On the isoperimetric problem with perimeter density r^p
OOn the isoperimetric problem with perimeter density r p . Gyula Csat´o
Facultad de Ciencias F´ısicas y Matem´aticas, Universidad de Concepci´on, Concepci´on, [email protected]
Abstract
In this paper the author studies the isoperimetric problem in R n with perimeterdensity | x | p and volume density 1 . We settle completely the case n = 2 , completinga previous work by the author: we characterize the case of equality if 0 ≤ p ≤ −∞ < p < − ∈ Ω). In thecase n ≥ −∞ < p < , showing among others thatthe results in 2 dimensions do not generalize for the range − n + 1 < p < . Let Ω ⊂ R n be a bounded open set with Lipschitz boundary ∂ Ω . We study the inequality (cid:18) n L n (Ω) ω n − (cid:19) n + p − n ≤ ω n − (cid:90) ∂ Ω | x | p d H n − ( x ) . (1)where L n (Ω) denotes the n dimensional Lebesgue measure of Ω , H n − is the n − ω n − = H n − ( S n − ) is the surface area of the unit n − p = 0 this is the classical isoperimetric inequality. Note that for ballscentered at the origin there is always equality.For p < , we introduce a new condition: Ω shall contain the origin.The inequality (1) is a particular case among a broader class of problems called isoperi-metric problems with densities. Given two positive functions f, g : R n → R one studiesthe existence of minimizers of I ( C ) = inf (cid:26)(cid:90) ∂ Ω g d H n − : Ω ⊂ R n and (cid:90) Ω f d L n = C (cid:27) . (2) (cid:82) ∂ Ω gd H n − is called the weighted perimeter. There are an increasing number of worksdealing with different types of weights f and g, see for instance [5], [6], [7], [10], [17],[19], [21], [23]. In the case where f ( x ) = g ( x ) = | x | p and p > , we must mention resultsappearing in [4], [8], [9] and [14], which, among other results, led to the final surprising factthat the minimizers are hyperspheres passing through the origin. Some other interesting Primary 49Q10, 49Q20, Secondary 26D10.
Key words and phrases.
Isoperimetric problem with density, radial weight a r X i v : . [ m a t h . DG ] J un esults for the case f ( x ) = | x | q and g ( x ) = | x | p can be found in [15], respectively [1] and[20].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 ≥ . Anotherproof of the same result can be found in [15], Section 7. Later (1) was shown by the author[11] for 0 ≤ p ≤ n = 2 , and then by [1] for any n ≥ . See also the very recent paper[20], which uses variational methods and the study of the Euler equations satisfied by aminimizer. The method of [1] contains a very original interpolation argument with which(1) is deduced from the classical isoperimetric inequality. For the sake of completenesswe repeat this proof in the special case of starshaped domains, see Proposition 15. Thegeneral case follows from a more standard, but weighted, symmetrization argument, see[1]. In the case p < n = 2 and − ≤ p ≤ p and adding theseadditional assumptions come from the singular Moser-Trudinger functional [2]. Theseassumptions 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: oneuses (1) for the level sets of the Greens function G Ω , with singularity at 0 . Obviouslythese level sets will always contain the origin and will be connected by the maximumprinciple.In the present paper we make the following further contributions. It turns out thatthere are big differences depending on the dimension n. The case n = 1 . The inequality (1) is elementary, but we have included it for com-pleteness. The inequality hols for all p ∈ R \ (0 , . If p ∈ (0 , , then minimzers of theweighted 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 < p < − The case n ≥ . We completely tackle the case p < . We prove that (1) remains trueif p < − n + 1 . This proof is basically the same as that of [3], with a slight difference inthe proof of the unicity result. However, the rather surprising result is that the inequalitydoes 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 centeredat the origin are stationary and stable, but nevertheless they are not global minimizers.We actually prove something stronger: there are no minimizers of the correspondingvariational problem (2) if − n + 1 < p < In this case H is the counting measure, ω = 2 and the inequality (1) becomes (cid:18) L (Ω)2 (cid:19) p ≤ (cid:90) ∂ Ω | x | p d H ( y ) . (3)2e assume that Ω is the union of disjoint open intervals, such that the closed intervals donot intersect. If p = 0 , and there is just one interval, then the inequality trivially readsas 1 = 1 . We have the following proposition.
Proposition 1 (i) If p ≥ , then inequality (3) holds true for any Ω . In case of equality Ω must be single interval and if p > this interval has to be centered at the origin.(ii) If < p < , then inequality (3) does not hold true.(iii) If p < , then inequality (3) holds true for all Ω containing the origin. In case ofequality 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 forthe weighted perimeter, under the constraint L (Ω) = c. One can verify that the uniqueminimizers are the intervals (0 , c ) or ( − c, . Proof (i) Let a = min { x | x ∈ ∂ Ω } and b = max { x | x ∈ ∂ Ω } . Then L (Ω) ≤ ( b − a ) ≤| a | + | b | and the inequality follows from the fact that the map s (cid:55)→ s p is increasing andconvex.(ii) Take Ω = (0 , c ) , for some c > . (iii) Since 0 ∈ Ω there exists a, b ∈ ∂ Ω such that a < , b > a, b ) ⊂ Ω . We have L (Ω) ≥ ( b − a ) = | a | + | b | . Using that s (cid:55)→ s p is decreasing and convex one obtains easilythe result. The following theorem summarizes the works by Betta-Brock-Mercaldo-Posteraro [3],Csat´o [11] and the results proven in the present paper. In this section | Ω | = L (Ω) shalldenote the area of a set and σ is the 1-dimensional Hausdorff measure. We will often justsay that a set Ω is C k meaning that its boundary ∂ Ω is a C k curve. Theorem 3
Let Ω ⊂ R be a bounded open Lipschitz set. Regarding the inequality (cid:18) | Ω | π (cid:19) p +12 ≤ π (cid:90) ∂ Ω | x | p dσ, (4) the following statements hold true:(i) If p ≥ , then (4) holds for all Ω . (ii) If − < p < , then (4) holds for all Ω connected and containing the origin.(iii) If p ≤ − then (4) holds for all Ω containing the origin.(iv) In case (i), if Ω is C and there is equality in (4) , then Ω must be a ball; centeredat the origin if p (cid:54) = 0 . If there is equality in case (ii) and (iii) then also Ω must be a ball. Remark 4
It follows from Morgan [21] that if one has equality in (4) (i.e. Ω is a min-imizer), then ∂ Ω \{ } is smooth. On the other hand if one shows the inequality (4) forsmooth sets, then it also holds for Lipschitz sets by approximation. So we can work withsmooth sets and have to be careful only if 0 ∈ ∂ Ω . roof (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 caseof Theorem 12. So it remains to deal with the case of equality in (i). This has also beendealt with in [11] as long as 0 / ∈ ∂ Ω or p > . The case 0 ∈ ∂ Ω is Proposition 5The rest of this section is devoted to the proof of the following Proposition 5. Itis based on variational methods and a careful analysis of the resulting Euler-Lagrangeequation.
