A weighted isoperimetric inequality on the hyperbolic plane
aa r X i v : . [ m a t h . A P ] D ec A weighted isoperimetric inequality on the hyperbolic plane
I. McGillivraySchool of Mathematics, University of BristolUniversity Walk, Bristol BS8 1TW, [email protected]
Abstract
We prove a counterpart of the log-convex density conjecture in the hyperbolic plane.
Key words: isoperimetric problem, log-convex density, hyperbolic planeMathematics Subject Classification 2010: 49K20
Let U stand for the open unit disc in R centred at the origin. Define ζ : [0 , → R ; t − t and φ : [0 , → [0 , + ∞ ); t Z t ζ dτ = 2 tanh − t. The hyperbolic metric on U is associated with the differential expression ds = ζ ( | z | ) | dz | and thehyperbolic distance between 0 and a point z in U is φ ( | z | ).Let h : [0 , → R be a non-decreasing φ -convex function (that is, h ◦ φ − is convex) and set ψ : [0 , → (0 , + ∞ ); t e h ( t ) . (1.1)Define densities f : U → R ; x ( ζ ψ )( | x | ); (1.2) g : U → R ; x ( ζψ )( | x | ); (1.3)on U . The weighted hyperbolic volume V = V f := f L is defined on the L -measurable sets in U . The weighted hyperbolic perimeter of a set E of locally finite perimeter in U is defined by P g ( E ) := Z U g d | Dχ E | ∈ [0 , + ∞ ] . (1.4)We study minimisers for the weighted isoperimetric problem I ( v ) := inf n P g ( E ) : E is a set with locally finite perimeter in U and V f ( E ) = v o (1.5)for v >
0. Our first main result is the following.
Theorem 1.1.
Let f and g be as in (1.2) and (1.3) where h : [0 , → R is a non-decreasing φ -convex function. Let v > and B a centred disc in U with V f ( B ) = v . Then B is a minimiserfor (1.5). x ∈ [0 ,
1) and v ≥ h in direction v by h ′ + ( x, v ) := lim t ↓ h ( x + tv ) − h ( x ) t ∈ R and define h ′− ( x, v ) similarly for x ∈ (0 ,
1) and v ≤
0. The directional derivatives exist in R as aconsequence of the fact that h is φ -convex. We introduce the notation ̺ + := h ′ + ( · , +1) , ̺ − := − h ′ + ( · , −
1) and ̺ := (1 / ̺ + + ̺ − )on (0 , h is locally of bounded variation and is differentiable a.e.with h ′ = ̺ a.e. on (0 , ̺ := ζ ′ /ζ = tζ noting that ζ ′ = tζ . Put e ̺ := ˆ ̺ + ̺ . Our secondmain result is a uniqueness theorem. Theorem 1.2.
Let f and g be as in (1.2) and (1.3) where h : [0 , → R is a non-decreasing φ -convex function. Suppose that R := inf { ̺ > } ∈ [0 , and set v := V ( B (0 , R )) . Let v > and E a minimiser for (1.5). The following hold:(i) if v ≤ v then E is a.e. equivalent to a ball B in B (0 , R ) with V ( B ) = V ( E ) ;(ii) if v > v then E is a.e. equivalent to a centred ball B with V ( B ) = V ( E ) . The above two results are the counterpart in the hyperbolic plane of the log-convex density con-jecture proved in [8] Theorems 1.1 and 1.2. A number of works have been devoted to this andsimilar questions in the Euclidean setting: as well as the last-mentioned, we refer to [3], [4], [14],[23], [26], [31]; in the hyperbolic setting we mention [11], [22], [28].We give a brief outline of the article. After a discussion of some preliminary material in Section2 we show that (1.5) admits an open relatively compact minimiser with C boundary in U . Theargument draws upon the regularity theory for almost minimal sets (cf. [34]; also [25]) and includesan adaptation of [27] Proposition 3.1. In Section 4 it is shown that the boundary is of class C , (and has weakly bounded curvature). This result is contained in [27] Corollary 3.7 (see also [9])but we include a proof for completeness. This Section also includes the result that a minimisermay be supposed to possess spherical cap symmetry (Theorem 4.5). Section 5 contains furtherresults on spherical cap symmetric sets useful in the sequel. The main result of Section 6 isTheorem 6.5 which shows that the generalised (mean) curvature is conserved along the boundaryof a minimiser in a weak sense (so long as the minimiser satisfies some additional requirements).In Section 7 it is shown that there exist convex minimisers of (1.5). Sections 8 and 9 comprise ananalytic interlude and are devoted to the study of solutions of the first-order differential equationthat appears in Theorem 6.6 subject to Dirichlet boundary conditions. Section 9 for examplecontains a comparison theorem for solutions to a Ricatti equation (Theorem 9.16 and Corollary9.17). These comparison theorems are new as far as the author is aware. Finally, Section 10concludes the proof of the two main theorems above. Geometric measure theory.
We use | · | to signify the Lebesgue measure on R (or occasionally L ). Let U be a non-empty open set in R . Let E be a L -measurable set in U . The set ofpoints in E with density t ∈ [0 ,
1] is given by E t := (cid:26) x ∈ R : lim ρ ↓ | E ∩ B ( x, ρ ) || B ( x, ρ ) | = t (cid:27) . As usual B ( x, ρ ) denotes the open ball in R with centre x ∈ R and radius ̺ >
0. The set E isthe measure-theoretic interior of E while E is the measure-theoretic exterior of E . The essentialboundary of E is the set ∂ ⋆ E := R \ ( E ∪ E ).2ecall that an integrable function u on U is said to have bounded variation if the distributionalderivative of u is representable by a finite Radon measure Du (cf. [1] Definition 3.1 for example)with total variation | Du | on U ; in this case, we write u ∈ BV( U ). The set E has (locally) finiteperimeter if χ E belongs to BV( U ) (resp. BV loc ( U )). The reduced boundary F E of E is definedby F E := n x ∈ supp | Dχ E | ∩ U : ν E ( x ) := lim ρ ↓ Dχ E ( B ( x, ρ )) | Dχ E | ( B ( x, ρ )) exists in R and | ν E ( x ) | = 1 o (cf. [1] Definition 3.54) and is a Borel set (cf. [1] Theorem 2.22 for example). We use H k ( k ∈ [0 , + ∞ )) to stand for k -dimensional Hausdorff measure. If E is a set of finite perimeter in U then F E ∩ U ⊂ E / ⊂ ∂ ∗ E and H ( ∂ ∗ E \ F E ) = 0 (2.1)by [1] Theorem 3.61.Let g be a positive locally Lipschitz density on U . Let E be a set of finite perimeter in U and V an open set in U . The weighted perimeter of E relative to V is defined by P g ( E, V ) := sup n Z V ∩ E div( gX ) dx : X ∈ C ∞ c ( V, R ) , k X k ∞ ≤ o . We often write P g ( E ) for P g ( E, U ). By the Gauss-Green formula ([1] Theorem 3.36 for example)and a convolution argument, P g ( E, V ) = sup n Z U g h ν E , X i d | Dχ E | : X ∈ C ∞ c ( U, R ) , supp[ X ] ⊂ V, k X k ∞ ≤ o = sup n Z U g h ν E , X i d | Dχ E | : X ∈ C c ( U, R ) , supp[ X ] ⊂ V, k X k ∞ ≤ o = Z V g d | Dχ E | (2.2)where we have also used [1] Propositions 1.47 and 1.23. Lemma 2.1.
Let ϕ be a C diffeomeorphism of R which coincides with the identity map on thecomplement of a compact set and E ⊂ R with χ E ∈ BV( R ) . Then(i) χ ϕ ( E ) ∈ BV( R ) ;(ii) ∂ ⋆ ϕ ( E ) = ϕ ( ∂ ⋆ E ) ;(iii) H ( F ϕ ( E )∆ ϕ ( F E )) = 0 .Proof. Part (i) follows from [1] Theorem 3.16 as ϕ is a proper Lipschitz function. Given x ∈ E we claim that y := ϕ ( x ) ∈ ϕ ( E ) . Let M stand for the Lipschitz constant of ϕ and L stand forthe Lipschitz constant of ϕ − . Note that B ( y, r ) ⊂ ϕ ( B ( x, Lr )) for each r >
0. As ϕ is a bijectionand using [1] Proposition 2.49, | ϕ ( E ) ∩ B ( y, r ) | ≤ | ϕ ( E ) ∩ ϕ ( B ( x, Lr ) | = | ϕ ( E ∩ B ( x, Lr )) | ≤ M | E ∩ B ( x, Lr ) | . This means that | ϕ ( E ) ∩ B ( y, r ) || B ( y, r ) | ≤ ( LM ) | E ∩ B ( x, Lr ) || B ( x, Lr ) | for r > ϕ ( E ) ⊂ [ ϕ ( E )] . The reverse inclusion canbe seen using the fact that ϕ is a bijection. In summary ϕ ( E ) = [ ϕ ( E )] . The corresponding3dentity for E can be seen in a similar way. These identities entail (ii) . From (2.1) and (ii) wemay write F ϕ ( E ) ∪ N = ϕ ( F E ) ∪ ϕ ( N ) for H -null sets N , N in R . Item (iii) follows. Curves with weakly bounded curvature.
Suppose the open set E in R has C boundary M .Denote by n : M → S the inner unit normal vector field. Given p ∈ M we choose a tangentvector t ( p ) ∈ S in such a way that the pair { t ( p ) , n ( p ) } forms a positively oriented basis for R .There exists a local parametrisation γ : I → M where I = ( − δ, δ ) for some δ > C with γ (0) = p . We always assume that γ is parametrised by arc-length and that ˙ γ (0) = t ( p ) wherethe dot signifies differentiation with respect to arc-length. Let X be a vector field defined in someneighbourhood of p in M . Then( D t X )( p ) := dds (cid:12)(cid:12)(cid:12) s =0 ( X ◦ γ )( s ) (2.3)if this limit exists and the divergence div M X of X along M at p is defined bydiv M X := h D t X, t i (2.4)evaluated at p . Suppose that X is a vector field in C ( U, R ) where U is an open neighbourhoodof p in R . Thendiv X = div M X + h D n X, n i (2.5)at p . If p ∈ M \ { } let σ ( p ) stand for the angle measured anti-clockwise from the position vector p to the tangent vector t ( p ); σ ( p ) is uniquely determined up to integer multiples of 2 π .Let E be an open set in R with C , boundary M . Let x ∈ M and γ : I → M a localparametrisation of M in a neighbourhood of x . There exists a constant c > | ˙ γ ( s ) − ˙ γ ( s ) | ≤ c | s − s | for s , s ∈ I ; a constraint on average curvature (cf. [13], [21]). That is, ˙ γ is Lipschitz on I . So˙ γ is absolutely continuous and differentiable a.e. on I with˙ γ ( s ) − ˙ γ ( s ) = Z s s ¨ γ ds (2.6)for any s , s ∈ I with s < s . Moreover, | ¨ γ | ≤ c a.e. on I (cf. [1] Corollary 2.23). As h ˙ γ , ˙ γ i = 1on I we see that h ˙ γ , ¨ γ i = 0 a.e. on I . The (geodesic) curvature k is then defined a.e. on I viathe relation¨ γ = k n (2.7)as in [21]. The curvature k of M is defined H -a.e. on M by k ( x ) := k ( s ) (2.8)whenever x = γ ( s ) for some s ∈ I and k ( s ) exists. We sometimes write H ( · , E ) = k .Let E be an open set in R with C boundary M . Let x ∈ M and γ : I → M a local parametrisa-tion of M in a neighbourhood of x . In case γ = 0 let θ stand for the angle measured anti-clockwisefrom e to the position vector γ and σ stand for the angle measured anti-clockwise from theposition vector γ to the tangent vector t = ˙ γ . Put r := | γ | on I . Then r , θ ∈ C ( I ) and˙ r = cos σ ; (2.9) r ˙ θ = sin σ ; (2.10)on I provided that γ = 0. Now suppose that M is of class C , . Let α stand for the anglemeasured anti-clockwise from the fixed vector e to the tangent vector t (uniquely determined4p to integer multiples of 2 π ). Then t = (cos α , sin α ) on I so α is absolutely continuous on I .In particular, α is differentiable a.e. on I with ˙ α = k a.e. on I . This means that α ∈ C , ( I ).In virtue of the identities r cos σ = h γ , t i and r sin σ = −h γ , n i we see that σ is absolutelycontinuous on I and σ ∈ C , ( I ). By choosing an appropriate branch we may assume that α = θ + σ (2.11)on I . We may choose σ in such a way that σ ◦ γ = σ on I . Flows.
Recall that a diffeomorphism ϕ : R → R is said to be proper if ϕ − ( K ) is compactwhenever K ⊂ R is compact. Given X ∈ C ∞ c ( R , R ) there exists a 1-parameter group of proper C ∞ diffeomorphisms ϕ : R × R → R as in [24] Lemma 2.99 that satisfy ∂ t ϕ ( t, x ) = X ( ϕ ( t, x )) for each ( t, x ) ∈ R × R ; ϕ (0 , x ) = x for each x ∈ R . (2.12)We often use ϕ t to refer to the diffeomorphism ϕ ( t, · ) : R → R . Lemma 2.2.
Let X ∈ C ∞ c ( R , R ) and ϕ be the corresponding flow as above. Then(i) there exists R ∈ C ∞ ( R × R , R ) and K > such that ϕ ( t, x ) = (cid:26) x + tX ( x ) + R ( t, x ) for x ∈ supp[ X ]; x for x supp[ X ]; where | R ( t, x ) | ≤ Kt for ( t, x ) ∈ R × R ;(ii) there exists R (1) ∈ C ∞ ( R × R , M ( R )) and K > such that dϕ ( t, x ) = (cid:26) I + tdX ( x ) + R (1) ( t, x ) for x ∈ supp[ X ]; I for x supp[ X ]; where | R (1) ( t, x ) | ≤ K t for ( t, x ) ∈ R × R ;(iii) there exists R (2) ∈ C ∞ ( R × R , R ) and K > such that J dϕ ( t, x ) = (cid:26) t div X ( x ) + R (2) ( t, x ) for x ∈ supp[ X ];1 for x supp[ X ]; where | R (2) ( t, x ) | ≤ K t for ( t, x ) ∈ R × R .Let x ∈ R , v a unit vector in R and M the line though x perpendicular to v . Then(iv) there exists R (3) ∈ C ∞ ( R × R , R ) and K > such that J d M ϕ ( t, x ) = (cid:26) t (div M X )( x ) + R (3) ( t, x ) for x ∈ supp[ X ];1 for x supp[ X ]; where | R (3) ( t, x ) | ≤ K t for ( t, x ) ∈ R × R .Proof. (i) First notice that ϕ ∈ C ∞ ( R × R ) by [19] Theorem 3.3 and Exercise 3.4. The statementfor x supp[ X ] follows by uniqueness (cf. [19] Theorem 3.1); the assertion for x ∈ supp[ X ] followsfrom Taylor’s theorem. (ii) follows likewise: note, for example, that[ ∂ tt dϕ ] αβ | t =0 = X α,βδ X δ + X α,γ X γ,β where the subscript , signifies partial differentiation. (iii) follows from (ii) and the definition ofthe 2-dimensional Jacobian (cf. [1] Definition 2.68). (iv) Using [1] Definition 2.68 together withthe Cauchy-Binet formula [1] Proposition 2.69, J d M ϕ ( t, x ) = | dϕ ( t, x ) v | for t ∈ R and the resultfollows from (ii) .Let U be an open set in R . Let I be an open interval in R containing 0. Let Z : I × U → R ; ( t, x ) Z ( t, x ) be a continuous time-dependent vector field on U with the properties5Z.1) Z ( t, · ) ∈ C c ( U, R ) for each t ∈ I ;(Z.2) supp[ Z ( t, · )] ⊂ K for each t ∈ I for some compact set K ⊂ U .By [19] Theorems I.1.1, I.2.1, I.3.1, I.3.3 there exists a unique flow ϕ : I × U → U such that(F.1) ϕ : I × U → U is of class C ;(F.2) ϕ (0 , x ) = x for each x ∈ U ;(F.3) ∂ t ϕ ( t, x ) = Z ( t, ϕ ( x, t )) for each ( t, x ) ∈ I × U ;(F.4) ϕ t := ϕ ( t, · ) : U → U is a proper diffeomorphism for each t ∈ I . Lemma 2.3.
Let Z be a time-dependent vector field with the properties ( Z. - ( Z. and ϕ be thecorresponding flow. Then(i) for ( t, x ) ∈ I × R , dϕ ( t, x ) = (cid:26) I + tdZ ( x ) + tR ( t, x ) for x ∈ K ; I for x K ; where sup K | R ( t, · ) | → as t → .Let x ∈ R , v a unit vector in R and M the line though x perpendicular to v . Then(ii) for ( t, x ) ∈ I × R , J d M ϕ ( t, x ) = (cid:26) t (div M Z )( x ) + tR (1) ( t, x ) for x ∈ K ;1 for x K. where sup K | R (1) ( t, · ) | → as t → .Proof. (i) We first remark that the flow ϕ : I × R → R associated to Z is continuouslydifferentiable in t, x in virtue of (Z.1) by [19] Theorem I.3.3. Put y ( t, x ) := dϕ ( t, x ) for ( t, x ) ∈ I × R . By [19] Theorem I.3.3,˙ y ( t, x ) = dZ ( t, ϕ ( t, x )) y ( t, x )for each ( t, x ) ∈ I × R and y (0 , x ) = I for each x ∈ R where I stands for the 2 × x ∈ K and t ∈ I , dϕ ( t, x ) = I + dϕ ( t, x ) − dϕ (0 , x )= I + t ˙ y (0 , x ) + t n dϕ ( t, x ) − dϕ (0 , x ) t − ˙ y (0 , x ) o = I + tdZ (0 , x ) + t n y ( t, x ) − y (0 , x ) t − ˙ y (0 , x ) o = I + tdZ ( x ) + t n y ( t, x ) − y (0 , x ) t − ˙ y (0 , x ) o . Applying the mean-value theorem component-wise and using uniform continuity of the matrix ˙ y in its arguments we see that y ( t, · ) − y (0 , · ) t − ˙ y (0 , · ) → K as t →
0. This leads to (i) . Part (ii) follows as in Lemma 2.2.6et f and g be positive locally Lipschitz densities on U . Let E be a set of finite perimeter in U with V f ( E ) < + ∞ . The first variation of weighted volume resp. perimeter along X ∈ C ∞ c ( U, R )is defined by δV f ( X ) := ddt (cid:12)(cid:12)(cid:12) t =0 V f ( ϕ t ( E )) , (2.13) δP + g ( X ) := lim t ↓ P g ( ϕ t ( E )) − P g ( E ) t , (2.14)whenever the limit exists. By Lemma 2.1 the g -perimeter in (2.14) is well-defined. Convex functions.
Suppose that h : [0 , → R is a φ -convex function. For x ∈ [0 ,
1) and v ≥ h ′ + ( x, v ) := lim t ↓ h ( x + tv ) − h ( x ) t ∈ R and define h ′− ( x, v ) similarly for x ∈ (0 ,
1) and v ≤
0. For future use we introduce the notation ̺ + := h ′ ( · , +1) ζ , ̺ − := − h ′ ( · , − ζ and ̺ := (1 / ̺ + + ̺ − )on (0 , h is differentiable a.e. and h ′ = ζ̺ a.e. on (0 , Lemma 2.4.
