω | is constant and Ω still satisfies the second limit in (28).
Let us now show (28). Again by a rescaling argument, we can fix one C and assume without loss of generality that C = α n R n for some R > 0. The idea is to choose Ω as a ball of radius R and center going away to infinity, with a long and narrow cylinder attached to it so that it contains the origing, see Figure 2. One has to choose the length and radius of the cylinder carefully. More precisely: let us fix some R > 0 define Then Ω shall be defined as the union of the following three sets: So we obtain that lim →0 |Ω | = lim →0 |D | = α n R n .
It remains to estimate the contribution coming from M . It is equal to To estimate L 1 we just use that x 2 1 + 4 ≥ 4 , and that p < 0 to get L 1 ≤ ω n−2 2(n+p−1) → 0. To estimate L 2 we use that x 2 1 + 4 ≥ x 2 1 and therefore we distinguish 3 cases. Case 1, p > −1. In this case p + 1 > 0 and we obtain by explicit integration So L 2 tends to zero as → 0 if and only if 2(n−2)−(n−1)(p+ 1) > 0, which is equivalent to p < n − 3 n − 1 .
Remark 14 Note that also Case 3 cannot occur if n = 2.
We give here a proof of Theorem 9 (i) in the special case of starshaped domains, since it is quite short and contains a new and interesting interpolation argument due to Alvino et alt. [1]. The general case follows from a more standard, but weighted, symmetrization argument, see also [1]. Moreover, if there is equality then Ω has to be a ball centered at the origin.

Proof
Step 1. Let us define by P p (Ω) the right hand side of the inequality in the proposition. We abbreviate S n−1 = S n−1

1
, see (14) for notation. From Lemma 18 in the appendix we obtain that Note that for any n ≥ 3 it holds that From Lemma 19 and the previous inequality we obtain Now use that for any a, b ≥ 0 the mapping t → log √ a + bt is concave, so that Using this it follows from (30) (with t 1 = 1, t 2 = 0) Define a new domain, also starshaped with respect to the origin, It follows from Lemma 18, the standard isoperimetric inequality and (36) that We plug this into the (31) and use the definition of Z to conclude that The idea is to use now Hölder inequality in the form We take (33) to the power (p + n − 1)/n and multiply by (nα n ) (1−p)/n to get (using (32) in the last step) Thus it follows from (36) that which proves the first part of the proposition.
Step 2. Let us consider the case of equality. In that case there must be equality in (33), which is only possible if for some constant c ∈ R R(θ) p+n−1 = cR(θ) This is only possible if R is constant.

Remark 16
The present proof does not work for n = 2, since in that case (29) is not satisfied for all p ∈ (0, 1).
5 Appendix: hyperspherical coordinates in R n .
We have used in Section 4 the explicit form of hyperspherical coordinates and their properties. Let us define Q ⊂ R n−1 by Q = (0, π) n−2 × (0, 2π).
For our purpose the following lemma will be useful.
Lemma 17 Suppose T ⊂ Q is an open set and a hypersurface Γ ⊂ R n is given by the parametrization F : where r is some smooth function r : T → (0, ∞). The surface element in this parametrization calculates as where d i means that d i should be omitted in the product. In particular, since d i > 0, the surface element is bigger than r n−1 g(ϕ).
Proof Using the relations H; H = 1 and H; ∂H ∂ϕi = 0, one obtains that Thus the matrix M with entries M ij is of the form M = A + r 2 D where A has rank 1 and D is diagonal with entries d i . Thus using the linearity of the determinant in the columns one obtains that (since no matrix with two columns of A survives when developing the determinant succesively with respect to the columns) where A i is the matrix obtained from r 2 D by replacing the i-th column of r 2 D by the i-th column of A. From this the lemma follows. We now use the hypershperical coordinates to deal with domains starshaped with respect to the origin. By definition, a bounded open Lipschitz domain Ω ∈ R n is starshaped with respect to the origin if there exists a function R : S n−1 → (0, ∞) such that Ω = {0} ∪ x ∈ R n \{0} : x |x| = θ ∈ S n−1 , 0 < |x| < R(θ) We shall call R the defining function of Ω. Note that R is not necessarily Lipschitz, even if Ω is, and it might even be discontinuous (for example Ω = (B r2 (0) ∩ {x 2 > 0}) ∪ (B r1 (0) ∩ {x 2 < 0}) ⊂ R 2 , with r 1 < r 2 ). But we will alway assume that R is also Lipschitz. It follows from the relation (34) that the volume of a starshaped domain calculates as L n (Ω) = 1 n S n−1 R n dH n−1 For an almost everywhere differentiable function f : S n−1 → R we recall that the norm of the covariant gradient of f at p on the manifold S n−1 can be calculated as where {E 1 , . . . , E n−1 } is any orthonormal basis of T p S n−1 , and ∇ E f is the derivative in direction E.

Lemma 18
Let Ω ⊂ R n be a bounded open Lipschitz set, starshaped with respect to the origin and defining function R as in (35). Assume also that R is Lipschitz. Then for any continuous function g : (0, ∞) → R it holds that ∂Ω g(|x|)dH n−1 (x) = S n−1 (g • R)R n−1 1 + 1 R 2 |∇ S n−1 R| 2 dH n−1 .
The assumption that g is continuous can be reduced, but we will need the lemma only for g(t) = t p . Proof We use the hyperspherical coordinates H, definitions of d i and g, respectively their properties (summarized in the beginning of this section). Let ϕ ∈ Q, p = H(ϕ) and ∂H ∂ϕ i , i = 1, . . . , n − 1 form an orthonormal basis of T p S n−1 . Thus we obtain from (37) (we can assume that R has been extended to a neighborhood of S n−1 ) that at p = H(ϕ) ∇R(H(ϕ)); where we define for ϕ ∈ Q r(ϕ) := R(H(ϕ)) and F (ϕ) = r(ϕ)H(ϕ).
Hence F : Q → ∂Ω is a parametrization of ∂Ω. Using Lemma 17 we obtain that det ∂F ∂ϕ i , ∂F ∂ϕ j =r n−1 d 1 d 2 · · · d n−1 1 + 1 r 2 Therefore, using the parametrization F to calculate the left side of (38), respectively the parametrization H to calculate the right side and (39), the lemma follows.