Proposition 5
Suppose < p ≤ and Ω ⊂ R is a bounded open set with C boundary ∂ Ω . If (cid:18) | Ω | π (cid:19) p +12 = 12 π (cid:90) ∂ Ω | x | p dσ ( x ) , then cannot lie on the boundary ∂ Ω . The proof of Proposition 5 is based on two lemmas. The first one, Lemma 6, is thevariational formula that we need, establishing the Euler-Lagrange equation satisfied bya minimizer. It is the generalization of the statement that minimizers of the classicalisoperimetric problem have constant curvature, with the difference that one introduces ageneralized curvature. Such formulae are broadly used to deal with isoperimetric problemswith densities, see for instance [5], [10], [14] [15], respectively [22] for a summary. Sincein the present case the derivation is very short and elementary, we provide the proof tomake the presentation self-contained. Afterwards, in Lemma 8, one uses the symmetryproperties of the Euler-Lagrange equation to show that minimizers are symmetric withrespect to any line through the origin and a point P on ∂ Ω with maximal distance fromthe origin. Such symmetrization arguments are also well known, see for instance Lemma2.1 in [14]. Then to conclude the proof of Proposition 5 one essentially compares thegeneralized curvature, which has to be constant, at the point P and at the origin, whichwill lead to a contradiction.We shall use the following notation: if Ω is simply connected, then γ : [0 , L ] → ∂ Ωshall always denote a simple closed curve bounding Ω . Mostly we will also assume that | γ (cid:48) | = 1 and ν = ( γ (cid:48) , − γ (cid:48) ) is the outward unit normal to ∂ Ω . (5)This can always be achieved by reparametrization and chosing the orientation properly.The prime ( · ) (cid:48) shall always denote the derivative with respect to the argument of γ. If | γ (cid:48) | = 1 then the curvature κ of ∂ Ω calculates as κ ( t ) = (cid:104) γ (cid:48) ( t ) , ν (cid:48) ( t ) (cid:105) . Lemma 6
Let p ∈ R , C > and suppose that Ω is a C minimizer of inf (cid:26)(cid:90) ∂ Ω | x | p dσ ; | Ω | = C (cid:27) . (6) Assume γ : [0 , L ] → ∂ Ω is a simple closed curve. Then p | γ ( t ) | p − (cid:104) γ ( t ) , ν ( t ) (cid:105) + | γ ( t ) | p κ ( t ) = constant ∀ t ∈ [0 , L ] (7) if / ∈ ∂ Ω . Whereas if γ ( t ) = 0 for some t ∈ [0 , L ] , then (7) holds for all t ∈ [0 , L ] \{ t } . emark 7 The function k = p | γ | p − (cid:104) γ, ν (cid:105) + | γ | p κ is usually called the generalized cur-vature of ∂ Ω . Proof
Step 1.
Without loss of generality we assume the second case 0 ∈ ∂ Ω and assumethat γ (0) = γ ( L ) = 0 . We can also assume that (5) holds, since (7) is independentof 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 adomain Ω s . Because Ω is a minimizer, we claim thatIf dds [ | Ω s | ] s =0 = 0 then dds (cid:20)(cid:90) ∂ Ω s | x | p dσ (cid:21) s =0 = 0 . (8)Let us show (8). If this is not the case, this means that the function h is such that thefirst equality in (8) holds true but dds (cid:20)(cid:90) ∂ Ω s | x | p dσ (cid:21) s =0 (cid:54) = 0 . Then define (cid:98) Ω s = λ s Ω s where λ s = (cid:112) | Ω | / (cid:112) | Ω s | . Note that | (cid:98) 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 dds (cid:90) ∂ (cid:98) Ω s | x | p dσ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) s =0 = dds (cid:20) λ p +1 s (cid:90) ∂ Ω s | x | p dσ (cid:21) s =0 = dds (cid:20)(cid:90) ∂ Ω s | x | p dσ (cid:21) s =0 (cid:54) = 0 . Thus for sufficiently small s (negative or positive depending on the sign of the last in-equality) the weighted perimeter of (cid:98) Ω s is strictly smaller than that of Ω . This contradictsthe fact that Ω is a minimizer.
Step 2.
We now calculate the derivatives in (8) explcitly. First one gets g s = γ + shν, g (cid:48) s = γ (cid:48) + sh (cid:48) ν + shν (cid:48) . We therefore obtain | Ω s | = (cid:90) L ( g s ) ( g s ) (cid:48) dt = (cid:90) L ( γ + shν )( γ (cid:48) + sh (cid:48) ν + shν (cid:48) ) , which leads to, using partial integration to get rid of derivatives of h , dds | Ω s | (cid:12)(cid:12)(cid:12) s =0 = (cid:90) L ( h (cid:48) γ ν + hγ ν (cid:48) + hν γ (cid:48) ) = (cid:90) L h | γ (cid:48) | = (cid:90) L h. On the other hand we have (cid:90) ∂ Ω s | x | p dσ = (cid:90) L | g s ( t ) | p | g (cid:48) s ( t ) | dt. We therefor obtain that dds (cid:20)(cid:90) ∂ Ω s | x | p dσ (cid:21) s =0 = A + B, A = (cid:34)(cid:90) L p | g s ( t ) | p − (cid:18) dds | g s ( t ) | (cid:19) | g (cid:48) s ( t ) | dt (cid:35) s =0 B = (cid:34)(cid:90) L | g s ( t ) | p dds | g (cid:48) s ( t ) | dt (cid:35) s =0 . Note that | g s | = (cid:112) (cid:104) γ + shν, γ + shν (cid:105) and dds | g s | (cid:12)(cid:12)(cid:12) s =0 = h (cid:104) γ, ν (cid:105)| γ | . Thus we get A = (cid:90) L p | γ | p − h (cid:104) γ, ν (cid:105) dt. Let us now calculate B. As above | g (cid:48) s | = (cid:112) (cid:104) γ (cid:48) + sh (cid:48) ν + shν (cid:48) , γ (cid:48) + sh (cid:48) ν + shν (cid:48) (cid:105) . This leads to dds | g (cid:48) s | (cid:12)(cid:12)(cid:12) s =0 = 1 | γ (cid:48) | ( h (cid:48) (cid:104) γ (cid:48) , ν (cid:105) + h (cid:104) γ (cid:48) , ν (cid:48) (cid:105) ) = h (cid:104) γ (cid:48) , ν (cid:48) (cid:105) . Setting this into B and adding A + B we finally obtain that (8) implies that (cid:90) L (cid:0) p | γ | p − (cid:104) γ, ν (cid:105) + | γ | p (cid:104) γ (cid:48) , ν (cid:48) (cid:105) (cid:1) h dt = 0 ∀ h ∈ C ∞ c (0 , L ) with (cid:90) L hdt = 0 . This implies the claim of the lemma.