Suppose that the function g takes the form (1.3). Then(i) the directional derivative g ′ + ( x, v ) exists in R for each x ∈ U and v ∈ R ;(ii) for v ∈ R , g ′ + ( x, v ) = ( g ( x ) n ζ | x |h x | x | , v i + h ′ + ( | x | , sgn h x, v i ) |h x | x | , v i| o for x ∈ U \ { } ; g (0) h ′ + (0 , +1) | v | for x = 0; (iii) if M is a C hypersurface in R such that cos σ = 0 on M then f is differentiable H -a.e.on M and ( ∇ f )( x ) = f ( x ) ̺ ( | x | ) h x, ·i| x | for H -a.e. x ∈ M .Proof. The assertion in (i) follows from the monotonicity of chords property while (ii) is straight-forward. (iii)
Let x ∈ M and γ : I → M be a C -parametrisation of M near x as above. Now r ∈ C ( I ) and ˙ r (0) = cos σ ( x ) = 0 so we may assume that r : I → r ( I ) ⊂ (0 , + ∞ ) is a C dif-feomorphism. The differentiability set D ( h ) of h has full Lebesgue measure in [0 , + ∞ ). It followsby [1] Proposition 2.49 that r − ( D ( h )) has full measure in I . This entails that f is differentiable H -a.e. on γ ( I ) ⊂ M . C regularity and boundedness We start with an existence theorem. Let U := B (0 ,
1) stand for the open centred unit disc in R .The below follows as in [29] Theorem 3.3. Theorem 3.1.
Suppose that h : [0 , → (0 , + ∞ ) is non-decreasing, lower semi-continuous anddiverges to infinity as r ↑ . Define f and g as in (1.2) and (1.3). Then (1.5) admits a minimiserfor each v > . U be an open set in R . Given positive locally Lipschitz densities f, g on U we consider thevariational problem I ( v ) := inf n P g ( E ) : E is a set of finite perimeter in U and V f ( E ) = v o (3.1)for v > Proposition 3.2.
Let f be a positive locally Lipschitz density on U . Let E be a set with finiteperimeter in U . Let X ∈ C ∞ c ( U, R ) . Then δV f ( X ) = Z E div( f X ) dx = − Z F E f h ν E , X i d H . Proof.
Let t ∈ R . By the area formula ([1] Theorem 2.71 and (2.74)), V f ( ϕ t ( E )) = Z ϕ t ( E ) f dx = Z E ( f ◦ ϕ t ) J d ( ϕ t ) x dx (3.2)and V f ( ϕ t ( E )) − V f ( E ) = Z E ( f ◦ ϕ t ) J dϕ t − f dx = Z E ( f ◦ ϕ t )( J dϕ t − dx + Z E f ◦ ϕ t − f dx. The density f is locally Lipschitz and in particular differentiable a.e. on U (see [1] 2.3 for example).By the dominated convergence theorem and Lemma 2.2, δV f ( X ) = Z E n f div( X ) + h∇ f, X i o dx = Z E div( f X ) dx = − Z F E f h ν E , X i d H by the generalised Gauss-Green formula [1] Theorem 3.36. Proposition 3.3.
Let g be a positive locally Lipschitz density on U . Let E be a set with finiteperimeter in U . Let X ∈ C ∞ c ( U, R ) . Then there exist constants C > and δ > such that | P g ( ϕ t ( E )) − P g ( E ) | ≤ C | t | for | t | < δ .Proof. Let t ∈ R . By Lemma 2.1 and [1] Theorem 3.59, P g ( ϕ t ( E )) = Z U g d | Dχ ϕ t ( E ) | = Z F ϕ t ( E ) ∩ U g d H = Z ϕ t ( F E ∩ U ) g d H . As F E is countably 1-rectifiable ([1] Theorem 3.59) we may use the generalised area formula [1]Theorem 2.91 to write P g ( ϕ t ( E )) = Z F E ∩ U ( g ◦ ϕ t ) J d F E ( ϕ t ) x d H . For each x ∈ F E ∩ U and any t ∈ R , | ( g ◦ ϕ t )( x ) − g ( x ) | ≤ K | ϕ ( t, x ) − x | ≤ K k X k ∞ | t | where K is the Lipschitz constant of f on supp[ X ]. The result follows upon writing P g ( ϕ t ( E )) − P g ( E ) = Z F E ∩ U g ◦ ϕ t ( J d F E ( ϕ t ) x −
1) + [ g ◦ ϕ t − g ] d H (3.3)and using Lemma 2.2. 8 emma 3.4. Let f be a positive locally Lipschitz density on U . Let E be a set with finite perimeterin U and p ∈ ( F E ) ∩ U . For any r ∈ (0 , d ( x, ∂U ) there exists X ∈ C ∞ c ( U, R ) with supp[ X ] ⊂ B ( p, r ) such that δV f ( X ) = 1 .Proof. By (2.2) and [1] Theorem 3.59 and (3.57) in particular, P f ( E, B ( p, r )) = Z B ( p,r ) ∩ F E f d H > . By the variational characterisation of the f -perimeter relative to B ( p, r ) we can find Y ∈ C ∞ c ( U, R )with supp[ Y ] ⊂ B ( p, r ) such that0 < Z E ∩ B ( p,r ) div( f Y ) dx = − Z F E ∩ B ( p,r ) f h ν E , Y i d H =: c where we make use of the generalised Gauss-Green formula (cf. [1] Theorem 3.36). Put X :=(1 /c ) Y . Then X ∈ C ∞ c ( U, R ) with supp[ X ] ⊂ B ( p, r ) and δV f ( X ) = 1 according to Proposition3.2. Proposition 3.5.
Let g be a positive lower semi-continuous density on U . Let V be a relativelycompact open set in U with Lipschitz boundary. Let E, F , F be sets with finite perimeter in U .Assume that E ∆ F ⊂⊂ V and E ∆ F ⊂⊂ U \ V . Define F := h F ∩ V i ∪ h F \ V i . Then F is a set of finite perimeter in U and P g ( E ) + P g ( F ) = P g ( F ) + P g ( F ) . Proof.
The function χ E | V ∈ BV( V ) and D ( χ E | V ) = ( Dχ E ) | V . We write χ VE for the boundarytrace of χ E | V (see [1] Theorem 3.87); then χ VE ∈ L ( ∂V, H ∂V ) (cf. [1] Theorem 3.88). We usesimilar notation elsewhere. By [1] Corollary 3.89, Dχ E = Dχ E V + ( χ VE − χ U \ VE ) ν V H ∂V + Dχ E ( U \ V ); Dχ F = Dχ F V + ( χ VF − χ U \ VF ) ν V H ∂V + Dχ F ( U \ V ); Dχ F = Dχ F V + ( χ VF − χ U \ VE ) ν V H ∂V + Dχ E ( U \ V ); Dχ F = Dχ E V + ( χ VE − χ U \ VF ) ν V H ∂V + Dχ F ( U \ V ) . From the definition of the total variation measure ([1] Definition 1.4), | Dχ E | = | Dχ E | V + | χ VE − χ U \ VE | H ∂V + | Dχ E | ( U \ V ); | Dχ F | = | Dχ F | V + | χ VE − χ U \ VE | H ∂V + | Dχ F | ( U \ V ); | Dχ F | = | Dχ F | V + | χ VE − χ U \ VE | H ∂V + | Dχ E | ( U \ V ); | Dχ F | = | Dχ E | V + | χ VE − χ U \ VE | H ∂V + | Dχ F | ( U \ V );where we also use the fact that χ VF = χ VE as E ∆ F ⊂⊂ V and similarly for F . The result nowfollows. Proposition 3.6.
Assume that f and g are positive locally Lipschitz densities on U . Let v > and suppose that the set E is a minimiser of (3.1). Let V be a relatively compact open set in U .There exist constants C > and δ ∈ (0 , d ( V, ∂U )) with the following property. For any x ∈ V and < r < δ , P g ( E ) − P g ( F ) ≤ C (cid:12)(cid:12) V f ( E ) − V f ( F ) (cid:12)(cid:12) (3.4) where F is any set with finite perimeter in U such that E ∆ F ⊂⊂ B ( x, r ) . roof. The proof follows that of [27] Proposition 3.1. We assume to the contrary that( ∀ C > ∀ δ ∈ (0 , d ( V, ∂U ))( ∃ x ∈ V )( ∃ r ∈ (0 , δ ))( ∃ F ⊂ U ) h F ∆ E ⊂⊂ B ( x, r ) ∧ ∆ P g > C | ∆ V f | i (3.5)in the language of quantifiers where we have taken some liberties with notation.Choose p , p ∈ F E ∩ U with p = p . Choose r > B ( p , r ) and B ( p , r ) are disjoint and lie in U . Choose vector fields X j ∈ C ∞ c ( U, R ) with supp[ X j ] ⊂ B ( p j , r )such that δV f ( X j ) = 1 and | P g ( ϕ ( j ) t ( E )) − P g ( E ) | ≤ a j | t | for | t | < δ j and j = 1 , a := max { a , a } . By (3.6), V f ( ϕ ( j ) t ( E )) − V f ( E ) = t + o ( t ) as t → j = 1 , . So there exist ε > > η > t − η | t | < V f ( ϕ ( j ) t ( E )) − V f ( E ) < t + η | t | ; (3.7) | P g ( ϕ ( j ) t ( E )) − P g ( E ) | < ( a + 1) | t | ;for | t | < ε and j = 1 ,
2. In particular, | V f ( ϕ ( j ) t ( E )) − V f ( E ) | > (1 − η ) | t | ; | P g ( ϕ ( j ) t ( E )) − P g ( E ) | < a − η | V f ( ϕ ( j ) t ( E )) − V f ( E ) | ; (3.8)for | t | < ε and j = 1 , C = (1 + a ) / (1 − η ) and δ ∈ (0 , d ( V, ∂U )) such that(a) 0 < δ < dist( B ( p , r ) , B ( p , r )),(b) sup { V f ( B ( x, δ )) : x ∈ V } < (1 − η ) ε .Choose x, r and F as in (3.5). In light of (a) we may assume that B ( x, r ) ∩ B ( p , r ) = ∅ . By (b), | V f ( F ) − V f ( E ) | ≤ V f ( B ( x, r )) ≤ V f ( B ( x, δ )) < (1 − η ) ε. (3.9)From (3.7) and (3.9) we can find t ∈ ( − ε, ε ) such that with F := ϕ (1) t ( E ), V f ( F ) − V f ( E ) = − n V f ( F ) − V f ( E ) o (3.10)by the intermediate value theorem. From (3.5), P g ( F ) < P g ( E ) − C | V f ( F ) − V f ( E ) | (3.11)while from (3.8), P g ( F ) < P g ( E ) + C | V f ( F ) − V f ( E ) | . (3.12)Let F be the set F := h F \ B ( p , r )) i ∪ h B ( p , r ) ∩ F i . E ∆ F ⊂⊂ B ( p , r ). By Proposition 3.5, F is a set of finite perimeter in U and P g ( E ) + P g ( F ) = P g ( F ) + P g ( F ) . We then infer from (3.11), (3.12) and (3.10) that P g ( F ) = P g ( F ) + P g ( F ) − P g ( E ) < P g ( E ) − C | V f ( F ) − V f ( E ) | + P g ( E ) + C | V f ( F ) − V f ( E ) | − P g ( E ) = P g ( E ) . On the other hand, V f ( F ) = V f ( F ) + V f ( F ) − V f ( E ) = V f ( E ) by (3.10). We therefore obtain acontradiction to the isoperimetric property of E .Let E be a set of finite perimeter in U and V a relatively compact open set in U . The minimalityexcess is the function ψ defined by ψ ( E, V ) := P ( E, V ) − ν ( E, V ) (3.13)where ν ( E, V ) := inf { P ( F, V ) : F is a set of finite perimeter in U with F ∆ E ⊂⊂ V } as in [34] (1.9). We recall that the boundary of E is said to be almost minimal in U if for eachrelatively compact open set V in U there exists T ∈ (0 , d ( V, ∂U )) and a positive constant K suchthat for every x ∈ V and r ∈ (0 , T ), ψ ( E, B ( x, r )) ≤ Kr . (3.14)This definition corresponds to [34] Definition 1.5. Theorem 3.7.
Assume that f and g are positive locally Lipschitz densities on U . Let v > andassume that the set E is a minimiser of (3.1). Then the boundary of E is almost minimal in U .Proof. Let V be a relatively compact open set in U and C > δ > δ -neighbourhood of V is denoted I δ ( V ); put W := I δ ( V ). Let x ∈ V and r ∈ (0 , δ ). Forthe sake of brevity write m := inf B ( x,r ) g and M := sup B ( x,r ) g . Let F be a set of finite perimeterin U such that F ∆ E ⊂⊂ B ( x, r ). By Proposition 3.6, P ( E, B ( x, r )) − P ( F, B ( x, r )) ≤ m P g ( E, B ( x, r )) − M P g ( F, B ( x, r ))= 1 m (cid:16) P g ( E, B ( x, r )) − P g ( F, B ( x, r )) (cid:17) + (cid:16) m − M (cid:17) P g ( F, B ( x, r )) ≤ m (cid:16) P g ( E, B ( x, r )) − P g ( F, B ( x, r )) (cid:17) + M − mm P g ( F, B ( x, r )) ≤ C inf W f | V f ( E ) − V f ( F ) | + (2 Lr ) sup W f (inf W f ) P ( F, B ( x, r )) ≤ Cπr sup W f inf W f + (2 Lr ) sup W f (inf W f ) P ( F, B ( x, r ))where L stands for the Lipschitz constant of the restriction of f to W . We then derive that ψ ( E, B ( x, r )) ≤ Cπr sup W f inf W f + (2 Lr ) sup W f (inf W f ) ν ( E, B ( x, r )) . By [16] (5.14), ν ( E, B ( x, r )) ≤ πr . The inequality in (3.14) now follows. Theorem 3.8.
Assume that f and g are positive locally Lipschitz densities on U . Let v > andsuppose that E is a minimiser of (3.1). Then there exists a set e E ⊂ U such that i) e E is a minimiser of (3.1);(ii) e E is equivalent to E ;(iii) e E is open and ∂ e E ∩ U is a C hypersurface in U .Proof. By [16] Proposition 3.1 there exists a Borel set F equivalent to E with the property that ∂F ∩ U = { x ∈ U : 0 < | F ∩ B ( x, ρ ) | < πρ for each ρ > } . By Theorem 3.7 and [34] Theorem 1.9, ∂F ∩ U is a C hypersurface in U (taking note of differencesin notation). The set e E := { x ∈ U : | F ∩ B ( x, ρ ) | = πρ for some ρ > } satisfies (i) - (iii) .Again, let U := B (0 ,
1) stand for the open centred unit disc in R . The below follows as in [29]Theorem 5.9. Theorem 3.9.
Suppose that h : [0 , → (0 , + ∞ ) is lower semi-continuous and non-decreasing.Define f and g as in (1.2) and (1.3). Then any minimiser of (1.5) is essentially relatively compactin U for each v > . Theorem 4.1.
Assume that f and g are positive locally Lipschitz densities on U . Let v > andsuppose that E is a minimiser of (3.1). Then there exists a set e E ⊂ U such that(i) e E is a minimiser of (3.1);(ii) e E is equivalent to E ;(iii) e E is open in U and ∂ e E ∩ U is a C , hypersurface in U .Proof. We may assume that E has the properties listed in Theorem 3.8. Put M := ∂E ∩ U .Let x ∈ M and V be a relatively compact open set in U containing x . Choose C > δ ∈ (0 , d ( V, ∂U )) as in Proposition 3.6. Let 0 < r < δ and X ∈ C ∞ c ( U, R ) with supp[ X ] ⊂ B ( x, r ).Then P g ( E ) − P g ( ϕ t ( E )) ≤ C | V f ( E ) − V f ( ϕ t ( E )) | for each t ∈ R . From the identity (3.3), − Z M ( g ◦ ϕ t )( J d M ( ϕ t ) x − d H ≤ C | V f ( E ) − V f ( ϕ t ( E )) | + Z M [ g ◦ ϕ t − g ] d H ≤ C | V f ( E ) − V f ( ϕ t ( E )) | + √ K k X k ∞ H ( M ∩ supp[ X ]) t where K stands for the Lipschitz constant of g restricted to I δ ( V ). On dividing by t and takingthe limit t → − Z M g div M X d H ≤ C (cid:12)(cid:12)(cid:12) Z M f h n, X i d H (cid:12)(cid:12)(cid:12) + √ K k X k ∞ H ( M ∩ supp[ X ])upon using Lemma 2.2 and Proposition 3.2. Replacing X by − X we derive that (cid:12)(cid:12)(cid:12) Z M g div M X d H (cid:12)(cid:12)(cid:12) ≤ C k X k ∞ H ( M ∩ supp[ X ])12here C = C k f k L ∞ ( I δ ( V )) + √ K .Let γ : I → M be a local C parametrisation of M near x . We first observe that the collection Y := { Y = X ◦ γ : X ∈ C c ( U, R ) } ∩ C c ( I, R )is dense in C c ( I, R ). To see this, note that the mapping φ : ( − δ, δ ) × R → R ; ( s, t ) γ ( s )+ tn (0)is a continuous bijection onto its range for small δ >
0. Let Y ∈ C c ( I, R ). By a partition ofunity argument there exists X ∈ C c ( U, R ) with the property that Y = X ◦ γ on I . Moreover,it holds that k Y k ∞ = k X k ∞ . Finally, use the fact that C c ( U, R ) is dense in C c ( U, R ) with theusual topology.Suppose that γ ( I ) ⊂ M ∩ B ( x, r ). Let Y = X ◦ γ ∈ Y for some X ∈ C c ( U, R ) with supp[ X ] ⊂ B ( x, r ). The above estimate entails that (cid:12)(cid:12)(cid:12) Z I ( g ◦ γ ) h ˙ Y , t i ds (cid:12)(cid:12)(cid:12) ≤ C (cid:12)(cid:12)(cid:12) supp[ Y ] (cid:12)(cid:12)(cid:12) k Y k ∞ . This means that the function w := ht belongs to [BV( I )] by [1] Proposition 3.6 where h := g ◦ γ .Note that h is Lipschitz on I . For s , s ∈ I with s < s , | w ( s ) − w ( s ) | = | Dw ([ s , s )) | ≤ | Dw | (( s , s ))= sup n Z ( s ,s ) ( g ◦ γ ) h t, ˙ Y i ds : Y ∈ C c (( s , s )) and k Y k ∞ ≤ o ≤ C | s − s | . Using the identity t ( s ) − t ( s ) = 1 h ( s ) ( w ( s ) − w ( s )) − h ( s ) − h ( s ) h ( s ) t ( s )we conclude that there exists C > | t ( s ) − t ( s ) | ≤ C | s − s | for any s , s ∈ I .We turn to the topic of spherical cap symmetrisation. Let U be the open unit ball in R centredat the origin. Denote by S τ the centred circle in R with radius τ >
0. We sometimes write S for S . Given x ∈ R , v ∈ S and α ∈ (0 , π ] the open cone with vertex x , axis v and opening angle 2 α is the set C ( x, v, α ) := n y ∈ R : h y − x, v i > | y − x | cos α o . Let E be an L -measurable set in U and τ >
0. The τ -section E τ of E is the set E τ := E ∩ S τ .Put L ( τ ) = L E ( τ ) := H ( E τ ) for τ > p ( E ) := { τ ∈ (0 ,
1) : L ( τ ) > } . The function L is L -measurable by [1] Theorem 2.93.Given τ ∈ (0 ,
1) and 0 < α ≤ π the spherical cap C ( τ, α ) is the set C ( τ, α ) := (cid:26) S τ ∩ C (0 , e , α ) if 0 < α < π ; S τ if α = π ;and has H -measure s ( τ, α ) := 2 ατ . The spherical cap symmetral E sc of the set E is defined by E sc := [ τ ∈ p ( E ) C ( τ, α ) (4.2)where α ∈ (0 , π ] is determined by s ( τ, α ) = L ( τ ). Observe that E sc is a L -measurable set in U and V f ( E sc ) = V f ( E ). Note also that if B is a centred open ball in U then B sc = B \ { } . We13ay that E is spherical cap symmetric if H (( E ∆ E sc ) τ ) = 0 for each τ ∈ (0 , B be a Borel set in (0 , A ( B )over B is the set A ( B ) := { x ∈ U : | x | ∈ B } . Theorem 4.2.
Let E be a set of finite perimeter in U . Then E sc is a set of finite perimeter in U and P ( E sc , A ( B )) ≤ P ( E, A ( B )) (4.3) for any Borel set B ⊂ (0 , and the same inequality holds with E sc replaced by any set F that is L -equivalent to E sc . Corollary 4.3.