Lemma 8
Let p ∈ R , C > and suppose that Ω is a C minimizer of (6) . Assume γ : [0 , L ] → ∂ Ω is a simple closed curve with | γ (cid:48) | = 1 . Suppose there exists t ∈ [0 , L ] suchthat γ ( t ) (cid:54) = 0 and ddt | γ ( t ) | (cid:12)(cid:12)(cid:12) t = t = 0 . Then γ is symmetric with respect to the line through γ ( t ) and the origin. Proof
Without loss of generality we can assume (by rotation and reparametrization) that t = 0 and γ (0) = ( γ (0) ,
0) and γ (0) > . By hypothesis (cid:104) γ (0) , γ (cid:48) (0) (cid:105) = 0 and hence γ (cid:48) (0) = 0 . So using Lemma 6, γ satisfies the two equations | γ (cid:48) | = 1 and p | γ | p − (cid:104) γ, ν (cid:105) + | γ | p (cid:104) γ (cid:48) , ν (cid:48) (cid:105) = k, (9)for some constant k. Deriving the first equation with respect to t and multiplying with | γ | p we get that | γ | p ( γ (cid:48) γ (cid:48)(cid:48) + γ (cid:48) γ (cid:48)(cid:48) ) = 0 . Finally, multiplying this equation, respectivelythe second in (9), with γ (cid:48) or γ (cid:48) and combining properly one easily gets that (using oncemore | γ (cid:48) | = 1) γ (cid:48)(cid:48) = γ (cid:48) | γ | p (cid:0) − k + p | γ | p − ( γ γ (cid:48) − γ γ (cid:48) ) (cid:1) =: F ( γ, γ (cid:48) ) γ (cid:48)(cid:48) = γ (cid:48) | γ | p (cid:0) k − p | γ | p − ( γ γ (cid:48) − γ γ (cid:48) ) (cid:1) =: F ( γ, γ (cid:48) ) . α = γ (cid:48) , we see that ( γ, α ) satisfies the initial value problem γ (0) =( γ (0) ,
0) and α (0) = (0 , γ (cid:48) (0)) (cid:18) γα (cid:19) (cid:48) = (cid:18) α ( F ( γ, α ) , F ( γ, α )) (cid:19) . The functions F and F are C and hence Lipschitz as long as γ (cid:54) = 0 . They have theproperties F ( γ , − γ , − α , α ) = F ( γ , γ , α , α ) , F ( γ , − γ , − α , α ) = − F ( γ , γ , α , α )Using these properties, it can be easily verified, by evaluating the differential equation at − t, that also the curve ω ( t ) = ( γ ( − t ) , − γ ( − t )) satisfies the initial value problem. Byuniqueness we obtain that ω ( t ) = γ ( t ) . Since ( F , F ) is locally Lipschitz, by the theoreyof ordinary differential equations (see for instance [24] page 68) the solution exists eitherfor all times t, it blows up or goes out of the region of definition of ( F , F ) . In the presentcase this means that the solution γ exists and is unique for all times t, unless | γ | goes to0 or to ∞ . But it cannot go to infinity, because then the weighted perimiter would go toinfinity and then Ω cannot be a minimizer. So we conclude, using that γ is continuousby 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 connectedset. If Ω is not connected, let us say Ω = Ω ∪ Ω and Ω ∩ Ω = ∅ for two nonempty sets,then choose R , R > | Ω i | = πR i . Using the hypothesis that there is equalityin Proposition 4 and Theorem 3 part (i) for Ω i we get( R + R ) p +12 = (cid:18) | Ω | π (cid:19) p +12 = 12 π (cid:18)(cid:90) ∂ Ω | x | p dσ + (cid:90) ∂ Ω | x | p dσ (cid:19) ≥ R p +11 + R p +12 > (cid:0) R + R (cid:1) p +12 In the last inequality we have assumed that 0 < p < a s + b s > ( a + b ) s , if 0 < s < a, b > . If p = 1 there is still strict inequality because for one of the i = 1 , (cid:82) ∂ Ω i | x | p > πR ( p +1) / i . If not, assume that we have shown Theorem3 (iv) first for connected sets. Then both Ω and Ω would have to be balls centered atthe origin, a contradiction to Ω ∩ Ω = ∅ . Assume now that Ω is not simply connected and is of the form Ω = Ω \ Ω for someΩ ⊂ Ω . Then we obtain using Theorem 3 (i) that (cid:18) | Ω | π (cid:19) p +12 < (cid:18) | Ω | π (cid:19) p +12 ≤ π (cid:90) ∂ Ω | x | p dσ < π (cid:90) ∂ Ω | x | p dσ = (cid:18) | Ω | π (cid:19) p +12 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, byrotating the domain, we know by Lemma 8 that the maximum of | γ | has to be achievedat t = L/ , that for some d > γ (cid:18) L (cid:19) = ( d,
0) and d = max t ∈ [0 ,L ] | γ ( t ) | , γ is symmetic with respect to the x axis { ( x , x ) ∈ R | x = 0 } . Using thissymmetry and the chosen orientation of γ, one obtains that γ (cid:48) (cid:18) L (cid:19) = (0 , , γ (cid:48)(cid:48) (cid:18) L (cid:19) ≤ κ (cid:18) L (cid:19) ≥ . Note also that ν ( L/
2) = (1 , . Hence we obtain for the generalized curvature at L/ k = p | γ | p − (cid:104) γ ; ν (cid:105) + | γ | p κ = pd p − d + d p κ ≥ pd p − > . (10) Step 3.
Using again Lemma 8 we obtain that γ ( − t ) = γ ( t ) and therefore γ (cid:48) (0) = 0and γ (cid:48) (0) = − . Therefore γ is invertible near zero and f = γ ◦ γ − ∈ C (( − (cid:15), (cid:15) )) forsome (cid:15) > . We use the new parametrization α of ∂ Ω near 0 , given by t (cid:55)→ α ( t ) = ( f ( t ) , t ) ,t ∈ ( − (cid:15), (cid:15) ) . This new parametrization allows us to reduce the problem to a 1-dimensionalone, analyzing the function f. Since α (cid:48) = ( f (cid:48) , , α (cid:48)(cid:48) = ( f (cid:48)(cid:48) ,
0) and keeping the reversedorientation in mind, we have the following formulas for the outer normal ν and curvature κ ν = ( − , f (cid:48) ) (cid:112) f (cid:48) , κ = − α (cid:48) α (cid:48)(cid:48) − α (cid:48) α (cid:48)(cid:48) | α (cid:48) | = f (cid:48)(cid:48) (1 + f (cid:48) ) / near t = 0 . The generalized curvature k = p | α | p − (cid:104) α, ν (cid:105) + | α | p κ near t = 0 is k = p | α ( t ) | p (cid:112) f (cid:48) ( t ) tf (cid:48) ( t ) − f ( t ) | α ( t ) | + | α ( t ) | p f (cid:48)(cid:48) ( t )(1 + f (cid:48) ( t ) ) / = m ( t ) , (11)where m ( t ) is just an abbreviation for the right hand side. Step 4.
We will show that lim t → m ( t ) = 0 . (12)This will be a contradiction to (10) and the fact that k has to be constant, by Lemma 6.Using the facts that p > , lim t → | α ( t ) | = 0 , lim t → (cid:112) f (cid:48) = 1 , and f ∈ C , to prove (12), it suffices to show that | tf (cid:48) ( t ) − f ( t ) || α ( t ) | = | tf (cid:48) ( t ) − f ( t ) | ( t + f ( t ) ) is bounded on ( − (cid:15), (cid:15) ) . (13)(13) is true for any f ∈ C ([ − (cid:15), (cid:15) ]) with f (0) = 0 . This can be seen in the following way.Since f (0) = 0 , one has lim t → f ( t ) /t = f (cid:48) (0) andlim sup t → | tf (cid:48) ( t ) − f ( t ) || t + f ( t ) | = lim t → | f ( t ) t | lim sup t → (cid:12)(cid:12)(cid:12)(cid:12) tf (cid:48) ( t ) − f ( t ) t (cid:12)(cid:12)(cid:12)(cid:12) = 11 + f (cid:48) (0) lim sup t → (cid:12)(cid:12)(cid:12)(cid:12) tf (cid:48) ( t ) − f ( t ) t (cid:12)(cid:12)(cid:12)(cid:12) . It follows from de l’ Hopital rule thatlim t → tf (cid:48) ( t ) − f ( t ) t = f (cid:48)(cid:48) (0)2 , which proves that (13) holds and the claim of Step 4.8 Some results in general dimensions n ≥ In this section we shall use the following notations and abbreviations: let n ∈ N B nδ ( y ) = { x ∈ R n | | x − y | < δ } , B nδ = B nδ (0) , α n = L n ( B n ) S n − δ = ∂B nδ = { x ∈ R n | | x | = δ } , ω n − = H n − ( S n − ) . (14)Recall that α n = ω n − /n. If the dimension is obvious we omit n in the superscript e.g. B δ = B nδ and the same for S nδ . We also set | Ω | := L n (Ω) . The following is the maintheorem, summarizing the results of [1], [3] and the present paper.
Theorem 9
Let n ≥ and Ω ⊂ R n be a bounded open Lipschitz set. Regarding theinequality (cid:18) n | Ω | ω n − (cid:19) n + p − n ≤ ω n − (cid:90) ∂ Ω | x | p d H n − . (15) the following statements hold true:(i) If p ≥ then (15) holds true for all Ω . (ii) If − n + 1 < p < , then we have for any C > (cid:26)(cid:90) ∂ Ω | x | p d H n − : 0 ∈ Ω ⊂ R n connected and | Ω | = C (cid:27) = 0 , where the infimum is taken over all bounded open smooth sets. In particular (15) cannothold for all Ω containing the origin.(iii) If p ≤ − n + 1 , then (15) holds true 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 originif p (cid:54) = 0 . Remark 10
In (ii), if one omits the condition that Ω has to contain the origin, then it istrivial that the infimum is zero, by taking any domain with constant volume and shiftingit to infinity. This is true in any dimension, in particular also if n = 2 , compare withTheorem 3 (ii). Proof
Part (i) and its sharp form (iv) have been proven in [1] and [3], respectively theclassical isoperimetric inequality. See also Proposition 15 for the case 0 < p < 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
Fix r > s = B r (0) + ( s, , where we will understand from nowon ( s,
0) = ( s, , . . . , ∈ R n and s ∈ R . Obviously | Ω s | = α n r n for all s and Ω s containsthe origin for all s < r. We shall denote the weighted perimeter as P (Ω s ) = (cid:90) ∂ Ω s | x | p d H n − . Assume that F = ( F , F , . . . , F n ) = ( F , ˜ F ) : U ⊂ R n − → ∂B r (0) is a parametrizationof ∂B r (0) up to a set of H n − 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 + s, ˜ F ) is a parametrizationof ∂ Ω s and one obtains that dds P (Ω s ) = p (cid:90) U (cid:16) ( F ( u ) + s )) + | ˜ F ( u ) | (cid:17) p − ( F ( u ) + s ) g ( u ) du. (16)This gives dds P (Ω s ) (cid:12)(cid:12)(cid:12) s =0 = p (cid:90) ∂B r | x | p − x d H n − = 0 for any p ∈ R . (17)Deriving (16) once more leads to d ds P (Ω s ) (cid:12)(cid:12)(cid:12) s =0 = p (cid:18) ( p − r p − (cid:90) ∂B r x d H n − + H n − ( ∂B r ) r p − (cid:19) Using that (cid:82) ∂B r x = 1 /n (cid:82) ∂B r ( x + · · · + x n ) = r /n H n − ( ∂B r ) the second variationsimplifies to d ds P (Ω s ) (cid:12)(cid:12)(cid:12) s =0 = r p − n ω n − r n − p ( p − n ) . One can now easily verify that if n ≥ p ( p − n ) is > − ∞ < p < − n + 2= 0 if p = − n + 2 < − n + 2 < p <
0= 0 if p = 0 > p > . (18)In view of (17) and the third inequality in (18), one obtains that P (Ω s ) decreases forincreasing s near the origin. This shows that (15) cannot hold true if − n + 2 < p < . The equations (18) remain true for much more general variations. We will not usethem but nevertheless summarize the result in the present case. For the methods andproofs 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 (cid:90) ∂B r ud H n − = 0 . (19)10ssume we are given a radial perimeter density G : (0 , ∞ ) → (0 , ∞ ) , smooth in a neigh-borhood of r, and P G is defined by P G (Ω) = (cid:90) ∂ Ω G ( | x | ) d H n − . Then the second variation for P G is given by d ds P G (Ω s ) (cid:12)(cid:12)(cid:12) s =0 = M + M where M = G ( r ) (cid:18)(cid:90) ∂B r |∇ (cid:80) u | − n − r (cid:90) ∂B r u (cid:19) , M = (cid:18) ( n − G (cid:48) ( r ) r + G (cid:48)(cid:48) ( r ) (cid:19) (cid:90) ∂B r u where ∇ (cid:80) u is the covariant derivative along ∂B r . By the standard isoperimetric inequal-ity one has that M ≥ u satisfying (19). This follows again by variationalmethods (taking G = 1), or equivalently, it is the Poincar´e inequality on the sphere withoptimal constant, see [16] Example (2.13). Moreover one can show by the first variationof P G that balls centered at the origin are always critical points of P G under the volumeconstraint. So this implies that B r can be a minimizer of P G if and only if M ≥ , i.e.( n − G (cid:48) ( r ) r + G (cid:48)(cid:48) ( r ) ≥ G ( r ) = r p , so the last inequality holds true if and only if p ( n + p − ≥ . This is precisely again (18). Note that if − n + 1 < p < − n + 2 , then p ( n + p − > , but nevertheless Theorem 9 (ii) shows that balls centered at the origin are not globalminimizers.We shall now prove Part (iii) of Theorem 9. However this can be done for much moregeneral densities and we state the result in that form. It follows the same idea as in theproof of the corresponding result when p ≥ Theorem 12