Let g be a positive lower semi-continuous radial function on U . Let E be a set offinite perimeter in U . Then P g ( E sc ) ≤ P g ( E ) .Proof. Assume that P g ( E ) < + ∞ . We remark that g is Borel measurable as f is lower semi-continuous. Let ( g h ) be a sequence of simple Borel measurable radial functions on U such that0 ≤ g h ≤ g and g h ↑ g on U as h → ∞ . By Theorem 4.2, P g h ( E sc ) = Z U g h d | Dχ E sc | ≤ Z U g h d | Dχ E | = P g h ( E )for each h . Taking the limit h → ∞ the monotone convergence theorem gives P g ( E sc ) ≤ P g ( E ). Lemma 4.4.
Let E be an L -measurable set in U such that E \ { } = E sc . Then there exists an L -measurable set F in U equivalent to E such that(i) ∂F ∩ U = { x ∈ U : 0 < | F ∩ B ( x, ρ ) | < | B ( x, ρ ) | for small ρ > } ;(ii) F = F ∪ Y where F := { x ∈ U : | F ∩ B ( x, ρ ) | = | B ( x, ρ ) | for some ρ > } and Y ⊂ ∂F ∩ U ;(iii) F is spherical cap symmetric.Proof. Put E := { x ∈ U : | E ∩ B ( x, ρ ) | = | B ( x, ρ ) | for some ρ > } ; E := { x ∈ U : | E ∩ B ( x, ρ ) | = 0 for some ρ > } . We claim that E is spherical cap symmetric. For take x ∈ E with τ = | x | ∈ (0 ,
1) and θ ( x ) ∈ (0 , π ). Then | E ∩ B ( x, ρ ) | = | B ( x, ρ ) | for some ρ >
0. Let y ∈ U with | y | = τ and 0 < θ ( y ) < θ ( x ).Choose a rotation O ∈ SO(2) such that OB ( x, ρ ) = B ( y, ρ ). By decreasing ρ if necessary we mayassume that B ( y, ρ ) ⊂ H . As E \ { } = E sc , O ( E ∩ B ( x, ρ )) ⊂ E ∩ B ( y, ρ ) and consequently | E ∩ B ( y, ρ ) | ≥ | E ∩ B ( x, ρ ) | = | B ( x, ρ ) | so that | E ∩ B ( y, ρ ) | = | B ( y, ρ ) | and y ∈ E . The claimfollows. It follows in a similar way that U \ E is spherical cap symmetric. It can then be seen thatthe set F := ( E ∪ E ) \ E inherits this property. As in [16] Proposition 3.1 the set F is equivalentto E and enjoys the property in (i) . Note that F = E ∪ ( E \ E ) = E ∪ [ E \ ( E ∪ E )] = F ∪ Y where Y := E \ ( E ∪ E ) ⊂ ∂F ∩ U . This leads to (ii) . Theorem 4.5.
Assume that f and g are positive locally Lipschitz densities on U . Let v > andsuppose that E is a minimiser for (3.1). Then there exists an L -measurable set e E in U with theproperties(i) e E is a minimiser of (3.1); ii) L e E = L a.e. on (0 , ;(iii) e E is open and has C , boundary in U ;(iv) e E \ { } = e E sc .Proof. By Corollary 4.3, E := E sc is a minimiser of (3.1) and L E = L E on (0 , E := F with F as in Lemma 4.4. Then L E = L a.e. on (0 ,
1) as E is equivalent to E , E is a minimiser of (3.1) and E is spherical cap symmetric. Moreover, ∂E ∩ U = { x ∈ U : 0 < | E ∩ B ( x, ρ ) | < | B ( x, ρ ) | for small ρ > } . As in the proof of Theorem 3.8, ∂E ∩ U is a C hypersurface in U . In fact, ∂E ∩ U is of class C , by Theorem 4.1. Put e E := ( E ) = { x ∈ U : | E ∩ B ( x, ρ ) | = | B ( x, ρ ) | for some ρ > } . Then e E is open and equivalent to E by Lemma 4.4 so that (i) and (ii) hold. Note that ∂ e E ∩ U = ∂E ∩ U so (iii) holds. As E is spherical cap symmetric the same is true of e E . But e E is openwhich entails that e E \ { } = e E sc . Let H := { x = ( x , x ) ∈ U : x > } stand for the upper hemidisc and S : U → U ; x = ( x , x ) ( x , − x )for reflection in the x -axis. Let O ∈ SO(2) represent rotation anti-clockwise through π/ Lemma 5.1.
Let E be an open set in U with C boundary M = ∂E ∩ U in U and assume that E \ { } = E sc . Let x ∈ M \ { } . Then(i) Sx ∈ M \ { } ;(ii) n ( Sx ) = Sn ( x ) ;(iii) cos σ ( Sx ) = − cos σ ( x ) .Proof. (i) The closure E of E is spherical cap symmetric. The spherical cap symmetral E isinvariant under S from the representation (4.2) and this entails (i) . (ii) is a consequence of thislast observation. (iii) Note that t ( Sx ) = O ⋆ n ( Sx ) = O ⋆ Sn ( x ). Then | x | cos σ ( Sx ) = h Sx, t ( Sx ) i = h Sx, O ⋆ Sn ( x ) i = h x, SO ⋆ Sn ( x ) i = h x, On ( x ) i = −h x, O ⋆ n ( x ) i = −h x, t ( x ) i = −| x | cos σ ( x )as SO ⋆ S = O and O = − O ⋆ .We introduce the projection π : U → [0 , x
7→ | x | . Lemma 5.2.
Let E be an open set in U with boundary M = ∂E ∩ U in U and assume that E \ { } = E sc .(i) Suppose = x ∈ U \ E and θ ( x ) ∈ (0 , π ] . Then there exists an open interval I in (0 , containing τ and α ∈ (0 , θ ( x )) such that A ( I ) \ S ( α ) ⊂ U \ E .(ii) Suppose = x ∈ E and θ ( x ) ∈ [0 , π ) . Then there exists an open interval I in (0 , containing τ and α ∈ ( θ ( x ) , π ) such that A ( I ) ∩ S ( α ) ⊂ E . iii) For each < τ ∈ π ( M ) , M τ is the union of two closed spherical arcs in S τ symmetric aboutthe x -axis.Proof. (i) We can find α ∈ (0 , θ ( x )) such that S τ \ S ( α ) ⊂ U \ E as can be seen from definition(4.2). This latter set is compact so dist( S τ \ S ( α ) , E ) >
0. This means that the ε -neighbourhoodof S τ \ S ( α ) is contained in U \ E for ε > (ii) Again from (4.2) we canfind α ∈ ( θ ( x ) , π ) such that S τ ∩ S ( α ) ⊂ E and the assertion follows as before. (iii) Suppose x , x are distinct points in M τ with 0 ≤ θ ( x ) < θ ( x ) ≤ π . Suppose y lies in theinterior of the spherical arc joining x and x . If y ∈ U \ E then x ∈ U \ E by (i) and hence x M . If y ∈ E we obtain the contradiction that x ∈ E by (ii) . Therefore y ∈ M . We inferthat the closed spherical arc joining x and x lies in M τ . The claim follows noting that M τ isclosed. Lemma 5.3.
Let E be an open set in U with C boundary M in U . Let x ∈ M . Then lim inf E ∋ y → x D y − x | y − x | , n ( x ) E ≥ . Proof.
Assume for a contradiction thatlim inf E ∋ y → x D y − x | y − x | , n ( x ) E ∈ [ − , . There exists η ∈ (0 ,
1) and a sequence ( y h ) in E such that y h → x as h → ∞ and D y h − x | y h − x | , n ( x ) E < − η (5.1)for each h ∈ N . Choose α ∈ (0 , π/
2) such that cos α = η . As M is C there exists r > B ( x, r ) ∩ C ( x, − n ( x ) , α ) ∩ E = ∅ . By choosing h sufficiently large we can find y h ∈ B ( x, r ) with the additional property that y h ∈ C ( x, − n ( x ) , α ) by (5.1). We are thus led to a contradiction. Lemma 5.4.
Let E be an open set in U with C boundary M in U and assume that E \{ } = E sc .For each < τ ∈ π ( M ) ,(i) | cos σ | is constant on M τ ;(ii) cos σ = 0 on M τ ∩ { x = 0 } ;(iii) h Ox, n ( x ) i ≤ for x ∈ M τ ∩ H (iv) cos σ ≤ on M τ ∩ H ;and if cos σ on M τ then(v) τ ∈ p ( E ) ;(vi) M τ consists of two disjoint singletons in S τ symmetric about the x -axis;(vii) L ( τ ) ∈ (0 , πτ ) ;(viii) M τ = { ( τ cos( L ( τ ) / τ ) , ± τ sin( L ( τ ) / τ ) } . roof. (i) By Lemma 5.2, M τ is the union of two closed spherical arcs in S τ symmetric aboutthe x -axis. In case M τ ∩ H consists of a singleton the assertion follows from Lemma 5.1. Nowsuppose that M τ ∩ H consists of a spherical arc in S τ with non-empty interior. It can be seen thatcos σ vanishes on the interior of this arc as 0 = r ′ = cos σ in a local parametrisation by (2.9). Bycontinuity cos σ = 0 on M τ . (ii) follows from Lemma 5.1. (iii) Let x ∈ M τ ∩ H so θ ( x ) ∈ (0 , π ).Then S ( θ ( x )) ∩ S τ ⊂ E as E is spherical cap symmetric. Then0 ≤ lim S ( θ ( x )) ∩ S τ ∋ y → x D y − x | y − x | , n ( x ) E = −h Ox, n ( x ) i by Lemma 5.3. (iv) The adjoint transformation O ⋆ represents rotation clockwise through π/ x ∈ M τ ∩ H . By (iii) ,0 ≥ h Ox, n ( x ) i = h x, O ⋆ n ( x ) i = h x, t ( x ) i = τ cos σ ( x )and this leads to the result. (v) As cos σ M τ we can find x ∈ M τ ∩ H . We claimthat S τ ∩ S ( θ ( x )) ⊂ E . For suppose that y ∈ S τ ∩ S ( θ ( x )) but y E . We may suppose that0 ≤ θ ( y ) < θ ( x ) < π . If y ∈ R \ E then x ∈ R \ E by Lemma 5.2. On the other hand, if y ∈ M then the spherical arc in H joining y to x is contained in M again by Lemma 5.2. This arc alsohas non-empty interior in S τ . Now cos σ = 0 on its interior so cos( σ ( x )) = 0 by (i) contradictingthe hypothesis. A similar argument deals with (vi) and this together with (v) in turn entails (vii) and (viii) . Lemma 5.5.
Let E be an open set in U with C boundary M in U and assume that E \{ } = E sc .Suppose that ∈ M . Then(i) (sin σ )(0+) = 0 ;(ii) (cos σ )(0+) = − .Proof. (i) Let γ be a C parametrisation of M in a neighbourhood of 0 with γ (0) = 0 asabove. Then n (0) = n (0) = e and hence t (0) = t (0) = − e . By Taylor’s Theorem γ ( s ) = γ (0) + t (0) s + o ( s ) = − e s + o ( s ) for s ∈ I . This means that r ( s ) = | γ ( s ) | = s + o ( s ) andcos θ = h e , γ i r = h e , γ i s sr → s → θ )(0 − ) = 0. Now t is continuous on I so t = − e + o (1)and cos α = h e , t i = o (1). We infer that (cos α )(0 − ) = 0. By (2.11), cos α = cos σ cos θ − sin σ sin θ on I and hence (sin σ )(0 − ) = 0. We deduce that (sin σ )(0+) = 0. Item (ii) followsfrom (i) and Lemma 5.4.The setΩ := π h ( M \ { } ) ∩ { cos σ = 0 } i (5.2)plays an important rˆole in the sequel. Lemma 5.6.
Let E be an open set in U with C boundary M in U and assume that E \{ } = E sc .Then Ω is an open set in (0 , .Proof. Suppose 0 < τ ∈ Ω. Choose x ∈ M τ ∩ { cos σ = 0 } . Let γ : I → M be a local C parametrisation of M in a neighbourhood of x such that γ (0) = x as before. By shrinking I ifnecessary we may assume that r = 0 and cos σ = 0 on I . Then the set { r ( s ) : s ∈ I } ⊂ Ωis connected and so an interval in R (see for example [32] Theorems 6.A and 6.B). By (2.9),˙ r (0) = cos σ (0) = cos σ ( p ) = 0. This means that the set { r ( s ) : s ∈ I } contains an open intervalabout τ . 17 Generalised (mean) curvature
Given a relatively compact set E of finite perimeter in U the first variation δV f ( Z ) resp. δP + g ( Z )of weighted volume and perimeter along a time-dependent vector field Z are defined as in (2.13)and (2.14). Proposition 6.1.
Let g be as in (1.3). Let E be a relatively compact open set in U with C boundary M = ∂E ∩ U in U . Let Z be a time-dependent vector field satisfying (Z.1) and (Z.2).Then δP + g ( Z ) = Z M g ′ + ( · , Z ) + g div M Z d H where Z := Z (0 , · ) ∈ C c ( U, R ) .Proof. The identity (3.3) holds for each t ∈ I with M in place of F E . The assertion follows onappealing to Lemma 2.3 and Lemma 2.4 with the help of the dominated convergence theorem.Given X, Y ∈ C ∞ c ( U, R ) let ψ resp. χ stand for the 1-parameter group of C ∞ diffeomorphisms of U associated to the vector fields X resp. Y as in (2.12). Let I be an open interval in R containingthe point 0. Suppose that the function σ : I → R is C . Define a flow via ϕ : I × U → U ; ( t, x ) χ ( σ ( t ) , ψ ( t, x )) . Lemma 6.2.
The time-dependent vector field Z associated with the flow ϕ is given by Z ( t, x ) = σ ′ ( t ) Y ( χ ( σ ( t ) , ψ ( t, x ))) + dχ ( σ ( t ) , ψ ( t, x )) X ( ψ ( t, x )) (6.1) for ( t, x ) ∈ I × U and satisfies (Z.1) and (Z.2).Proof. For t ∈ I and x ∈ U we compute using (2.12), ∂ t ϕ ( t, x ) = ( ∂ t χ )( σ ( t ) , ψ ( t, x )) σ ′ ( t ) + dχ ( σ ( t ) , ψ ( t, x )) ∂ t ψ ( t, x )and this gives (6.1). Put K := supp[ X ], K := supp[ Y ] and K := K ∪ K . Then ( Z.
2) holdswith this choice of K .Let E be a relatively compact open set in U with C boundary M in U . Define Λ := ( M \ { } ) ∩{ cos σ = 0 } andΛ := { x ∈ M : H (Λ ∩ B ( x, ρ )) = H ( M ∩ B ( x, ρ )) for some ρ > } . (6.2)For future reference put Λ ± := Λ ∩ { x ∈ M : ±h x, n i > } . Lemma 6.3.
Let E be a relatively compact open set in U with C , boundary M in U and supposethat E \ { } = E sc . Then(i) Λ is a countable disjoint union of well-separated open circular arcs centred at ;(ii) H (Λ \ Λ ) = 0 ;(iii) g is differentiable H -a.e. on M \ Λ with g as in (1.3). The term well-separated in (i) means the following: if Γ is an open circular arc in Λ withΓ ∩ (Λ \ Γ) = ∅ then d (Γ , Λ \ Γ) > Proof. (i)
Let x ∈ Λ and γ : I → M a C , parametrisation of M near x . By shrinking I ifnecessary we may assume that γ ( I ) ⊂ M ∩ B ( x, ρ ) with ρ as in (6.2). So cos σ = 0 H -a.e. on γ ( I ) and hence cos σ = 0 a.e. on I . This means that cos σ = 0 on I as σ ∈ C , ( I ) and that18 is constant on I by (2.9). Using (2.10) it can be seen that γ ( I ) is an open circular arc centredat 0. By compactness of M it follows that Λ is a countable disjoint union of open circular arcscentred on 0. The well-separated property flows from the fact that M is C . (ii) follows as aconsequence of this property. (iii) Let x ∈ M \ Λ and γ : I → M a C , parametrisation of M near x with properties as before. We assume that x lies in the upper half-plane H . By shrinking I if necessary we may assume that γ ( I ) ⊂ ( M \ Λ ) ∩ H . Let s , s , s ∈ I with s < s < s . Then y := γ ( s ) ∈ M \ Λ . So H ( M ∩ { cos σ = 0 } ∩ B ( y, ρ )) > ρ >
0. This means thatfor small η > γ (( s − η, s + η )) ∩ { cos σ = 0 } has positive H -measure. Consequently, r ( s ) − r ( s ) = R s s cos σ ds < r is strictlydecreasing on I . So h is differentiable a.e. on r ( I ) ⊂ (0 , + ∞ ) in virtue of the fact that h is φ -convex and hence locally Lipschitz. This entails (iii) . Proposition 6.4.
Let f and g be as in (1.2) and (1.3). Given v > let E be a minimiserof (3.1). Assume that E is a relatively compact open set in U with C boundary M in U andsuppose that E \ { } = E sc . Suppose that M \ Λ = ∅ . Then there exists λ ∈ R such that for any X ∈ C c ( U, R ) , ≤ Z M n g ′ + ( · , X ) + g div M X − λf h n, X i o d H . Proof.
Let X ∈ C ∞ c ( U, R ). Let x ∈ M and r > M ∩ B ( x, r ) ⊂ M \ Λ . Choose Y ∈ C ∞ c ( U, R ) with supp[ Y ] ⊂ B ( x, r ) as in Lemma 3.4. Let ψ resp. χ stand for the 1-parametergroup of C ∞ diffeomorphisms of U associated to the vector fields X resp. Y as in (2.12). For each( s, t ) ∈ R the set χ s ( ψ t ( E )) is an open set in U with C boundary and ∂ ( χ s ◦ ψ t )( E ) = ( χ s ◦ ψ t )( M )by Lemma 2.1. Define V ( s, t ) := V f ( χ t ( ψ s ( E ))) − V f ( E ) ,P ( s, t ) := P g ( χ t ( ψ s ( E ))) , for ( s, t ) ∈ R . We write F = ( χ t ◦ ψ s )( E ). Arguing as in Proposition 3.2, ∂ t V ( s, t ) = lim h → (1 /h ) { V f ( χ h ( F )) − V f ( F ) } = Z F div( f Y ) dx = Z E (div( f Y ) ◦ χ t ◦ ψ s ) J d ( χ t ◦ ψ s ) x dx with an application of the area formula (cf. [1] Theorem 2.71). This last varies continuously in( s, t ). The same holds for partial differentiation with respect to s . Indeed, put η := χ t ◦ ψ s . Thennoting that J d ( η ◦ ψ h ) = ( J dη ) ◦ ψ h J dψ h and using the dominated convergence theorem, ∂ s V ( s, t ) = lim h → (1 /h ) n V f ( η ( ψ h ( E ))) − V f ( η ( E )) o = lim h → (1 /h ) n Z E ( f ◦ η ◦ ψ h ) J d ( η ◦ ψ h ) x dx − Z E ( f ◦ η ) J dη x dx o = lim h → (1 /h ) n Z E [( f ◦ η ◦ ψ h ) − ( f ◦ η )] J d ( η ◦ ψ h ) x dx + Z E ( f ◦ η )[( J dη ◦ ψ h − J dη ] J dψ h dx + Z E ( f ◦ η ) J dη [ J dψ h − dx o = Z E h∇ ( f ◦ η ) , X i J dη x dx + Z E ( f ◦ η ) h∇ J dη, X i dx + Z E ( f ◦ η ) J dη div X dx where the explanation for the last term can be found in the proof of Proposition 3.2. In this regardwe note that d ( dχ t ) (for example) is continuous on I × R (cf. [1] Theorem 3.3 and Exercise 3.2)and in particular ∇ J dχ t is continuous on I × U . The expression above also varies continuously in19 s, t ) as can be seen with the help of the dominated convergence theorem. This means that V ( · , · )is continuously differentiable on R . Note that ∂ t V (0 ,
0) = Z E div( f Y ) dx = 1by choice of Y . By the implicit function theorem there exists η > C function σ : ( − η, η ) → R such that σ (0) = 0 and V ( s, σ ( s )) = 0 for s ∈ ( − η, η ); moreover, σ ′ (0) = − ∂ s V (0 ,
0) = − Z E n h∇ f, X i + f div X o dx = − Z E div( f X ) dx = Z M f h n, X i d H by the Gauss-Green formula (cf. [1] Theorem 3.36).The mapping ϕ : ( − η, η ) × U → U ; t χ ( σ ( t ) , ψ ( t, x ))satisfies conditions (F.1)-(F.4) above with I = ( − η, η ) where the associated time-dependent vectorfield Z is given as in (6.1) and satisfies (Z.1) and (Z.2); moreover, Z = Z (0 , · ) = σ ′ (0) Y + X .Note that Z = X on M \ B ( x, r ).The mapping I → R ; t P g ( ϕ t ( E )) is right-differentiable at t = 0 as can be seen from Proposition6.1 and has non-negative right-derivative there. By Proposition 6.1 and Lemma 6.3,0 ≤ δP + g ( Z ) = Z M g ′ + ( · , Z ) + g div M Z d H = Z M \ Λ g ′ + ( · , Z ) + g div M Z d H + Z Λ g ′ + ( · , X ) + g div M X d H = Z M \ Λ σ ′ (0) h∇ g, Y i + h∇ g, X i + σ ′ (0) g div M Y + g div M X d H + Z Λ g ′ + ( · , X ) + g div M X d H = Z M g ′ + ( · , X ) + g div M X d H + σ ′ (0) Z M g ′ + ( · , Y ) + g div M Y d H . (6.3)The identity then follows upon inserting the expression for σ ′ (0) above with λ = − R M g ′ + ( · , Y ) + g div M Y d H . The claim follows for X ∈ C c ( R , R ) by a density argument. Theorem 6.5.