Let n ≥ and a : (0 , ∞ ) → [0 , ∞ ) be a continuous function such that t (cid:55)→ a (cid:16) t n (cid:17) t n − n is non-increasing and convex (20) Then any bounded open Lipschitz set Ω ⊂ R n containing the origin satisfies (cid:90) ∂B R a ( | x | ) d H n − ( x ) ≤ (cid:90) ∂ Ω a ( | x | ) d H n − ( x ) , (21) 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 ) > for all t ∈ (0 , ∞ ) , then Ω must be a ball centeredat 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 − × (0 , π ) and H : Q → S n − denote the hypershpericalcoordinates and g ( ϕ ) dϕ the surface element of S n − in these coordinates, see Appendix,Section 5. By an approximation argument, using that 0 / ∈ ∂ Ω and a continuous, we canassume that Ω is of the following form (this is exactly the same as in [3] proof of Theorem11igure 1: Construction of Ω i and r i,j i , i = 1 , . . . , l such that Ω = (cid:83) li =1 Ω i , and each Ω i is givenin the following way (see Figure 1 giving an idea how to construct Ω i ). We assume thatthere exists disjoint open subsets T i ⊂ Q, such that Q = l (cid:91) i =1 T i , and for each i = 1 , . . . , l there exists K i ∈ N and functions r i,s for s = 1 , . . . , K i r i,s ∈ C (cid:0) T i (cid:1) , r i, < r i, < . . . < r i, K i , such thatΩ i = (cid:8) rH ( ϕ ) | ϕ ∈ T i , r i, s − ( ϕ ) ≤ r ≤ r i, s ( ϕ ) for s = 1 , . . . , K i (cid:9) . We can assume that r i, = 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 − g ( ϕ ) . Therefore weobtain that | Ω i | = (cid:90) T i K i (cid:88) s =1 (cid:90) r i, s ( ϕ ) r i, s − ( ϕ ) t n − g ( ϕ ) dt dϕ = (cid:90) T i K i (cid:88) s =1 n (cid:0) r ni, s − r ni, s − (cid:1) g ( ϕ ) dϕ ≥ (cid:90) T i n r ni, ( ϕ ) g ( ϕ ) dϕ. Let us define a function r : Q → R , H n − almost eveywhere in Q, by r : l (cid:91) i =1 T i → R , r ( ϕ ) = r i, ( ϕ ) if ϕ ∈ T i . p ∈ S n − we set ˜ r ( p ) = r ( H − ( p )) , and thus ˜ r ( H ( ϕ )) = r ( ϕ ) almost everywhere in Q. In this way we obtain that | Ω | = l (cid:88) i =1 | Ω i | ≥ n (cid:90) Q r ( ϕ ) n g ( ϕ ) dϕ = 1 n (cid:90) S n − ˜ r n d H n − . (22) Step 2.
We now estimate the weighted perimeter. For i = 1 , . . . , l us define Γ i byΓ i = K i (cid:91) s =2 (cid:8) r i,s ( ϕ ) H ( ϕ ) | ϕ ∈ T i (cid:9) . In this way (cid:83) li =1 Γ i ⊂ ∂ Ω and we obtain that (cid:90) ∂ Ω a ( | x | ) d H n − ( x ) ≥ l (cid:88) i =1 (cid:90) Γ i a ( | x | ) d H n − ( x ) = l (cid:88) i =1 2 K i (cid:88) s =2 (cid:90) T i a ( r i,s ( ϕ ))˜ g i,s ( ϕ ) dϕ, where ˜ g i,s is given by ˜ g i,s ( ϕ ) = (cid:115) det (cid:18)(cid:28) ∂F i,s ∂ϕ j , ∂F i,s ∂ϕ k (cid:29)(cid:19) j,k =1 ,...,n − (23)with the parametrizations F i,s ( ϕ ) = r i,s ( ϕ ) H ( ϕ ) . Using Lemma 17 and the fact that a ≥ , we obtain that a ˜ g i,s ≥ a r n − i,s g for all i, s. (24)It follows that (cid:90) ∂ Ω a ( | x | ) d H n − ( x ) ≥ l (cid:88) i =1 2 K i (cid:88) s =2 (cid:90) T i a ( r i,s ( ϕ )) r n − i,s ( ϕ ) g ( ϕ ) dϕ ≥ l (cid:88) i =1 (cid:90) T i a ( r i, ( ϕ )) r n − i, ( ϕ ) g ( ϕ ) dϕ Using again the definition of r and ˜ r introduced at the end of Step 1 we obtain (cid:90) ∂ Ω a ( | x | ) d H n − ( x ) ≥ (cid:90) S n − a (˜ r )˜ r n − d H n − . (25) Step 3.
Let us define h by h ( t ) = a (cid:16) t n (cid:17) t n − n . It now follows from (22), from the fact that h is decreasing, from Jensen inequality andfinally (25), that h (cid:18) n | Ω | ω n − (cid:19) ≤ h (cid:18) ω n − (cid:90) S n − ˜ r n d H n − (cid:19) ≤ ω n − (cid:90) S n − h (˜ r n ) d H n − = 1 ω n − (cid:90) S n − a (˜ r )˜ r n − d H n − ≤ ω n − (cid:90) ∂ Ω a ( | x | ) d H n − ( x ) . (26)Since | Ω | = | B R | one easily verifies that R = (cid:18) n | Ω | ω n − (cid:19) n and h (cid:18) n | Ω | ω n − (cid:19) = a ( R ) R n − = 1 ω n − (cid:90) ∂B R a ( | x | ) d H n − ( x ) . (27)13lugging this identity into (26) concludes the proof inequality (21). Step 4.
We now deal with the case of equality in (21), first with the additionalassumption that Ω is a C set starshaped with respect to the origin: i.e. ∂ Ω can beparametrized (up to a set of H n − measure zero) as ∂ Ω = { r ( ϕ ) H ( ϕ ) | ϕ ∈ Q } . In otherwords l = 1 , K = 1 and r = r , ∈ C ( 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 and a neighborhood of U in Q containing ϕ such that, recalling (23),˜ g , > r n − g in U. Thus we obtain a strict inequality in (25) and we cannot have equality in (21). (Notethat if we additionally assume that h is strictly convex, then equality in Jensen inequalityalso implies that ˜ r must be constant.) We thus obtain that Ω is a ball centered at theorigin with radius r. Step 5.
Let us treat now the general case of equality in (21). Since Ω is bounded thereexsist
M > | x | ≤ M for all x ∈ ∂ Ω and all x ∈ ∂B R . Let us define ˜ a by ˜ a ( t ) = (cid:40) a ( t ) − a ( M ) if 0 < t ≤ M t ≥ M. Note that ˜ a is continuous and ˜ a ≥ , because a is non-increasing (Remark 13). Moreover˜ h defined by˜ h ( t ) = ˜ a (cid:0) t n (cid:1) t n − n = (cid:40) a (cid:0) t n (cid:1) t n − n − a ( M ) t n − n if 0 < t n ≤ M t n ≥ M is non-increasing and convex, because it is the sum of two nonincreasing and convexfunctions. Therefore we can apply (21) to obtain that (cid:90) ∂ Ω ˜ a ( | x | ) d H n − ( x ) ≥ (cid:90) ∂B R ˜ a ( | x | ) d H n − ( x ) . By the choice of M and the definition of ˜ a we have that ˜ a ( | x | ) = a ( | x | ) − a ( M ) for any x ∈ ∂ Ω ∪ ∂B R . So the inequality becomes (cid:90) ∂ Ω a ( | x | ) d H n − ( x ) − a ( M ) H n − ( ∂ Ω) ≥ (cid:90) ∂B R a ( | x | ) d H n − ( x ) − a ( M ) H n − ( ∂B R ) . Using now the assumption that we have equality in (21) and that a ( M ) >
0, we obtain that H n − ( ∂ Ω) ≤ H n − ( ∂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 andthe result follows.We will now prove Part (ii) of the main theorem. We will use the notation: if x ∈ R n write x = ( x , x (cid:48) ) , where x (cid:48) ∈ R n − . roof of Theorem 9 Part (ii). Step 1.