Let f and g be as in (1.2) and (1.3). Given v > let E be a minimiser of (1.5).Assume that E is a relatively compact open set in U with C , boundary M in U and suppose that E \ { } = E sc . Suppose that M \ Λ = ∅ . Then there exists λ ∈ R such that(i) k + e ̺ sin σ + λζ = 0 H -a.e. on M \ Λ ;(ii) e ̺ − − λζ ≤ k ≤ e ̺ + − λζ on Λ +1 ;(iii) − e ̺ + − λζ ≤ k ≤ − e ̺ − − λζ on Λ − . The expression ( k + e ̺ sin σ ) /ζ is called the generalised (mean) curvature. Proof. (i)
Let x ∈ M and r > M ∩ B ( x, r ) ⊂ M \ Λ . Choose X ∈ C c ( U, R ) withsupp[ X ] ⊂ B ( x, r ). We know from Lemma 6.3 that g is differentiable H -a.e. on supp[ X ]. Let λ be as in Proposition 6.4. Replacing X by − X we deduce from Proposition 6.4 that0 = Z M n h∇ g, X i + g div M X − λf h n, X i o d H . C , manifolds. So Z M h∇ g, X i + g div M X d H = Z M ∂ n g h n, X i + h∇ M g, X i + g div M X d H = Z M ∂ n g h n, X i + div M ( gX ) d H = Z M ∂ n g h n, X i − H h n, X i d H = Z M gu { ∂ n log g − H } d H where u = h n, X i . Combining this with the equality above we see that Z M ug { ∂ n log g − H − λζ } d H = 0for all X ∈ C c ( U, R ) with support in B ( x, r ). This leads to the result. (ii) Let x ∈ M and r > M ∩ B ( x, r ) ⊂ Λ +1 . Let φ ∈ C ( S τ ) with support in S τ ∩ B ( x, r )where τ = | x | . We can construct X ∈ C c ( U, R ) with the property that X = φn on M ∩ B ( x, r ).By Lemma 2.4, g ′ + ( · , X ) = g n ζ | x | φ + h ′ + ( | x | , sgn φ ) | φ | o on Λ +1 . Let us assume that φ ≥
0. As h· , n i > +1 we have that g ′ + ( · , X ) = gφ n ζ | x | + h ′ + ( | x | , +1) o = gφ e ̺ + so by Proposition 6.4,0 ≤ Z M n g ′ + ( · , X ) + g div M X − λf h n, X i o d H = Z M gφ ne ̺ + − k − λζ o d H . We conclude that e ̺ + − k − λζ ≥ M ∩ B ( x, r ). Now assume that φ ≤
0. Then g ′ + ( · , X ) = gφ n ζ | x | − h ′ + ( | x | , − o = gφ e ̺ − so 0 ≤ Z M gφ ne ̺ − − k − λζ o d H and hence e ̺ − − k − λζ ≤ M ∩ B ( x, r ). This shows (ii) . (iii) The argument is similar. Then g ′ + ( · , X ) = g {− ζ | x | φ + h ′ + ( | x | , − sgn φ ) | φ |} on Λ − . Assume in the first instance that φ ≥
0. Then g ′ + ( · , X ) = − gφ e ̺ − so0 ≤ Z M gφ n − e ̺ − − k − λζ o d H . We conclude that − e ̺ − − k − λζ ≥ M ∩ B ( x, r ). Next suppose that φ ≤
0. Then g ′ + ( · , X ) = − gφ e ̺ + so0 ≤ Z M gφ n − e ̺ + − k − λζ o d H and − e ̺ + − k − λζ ≤ M ∩ B ( x, r ). 21et E be an open set in R with C boundary M and assume that E \ { } = E sc and that Ω isas in (5.2). Bearing in mind Lemma 5.4 we may define θ : Ω → (0 , π ); τ L ( τ ) / τ ; (6.4) γ : Ω → M ; τ ( τ cos θ ( τ ) , τ sin θ ( τ )) . (6.5)The function u : Ω → [ − , τ sin( σ ( γ ( τ ))) . (6.6)plays a key role. Theorem 6.6.
Let f and g be as in (1.2) and (1.3). Given v > let E be a minimiser of (1.5).Assume that E is a relatively compact open set in U with C , boundary M in U and suppose that E \ { } = E sc . Suppose that M \ Λ = ∅ and let λ be as in Theorem 6.5. Then u ∈ C , (Ω) and u ′ + (1 /τ + e ̺ ) u + λζ = 0 a.e. on Ω .Proof. Let τ ∈ Ω and x a point in the open upper half-plane such that x ∈ M τ . There exists a C , parametrisation γ : I → M of M in a neighbourhood of x with γ (0) = x as above. Put u := sin σ on I . By shrinking the open interval I if necessary we may assume that r : I → r ( I )is a diffeomorphism and that r ( I ) ⊂⊂ Ω. Note that γ = γ ◦ r − and u = u ◦ r − on r ( I ). Itfollows that u ∈ C , (Ω). By (2.9), u ′ = ˙ u ˙ r ◦ r − = ˙ σ ◦ r − a.e. on r ( I ). As ˙ α = k a.e. on I and using the identity (2.10) we see that ˙ σ = ˙ α − ˙ θ = k − (1 /r ) sin σ a.e on I . Thus, u ′ = k − (1 /τ ) sin( σ ◦ γ ) = k − (1 /τ ) u a.e. on r ( I ). By Theorem 6.5, k + e ̺ sin σ + λζ = 0 H -a.e. on M \ Λ . So u ′ = − e ̺ ( τ ) u − λζ − (1 /τ ) u = − (1 /τ + e ̺ ( τ )) u − λζ a.e. on r ( I ). The result follows. Lemma 6.7.
Suppose that E is a bounded open set in U with C boundary M in U and that E \ { } = E sc . Then(i) θ ∈ C (Ω) ;(ii) θ ′ = − τ u √ − u on Ω .Proof. Let τ ∈ Ω and x a point in the open upper half-plane such that x ∈ M τ . There exists a C parametrisation γ : I → M of M in a neighbourhood of x with γ (0) = x as above. By shrinkingthe open interval I if necessary we may assume that r : I → r ( I ) is a diffeomorphism and that r ( I ) ⊂⊂ Ω. It then holds that θ = θ ◦ r − and σ ◦ γ = σ ◦ r − on r ( I ) by choosing an appropriate branch of θ . It follows that θ ∈ C (Ω). By the chain-rule,(2.10) and (2.9), θ ′ = ˙ θ ˙ r ◦ r − = ( 1 r tan σ ) ◦ r − = (1 /τ ) tan( σ ◦ γ )on r ( I ). By Lemma 5.4, cos( σ ◦ γ ) = −√ − u on Ω. This entails (ii) .22 Convexity
Lemma 7.1.
Let f and g be as in (1.2) and (1.3). Given v > let E be a minimiser of (1.5).Assume that E is a relatively compact open set in U with C , boundary M in U and suppose that E \ { } = E sc . Let x ∈ ( M \ Λ ) ∪ Λ . Suppose that γ : I → M is a local parametrisation of M near x and assume that γ ( I ) ⊂⊂ ( M \ Λ ) ∪ Λ . Then k ∈ BV( I ) .Proof. For x ∈ Λ the assertion follows from Lemma 6.3. Suppose x ∈ M \ Λ and that x lies inthe upper half-plane. By Theorem 6.5, k + e ̺ ( r ) sin σ + λζ ( r ) = 0a.e. on I . As remarked in Section 2, σ ∈ C , ( I ) so the function I R ; s sin σ ( s ) is aLipschitz function on I . As in the proof of Lemma 6.3 the function s r ( s ) is strictly decreasingon I so that the function s e ̺ ( r ( s )) is non-increasing. The product e ̺ ( r ) sin σ belongs toBV( I ) by [1] Proposition 3.2. The result follows.Let E be a relatively compact open set in U with C , boundary M in U and assume that E \ { } = E sc . Put d := sup {| x | : x ∈ M } ∈ (0 ,
1) and b := ( d, γ : I → M be a C , parametrisation of M near b with γ (0) = b . In light of Lemma 7.1 we may assume that k is right-continuous on I (see [1] Theorem 3.28 for example). Define k ( d − ) := k (0) and notethat this quantity does not depend on the particular parametrisation chosen subject to the aboveconventions. Lemma 7.2.
Let E be a relatively compact open set in U with C , boundary M in U and assumethat E \ { } = E sc . Put d := sup {| x | : x ∈ M } ∈ (0 , and b := ( d, . Then k ( d − ) ≥ /d. Proof.
Let γ : I → M be a C , parametrisation of M near b with γ (0) = b and assume that k is right-continuous on I . For s ∈ I , γ ( s ) = de + se + Z s n ˙ γ ( u ) − ˙ γ (0) o du and ˙ γ ( u ) − ˙ γ (0) = Z u k n dv by (2.6). By the Fubini-Tonelli Theorem, γ ( s ) = de + se + Z s ( s − u ) k ( u ) n ( u ) du = de + se + R ( s )for s ∈ I . Assume for a contradiction that k ( d − ) < l < /d for some l ∈ R . By right-continuitywe can find δ > k < l a.e. on [0 , δ ]. So h R ( s ) , e i = Z s ( s − u ) k ( u ) h n ( u ) , e i du > − (1 / s l as s ↓ r ( s ) − d = 2 d h R ( s ) , e i + s + o ( s ) > − dls + s + o ( s )as s ↓
0. Alternatively, r ( s ) − d s > − dl + o (1) . As 1 − dl > s ∈ I with r ( s ) > d , contradicting the definition of d .23 emma 7.3. Let f and g be as in (1.2) and (1.3). Given v > let E be a minimiser of (1.5).Assume that E is a relatively compact open set in U with C , boundary M in U and suppose that E \ { } = E sc . Suppose that M \ Λ = ∅ . Then λζ ( d ) ≤ − /d − e ̺ − ( d ) < with λ as in Theorem6.5.Proof. We write M as the disjoint union M = ( M \ Λ ) ∪ Λ . With b as above, suppose that b ∈ Λ . Then b ∈ Λ ; in fact, b ∈ Λ − . By Theorem 6.5, λζ ≤ − e ̺ − − k at b . By Lemma 7.2, λζ ≤ − /d − e ̺ − ( d ). Now suppose that b lies in the open set M \ Λ in M . Let γ : I → M be a C , parametrisation of M near b with γ ( I ) ⊂ M \ Λ . By Theorem 6.5, k + e ̺ ( r ) sin σ + λζ ( r ) = 0a.e. on I . Now sin σ ( s ) → s →
0. On taking the limit s ↓ k ( d − ) + e ̺ − ( d ) + λζ ( d ) = 0 and1 /d + e ̺ ( d − ) + λζ ( d ) ≤ λζ ( d ) ≤ − /d − e ̺ − ( d ). Theorem 7.4.
Let f and g be as in (1.2) and (1.3). Given v > let E be a minimiser of (1.5).Assume that E is a relatively compact open set in U with C , boundary M in U and suppose that E \ { } = E sc . Suppose that M \ Λ = ∅ . Then E is convex.Proof. The proof runs along similar lines as [29] Theorem 6.5. By Theorem 6.5, k + e ̺ sin σ + λζ = 0 H -a.e. on M \ Λ . By Lemma 7.3,1 dζ ( d ) + e ̺ − ( d ) ζ ( d ) ≤ − λ = k + e ̺ sin σζ = kζ + e ̺ζ sin σ ≤ kζ + e ̺ − ( d ) ζ ( d ) H -a.e. on M \ Λ using the fact that h is φ -convex. This in turn implies that k is boundedbelow by ζ (0) /dζ ( d ) > H -a.e. on M \ Λ .On Λ +1 , k/ζ ≥ e ̺ − /ζ − λ > λ <
0; on the other hand, k < +1 . So in fact Λ +1 = ∅ . If b ∈ Λ − then k = 1 /d . On Λ − ∩ B (0 , d ), kζ ≥ − e ̺ + ζ − λ ≥ − ̺ + ζ + 1 dζ ( d ) + ̺ − ( d ) ζ ( d ) ≥ dζ ( d ) > . Therefore k > H -a.e. on M . The set E is then convex by a modification of [33] Theorem1.8 and Proposition 1.4. It is sufficient that the function f (here α ) in the proof of the formertheorem is non-decreasing. Let 0 ≤ a < b < ̺ ≥ a, b ] such that ̺/ζ is non-decreasing on[ a, b ]. Let h be a primitive of ̺ on [ a, b ] so that h ∈ C , ([ a, b ]) and introduce the functions ψ : [ a, b ] → R ; x e h ( x ) ; (8.1) f : [ a, b ] → R ; x x ( ζ ψ )( x ); (8.2) g : [ a, b ] → R ; x x ( ζψ )( x ) . (8.3)Then g ′ = (1 /x + e ̺ ) g = (1 /x + ˆ ̺ ) g + ( ̺/ζ ) f (8.4)a.e. on ( a, b ). Define m = m ( ̺/ζ, a, b ) := g ( b ) − g ( a ) R ba f dt . (8.5)If ̺/ζ takes the constant value R ∋ λ ≥ a, b ] we use the notation m ( λ, a, b ) and we write m = m ( a, b ) := m (0 , a, b ). A computation gives m = ˆ ̺ ( b ) − ˆ ̺ ( a ) ζ ( b ) − ζ ( a ) = R ba (1 + x ˆ ̺ ) ζ dx R ba xζ dx (8.6)noting that ˆ ̺ = ζ ′ /ζ = ζ . 24 emma 8.1. Let ≤ a < b < and ̺ ≥ be a non-decreasing bounded function on [ a, b ] . Then m ≤ m .Proof. Note that g is convex on [ a, b ] as can be seen from (8.4). By the Hermite-Hadamardinequality (cf. [20], [18]),1 b − a Z ba g dt ≤ g ( a ) + g ( b )2 . (8.7)The inequality ( b − a )( g ( a ) + g ( b )) ≤ ( a + b )( g ( b ) − g ( a )) entails Z ba g dt ≤ a + b g ( b ) − g ( a ))and the result follows on rearrangement. Lemma 8.2.
Let I = [ a, b ] with −∞ < a < b < + ∞ . Let p, q be non-decreasing positive continuousprobability densities on I with the property that p/q is non-increasing on I . Let θ be a positivenon-decreasing function on I . Then(i) R I pθ dx ≤ R I qθ dx ;(ii) if p/q is strictly decreasing on [ a, b ] then equality holds if and only if θ is a positive constant.Proof. (i) As p, q are probability densities on I there exists c ∈ ( a, b ) with p ( c ) = q ( c ) by theintermediate value theorem. Put I := [ a, c ] and I := [ c, b ]. Then Z I p − q dx = Z I q − p dx. As p/q is non-increasing on I and ( p/q )( c ) = 1, p ≥ q on I and p ≤ q on I . Note that Z I pθ dx − Z I qθ dx = Z I ( p − q ) θ dx − Z I ( q − p ) θ dx. Now, Z I ( p − q ) θ dx ≤ θ ( c ) Z I ( p − q ) dx = θ ( c ) Z I ( q − p ) dx ≤ Z I ( q − p ) θ dx and this leads to the result. (ii) Suppose that |{ θ = θ ( c ) }| >
0. Then either | I ∩ { θ = θ ( c ) }| > | I ∩ { θ = θ ( c ) }| >
0. Assume the former. Then | I ∩ { θ < θ ( c ) }| > Lemma 8.3.
The function (0 , → R ; x x ˆ ̺xζ is strictly decreasing.Proof. Note that 1 + x ˆ ̺ = 1 + x ζ = ζ − x ∈ (0 ,
1) because ζ ′ = xζ , ˆ ̺ = xζ and ζ − x ζ = 2.So 1 + x ˆ ̺xζ = ζ − xζ = 2 − (1 − x )2 x = 12 (cid:16) x + x (cid:17) =: q ( x )for x ∈ (0 ,
1) and q is strictly decreasing on (0 ,
1) by calculus.25 emma 8.4.
Let ≤ a < b < and λ ≥ . Then(i) m ( λ, a, b ) ≤ λ + m ( a, b ) ;(ii) equality holds if and only if λ = 0 .Proof. (i) We may assume that λ >
0. Note that ψ = e λφ on I by choosing an appropriateprimitive. Integrating (8.4) we obtain, g ( b ) − g ( a ) = Z ba (1 + x ˆ ̺ ) ζψ dx + λ Z ba f dx hence m ( λ, a, b ) − λ = R ba (1 + x ˆ ̺ ) ζψ dx R ba xζ ψ dx . To obtain the inequality in (i) we show that R ba (1 + x ˆ ̺ ) ζψ dx R ba (1 + x ˆ ̺ ) ζ dx ≤ R ba xζ ψ dx R ba xζ dx . Define positive continuous probability densities p := (1 + x ˆ ̺ ) ζ R ba (1 + x ˆ ̺ ) ζψ dx and q := xζ R ba xζ ψ dx on [ a, b ]. Then p and q are strictly increasing on [ a, b ] and pq = R ba xζ ψ dx R ba (1 + x ˆ ̺ ) ζψ dx x ˆ ̺xζ is strictly decreasing on [ a, b ] by Lemma 8.3. The inequality then follows from Lemma 8.2 with θ := ψ = e λφ . (ii) Suppose that equality holds in (i) . Then ψ is constant on [ a, b ] by Lemma 8.2 which entailsthat λ = 0. Theorem 8.5.
Let ≤ a < b < and ̺ ≥ be a bounded function on [ a, b ] such that ̺/ζ isnon-decreasing on [ a, b ] . Then(i) m ( ̺/ζ, a, b ) ≤ ( ̺/ζ )( b − ) + m ( a, b ) ;(ii) equality holds if and only if ̺ ≡ on [ a, b ) .Proof. (i) Define h := R · a ̺ dτ on [ a, b ] so that h ′ = ̺ a.e. on ( a, b ). Note that h = Z · a ( ̺/ζ ) ζ dτ ≤ λφ on [ a, b ] with λ := ( ̺/ζ )( b − ). Define h : [ a, b ] → R ; t h ( b ) − λ ( φ ( b ) − φ ). Then h ( b ) = h ( b ), h ′ = λζ so that h ′ = ̺ = ( ̺/ζ ) ζ ≤ λζ = h ′ a.e. on ( a, b ) and hence h ≥ h on [ a, b ]. We derive Z ba f dt = Z ba tζ ψ dt = Z ba tζ e h dt ≥ Z ba tζ e h dt = Z ba f dt and g ( b ) − g ( a ) = bζ ( b ) e h ( b ) − aζ ( a ) e h ( a ) = bζ ( b ) e h ( b ) − aζ ( a ) e h ( a ) ≤ bζ ( b ) e h ( b ) − aζ ( a ) e h ( a ) = g ( b ) − g ( a )26n obvious notation. This entails that m ( ̺/ζ, a, b ) ≤ m ( λ, a, b ) and the result follows with the helpof Lemma 8.4. (ii) Suppose that ̺ a, b ). If ̺/ζ is constant on [ a, b ) the assertion follows from Lemma 8.4.Assume then that ̺/ζ is not constant on [ a, b ). Then h h on [ a, b ] in the above notation and R ba tζ e h dt > R ba tζ e h dt which entails strict inequality in (i) .With the above notation defineˆ m = ˆ m ( ̺/ζ, a, b ) := g ( a ) + g ( b ) R ba f dt . (8.8)A computation givesˆ m := ˆ m (0 , a, b ) = 2 b − a . (8.9) Lemma 8.6.
Let ≤ a < b < + ∞ and ̺ ≥ be a non-decreasing bounded function on [ a, b ] .Then ˆ m ≥ ˆ m .Proof. This follows by the Hermite-Hadamard inequality (8.7).Define N := ( ζ − ζ ( a ))( g + g ( a )); N := ζ − ζ ( a ) ; D := 2 + a e ̺ ( a ) + t e ̺ ; D := 2 + a ˆ ̺ ( a ) + t ˆ ̺ ; e D := a̺ ( a ) + t̺ ; R := ζ ( a )( ζ − ζ ( a ))( ψ − ψ ( a ));with ζ := tζ on [ a, b ]. Note that N = ( ζ − ζ ( a ))( ψζ + ψ ( a ) ζ ( a ))= ( ζ − ζ ( a ))( ψ [ ζ + ζ ( a )] + ζ ( a )[ ψ ( a ) − ψ ])= ψ [ ζ − ζ ( a ) ] − ζ ( a )( ζ − ζ ( a ))( ψ − ψ ( a )) = ψN − R and D = ( D + e D ) = D + e D ( D + e D ) + D e D = D + e DD + D e D = D + ( a̺ ( a ) + t̺ ) D + e D (2 + a ˆ ̺ ( a ) + t ˆ ̺ )= D + t̺D + (2 + t ˆ ̺ ) e D + a̺ ( a ) D + a ˆ ̺ ( a ) e D (8.10) Lemma 8.7.
It holds that(i) R (1 + t ˆ ̺ ) − tR ′ ≤ a.e. on [ a, b ] ;(ii) RD ′ − R ′ (2 + t ˆ ̺ ) ≤ a.e. on [ a, b ] .Proof. (i) This follows from the fact that the function Rζ − ζ ( a ) = ζ ( a )( ψ − ψ ( a ))is non-decreasing as ̺ ≥ a, b ]. (ii) We compute RD ′ − R ′ (2 + t ˆ ̺ ) = ζ ( a )( ζ − ζ ( a ))( ψ − ψ ( a ))( t ˆ ̺ ) ′ − ζ ( a ) n ζ ′ ( ψ − ψ ( a )) + ( ζ − ζ ( a )) ψ ′ o (2 + t ˆ ̺ )= ζ ( a ) n − ζ ( a )( ψ − ψ ( a ))( t ˆ ̺ ) ′ + ( ψ − ψ ( a )) h ζ ( t ˆ ̺ ) ′ − ζ ′ (1 + t ˆ ̺ ) i − ζ ′ ( ψ − ψ ( a )) − ( ζ − ζ ( a )) ψ e ̺ (2 + t ˆ ̺ ) o a.e. on [ a, b ]. The expression ζ ( t ˆ ̺ ) ′ − ζ ′ (1 + t ˆ ̺ ) < emma 8.8. It holds that N ′ D − N D ′ = tζ D on (0 , .Proof. First note that t ζ = ζ − ζ = ζ − ζ and 1 + t ˆ ̺ = 1 + t ζ = ζ −
1. Hence ζ − ζ ( a ) = ζ − ζ ( a ) − ζ − ζ ( a )) = ( ζ − ζ ( a ))( ζ + ζ ( a ) − a ˆ ̺ ( a ) + t ˆ ̺ = ζ + ζ ( a ) − . It can then be seen that N D = ζ − ζ ( a ) a ˆ ̺ ( a ) + t ˆ ̺ = ζ − ζ ( a )on [0 ,
1) and the statement follows.
Lemma 8.9.
Let ≤ a < b < and ̺ ≥ be a function on [ a, b ] such that ̺/ζ is non-decreasingand bounded on [ a, b ] . Assume that ̺ ∈ C ([ a, b ]) . Then N ′ D − N D ′ = tζ ψ [ D + t̺D + (2 + t ˆ ̺ ) e D + a̺ ( a ) t ˆ ̺ ] + [ ζ ( a ) − ζ ] ψ̺ + ζ ( a ) ψ̺ [ t ˆ ̺ − D ]+ n R (1 + t ˆ ̺ ) − R ′ t o ̺ + RD ′ − R ′ [ D − t̺ ] + ( R − ψN ) t ( ̺ ′ − ˆ ̺̺ ) on [ a, b ] .Proof. Note that N ′ = 2 ζ ζ ′ = 2 tζ (1 + t ˆ ̺ ). We compute using Lemma 8.8, N ′ D − N D ′ = ( ψN ′ + ψ̺N − R ′ )( D + e D ) − ( ψN − R )( D ′ + e D ′ )= ψN ′ D + ψN ′ e D + ψ̺N D − R ′ D − ψN D ′ − ψN e D ′ + RD ′ = ψ [ N ′ D − N D ′ ] + ψN ′ e D + ψ̺N D − R ′ D − ψN e D ′ + RD ′ = tζ ψD + ψN ′ e D + ψ̺N D − R ′ D − ψN e D ′ + RD ′ = tζ ψD + tζ ψt̺D + ψN ′ e D + R e D ′ − R ′ D − ψN e D ′ + RD ′ − ζ ( a ) ψ̺D = tζ ψD + tζ ψt̺D + ( R − ψN ) e D ′ + ψN ′ e D − R ′ D + RD ′ − ζ ( a ) ψ̺D = tζ ψD + tζ ψt̺D + tζ ψ (2[1 + t ˆ ̺ ]) e D + ( R − ψN ) e D ′ − R ′ D + RD ′ − ζ ( a ) ψ̺D = tζ ψ [ D + t̺D + (2 + t ˆ ̺ ) e D ] + tζ ψt ˆ ̺ e D + ( R − ψN ) e D ′ − R ′ D + RD ′ − ζ ( a ) ψ̺D = tζ ψ [ D + t̺D + (2 + t ˆ ̺ ) e D ] + tζ ψt ˆ ̺ ( a̺ ( a ) + t̺ ) + ( R − ψN )[ t ( ̺ ′ − ˆ ̺̺ ) + (1 + t ˆ ̺ ) ̺ ] − R ′ D + RD ′ − ζ ( a ) ψ̺D = tζ ψ [ D + t̺D + (2 + t ˆ ̺ ) e D + a̺ ( a ) t ˆ ̺ ] + tζ ψt ˆ ̺t̺ + ( R − ψN ) t ( ̺ ′ − ˆ ̺̺ ) + R (1 + t ˆ ̺ ) ̺ − R ′ D + RD ′ − tζ ψ (1 + t ˆ ̺ ) t̺ − ζ ( a ) ψ̺D + ζ ( a ) ψ (1 + t ˆ ̺ ) ̺ = tζ ψ [ D + t̺D + (2 + t ˆ ̺ ) e D + a̺ ( a ) t ˆ ̺ ] − tζ ψt̺ + ζ ( a ) ψ (1 + t ˆ ̺ ) ̺ + R (1 + t ˆ ̺ ) ̺ − R ′ ( t̺ ) − R ′ [ D − t̺ ] − ζ ( a ) ψ̺D + RD ′ + ( R − ψN ) t ( ̺ ′ − ˆ ̺̺ )= tζ ψ [ D + t̺D + (2 + t ˆ ̺ ) e D + a̺ ( a ) t ˆ ̺ ] − [ ζ − ζ ( a ) ] ψ̺ + ζ ( a ) ψ̺ [ t ˆ ̺ − D ] + R (1 + t ˆ ̺ ) ̺ − R ′ ( t̺ ) + RD ′ − R ′ [ D − t̺ ] + ( R − ψN ) t ( ̺ ′ − ˆ ̺̺ )and some rearrangement produces the identity.We prove a reverse Hermite-Hadamard inequality. Theorem 8.10.
Let ≤ a < b < and ̺ ≥ be a function on [ a, b ] such that ̺/ζ is non-decreasingand bounded on [ a, b ] . Then i) ( ζ ( b ) − ζ ( a )) ˆ m ( ̺/ζ, a, b ) ≤ a e ̺ ( a +) + b e ̺ ( b − ) ;(ii) equality holds if and only if ̺ ≡ on [ a, b ) . This last inequality can be written in the form g ( a ) + g ( b )2 + a̺ ( a +) + b̺ ( b − ) ≤ b − a Z ba g dt ;comparing with (8.7) justifies naming this a reverse Hermite-Hadamard inequality. Proof. (i)
We assume in the first instance that ̺ ∈ C (( a, b )). We prove the above result in theform Z ba f dt ≥ ( ζ ( b ) − ζ ( a ))( g ( a ) + g ( b ))2 + a e ̺ ( a ) + b e ̺ ( b ) . (8.11)Put w := ( ζ − ζ ( a ))( g ( a ) + g )2 + a e ̺ ( a ) + t e ̺ = ND for t ∈ [ a, b ] so that Z ba w ′ dt = ( ζ ( b ) − ζ ( a ))( g ( a ) + g ( b ))2 + a e ̺ ( a ) + b e ̺ ( b ) . By Lemma 8.9, w ′ D = N ′ D − N D ′ = tζ ψ [ D + t̺D + (2 + t ˆ ̺ ) e D + a̺ ( a ) t ˆ ̺ ] + [ ζ ( a ) − ζ ] ψ̺ + ζ ( a ) ψ̺ [ t ˆ ̺ − D ]+ n R (1 + t ˆ ̺ ) − R ′ t o ̺ + RD ′ − R ′ [ D − t̺ ] + ( R − ψN ) t ( ̺ ′ − ˆ ̺̺ ) . From a comparison with (8.10) it can be seen that D + t̺D + (2 + t ˆ ̺ ) e D + a̺ ( a ) t ˆ ̺ ≤ D . By assumption ̺/ζ is non-decreasing on [ a, b ] which entails that ̺ ′ − ˆ ̺̺ ≥ a, b ]. Theseobservations together with Lemma 8.7 yield the inequality.Let us now assume that ̺ ≥ a, b ] such that ̺/ζ is non-decreasing and boundedon [ a, b ]. Extend ̺ to R via ̺ ( t ) := ̺ ( a +) for t ∈ ( −∞ , a ]; ̺ ( t ) for t ∈ ( a, b ]; ̺ ( b − ) for t ∈ ( b, + ∞ );for t ∈ R . Let ( ψ ε ) ε> be a family of mollifiers (see e.g. [1] 2.1) and set ̺ ε := ̺ ⋆ ψ ε on R foreach ε >
0. Then ̺ ε ∈ C ∞ ( R ) and is non-decreasing on R for each ε >
0. Put ̺ ε := ̺ ε | [ a,b ] for each ε >
0. Then ( ̺ ε ) ε> converges to ̺ in L (( a, b )) by [1] 2.1 for example. Note that h ε := R · a ̺ ε dt → h pointwise on [ a, b ] as ε ↓ h ε ) is uniformly bounded on [ a, b ].Moreover, ̺ ε ( a ) → ̺ ( a +) and ̺ ε ( b ) → ̺ ( b − ) as ε ↓
0. By the above result,( ζ ( b ) − ζ ( a )) ˆ m ( ̺ ε /ζ, a, b ) ≤ a e ̺ ε ( a ) + b e ̺ ε ( b )for each ε >
0. The inequality follows on taking the limit ε ↓ ii) We now consider the equality case. We claim that( ζ ( b ) − ζ ( a ))( g ( a ) + g ( b ))2 + a e ̺ ( a ) + b e ̺ ( b ) ≤ Z ba f dt −
12 + a e ̺ ( a ) + b e ̺ ( b ) Z ba ( ζ − ζ ( a ) ) ψ̺ dt ; (8.12)this entails the equality condition in (ii) . First suppose that ̺ ∈ C (( a, b )). From the equalityabove, w ′ ≤ tζ ψ − ( ζ − ζ ( a ) ) ψ̺D ≤ tζ ψ − ( ζ − ζ ( a ) ) ψ̺ a e ̺ ( a ) + b e ̺ ( b )as D ≤ a e ̺ ( a ) + b e ̺ ( b ) on [ a, b ]. This establishes the inequality in this case. Now supposethat ̺ ≥ a, b ] such that ̺/ζ is non-decreasing and bounded on [ a, b ]. Then(8.12) holds with ̺ ε in place of ̺ for each ε >
0. The inequality for ̺ follows by the dominatedconvergence theorem. Let L stand for the collection of Lebesgue measurable sets in [0 , + ∞ ). Define a measure µ on([0 , + ∞ ) , L ) by µ ( dx ) := (1 /x ) dx . Let 0 ≤ a < b < + ∞ . Suppose that u : [ a, b ] → R is an L -measurable function with the property that µ ( { u > t } ) < + ∞ for each t > . (9.1)The distribution function µ u : (0 , + ∞ ) → [0 , + ∞ ) of u with respect to µ is given by µ u ( t ) := µ ( { u > t } ) for t > . Note that µ u is right-continuous and non-increasing on (0 , ∞ ) and µ u ( t ) → t → ∞ .Let u be a Lipschitz function on [ a, b ]. Define Z := { u differentiable and u ′ = 0 } , Z := { u not differentiable } and Z := Z ∪ Z . By [1] Lemma 2.96, Z ∩ { u = t } = ∅ for L -a.e. t ∈ R and hence N := u ( Z ) ⊂ R is L -negligible.We make use of the coarea formula ([1] Theorem 2.93 and (2.74)), Z [ a,b ] φ | u ′ | dx = Z ∞−∞ Z { u = t } φ d H dt (9.2)for any L -measurable function φ : [ a, b ] → [0 , ∞ ]. Lemma 9.1.
Let ≤ a < b < + ∞ and u a Lipschitz function on [ a, b ] . Then(i) µ u ∈ BV loc ((0 , + ∞ )) ;(ii) Dµ u = − u ♯ µ ;(iii) Dµ au = Dµ u ((0 , + ∞ ) \ N ) ;(iv) Dµ su = Dµ u N ;(v) A := n t ∈ (0 , + ∞ ) : L ( Z ∩ { u = t } ) > o is the set of atoms of Dµ u and Dµ ju = Dµ u A ;(vi) µ u is differentiable L -a.e. on (0 , + ∞ ) with derivative given by µ ′ u ( t ) = − Z { u = t }\ Z | u ′ | d H τ for L -a.e. t ∈ (0 , + ∞ ) ; vii) Ran( u ) ∩ [0 , + ∞ ) = supp( Dµ u ) . The notation above Dµ au , Dµ su , Dµ ju stands for the absolutely continuous resp. singular resp.jump part of the measure Dµ u (see [1] 3.2 for example). Proof.
For any ϕ ∈ C ∞ c ((0 , + ∞ )) with supp[ ϕ ] ⊂ ( τ, + ∞ ) for some τ > Z ∞ µ u ϕ ′ dt = Z [ a,b ] ϕ ◦ u dµ = Z [ a,b ] χ { u>τ } ϕ ◦ u dµ (9.3)by Fubini’s theorem; so µ u ∈ BV loc ((0 , + ∞ )) and Dµ u is the push-forward of µ under u , Dµ u = − u ♯ µ (cf. [1] 1.70). By (9.2), Dµ u ((0 , + ∞ ) \ N )( A ) = − µ ( { u ∈ A } \ Z ) = − Z A Z { u = t }\ Z | u ′ | d H τ dt for any L -measurable set A in (0 , + ∞ ). In light of the above, we may identify Dµ au = Dµ u ((0 , + ∞ ) \ N ) and Dµ su = Dµ u N . The set of atoms of Dµ u is defined by A := { t ∈ (0 , + ∞ ) : Dµ u ( { t } ) = 0 } . For t > Dµ u ( { t } ) = Dµ su ( { t } ) = ( Dµ u N )(( { t } ) = − u ♯ µ ( N ∩ { t } ) = − µ ( Z ∩ { u = t } )and this entails (v) . The monotone function µ u is a good representative within its equivalenceclass and is differentiable L -a.e. on (0 , + ∞ ) with derivative given by the density of Dµ u withrespect to L by [1] Theorem 3.28. Item (vi) follows from (9.2) and (iii) . Item (vii) follows from (ii) .Let 0 < a < b < ̺ ≥ a, b ] such that ̺/ζ is non-decreasing and boundedon [ a, b ]. Let η ∈ {± } . We study solutions to the first-order linear ordinary differential equation u ′ + (1 /x + e ̺ ) u + λζ = 0 a.e. on ( a, b ) with u ( a ) = η and u ( b ) = η (9.4)where u ∈ C , ([ a, b ]) and λ ∈ R . In case ̺ ≡ a, b ] we use the notation u . Lemma 9.2.
Let < a < b < and ̺ ≥ be a function on [ a, b ] such that ̺/ζ is non-decreasingand bounded on [ a, b ] . Let η ∈ {± } . Then(i) there exists a solution ( u, λ ) of (9.4) with u ∈ C , ([ a, b ]) and λ = λ η ∈ R ;(ii) the pair ( u, λ ) in (i) is unique;(iii) λ η is given by − λ (1 , = λ ( − , − = m ; λ (1 , − = − λ ( − , = ˆ m ; (iv) if η = (1 , or η = ( − , − then u is uniformly bounded away from zero on [ a, b ] .Proof. (i) For η = (1 ,
1) define u : [ a, b ] → R by u := m R · a f ds + g ( a ) g (9.5)with m as in (8.5). Then u ∈ C , ([ a, b ]) and satisfies (9.4) with λ = − m . For η = (1 , −
1) set u = ( − ˆ m R · a f ds + g ( a )) / g with λ = ˆ m . The cases η = ( − , −
1) and η = ( − ,
1) can be dealt withusing linearity. (ii)
We consider the case η = (1 , u , λ ) resp. ( u , λ ) solve(9.4). By linearity u := u − u solves u ′ + (1 /x + e ̺ ) u + λζ = 0 a.e. on ( a, b ) with u ( a ) = u ( b ) = 0where λ = λ − λ . An integration gives that u = ( − λ R · a f ds + c ) / g for some constant c ∈ R andthe boundary conditions entail that λ = c = 0. The other cases are similar. (iii) follows as in (i) . (iv) If η = (1 ,
1) then u > a, b ] from (9.5) as m > The boundary condition η η = − . emma 9.3. Let < a < b < and ̺ ≥ be a function on [ a, b ] such that ̺/ζ is non-decreasingand bounded on [ a, b ] . Let ( u, λ ) solve (9.4) with η = (1 , − . Then(i) there exists a unique c ∈ ( a, b ) with u ( c ) = 0 ;(ii) u ′ < a.e. on [ a, c ] and u is strictly decreasing on [ a, c ] ;(iii) Dµ su = 0 .Proof. (i) We first observe that u ′ ≤ − ˆ mζ < { u ≥ } in view of (9.4). Suppose u ( c ) = u ( c ) = 0 for some c , c ∈ ( a, b ) with c < c . We may assume that u ≥ c , c ].This contradicts the above observation. Item (ii) is plain. For any L -measurable set B in(0 , + ∞ ), Dµ su ( B ) = µ ( { u ∈ B } ∩ Z ) = 0 using Lemma 9.1 and (ii) . Lemma 9.4.