It is sufficient to find a sequence of sets Ω (cid:15) such that each Ω (cid:15) is Lipschitz, bounded, connected, 0 ∈ Ω (cid:15) andlim (cid:15) → | Ω (cid:15) | → C > (cid:15) → (cid:90) ∂ Ω (cid:15) | x | p d H n − = 0 . (28)Then, by approximation, (28) holds also for a sequence of smooth sets. Then this sequencecan be rescaled, defining a new sequence (cid:98) Ω (cid:15) = λ (cid:15) Ω (cid:15) with λ (cid:15) = C /n | Ω (cid:15) | − /n so that | (cid:98) Ω (cid:15) | is constant and (cid:98) Ω (cid:15) still satisfies the second limit in (28).Let us now show (28). Again by a rescaling argument, we can fix one C and assumewithout loss of generality that C = α n R n for some R > . The idea is to choose Ω (cid:15) asa ball of radius R and center going away to infinity, with a long and narrow cylinderattached to it so that it contains the origing, see Figure 2. One has to choose the lengthand radius of the cylinder carefully. More precisely: let us fix some R > c (cid:15) = (cid:112) R − (cid:15) + (cid:15) − n +1 . Then Ω (cid:15) shall be defined as the union of the following three sets: Ω (cid:15) = A (cid:15) ∪ M (cid:15) ∪ D (cid:15) ,A (cid:15) = { x ∈ B n(cid:15) : x ≤ } M (cid:15) = (cid:0) , (cid:15) − n +1 (cid:1) × B n − (cid:15) = (cid:8) x ∈ R n : 0 ≤ x ≤ (cid:15) − n +1 , | x (cid:48) | < (cid:15) (cid:9) D (cid:15) = (cid:8) x ∈ R n : | x − ( c (cid:15) , | < R, x ≥ (cid:15) − n +1 (cid:9) Figure 2: the domain Ω (cid:15)
By construction Ω (cid:15) is a bounded connected Lipschitz set containing the origin. Notethat | A (cid:15) | = 12 α n (cid:15) n , | M (cid:15) | = α n − (cid:15) n − (cid:15) n − = α n − (cid:15) n − . So we obtain that lim (cid:15) → | Ω (cid:15) | = lim (cid:15) → | D (cid:15) | = α n R n . Step 2.
It remains to estimate the weighted perimeter of ∂ Ω (cid:15) . The contribution comingfrom A (cid:15) is 12 (cid:90) ∂B n(cid:15) | x | p d H n − = ω n − (cid:15) n − p → (cid:15) → , n + p − > . Note that for any x ∈ ∂D (cid:15) , we have the estimates, (since p < | x | ≥ (cid:15) − n +1 and hence | x | p ≤ (cid:15) − p ( n − . This gives the estimate (cid:90) ∂D (cid:15) | x | p d H n − ≤ (cid:15) − p ( n − ω n − R n − → (cid:15) → . It remains to estimate the contribution coming from M (cid:15) . It is equal to (cid:90) (cid:15) − n +1 (cid:90) S n − (cid:15) (cid:0) x + | x (cid:48) | (cid:1) p d H n − ( x (cid:48) ) dx = ω n − (cid:15) n − (cid:90) (cid:15) − n +1 (cid:0) x + (cid:15) (cid:1) p dx = L + L , where L = ω n − (cid:15) n − (cid:90) (cid:15) (cid:0) x + (cid:15) (cid:1) p dx and L = ω n − (cid:15) n − (cid:90) (cid:15) − n +1 (cid:15) (cid:0) x + (cid:15) (cid:1) p dx . To estimate L we just use that x + (cid:15) ≥ (cid:15) , and that p < L ≤ ω n − (cid:15) n + p − → . To estimate L we use that x + (cid:15) ≥ x and therefore L ≤ ω n − (cid:15) n − (cid:90) (cid:15) − n +1 (cid:15) x p dx . we distinguish 3 cases. Case 1, p > − . In this case p + 1 > L ≤ ω n − p + 1 (cid:15) n − (cid:16) (cid:15) − ( n − p +1) − (cid:15) p +1) (cid:17) ≤ ω n − p + 1 (cid:15) n − − ( n − p +1) . So L tends to zero as (cid:15) → n − − ( n − p +1) > , which is equivalentto p < n − n − . This is true, because p < n ≥ . Case 2, p = − . Explicit integration as in Case 1 gives that L ≤ − ω n − ( n + 1) (cid:15) n − log( (cid:15) ) → , because n ≥ . Case 3, p < − . Hence p + 1 < L ≤ − ω n − p + 1 (cid:15) n − (cid:16) (cid:15) p +1) − (cid:15) − ( n − p +1) (cid:17) ≤ − ω n − p + 1 (cid:15) n − p +1) → , using again that n + p − > . 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, sinceit is quite short and contains a new and interesting interpolation argument due to Alvinoet alt. [1]. The general case follows from a more standard, but weighted, symmetrizationargument, see also [1]. 16 roposition 15
Let n ≥ , < p < , Ω be a bounded open domain, starshaped withrespect to the origin and with defining function R (see (35) in Section 5) such that R isLipschitz. Then it holds that nα − pn n | Ω | p + n − n ≤ (cid:90) ∂ Ω | x | p d H n − . 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 theproposition. We abbreviate S n − = S n − , see (14) for notation. From Lemma 18 in theappendix we obtain that P p (Ω) := (cid:90) ∂ Ω | x | p d H n − = (cid:90) S n − R p + n − (cid:114) R |∇ S n − R | d H n − . Define Z = R p + n − n − and A = (cid:18) n − p + n − (cid:19) ≤ , Note that for any n ≥ ≥ A = (cid:18) n − p + n − (cid:19) ≥ (cid:18) − p + 3 − (cid:19) ≥ − p ≥ p ∈ [0 , . (29)From Lemma 19 and the previous inequality we obtain P p (Ω) = (cid:90) S n − Z n − (cid:114) Z |∇ S n − Z | A d H n − ≥ (cid:90) S n − Z n − (cid:114) Z |∇ S n − Z | (1 − p ) d H n − , (30)Now use that for any a, b ≥ t (cid:55)→ log (cid:2)(cid:82) √ a + bt (cid:3) is concave, so that (cid:90) (cid:112) a + b ((1 − p ) t + pt ) ≥ (cid:18)(cid:90) (cid:112) a + bt (cid:19) p (cid:18)(cid:90) (cid:112) a + bt (cid:19) − p , t , t ≥ . Using this it follows from (30) (with t = 1 , t = 0) P p (Ω) ≥ (cid:18)(cid:90) S n − Z n − d H n − (cid:19) p (cid:32)(cid:90) S n − Z n − (cid:114) Z |∇ S n − Z | d H n − (cid:33) − p (31)Define a new domain, also starshaped with respect to the origin,Ω Z = { } ∪ (cid:26) x ∈ R n \{ } : x | x | = θ ∈ S n − , < | x | < Z ( θ ) (cid:27) . It follows from Lemma 18, the standard isoperimetric inequality and (36) that (cid:90) S n − Z n − (cid:114) Z |∇ S n − Z | d H n − = H n − ( ∂ Ω Z ) ≥ L n (Ω Z ) n − n nα n n = (cid:18) n (cid:90) S n − Z n d H n − (cid:19) n − n α n n n = (cid:18)(cid:90) S n − Z n d H n − (cid:19) n − n ( nα n ) n .