Let < a < b < and ̺ ≥ be a function on [ a, b ] such that ̺/ζ is non-decreasingand bounded on [ a, b ] . Let ( u, λ ) solve (9.4) with η = (1 , − . Assume that(a) u is differentiable at both a and b and that (9.4) holds there;(b) u ′ ( a ) < and u ′ ( b ) < ;(c) ̺ is differentiable at a and b .Put v := − u . Then(i) R { v =1 }\ Z v | v ′ | d H τ ≥ R { u =1 }\ Z u | u ′ | d H τ ;(ii) equality holds if and only if ̺ ≡ on [ a, b ) .Proof. First, { u = 1 } = { a } by Lemma 9.3. Further 0 < − au ′ ( a ) = 1 + a [ ζ ( a ) ˆ m + e ̺ ( a )] from(9.4). On the other hand { v = 1 } ⊃ { b } and 0 < bv ′ ( b ) = − b [ ζ ( b ) ˆ m − e ̺ ( b )]. Thus Z { v =1 }\ Z v | v ′ | d H τ − Z { u =1 }\ Z u | u ′ | d H τ ≥ − b [ ζ ( b ) ˆ m − e ̺ ( b )] −
11 + a [ ζ ( a ) ˆ m + e ̺ ( a )] . By Theorem 8.10, 0 ≤ [ aζ ( a ) − bζ ( b )] ˆ m + 2 + a e ̺ ( a ) + b e ̺ ( b ), noting that ̺ ( a ) = ̺ ( a +) in virtueof (c) and similarly at b . A rearrangement leads to the inequality. The equality assertion followsfrom Theorem 8.10. Theorem 9.5.
Let < a < b < and ̺ ≥ be a function on [ a, b ] such that ̺/ζ is non-decreasingand bounded on [ a, b ] . Let ( u, λ ) solve (9.4) with η = (1 , − and set v := − u . Assume that u > − on [ a, b ) . Then(i) − µ ′ v ≥ − µ ′ u for L -a.e. t ∈ (0 , ;(ii) if ̺ on [ a, b ) then there exists t ∈ (0 , such that − µ ′ v > − µ ′ u for L -a.e. t ∈ ( t , ;(iii) for t ∈ [ − , , µ u ( t ) = log n − ( b − a ) t + p ( b − a ) t + 4 ab a o and µ v = µ u on [ − , ;in obvious notation. roof. (i) The set Y u := Z ,u ∪ (cid:16) { u ′ + (1 /x + e ̺ ) u + λζ = 0 } \ Z ,u (cid:17) ∪ { ̺ not differentiable } ⊂ [ a, b ](in obvious notation) is a null set in [ a, b ] and likewise for Y v . By [1] Lemma 2.95 and Lemma2.96, { u = t } ∩ ( Y u ∪ Z ,u ) = ∅ for a.e. t ∈ (0 ,
1) and likewise for the function v . Let t ∈ (0 , { u = t } ∩ ( Y u ∪ Z ,u ) = ∅ and { v = t } ∩ ( Y v ∪ Z ,v ) = ∅ . Put c := max { u ≥ t } .Then c ∈ ( a, b ), { u > t } = [ a, c ) by Lemma 9.3 and u is differentiable at c with u ′ ( c ) <
0. Put d := max { v ≤ t } = max { u ≥ − t } . As u is continuous on [ a, b ] it holds that a < c < d < b .Moreover, u ′ ( d ) < v ( d ) = t and d Z v . Put e u := u/t and e v := v/t on [ c, d ]. Then e u ′ + (1 /τ + e ̺ ) e u + ( ˆ m/t ) ζ = 0 a.e. on ( c, d ) and e u ( c ) = − e u ( d ) = 1; e v ′ + (1 /τ + e ̺ ) e v − ( ˆ m/t ) ζ = 0 a.e. on ( c, d ) and − e v ( c ) = e v ( d ) = 1 . By Lemma 9.4, Z { v = t }\ Z v | v ′ | d H τ ≥ Z [ c,d ] ∩{ v = t }\ Z v | v ′ | d H τ = (1 /t ) Z [ c,d ] ∩{ e v =1 }\ Z v | e v ′ | d H τ ≥ (1 /t ) Z [ c,d ] ∩{ e u =1 }\ Z u | e u ′ | d H τ = Z { u = t }\ Z u | u ′ | d H τ . By Lemma 9.1, − µ ′ u ( t ) = Z { u = t }\ Z u | u ′ | d H τ for L -a.e. t ∈ (0 ,
1) and a similar formula holds for v . The assertion in (i) follows. (ii) Assume that ̺ a, b ). Put α := inf { ̺ > } ∈ [ a, b ). Note that max { v ≤ t } → b as t ↑ v < a, b ) by assumption. Choose t ∈ (0 ,
1) such that max { v ≤ t } > α . Then for t > t , a < max { u ≥ t } < max { u ≥ − t } = max { v ≤ t } < max { v ≤ t } < b ;that is, the interval [ c, d ] with c, d as described above intersects ( α, b ]. So for L -a.e. t ∈ ( t , Z { v = t }\ Z v | v ′ | d H τ > Z { u = t }\ Z u | u ′ | d H τ . by the equality condition in Lemma 9.4. The conclusion follows from the representation of µ u resp. µ v in Lemma 9.1. (iii) A computation givesˆ ̺u + ˆ m ζ = ˆ m ζ ( a ) + ζ ( a ) = 2 b − a on [ a, b ] andˆ m = 1 − abb − a . We then derive that u ( τ ) = 1 b − a n − τ + abτ o for τ ∈ [ a, b ]; u is strictly decreasing on its domain. This leads to the formula in (iii) . A similarcomputation gives µ v ( t ) = log n b ( b − a ) t + p ( b − a ) t + 4 ab o for t ∈ [ − , orollary 9.6. Let < a < b < and ̺ ≥ be a function on [ a, b ] such that ̺/ζ is non-decreasingand bounded on [ a, b ] . Suppose that ( u, λ ) solves (9.4) with η = (1 , − and set v := − u . Assumethat u > − on [ a, b ) . Then(i) µ u ( t ) ≤ µ v ( t ) for each t ∈ (0 , ;(ii) if ̺ on [ a, b ) then µ u ( t ) < µ v ( t ) for each t ∈ (0 , .Proof. (i) By [1] Theorem 3.28 and Lemma 9.3, µ u ( t ) = µ u ( t ) − µ u (1) = − Dµ u (( t, − Dµ au (( t, − Dµ su (( t, − Z ( t, µ ′ u ds for each t ∈ (0 ,
1) as µ u (1) = 0. On the other hand, µ v ( t ) = µ v (1) + ( µ v ( t ) − µ v (1)) = µ v (1) − Dµ v (( t, µ v (1) − Z ( t, µ ′ v ds − Dµ sv (( t, t ∈ (0 , Dµ sv (( t, ≤ (ii) follows from Theorem 9.5 (ii) . Corollary 9.7.
Let < a < b < and ̺ ≥ be a function on [ a, b ] such that ̺/ζ is non-decreasingand bounded on [ a, b ] . Suppose that ( u, λ ) solves (9.4) with η = (1 , − . Assume that u > − on [ a, b ) . Let ϕ ∈ C (( − , be an odd strictly increasing function with ϕ ∈ L (( − , . Then(i) R { u> } ϕ ( u ) dµ < + ∞ ;(ii) R ba ϕ ( u ) dµ ≤ ;(iii) equality holds in (ii) if and only if ̺ ≡ on [ a, b ) .In particular,(iv) R ba u √ − u dµ ≤ with equality if and only if ̺ ≡ on [ a, b ) .Proof. (i) Put I := { > u > } . The function u : I → (0 ,
1) is C , and u ′ ≤ − ˆ mζ a.e. on I byLemma 9.3. It has C , inverse v : (0 , → I ⊂ [ a, b ], v ′ = 1 / ( u ′ ◦ v ) and | v ′ | ≤ / ( ˆ m [ ζ ◦ v ]) a.e.on (0 , Z { u> } ϕ ( u ) dµ = − Z ϕ ( v ′ /v ) dt from which the claim is apparent. (ii) The integral is well-defined because ϕ ( u ) + = ϕ ( u ) χ { u> } ∈ L (( a, b ) , µ ) by (i) . By Lemma 9.3 the set { u = 0 } consists of a singleton and has µ -measure zero.So Z ba ϕ ( u ) dµ = Z { u> } ϕ ( u ) dµ + Z { u< } ϕ ( u ) dµ = Z { u> } ϕ ( u ) dµ − Z { v> } ϕ ( v ) dµ where v := − u as ϕ is an odd function. We remark that in a similar way to (9.3), Z ϕ ′ µ u dt = Z { u> } n ϕ ( u ) − ϕ (0) o dµ = Z { u> } ϕ ( u ) dµ using oddness of ϕ and an analogous formula holds with v in place of u . Thus we may write Z ba ϕ ( u ) dµ = Z ϕ ′ µ u dt − Z ϕ ′ µ v dt = Z ϕ ′ n µ u − µ v o dt ≤ ϕ ′ > , (iii) Suppose that ̺ a, b ). Then strict inequalityholds in the above by Corollary 9.6. If ̺ ≡ a, b ) the equality follows from Theorem 9.5. (iv) follows from (ii) and (iii) with the particular choice ϕ : ( − , → R ; t t/ √ − t . The boundary condition η η = 1 . Let 0 < a < b < ̺ ≥ a, b ] such that ̺/ζ is non-decreasing and bounded on [ a, b ]. We study solutions of the auxilliary Riccati equation w ′ + λζw = (1 /x + e ̺ ) w a.e. on ( a, b ) with w ( a ) = w ( b ) = 1; (9.6)with w ∈ C , ([ a, b ]) and λ ∈ R . If ̺ ≡ a, b ] then we write w instead of w . Suppose ( u, λ )solves (9.4) with η = (1 , u > a, b ] by Lemma 9.2 and we may set w := 1 /u . Then( w, − λ ) satisfies (9.6). Lemma 9.8.
Let < a < b < and ̺ ≥ be a function on [ a, b ] such that ̺/ζ is non-decreasingand bounded on [ a, b ] . Then(i) there exists a solution ( w, λ ) of (9.6) with w ∈ C , ([ a, b ]) and λ ∈ R ;(ii) the pair ( w, λ ) in (i) is unique;(iii) λ = m .Proof. (i) Define w : [ a, b ] → R by w := g m R · a f ds + g ( a ) . Then w ∈ C , ([ a, b ]) and ( w, m ) satisfies (9.6). (ii) We claim that w > a, b ] for any solution( w, λ ) of (9.6). For otherwise, c := min { w = 0 } ∈ ( a, b ). Then u := 1 /w on [ a, c ) satisfies u ′ + (1 /τ + e ̺ ) u − λζ = 0 a.e. on ( a, c ) and u ( a ) = 1 , u ( c − ) = + ∞ . Integrating, we obtain g u − g ( a ) − λ Z · a g ζ dt = 0 on [ a, c )and this entails the contradiction that u ( c − ) < + ∞ . We may now use the uniqueness statementin Lemma 9.2. (iii) follows from (ii) and the particular solution given in (i) .We introduce the mapping ω : (0 , ∞ ) × (0 , ∞ ) → R ; ( t, x )
7→ − (2 /t ) coth( x/ . For ξ > | ω ( t, x ) − ω ( t, y ) | ≤ cosech [ ξ/ /t ) | x − y | (9.7)for ( t, x ) , ( t, y ) ∈ (0 , ∞ ) × ( ξ, ∞ ) and ω is locally Lipschitzian in x on (0 , ∞ ) × (0 , ∞ ) in the sense of[19] I.3. Let 0 < a < b < + ∞ and set λ := A/G >
1. Here, A = A ( a, b ) stands for the arithmeticmean of a, b as introduced in the previous Section while G = G ( a, b ) := p | ab | stands for theirgeometric mean. We refer to the inital value problem z ′ = ω ( t, z ) on (0 , λ ) and z (1) = µ (( a, b )) . (9.8)Define z : (0 , λ ) → R ; t n λ + √ λ − t t o . emma 9.9. Let < a < b < + ∞ . Then(i) w ( τ ) = AτG + τ for τ ∈ [ a, b ] ;(ii) k w k ∞ = λ ;(iii) µ w = z on [1 , λ ) ;(iv) z satisfies (9.8) and this solution is unique;(v) R { w =1 } | w ′ | d H τ = 2 coth( µ (( a, b )) / ;(vi) R ba √ w − dxx = π .Proof. (i) Note that m = 1 + aba + b . (i) follows from the representation of w in Lemma 9.8 by direct computation. (ii) follows bycalculus. (iii) follows by solving the quadratic equation tτ − Aτ + G t = 0 for τ with t ∈ (0 , λ ).Uniqueness in (iv) follows from [19] Theorem 3.1 as ω is locally Lipschitzian with respect to x in(0 , ∞ ) × (0 , ∞ ). For (v) note that | aw ′ ( a ) | = 1 − a/A and | bw ′ ( b ) | = b/A − µ (( a, b )) /
2) = 2( a + b ) / ( b − a ) . (vi) We may write Z ba p w − dττ = Z ba ab + τ p ( a + b ) τ − ( ab + τ ) dττ = Z ba ab + τ p ( τ − a )( b − τ ) dττ . The substitution s = τ followed by the Euler substitution (cf. [17] 2.251) p ( s − a )( b − s ) = t ( s − a ) gives Z ba p w − dττ = Z ∞
11 + t + abb + a t dt = π. Lemma 9.10.
Let < a < b < + ∞ . Then(i) for y > a the function x by − ax ( y − a )( b − x ) is strictly increasing on ( −∞ , b ] ;(ii) the function y ( b − a ) y ( y − a )( b − y ) is strictly increasing on [ G, b ] ;(iii) for x < b the function y by − ax ( y − a )( b − x ) is strictly decreasing on [ a, + ∞ ) Proof.
The proof is an exercise in calculus.
Lemma 9.11.
Let < a < b < . Then m ζ − ˆ ̺ = 1 A at a, b . roof. Note that m = 1 + aba + b so that m − t = 1 + ab − ( a + b ) ta + b for t ∈ [ a, b ] and the result follows. Lemma 9.12.
Let < a < b < and ̺ ≥ be a function on [ a, b ] such that ̺/ζ is non-decreasingand bounded on [ a, b ] . Let ( w, λ ) solve (9.6). Assume(i) w is differentiable at both a and b and that (9.6) holds there;(ii) w ′ ( a ) > and w ′ ( b ) < ;(iii) w > on ( a, b ) ;(iv) ̺ is differentiable at a and b .Then Z { w =1 }\ Z w | w ′ | d H τ ≥ µ (( a, b )) / with equality if and only if ̺ ≡ on [ a, b ) .Proof. At the end-points x = a, b the condition (i) entails that w ′ + mζ − e ̺ = 1 /x = w ′ + m ζ − ˆ ̺ so that w ′ − w ′ = ( m − m ) ζ + ̺ at x = a, b. (9.9)We consider the four cases(a) w ′ ( a ) ≥ w ′ ( a ) and w ′ ( b ) ≥ w ′ ( b );(b) w ′ ( a ) ≥ w ′ ( a ) and w ′ ( b ) ≤ w ′ ( b );(c) w ′ ( a ) ≤ w ′ ( a ) and w ′ ( b ) ≥ w ′ ( b );(d) w ′ ( a ) ≤ w ′ ( a ) and w ′ ( b ) ≤ w ′ ( b );in turn.(a) Note that0 < aw ′ ( a ) = 1 − ( mζ ( a ) − e ̺ ( a )) a ; (9.10)0 > bw ′ ( b ) = 1 − ( mζ ( b ) − e ̺ ( b )) b. (9.11)Put x := 1 / ( mζ ( b ) − e ̺ ( b )) and y := 1 / ( mζ ( a ) − e ̺ ( a )). We claim that a < y ≤ x < b and A ≤ x. The inequality x < b follows from (9.11) and the inequality a < y follows from (9.10). Because ̺/ζ is non-decreasing, ̺ ( b ) ζ ( b ) ≤ ̺ ( b ) − ̺ ( a ) ζ ( b ) − ζ ( a ) .
37y Theorem 8.5, m = m ( ̺/ζ, a, b ) ≤ ( ̺/ζ )( b − ) + m ( a, b )= ̺ ( b − ) ζ ( b ) + ˆ ̺ ( b ) − ˆ ̺ ( a ) ζ ( b ) − ζ ( a ) ≤ ̺ ( b − ) − ̺ ( a ) ζ ( b ) − ζ ( a ) + ˆ ̺ ( b ) − ˆ ̺ ( a ) ζ ( b ) − ζ ( a ) = e ̺ ( b − ) − e ̺ ( a ) ζ ( b ) − ζ ( a ) . This leads to the inequality y ≤ x . By Lemma 8.5 and Lemma 9.11,1 x = mζ ( b ) − e ̺ ( b − ) ≤ ̺ ( b − ) + m ζ ( b ) − e ̺ ( b − ) = m ζ ( b ) − ˆ ̺ ( b ) = 1 A so that x ≥ A . Now Z { w =1 }\ Z w | w ′ | d H x = 1 aw ′ ( a ) − bw ′ ( b ) = 1 − (1 /y ) a + 1 − − (1 /x ) b + 1 = by − ax ( y − a )( b − x ) . If y ∈ ( a, A ] use Lemma 9.10 (iii) to replace y by A then (i) to derive the desired inequality;otherwise, use Lemma 9.10 (i) and (ii) .(b) Define x and y as above. Then y ≤ x < b as before. Condition (b) together with (9.9) meansthat ( m − m ) ζ ( a ) + ̺ ( a ) ≥
0. So mζ ( a ) − e ̺ ( a ) ≤ m ζ ( a ) − ˆ ̺ ( a ) = 1 /A by Lemma 9.11; that is, y ≥ A . Now argue as in (a).(c) In this case,1 aw ′ ( a ) − bw ′ ( b ) ≥ aw ′ ( a ) − bw ′ ( b ) = 2 coth( µ (( a, b )) / w ′ ( b ) = w ′ ( b ) so that ( m − m ) ζ ( b )+ ̺ ( b ) = 0 and ̺ vanisheson [ a, b ) by Theorem 8.5.(d) Condition (d) together with (9.9) means that ( m − m ) ζ ( b )+ ̺ ( b ) ≤
0; that is, m ≥ ( ̺/ζ )( b − )+ m . On the other hand, by Theorem 8.5, m ≤ ( ̺/ζ )( b − ) + m . In consequence, m = ( ̺/ζ )( b − ) + m . It then follows that ̺ ≡ a, b ) by Theorem 8.5. Now use Lemma 9.9. Lemma 9.13.
Let φ : (0 , + ∞ ) → (0 , + ∞ ) be a convex non-increasing function with inf (0 , + ∞ ) φ > . Let Λ be an at most countably infinite index set and ( x h ) h ∈ Λ a sequence of points in (0 , + ∞ ) with P h ∈ Λ x h < + ∞ . Then X h ∈ Λ φ ( x h ) ≥ φ ( X h ∈ Λ x h ) and the left-hand side takes the value + ∞ in case Λ is countably infinite and is otherwise finite.Proof. Suppose 0 < x < x < + ∞ . By convexity φ ( x ) + φ ( x ) ≥ φ ( x + x ) ≥ φ ( x + x ) as φ is non-increasing. The result for finite Λ follows by induction. Theorem 9.14.
Let < a < b < and ̺ ≥ be a function on [ a, b ] such that ̺/ζ is non-decreasingand bounded on [ a, b ] . Let ( w, λ ) solve (9.6). Assume that w > on ( a, b ) . Then(i) for L -a.e. t ∈ (1 , k w k ∞ ) , − µ ′ w ≥ (2 /t ) coth((1 / µ w ); (9.12) (ii) if ̺ on [ a, b ) then there exists t ∈ (1 , k w k ∞ ) such that strict inequality holds in (9.12)for L -a.e. t ∈ (1 , t ) . roof. (i) The set Y w := Z ,w ∪ (cid:16) { w ′ + mζw = (1 /x + e ̺ ) w } \ Z ,w (cid:17) ∪ { ̺ not differentiable } ⊂ [ a, b ]is a null set in [ a, b ]. By [1] Lemma 2.95 and Lemma 2.96, { w = t } ∩ ( Y w ∩ Z ,w ) = ∅ for a.e. t > t ∈ (1 , k w k ∞ ) and assume that { w = t } ∩ ( Y w ∩ Z ,w ) = ∅ . We write { w > t } = S h ∈ Λ I h whereΛ is an at most countably infinite index set and ( I h ) h ∈ Λ are disjoint non-empty well-separated openintervals in ( a, b ). The term well-separated means that for each h ∈ Λ, inf k ∈ Λ \{ h } d ( I h , I k ) > w ′ = 0 on ∂I h for each h ∈ Λ. Put e w := w/t on { w > t } so e w ′ + ( mt ) ζ e w = (1 /x + e ̺ ) e w a.e. on { w > t } and e w = 1 on { w = t } . We use the fact that the mapping φ : (0 , + ∞ ) → (0 , + ∞ ); t coth t satisfies the hypotheses ofLemma 9.13. By Lemmas 9.12 and 9.13,(0 , + ∞ ] ∋ Z { w = t }\ Z w | w ′ | d H x = (1 /t ) Z { e w =1 } | e w ′ | d H τ = (1 /t ) X h ∈ Λ Z ∂I h | e w ′ | d H τ ≥ (2 /t ) X h ∈ Λ coth((1 / µ ( I h )) ≥ (2 /t ) coth((1 / X h ∈ Λ µ ( I h ))= (2 /t ) coth((1 / µ ( { w > t } ))) = (2 /t ) coth((1 / µ w ( t )) . The statement now follows from Lemma 9.1. (ii)
Suppose that ̺ a, b ). Put α := min { ̺ > } ∈ [ a, b ). Now that { w > t } ↑ ( a, b ) as t ↓ w > a, b ). Choose t ∈ (1 , k w k ∞ ) such that { w > t } ∩ ( α, b ) = ∅ . Then for each t ∈ (1 , t )there exists h ∈ Λ such that ̺ I h . The statement then follows by Lemma 9.12. Lemma 9.15.
Let ∅ 6 = S ⊂ R be bounded and suppose S has the property that for each s ∈ S there exists δ > such that [ s, s + δ ) ⊂ S . Then S is L -measurable and | S | > .Proof. For each s ∈ S put t s := inf { t > s : t S } . Then s < t s < + ∞ , [ s, t s ) ⊂ S and t s S .Define C := n [ s, t ] : s ∈ S and t ∈ ( s, t s ) o . Then C is a Vitali cover of S (see [7] Chapter 16 for example). By Vitali’s Covering Theorem (cf.[7] Theorem 16.27) there exists an at most countably infinite subset Λ ⊂ C consisting of pairwisedisjoint intervals such that | S \ [ I ∈ Λ I | = 0 . Note that I ⊂ S for each I ∈ Λ. Consequently, S = S I ∈ Λ I ∪ N where N is an L -null set andhence S is L -measurable. The positivity assertion is clear. Theorem 9.16.
Let < a < b < and ̺ ≥ be a function on [ a, b ] such that ̺/ζ is non-decreasing and bounded on [ a, b ] . Let ( w, λ ) solve (9.6). Assume that w > on ( a, b ) . Put T := min {k w k ∞ , k w k ∞ } > . Then(i) µ w ( t ) ≤ µ w ( t ) for each t ∈ [1 , T ) ; ii) k w k ∞ ≤ k w k ∞ ;(iii) if ̺ on [ a, b ) then there exists t ∈ (1 , k w k ∞ ) such that µ w ( t ) < µ w ( t ) for each t ∈ (1 , t ) .Proof. (i) We adapt the proof of [19] Theorem I.6.1. The assumption entails that µ w (1) = µ w (1) = µ (( a, b )). Suppose for a contradiction that µ w ( t ) > µ w ( t ) for some t ∈ (1 , T ).For ε > z ′ = ω ( t, z ) + ε and z (1) = µ (( a, b )) + ε (9.13)on (0 , T ). Choose υ ∈ (0 ,
1) and τ ∈ ( t, T ). By [19] Lemma I.3.1 there exists ε > ≤ ε < ε (9.13) has a continuously differentiable solution z ε defined on [ υ, τ ] and thissolution is unique by [19] Theorem I.3.1. Moreover, the sequence ( z ε ) <ε<ε converges uniformlyto z on [ υ, τ ].Given 0 < ε < η < ε it holds that z ≤ z ε ≤ z η on [1 , τ ] by [19] Theorem I.6.1. Note for examplethat z ′ ≤ ω ( · , z ) + ε on (1 , τ ). In fact, ( z ε ) <ε<ε decreases strictly to z on (1 , τ ). For if, say, z ( s ) = z ε ( s ) for some s ∈ (1 , τ ) then z ′ ε ( s ) = ω ( s, z ε ( s )) + ε > ω ( s, z ( s )) = z ′ ( s ) by (9.13); whileon the other hand z ′ ε ( s ) ≤ z ′ ( s ) by considering the left-derivative at s and using the fact that z ε ≥ z on [1 , τ ]. This contradicts the strict inequality.Choose ε ∈ (0 , ε ) such that z ε ( t ) < µ w ( t ) for each 0 < ε < ε . Now µ w is right-continuous andstrictly decreasing as µ w ( t ) − µ w ( s ) = − µ ( { s < w ≤ t } ) < ≤ s < t < k w k ∞ by continuityof w . So the set { z ε < µ w } ∩ (1 , t ) is open and non-empty in (0 , + ∞ ) for each ε ∈ (0 , ε ). Thusthere exists a unique s ε ∈ [1 , t ) such that µ w > z ε on ( s ε , t ] and µ w ( s ε ) = z ε ( s ε )for each ε ∈ (0 , ε ). As z ε (1) > µ (( a, b )) it holds that each s ε >
1. Note that 1 < s ε < s η whenever0 < ε < η as ( z ε ) <ε<ε decreases strictly to z as ε ↓ S := n s ε : 0 < ε < ε o ⊂ (1 , t ) . We claim that for each s ∈ S there exists δ > s, s + δ ) ⊂ S . This entails that S is L -measurable with positive L -measure by Lemma 9.15.Suppose s = s ε ∈ S for some ε ∈ (0 , ε ) and put z := z ε ( s ) = µ w ( s ). Put k := cosech ( z ( t ) / ≤ ζ < η < ε defineΩ ζ,η := n ( u, y ) ∈ R : u ∈ (0 , t ) and z ζ ( u ) < y < z η ( u ) o and note that this is an open set in R . We remark that for each ( u, y ) ∈ Ω ζ,η there exists aunique ν ∈ ( ζ, η ) such that y = z ν ( u ). Given r > s + r < t set Q = Q r := n ( u, y ) ∈ R : s ≤ u < s + r and | y − z | < k z ε − z k C ([ s,s + r ]) o . Choose r ∈ (0 , t − s ) and ε ∈ ( ε, ε ) such that(a) Q r ⊂ Ω ,ε ;(b) k z ε − z k C ([ s,s + r ]) < sε/ (2 k );(c) sup η ∈ ( ε,ε ) k z η − z k C ([ s,s + r ]) ≤ k z ε − z k C ([ s,s + r ]) ;(d) z η < µ w on [ s + r, t ] for each η ∈ ( ε, ε ). 40e can find δ ∈ (0 , r ) such that z ε < µ w < z ε on ( s, s + δ ) as z ε ( s ) > z ; in other words, thegraph of µ w restricted to ( s, s + δ ) is contained in Ω ε,ε .Let u ∈ ( s, s + δ ). Then µ w ( u ) = z η ( u ) for some η ∈ ( ε, ε ) as above. We claim that u = s η sothat u ∈ S . This implies in turn that [ s, s + δ ) ⊂ S . Suppose for a contradiction that z η < µ w on( u, t ]. Then there exists v ∈ ( u, t ] such that µ w ( v ) = z η ( v ). In view of condition (d), v ∈ ( u, s + r ).By [1] Theorem 3.28 and Theorem 9.14, µ w ( v ) − µ w ( u ) = Dµ w (( u, v ]) = Dµ aw (( u, v ]) + Dµ sw (( u, v ]) ≤ Dµ aw (( u, v ]) = Z vu µ ′ w dτ ≤ Z vu ω ( · , µ w ) dτ. On the other hand, z η ( v ) − z η ( u ) = Z vu z ′ η dτ = Z vu ω ( · , z η ) dτ + η ( v − u ) . We derive that ε ( v − u ) ≤ η ( v − u ) ≤ Z vu n ω ( · , µ w ) − ω ( · , z η ) o dτ ≤ k Z vu | µ w − z η | dµ using the estimate (9.7). Thus ε ≤ k v − u Z vu | µ w − z η | dµ ≤ ( k/s ) k µ w − z η k C ([ u,v ]) ≤ ( k/s ) n k µ w − z k C ([ s,s + r ]) + k z η − z k C ([ s,s + r ]) o ≤ (2 k/s ) k z ε − z k C ([ s,s + r ]) < ε by (b) and (c) giving rise to the desired contradiction.By Theorem 9.14, µ ′ w ≤ ω ( · , µ w ) for L -a.e. t ∈ S . Choose s ∈ S such that µ w is differentiableat s and the latter inequality holds at s . Let ε ∈ (0 , ε ) such that s = s ε . For any u ∈ ( s, t ), µ w ( u ) − µ w ( s ) > z ε ( u ) − z ε ( s ) . We deduce that µ ′ w ( s ) ≥ z ′ ε ( s ). But then µ ′ w ( s ) ≥ z ′ ε ( s ) = ω ( s, z ǫ ( s )) + ε > ω ( s, µ w ( s )) . This strict inequality holds on a set of full measure in S . This contradicts Theorem 9.14. (ii) Use the fact that k w k ∞ = sup { t > µ w ( t ) > } . (iii) Assume that ̺ a, b ). Let t ∈ (1 , k w k ∞ ) be as in Lemma 9.14. Then for t ∈ (1 , t ), µ w ( t ) − µ w (1) = Dµ w ((1 , t ]) = Dµ aw ((1 , t ]) + Dµ sw ((1 , t ]) ≤ Dµ aw ((1 , t ])= Z (1 ,t ] µ ′ w ds < Z (1 ,t ] ω ( s, µ w ) ds ≤ Z (1 ,t ] ω ( s, µ w ) ds = µ w ( t ) − µ w (1)by Theorem 9.14, Lemma 9.9 and the inequality in (i) . Corollary 9.17.
Let < a < b < and ̺ ≥ be a function on [ a, b ] such that ̺/ζ is non-decreasing and bounded on [ a, b ] . Suppose that ( w, λ ) solves (9.6). Assume that w > on ( a, b ) .Let ≤ ϕ ∈ C ((1 , + ∞ )) be strictly decreasing with R ba ϕ ( w ) dµ < + ∞ . Then(i) R ba ϕ ( w ) dµ ≥ R ba ϕ ( w ) dµ ; ii) equality holds in (i) if and only if ̺ ≡ on [ a, b ) .In particular,(iii) R ba √ w − dµ ≥ π with equality if and only if ̺ ≡ on [ a, b ) .Proof. (i) Let ϕ ≥ , + ∞ ) which is piecewise C . Suppose that ϕ (1+) < + ∞ . By Tonelli’s Theorem, Z [1 , + ∞ ) ϕ ′ µ w ds = Z [1 , + ∞ ) ϕ ′ n Z ( a,b ) χ { w>s } dµ o ds = Z ( a,b ) n Z [1 , + ∞ ) ϕ ′ χ { w>s } ds o dµ = Z ( a,b ) n ϕ ( w ) − ϕ (1) o dµ = Z ( a,b ) ϕ ( w ) dµ − ϕ (1) µ (( a, b ))and a similar identity holds for µ w . By Theorem 9.16, R ba ϕ ( w ) dµ ≥ R ba ϕ ( w ) dµ . Now supposethat 0 ≤ ϕ ∈ C ((1 , + ∞ )) is strictly decreasing with R ba ϕ ( w ) dµ < + ∞ . The inequality holds forthe truncated function ϕ ∧ n for each n ∈ N . An application of the monotone convergence theoremestablishes the result for ϕ . (ii) Suppose that equality holds in (i) . For c ∈ (1 , + ∞ ) put ϕ := ϕ ∨ ϕ ( c ) − ϕ ( c ) and ϕ := ϕ ∧ ϕ ( c ).By (i) we deduce Z ba ϕ ( w ) dµ = Z ba ϕ ( w ) dµ ;and hence by the above that Z [ c, + ∞ ) ϕ ′ n µ w − µ w o ds = 0 . This means that µ w = µ w on ( c, + ∞ ) and hence on (1 , + ∞ ). By Theorem 9.16 we concludethat ̺ ≡ a, b ). (iii) flows from (i) and (ii) noting that the function ϕ : (1 , + ∞ ) → R ; t / √ t − The case a = 0 . Let 0 < b < ̺ ≥ , b ] such that ̺/ζ is non-decreasingand bounded on [0 , b ]. We study solutions to the first-order linear ordinary differential equation u ′ + (1 /x + e ̺ ) u + λζ = 0 a.e. on (0 , b ) with u (0) = 0 and u ( b ) = 1 (9.14)where u ∈ C , ([0 , b ]) and λ ∈ R . If ̺ ≡ , b ] then we write u instead of u . Lemma 9.18.
Let < b < and ̺ ≥ be a function on [0 , b ] such that ̺/ζ is non-decreasingand bounded on [0 , b ] . Then(i) there exists a solution ( u, λ ) of (9.14) with u ∈ C , ([0 , b ]) and λ ∈ R ;(ii) λ is given by λ = − g ( b ) /F ( b ) where F := R · f ds ;(iii) the pair ( u, λ ) in (i) is unique;(iv) u > on (0 , b ] .Proof. (i) The function u : [0 , b ] → R given by u = g ( b ) F ( b ) F g (9.15)42n [0 , b ] solves (9.14) with λ as in (ii) . (iii) Suppose that ( u , λ ) resp. ( u , λ ) solve (9.14). Bylinearity u := u − u solves u ′ + (1 /x + e ̺ ) u + λζ = 0 a.e. on (0 , b ) with u (0) = u ( b ) = 0where λ = λ − λ . An integration gives that u = ( − λF + c ) / g for some constant c ∈ R and theboundary conditions entail that λ = c = 0. (iv) follows from the formula (9.15) and unicity. Lemma 9.19.
Suppose −∞ < a < b < + ∞ and that φ : [ a, b ] → R is convex. Suppose that thereexists ξ ∈ ( a, b ) such that φ ( ξ ) = b − ξb − a φ ( a ) + ξ − ab − a φ ( b ) . Then φ ( c ) = b − cb − a φ ( a ) + c − ab − a φ ( b ) for each c ∈ [ a, b ] .Proof. Let c ∈ ( ξ, b ). By monotonicity of chords, φ ( ξ ) − φ ( a ) ξ − a ≤ φ ( c ) − φ ( ξ ) c − ξ so φ ( c ) ≥ c − aξ − a φ ( ξ ) − c − ξξ − a φ ( a )= c − aξ − a n b − ξb − a φ ( a ) + ξ − ab − a φ ( b ) o − c − ξξ − a φ ( a )= b − cb − a φ ( a ) + c − ab − a φ ( b )and equality follows. The case c ∈ ( a, ξ ) is similar. Lemma 9.20.
Let < b < and ̺ ≥ be a function on [0 , b ] such that ̺/ζ is non-decreasingand bounded on [0 , b ] . Let ( u, λ ) satisfy (9.14). Then(i) u ≥ u on [0 , b ] ;(ii) if ̺ on [0 , b ) then u > u on (0 , b ) .Proof. (i) The mapping F : [0 , b ] → [0 , F ( b )] is a bijection with inverse F − . Define η : [0 , F ( b )] → R via η := ( t g ) ◦ F − . Then η ′ = ( t g ) ′ f ◦ F − = 2 + t e ̺ζ ◦ F − = h t ˆ ̺ζ + t̺ζ i ◦ F − = (1 + t̺/ζ ) ◦ F − a.e. on (0 , F ( b )) so η ′ is non-decreasing there. This means that η is convex on [0 , F ( b )]. Inparticular, η ( s ) ≤ [ η ( F ( b )) /F ( b )] s for each s ∈ [0 , F ( b )]. For t ∈ [0 , b ] put s := F ( t ) to obtain t g ( t ) ≤ ( b g ( b ) /F ( b )) F ( t ). A rearrangement gives u ≥ u on [0 , b ] noting that u : [0 , b ] → R ; t t/b . (ii) Assume ̺ , b ). Suppose that u ( c ) = u ( c ) for some c ∈ (0 , b ). Then η ( F ( c )) = [ η ( F ( b )) /F ( b )] F ( c ). By Lemma 9.19, η ′ is constant on (0 , F ( b )). This implies that ̺ ≡ , b ). Lemma 9.21.
Let < b < + ∞ . Then R b u √ − u dµ = π/ .Proof. The integral is elementary as u ( t ) = t/b for t ∈ [0 , b ].43 Lemma 10.1.
Let x ∈ H and v be a unit vector in R such that the pair { x, v } forms a positivelyoriented orthogonal basis for R . Put b := ( τ, where | x | = τ and γ := θ ( x ) ∈ (0 , π ) . Let α ∈ (0 , π/ such that h v, x − b i| x − b | = cos α. Then(i) C ( x, v, α ) ∩ H ∩ C (0 , e , γ ) = ∅ ;(ii) for any y ∈ C ( x, v, α ) ∩ H \ B (0 , τ ) the line segment [ b, y ] intersects S τ outside the closedcone C (0 , e , γ ) . We point out that C (0 , e , γ ) is the open cone with vertex 0 and axis e which contains the point x on its boundary. We note that cos α ∈ (0 ,
1) because h v, x − b i = −h v, b i = −h (1 /τ ) Ox, b i = −h Op, e i = h x, O ⋆ e i = h x, e i > | x − b | = h v, x − b i then b = x − λv for some λ ∈ R and hence x = h e , x i = τ and x = 0. Proof. (i)
For ω ∈ S define the open half-space H ω := { y ∈ R : h y, ω i > } . We claim that C ( x, v, α ) ⊂ H v . For given y ∈ C ( x, v, α ), h y, v i = h y − x, v i > | y − x | cos α > . On the other hand, it holds that C (0 , e , γ ) ∩ H ⊂ H − v . This establishes (i) . (ii) By some trigonometry γ = 2 α . Suppose that ω is a unit vector in C ( b, − e , π/ − α ). Then λ := h ω, e i < cos α since upon rewriting the membership condition for C ( b, − e , π/ − α ) weobtain the quadratic inequality λ − αλ + cos γ > . For ω a unit vector in C (0 , e , γ ) the opposite inequality h ω, e i ≥ cos α holds. This shows that C ( b, − e , π/ − α ) ∩ C (0 , e , γ ) ∩ S τ = ∅ . The set C ( x, v, α ) ∩ H is contained in the open convex cone C ( b, − e , π/ − α ). Suppose y ∈ C ( x, v, α ) ∩ H \ B (0 , τ ). Then the line segment [ b, y ] is contained in C ( b, − e , π/ − α ) ∪ { b } .Now the set C ( b, − e , π/ − α ) ∩ S τ disconnects C ( b, − e , π/ − α ) ∪ { b } . This entails that( b, y ] ∩ C ( b, − e , π/ − α ) ∩ S τ = ∅ . The foregoing paragraph entails that ( b, y ] ∩ C (0 , e , γ ) ∩ S τ = ∅ .This establishes the result. Lemma 10.2.