17e plug this into the (31) and use the definition of Z to conclude that P p (Ω) ≥ (cid:18)(cid:90) S n − R p + n − d H n − (cid:19) p (cid:18)(cid:90) S n − R ( p + n − nn − d H n − (cid:19) (1 − p )( n − n ( nα n ) − pn . (32)The idea is to use now H¨older inequality in the form (cid:90) S n − R n = (cid:90) S n − R np R n − np ≤ (cid:18)(cid:90) S n − R npq (cid:19) q (cid:18)(cid:90) S n − R ( n − np ) q (cid:48) (cid:19) q (cid:48) (33)with 1 < q < ∞ , q = p + n − pn , q (cid:48) = p + n − n − − p ) , q + 1 q (cid:48) = 1 . We take (33) to the power ( p + n − /n and multiply by ( nα n ) (1 − p ) /n to get (using (32)in the last step)( nα n ) − pn (cid:18)(cid:90) S n − R n d H n − (cid:19) p + n − n ≤ ( nα n ) − pn (cid:18)(cid:90) S n − R p + n − d H n − (cid:19) p + n − nq (cid:18)(cid:90) S n − R n ( p + n − n − d H n − (cid:19) p + n − nq (cid:48) =( nα n ) − pn (cid:18)(cid:90) S n − R p + n − d H n − (cid:19) p (cid:18)(cid:90) S n − R n ( p + n − n − d H n − (cid:19) (1 − p )( n − n ≤ P p (Ω) . Thus it follows from (36) that nα − pn n | Ω | p + n − n = ( nα n ) − pn (cid:18)(cid:90) S n − R n d H n − (cid:19) p + n − n ≤ P p (Ω) , 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 − = cR ( θ ) n ( p + n − n − for all θ ∈ S n − . 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 notsatisfied for all p ∈ (0 , . R n . We have used in Section 4 the explicit form of hyperspherical coordinates and theirproperties. Let us define Q ⊂ R n − by Q = (0 , π ) n − × (0 , π ) . H = ( H , . . . , H n ) : Q → S n − ⊂ R n are defined as: for k = 1 , . . . , n and ϕ = ( ϕ , . . . , ϕ n − ) H k ( ϕ ) = cos ϕ k k − (cid:89) l =0 sin ϕ l with convention ϕ = π , ϕ n = 0 . A calculation shows that d i ( ϕ ) := (cid:68) ∂H∂ϕ i , ∂H∂ϕ i (cid:69) = (cid:81) i − l =0 sin ϕ l > . One verifies that themetric tensor S n − in these coordinates is diagonal g ij ( ϕ ) = (cid:68) ∂H∂ϕ i , ∂H∂ϕ j (cid:69) = δ ij d i ( ϕ ) , ( δ ij = 1 if i = j and 0 else) and hence the surface element g is given by g ( ϕ ) = (cid:113) det g ij ( ϕ ) = n − (cid:89) k =1 d k ( ϕ ) = n − (cid:89) k =1 (sin ϕ k ) n − k − . Let us also denote the polar coordinates in R n , denoted as G : (0 , ∞ ) × Q → R n , givenby G ( t, ϕ ) = tH ( ϕ ) . Its Jacobian determinant is then given bydet DG ( t, ϕ ) = t n − g ( ϕ ) . (34)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 theparametrization F : T → Γ F ( ϕ ) = r ( ϕ ) H ( ϕ ) , where r is some smooth function r : T → (0 , ∞ ) . The surface element in this parametriza-tion calculates as det (cid:18)(cid:28) ∂F∂ϕ i , ∂F∂ϕ j (cid:29)(cid:19) i,j =1 ,...,n − = r n − (cid:32) n − (cid:88) i =1 (cid:16) ∂r∂ϕ i (cid:17) d · · · (cid:98) d i · · · d n − + r n − (cid:89) i =1 d i (cid:33) , where (cid:98) d i means that d i should be omitted in the product. In particular, since d i > , thesurface element is bigger than r n − g ( ϕ ) . Proof
Using the relations (cid:104) H ; H (cid:105) = 1 and (cid:68) H ; ∂H∂ϕ i (cid:69) = 0 , one obtains that M ij := (cid:28) ∂F∂ϕ i , ∂F∂ϕ j (cid:29) = ∂r∂ϕ i ∂r∂ϕ j + r g ij = ∂r∂ϕ i ∂r∂ϕ j + r δ ij d i . Thus the matrix M with entries M ij is of the form M = A + r D where A has rank 1 and D is diagonal with entries d i . Thus using the linearity of the determinant in the columnsone obtains that (since no matrix with two columns of A survives when developing thedeterminant succesively with respect to the columns)det M = n − (cid:88) i =1 det (cid:99) A i + det r D, (cid:99) A i is the matrix obtained from r D by replacing the i -th column of r 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 re-spect to the origin. By definition, a bounded open Lipschitz domain Ω ∈ R n is starshapedwith respect to the origin if there exists a function R : S n − → (0 , ∞ ) such thatΩ = { } ∪ (cid:26) x ∈ R n \{ } : x | x | = θ ∈ S n − , < | x | < R ( θ ) (cid:27) (35)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 r (0) ∩ { x > } ) ∪ ( B r (0) ∩ { x < } ) ⊂ R , with r < r ). But we will alway assume that R is alsoLipschitz. It follows from the relation (34) that the volume of a starshaped domaincalculates as L n (Ω) = 1 n (cid:90) S n − R n d H n − (36)For an almost everywhere differentiable function f : S n − → R we recall that the normof the covariant gradient of f at p on the manifold S n − can be calculated as |∇ S n − f | = n − (cid:88) i =1 |∇ E i f | , (37)where { E , . . . , E n − } is any orthonormal basis of T p S n − , and ∇ E f is the derivative indirection E. Lemma 18
Let Ω ⊂ R n be a bounded open Lipschitz set, starshaped with respect to theorigin and defining function R as in (35) . Assume also that R is Lipschitz. Then for anycontinuous function g : (0 , ∞ ) → R it holds that (cid:90) ∂ Ω g ( | x | ) d H n − ( x ) = (cid:90) S n − ( g ◦ R ) R n − (cid:114) R |∇ S n − R | d H n − . (38)The assumption that g is continuous can be reduced, but we will need the lemma onlyfor g ( t ) = t p . Proof