Let E be an open set in U such that M := ∂E ∩ U is a C , hypersurface in U .Assume that E \ { } = E sc . Suppose(i) x ∈ ( M \ { } ) ∩ H ;(ii) sin( σ ( x )) = − .Then E is not convex. roof. Let γ : I → M be a C , parametrisation of M in a neighbourhood of x with γ (0) = x asabove. As sin( σ ( x )) = − n ( x ) and hence n (0) point in the direction of x . Put v := − t (0) = − t ( x ). We may write γ ( s ) = γ (0) + st (0) + R ( s ) = x − sv + R ( s )for s ∈ I where R ( s ) = s R ˙ γ ( ts ) − ˙ γ (0) dt and we can find a finite positive constant K suchthat | R ( s ) | ≤ Ks on a symmetric open interval I about 0 with I ⊂⊂ I . Then h γ ( s ) − x, v i| γ ( s ) − x | = h− sv + R , v i| − sv + R | = 1 − h ( R /s ) , v i| v − R /s | → s ↑
0. Let α be as in Lemma 10.1 with x and v as just mentioned. The above estimate entailsthat γ ( s ) ∈ C ( x, v, α ) for small s <
0. By (2.9) and Lemma 5.4 the function r is non-increasingon I . In particular, r ( s ) ≥ r (0) = | x | =: τ for I ∋ s < γ ( s ) B (0 , τ ).Choose δ > γ ( s ) ∈ C ( x, v, α ) ∩ H for each s ∈ [ − δ , β := inf { s ∈ [ − δ ,
0] : r ( s ) = τ } . Suppose first that β ∈ [ − δ , E is not convex (see Lemma 5.2). Now supposethat β = 0. Let γ be as in Lemma 10.1. Then the open circular arc S τ \ C (0 , e , γ ) does notintersect E : for otherwise, M intersects S τ \ C (0 , e , γ ) and β < s ∈ [ − δ , b and γ ( s ) lie in E . But by Lemma 10.1 the line segment[ b, γ ( s )] intersects S τ in S τ \ C (0 , e , γ ). Let c ∈ [ b, γ ( s )] ∩ S τ . Then c E . This shows that E is not convex. But if E is convex then E is convex. Therefore E is not convex. Theorem 10.3.
Let f and g be as in (1.2) and (1.3). Given v > let E be a minimiser of (1.5).Assume that E is a relatively compact open set in U with C , boundary M in U and suppose that E \ { } = E sc . Assume that R := inf { ̺ > } ∈ [0 , . (10.2) Then Ω ∩ ( R,
1) = ∅ with Ω as in (5.2).Proof. Suppose that Ω ∩ ( R, = ∅ . As Ω is open in (0 ,
1) by Lemma 5.6 we may write Ω as acountable union of disjoint open intervals in (0 , a, b ) for some 0 ≤ a < b < ∩ ( R, = ∅ . Let us assumefor the time being that a >
0. Note that [ a, b ] ⊂ π ( M ) and cos σ vanishes on M a ∪ M b . We alsoremark that M \ Λ = ∅ as cos σ = 0 on M ∩ A (( a, b )). Let λ ∈ R be as in Theorem 6.5.Let u : Ω → [ − ,
1] be as in (6.6). Then u has a continuous extension to [ a, b ] and u = ± τ = a, b . This may be seen as follows. For τ ∈ ( a, b ) the set M τ ∩ H consists of a singleton byLemma 5.4. The limit x := lim τ ↓ a M τ ∩ H ∈ S a ∩ H exists as M is C . There exists a C , parametrisation γ : I → M with γ (0) = x as above. By (2.9) and Lemma 5.4, r is decreasingon I . So r > a on I ∩ { s < } for otherwise the C property fails at x . It follows that γ = γ ◦ r and σ = σ ◦ γ ◦ r on I ∩ { s < } . Thus sin( σ ◦ γ ) ◦ r = sin σ on I ∩ { s < } . Now the functionsin σ is continuous on I . So u → sin σ (0) ∈ {± } as τ ↓ a . Put η := u ( a ) and η := u ( b ).Let us consider the case η = ( η , η ) = (1 , u < a, b ) for otherwise cos( σ ◦ γ )vanishes at some point in ( a, b ) contradicting the definition of Ω. By Theorem 6.6 the pair ( u, λ )satisfies (9.4) with η = (1 , u > a, b ]. Put w := 1 /u . Then ( w, − λ ) satisfies(9.6) and w > a, b ). By Lemma 6.7, θ ( b ) − θ ( a ) = Z ba θ ′ dτ = − Z ba u √ − u dττ = − Z ba √ w − dττ . By Corollary 9.17, | θ ( b ) − θ ( a ) | > π . But this contradicts the definition of θ in (6.4) as θ takesvalues in (0 , π ) on ( a, b ). If η = ( − , −
1) then λ > η = ( − , θ ( b ) − θ ( a ) < θ ( a ) ∈ (0 , π ]. As before the limit x := lim τ ↓ a M τ ∩ H ∈ S a ∩ H M is C . Using a local parametrisation it can be seen that θ ( a ) = θ ( x ) and sin( σ ( x )) = −
1. If θ ( a ) ∈ (0 , π ) then E is not convex by Lemma 10.2. This contradicts Theorem 7.4. Notethat we may assume that θ ( a ) ∈ (0 , π ). For otherwise, h γ, e i < τ > a near a , contradictingthe definition of γ (6.5). If η = (1 , −
1) then λ > a = 0. By Lemma 5.5, u (0) = 0 and u ( b ) = ±
1. Suppose u ( b ) = 1. Againemploying the formula above, θ ( b ) − θ (0) < − π/ φ : (0 , → R ; t t/ √ − t is strictly increasing and Lemma 9.21. This means that θ (0) > π/ C property at 0 ∈ M . If u ( b ) = − λ > Lemma 10.4.
Let E be an open set in U such that M := ∂E ∩ U is a C , hypersurface in U .Let r ∈ (0 , and assume that the hyperbolic geodesic curvature ˆ k := k + ˆ ̺ sin σζ = − λ H -a.e. on M ∩ B (0 , r ) for some λ ∈ R . Then k is locally constant on M ∩ B (0 , r ) .Proof. Let x ∈ M ∩ B (0 , r ) and γ : I → M a local parametrisation with the usual conventions.We may assume that γ ( I ) ⊂ M ∩ B ( x, r ). Then (cid:16) r (cid:17) k + r sin σ + λ = 0a.e. on I . So k is a.e. differentiable and ˙ k = 0 a.e. on I . This leads to the statement. Lemma 10.5.
Let f and g be as in (1.2) and (1.3). Let v > .(i) Let E be a relatively compact minimiser of (1.5) in U . Assume that E is open, M := ∂E ∩ U is a C , hypersurface in U and E \{ } = E sc . Then for any r ∈ (0 , with r ≥ R , M \ B (0 , r ) consists of a finite union of disjoint centred circles.(ii) There exists a relatively compact minimiser E of (1.5) such that ∂E ∩ U consists of a finiteunion of disjoint centred circles in U .Proof. (i) First observe that ∅ 6 = π ( M ) = h π ( M ) ∩ [0 , r ] i ∪ h π ( M ) ∩ ( r, i \ Ωby Lemma 10.3. We assume that the latter member is non-empty. By definition of Ω, cos σ = 0on M ∩ A (( r, τ ∈ π ( M ) ∩ ( r, M τ = S τ . Suppose for a contradiction that M τ = S τ . By Lemma 5.2, M τ is the union of two closed spherical arcs in S τ . Let x be a pointon the boundary of one of these spherical arcs relative to S τ . There exists a C , parametrisation γ : I → M of M in a neighbourhood of x with γ (0) = x as before. By shrinking I if necessarywe may assume that γ ( I ) ⊂ A (( r, τ > r . By (2.9), ˙ r = 0 on I as cos σ = 0 on I becausecos σ = 0 on M ∩ A (( r, r is constant on I . This means that γ ( I ) ⊂ S τ . As thefunction sin σ is continuous on I it takes the value ± r ˙ θ = sin σ = ± I .This means that θ is either strictly decreasing or strictly increasing on I . This entails that thepoint x is not a boundary point of M τ in S τ and this proves the claim.It follows from these considerations that M \ B (0 , r ) consists of a finite union of disjoint centredcircles. Note that g ≥ g (0) =: c > U . As a result, + ∞ > P g ( E ) ≥ cP ( E ) and in particularthe relative perimeter P ( E, U \ B (0 , r )) < + ∞ . This explains why M \ B (0 , r ) comprises onlyfinitely many circles. (ii) Let E be a relatively compact minimiser of (1.5) as in Theorem 3.9. By Theorem 4.5 we mayassume that E is open, M := ∂E ∩ U is a C , hypersurface in U and E \ { } = E sc .46ssume that R ∈ (0 , (i) , M \ B (0 , R ) consists of a finite union of disjoint centred circles.We claim that only one of the possibilities M R = ∅ , M R = S R , M R = { Re } or M R = {− Re } (10.3)holds. To prove this suppose that M R = ∅ and M R = S R . Bearing in mind Lemma 5.2 we maychoose x ∈ M R such that x lies on the boundary of M R relative to S R . Assume that x ∈ H .Let γ : I → M be a local parametrisation of M with γ (0) = x with the usual conventions. Wefirst notice that cos( σ ( x )) = 0 for otherwise we obtain a contradiction to Theorem 10.3. As r isdecreasing on I and x is a relative boundary point it holds that r < R on I + := I ∩ { s > } .As M \ Λ is open in M we may suppose that γ ( I + ) ⊂ M \ Λ (see Lemma 6.3). According toTheorem 6.5 the hyperbolic geodesic curvature ˆ k of γ ( I + ) ∩ B (0 , R ) is a.e. constant as ̺ vanisheson (0 , R ). This in turn implies after Lemma 10.4 that γ ( I + ) ∩ B (0 , R ) consists of a line or circulararc. The fact that cos( σ ( x )) = 0 means that γ ( I + ) ∩ B (0 , R ) cannot be a line. So γ ( I + ) ∩ B (0 , R )is an open arc of a circle C containing x in its closure with centre on the line-segment (0 , x ) andradius r ∈ (0 , R ). It can be argued using local parametrisation that C ∩ B (0 , R ) ∩ H ⊂ M . Thiscontradicts C -smoothness of M . In summary, M R ⊂ {± Re } . Finally note that if M R = {± Re } then M R = S R by Lemma 5.2. This establishes (10.3).Suppose that M R = ∅ . As both sets M and S R are compact, d ( M, S R ) >
0. Assume firstthat S R ⊂ E . Put F := B (0 , R ) \ E and suppose F = ∅ . Then F is a set of finite perimeter, F ⊂⊂ B (0 , R ) and P ( F ) = P ( E, B (0 , R )). Let B be a centred open ball with | B | = | F | . Bythe classical isoperimetric inequality, P ( B ) ≤ P ( F ). Define E := ( B (0 , R ) ∪ E ) \ B . Then V f ( E ) = V f ( E ) and P g ( E ) ≤ P g ( E ). That is, E is a minimiser of (1.5) such that ∂E consistsof a finite union of disjoint centred circles. Now suppose that S R ⊂ U \ E . In like fashion we mayredefine E via E := B ∪ ( E \ B (0 , R )) with B a centred open ball in B (0 , R ). The remainingcases in (10.3) can be dealt with in a similar way. The upshot of this argument is that there existsa minimiser of (1.5) whose boundary M in U consists of a finite union of disjoint centred circles.This works in case R ∈ (0 , R = 0. By (i) , M \ B (0 , r ) consists of a finite union of disjoint centred circlesfor any r ∈ (0 , M fails to be C at the origin. The assertionfollows. Lemma 10.6.
Suppose that the function J : [0 , + ∞ ) → [0 , + ∞ ) is continuous non-decreasing and J (0) = 0 . Let N ∈ N ∪ { + ∞} and { t h : h = 0 , . . . , N + 1 } a sequence of points in [0 , + ∞ ) with t > t > · · · > t h > t h +1 > · · · ≥ . Then + ∞ ≥ N +1 X h =0 J ( t h ) ≥ J ( N +1 X h =0 ( − h t h ) . Proof.
We suppose that N = + ∞ . The series P ∞ h =0 ( − h t h converges by the alternating seriestest. For each n ∈ N , n +1 X h =0 ( − h t h ≤ t and the same inequality holds for the infinite sum. As in Step 2 in [5] Theorem 2.1,+ ∞ ≥ ∞ X h =0 J ( t h ) ≥ J ( t ) ≥ J ( ∞ X h =0 ( − h t h )as J is non-decreasing. 47 roof of Theorem 1.1. There exists a relatively compact minimiser E of (1.5) with the propertythat ∂E ∩ U consists of a finite union of disjoint centred circles according to Lemma 10.5. So wemay write E = N [ h =0 A (( a h +1 , a h ))where N ∈ N ∪ { } and 1 > a > a > · · · > a N > a N +1 ≥
0. Define f : [0 , → R ; x x ( ζ ψ )( x ); g : [0 , → R ; x x ( ζψ )( x ); F : [0 , → R ; t Z t τ f dτ. Then F : [0 , → [0 , + ∞ ) is a bijection with inverse F − . Define the strictly increasing function J : [0 , → R ; t g ◦ F − . Put t h := F ( a h ) for h = 0 , . . . , N + 1. Then + ∞ > t > t > · · · > t N > t N +1 ≥
0. Put B := B (0 , r ) where r := F − ( v/ π ) so that V f ( B ) = v . Note that v = V f ( E ) = 2 π N X h =0 n F ( a h ) − F ( a h +1 ) o = 2 π N +1 X h =0 ( − h t h . By Lemma 10.6, P g ( E ) = 2 π N +1 X h =0 g ( a h ) = 2 π N +1 X h =0 J ( t h ) ≥ πJ ( N +1 X h =0 ( − h t h ) = 2 πJ ( v/ π ) = P g ( B ) . Proof of Theorem 1.2.
Let v > E be a minimiser for (1.5). Then E is essentially boundedby Theorem 3.9. By Theorem 4.5 there exists an L -measurable set e E with the properties(a) e E is a minimiser of (1.5);(b) L e E = L E a.e. on (0 , e E is open and has C , boundary in U ;(d) e E \ { } = e E sc . (i) Suppose that 0 < v ≤ v so that R >
0. Choose r ∈ (0 , R ] such that V ( B (0 , r )) = V ( E ) = v .Suppose that e E \ B (0 , R ) = ∅ . By Lemma 10.5 there exists t > R such that S t ⊂ M . As thefunction ζψ is strictly increasing, g | S t > g | S r . So P g ( E ) = P g ( e E ) ≥ πg | S t > πg | S r = P g ( B (0 , r )).This contradicts the fact that E is a minimiser for (1.5). So e E ⊂ B (0 , R ) and L e E = 0 on ( R, | E \ B (0 , R ) | = 0. By the uniqueness property in the classical isoperimetrictheorem (see for example [15] Theorem 4.11) the set E is equivalent to a ball B in B (0 , R ). (ii) With r ∈ (0 ,
1) as before, V ( B (0 , r )) = V ( E ) = v > v = V ( B (0 , R )) so r > R . If e E \ B (0 , r ) = ∅ we derive a contradiction in the same way as above. Consequently, e E = B := B (0 , r ). Thus, L E = L B a.e. on (0 , | E \ B | = 0. This entails that E is equivalent to B .48 eferences [1] Ambrosio, L., Fusco, N., Pallara, D., Functions of bounded variation and free discontinuityproblems , Oxford University Press, 2000.[2] Barchiesi, M., Cagnetti, F., Fusco, N.,
Stability of the Steiner symmetrization of convex sets ,J. Eur. Math. Soc. 15 (2013) 1245-1278.[3] Borell, C.,
The Ornstein-Uhlenbeck velocity process in backward time and isoperimetry ,Chalmers Tekniska H´ogskola, G¨oteborgs Universitet, 1986.[4] Boyer, W., Brown, B., Chambers, G., Loving, A., Tammen, S.,
Isoperimetric regions in R n with density r p , Analysis and Geometry in Metric Spaces, ISSN (Online) 2299-3274. (2016).[5] Betta, F.M., Brock, F., Mercaldo, A., Posteraro, M.R., Weighted isoperimetric inequalities on R n and applications to rearrangements , Math. Nachr. 281, 4 (2008) 466-498.[6] Buttazzo, G., Giaquinta, M., Hildebrandt, S., One-dimensional variational problems , OxfordUniversity Press, 1998.[7] Carothers, N.L.,
Real analysis , Cambridge University Press, 2000.[8] Chambers, G.R.,
Proof of the log-convex density conjecture , to appear in J. Eur. Math. Soc.(2016).[9] Cinti, E., Pratelli, A.,
Regularity of isoperimetric sets in R with density , to appear in Math.Ann. (2017).[10] De Giorgi, E., Sulla propriet´a isoperimetrica dell’ipersfera, nella classe degli insiemi aventifrontiera orientata di misura finita , Atti Accad. Naz. Lincei Mem. Cl. Sci. Fis. Mat. Mat. Sez.I,5 (1958) 33-44.[11] Di Giosia, L., Habib, J., Kenigsberg, L., Pittman, D., Zhu, W.,
The log convex densityconjecture in hyperbolic space , Rose-Hulman Undergrad. Math. J. 18, 1 (2017)[12] Do Carmo, M.,
Differential geometry of curves and surfaces , Prentice-Hall, Inc., 1976.[13] Dubins, L.E.,
On curves of minimal length with a constraint on average curvature, and withprescribed initial and terminal positions and tangents , Amer. J. Math. 79, 3 (1957) 497-516[14] Figalli, A., Maggi, F.,
On the isoperimetric problem for radial log-convex densities , Calc. Var.Partial Differential Equations 48, 3-4 (2013) 447-489.[15] Fusco, N.,
The classical isoperimetric theorem , Rend. Accad. Sci. Fis. Mat. Napoli (4) 71(2004) 63-107.[16] Giusti, E.,
Minimal surfaces and functions of bounded variation , Birkh¨auser, 1984.[17] Gradshteyn, I.S., Ryzhik, I.M.,
Tables of integrals, series and products , Academic Press, 1965.[18] Hadamard, J., ´Etude sur les propri´et´es des fonctions entieres et en particulier d’une fonctionconsid´er´ee par Riemann , J. Math. Pures et Appl., 58 (1893) 171-215.[19] Hale, J.,
Ordinary differential equations , Wiley, 1969.[20] Hermite, C.,
Sur deux limites d’une int´egrale d´efinie , Mathesis 3, 82 (1883).[21] Howard, H., Treibergs, A.,
A reverse isoperimetric inequality, stability and extremal theoremsfor plane curves with bounded curvature , Rocky Mountain J. Math. 25, 2 (1995) 635-684[22] Howe, S.,
The log-convex density conjecture and vertical surface area in warped products , Adv.Geom. 15, 4 (2015) 4923] Kolesnikov, A.V., Zhdanov, R.I.,
On isoperimetric sets of radially symmetric measures , Con-centration, functional inequalities and isoperimetry , 123-154, Contemp. Math., 545, AmericanMathematical Society, 2011.[24] Lee, J.M., Manifolds and differential geometry, American Mathematical Society, 2009.[25] Maggi, F.,
Sets of finite perimeter and geometric variational problems , Cambridge UniversityPress, 2012.[26] McGillivray, I.,
An isoperimetric inequality in the plane with a log-convex density , (submitted)(2017)[27] Morgan, F.,
Regularity of isoperimetric hypersurfaces in Riemannian manifolds , Trans. AMS,355, 12 (2003) 5041-5052.[28] Morgan, F., Hutchings, M., Howards, H.,
The isoperimetric problem on surfaces of revolutionof decreasing Gauss curvature , Trans. Amer. Math. Soc. 352 (2000) 11, 4889-4909[29] Morgan, F., Pratelli, A.,
Existence of isoperimetric regions in R n with density , Ann. GlobalAnal. Geom. 43, 4 (2013) 331-365.[30] Perugini, M., PhD Thesis (in preparation), Univ. Sussex, 2016.[31] Rosales, C., Ca˜nete, A., Bale, V., Morgan, F., On the isoperimetric problem in Euclideanspace with density , Calc. Var. Part. Diff. Eqns. 31, 1 (2008) 27-46.[32] Simmons, G.F.,
Introduction to topology and modern analysis , McGraw-Hill, 1963.[33] Spivak, M.,
A comprehensive introduction to differential geometry, volume 2 , Publish or Per-ish, 1999.[34] Tamanini, I.,
Regularity results for almost minimal oriented hypersurfaces in R nn