We use the hyperspherical coordinates H, definitions of d i and g, respectively theirproperties (summarized in the beginning of this section). Let ϕ ∈ Q, p = H ( ϕ ) and d i ( ϕ ) = (cid:12)(cid:12)(cid:12)(cid:12) ∂H∂ϕ i (cid:12)(cid:12)(cid:12)(cid:12) , then E i ( p ) = 1 (cid:112) d i ( ϕ ) ∂H∂ϕ i , i = 1 , . . . , n − T p S n − . Thus we obtain from (37) (we can assume that R has been extended to a neighborhood of S n − ) that at p = H ( ϕ ) |∇ S n − R | = n − (cid:88) i =1 d i ( ϕ ) (cid:12)(cid:12)(cid:12)(cid:68) ∇ R ( H ( ϕ )); ∂H∂ϕ i (cid:69)(cid:12)(cid:12)(cid:12) = n − (cid:88) i =1 d i ( ϕ ) (cid:104) ∂∂ϕ i ( R ◦ H ) (cid:105) = n − (cid:88) i =1 d i ( ϕ ) (cid:104) ∂r∂ϕ i (cid:105) (39)20here we define for ϕ ∈ Qr ( 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 (cid:115) det (cid:16)(cid:68) ∂F∂ϕ i , ∂F∂ϕ j (cid:69)(cid:17) = r n − (cid:112) d d · · · d n − (cid:118)(cid:117)(cid:117)(cid:116) r n − (cid:88) i =1 d i (cid:16) ∂r∂ϕ i (cid:17) = r n − (cid:118)(cid:117)(cid:117)(cid:116) r n − (cid:88) i =1 d i (cid:16) ∂r∂ϕ i (cid:17) g ( ϕ ) . Therefore, using the parametrization F to calculate the left side of (38), respectively theparametrization H to calculate the right side and (39), the lemma follows. Lemma 19
Let R be a Lipschitz function, mapping S n − to (0 , ∞ ) , α ∈ R and define Z : S n − → R by Z = R α . Then the following identity holds α R |∇ S n − R | = 1 Z |∇ S n − Z | . In particular if p ∈ R is such that p + n − (cid:54) = 0 and Z = R p + n − n − , then R p + n − (cid:114) R |∇ S n − R | = Z n − (cid:115) Z |∇ S n − Z | (cid:18) n − p + n − (cid:19) Proof
Let E ∈ T p S n − be a tangent vector. Then we get ∇ E Z = (cid:104)∇ Z, E (cid:105) = αR α − (cid:104)∇ R, E (cid:105) and it follows from (37) that |∇ S n − Z | = α R α − |∇ S n − R | . From this the lemmafollows easily.
Acknowledgements
The author was supported by the Chilean Fondecyt Iniciaci´on grantnr. 11150017. He would also like to thank Friedemann Brock for some helpful discussionsrelated to the simplification of the proofs in [1].
References [1] A. Alvino, F. Brock, F. Chiacchio, A. Mercaldo and M.R. Posteraro, Some isoperi-metric inequalities on R n with respect to weights | x | α , J. Math. Anal. Appl. (2017)http://dx.doi.org/10.1016/j.jmaa.2017.01.085.[2] Adimurthi A. and Sandeep K., A singular Moser-Trudinger embedding and its applications,
NoDEA Nonlinear Differential Equations Appl. , 13 (2007), no. 5-6, 585–603.
3] Betta M.F., Brock F., Mercaldo A. and Posteraro M.R., A weighted isoperimetric in-equality and applications to symmetrization,
J. of Inequal. and Appl. , 4 (1999), 215–240.[4] Boyer W., Brown B., Chambers G.R., Loving A. and Tammen S., Isoperimetric Regionsin R n with density r p , arXiv:1504.01720.[5] Bayle V., Ca˜nete A. , Morgan F. and Rosales C., On the isoperimetric problem in Euclideanspace with density, Calc. Var. Partial Differential Equations
31 (2008), no. 1, 27–46.[6] Cabr´e X. and Ros-Oton X., Sobolev and isoperimetric inequalities with monomial weights,
J. Differential Equations
255 (2013), no. 11, 4312–4336.[7] Cabr´e X., Ros-Oton X. and Serra J., Euclidean balls solve some isoperimetric problemswith nonradial weights,
C. R. Math. Acad. Sci. Paris
350 (2012), no. 21-22, 945–947.[8] Ca˜nete A., Miranda M. and Vittone D., Some isoperimetric problems in planes with den-sity,
J. Geom. Anal. , 20 (2010), no. 2, 243–290.[9] Carroll C., Jacob A. Adam, Quinn C. and Walters R., The isoperimetric problem on planeswith density,
Bull. Aust. Math. Soc. , 78 (2008), no. 2, 177–197.[10] Chambers G.R., Proof of the log-convex density conjecture, arXiv:1311.4012.[11] Csat´o G., An isoperimetric problem with density and the Hardy-Sobolev inequality in R , Differential Integral Equations , 28 (2015), no. 9/10, 971–988.[12] Csat´o G. and Roy P., Extremal functions for the singular Moser-Trudinger inequality in 2dimensions,
Calc. Var. Partial Differential Equations , 54 (2015), no. 2, 2341–2366.[13] Csat´o G. and Roy P., The singular Moser-Trudinger inequality on simply connected do-mains,
Communications in Partial Differential Equations , to appear.[14] Dahlberg J., Dubbs A., Newkirk E. and Tran H., Isoperimetric regions in the plane withdensity r p , New York J. Math. , 16 (2010), 31–51.[15] D´ıaz A., Harman N., Howe S. and Thompson D., Isoperimetric problems in sectors withdensity,
Adv. Geom. , 12 (2012), 589–619.[16] Barbosa J.L. and do Carmo M., Stability of hypersurfaces with constant mean curvature,
Math. Z. , 185 (1984), no. 3, 339–353.[17] Figalli A. and Maggi F., On the isoperimetric problem for radial log-convex densities,
Calc.Var. Partial Differential Equations , 48 (2013), no. 3-4, 447–489.[18] Flucher M., Extremal functions for the Trudinger-Moser inequality in 2 dimensions,
Com-ment. Math. Helvetici , 67 (1992), 471–497.[19] Fusco N., Maggi F. and Pratelli A., On the isoperimetric problem with respect to a mixedEuclidean-Gaussian density,
J. Funct. Anal. , 260 (2011), no. 12, 3678–3717.[20] Di Giosia L., Habib J., Kenigsberg L., Pittman D and Zhu W,
Balls Isoperimetric in R n with Volume and Perimeter Densities r m and r k , arXiv:1610.05830v1.[21] Morgan F., Regularity of isoperimetric hypersurfaces in Riemannian manifolds, Trans.Amer. Math. Soc. , 355 (2003), no. 12, 5041–5052.[22] Morgan F., http://sites.williams.edu/Morgan/2010/06/22/variation-formulae-for-perimeter-and-volume-densities/.[23] Morgan F. and Pratelli A., Existence of isoperimetric regions in R n with density, Ann.Global Anal. Geom. , 43 (2013), no. 4, 331–365.[24] Walter W.,
Ordinary differential equations , English translation, Springer, 1998., English translation, Springer, 1998.