AA NOTE ON THE SQUEEZING FUNCTION
ALEXANDER YU. SOLYNIN
Abstract.
The squeezing problem on C can be stated as follows. Supposethat Ω is a multiply connected domain in the unit disk D containing the origin z = 0. How far can the boundary of Ω be pushed from the origin by aninjective holomorphic function f : Ω → D keeping the origin fixed?In this note, we discuss recent results on this problem obtained by Ng, Tangand Tsai (Math. Anal. 2020) and by Gumenyuk and Roth (arXiv:2011.13734,2020) and also prove few new results using a method suggested in one of ourprevious papers (Zapiski Nauchn. Sem. POMI 1993). The squeezing problem.
The squeezing function S Ω ( z ) of a planar domain Ω isdefined as follows. Suppose that Ω ⊂ C is such that there is an injective holomorphicfunction f ( z ) from Ω to the unit disk D = { z : | z | < } . Let U (Ω) be the class ofall such functions. Then S Ω : Ω → R is defined by S Ω ( z ) = sup { dist(0 , ∂f (Ω)) : f ∈ U (Ω) , f ( z ) = 0 } . (1)According to [8], the squeezing function was first introduced in 2012 by Dong,Guan and Zhang [8] but the concept itself goes back to the work of Liu, Sun andYau [7]. These authors defined S Ω and used it in a more general setting, namely,for classes of injective holomorphic mappings defined on domains Ω in C n , n ≥ squeezing problem , i.e. the problem to find or characterize S Ω ( z ), is a difficulttask, even in complex dimension 1. Any function f ∈ U (Ω) such that f ( z ) = 0 and S Ω ( z ) = dist(0 , ∂f (Ω)) will be called extremal for the squeezing problem (for thepoint z ∈ Ω). In this case, the image f (Ω) will be called an extremal domain.For doubly connected domains Ω ⊂ C , the squeezing function was identified byNg, Tang and Tsai in 2020 [8]. To state the main result of [8], we first introducenecessary notations. As well known, see [4, Theorem 5.4], every domain Ω ⊂ C with a finite number of boundary continua γ , . . . , γ n , n ≥
2, can be mapped by w = ϕ a,k ( z ), a ∈ Ω, 1 ≤ k ≤ n , conformally on D slit along arcs C j,k = C j,k ( a ), j (cid:54) = k , on the circles of radii 0 < r j,k = r j,k ( a ) < ϕ a,k ( a ) = 0, anon-degenerate boundary continuum γ k corresponds to the unit circle T = ∂ D and,for j (cid:54) = k , γ j corresponds to C j,k . Under the additional normalization ϕ (cid:48) a,k ( a ) > ϕ a,k is uniquely determined.If γ k is a singleton, we put r j,k ( a ) = 0 for all a ∈ Ω and all j (cid:54) = k . Withoutthe uniqueness statement, the mapping property, discussed above, remains true fordomains of any connectivity. Theorem 1 ([8],[3]) . Let Ω ⊂ C be a doubly connected domain with boundarycomponents γ and γ , at least one of which is non-degenerate. Then S Ω ( z ) = max { r , ( z ) , r , ( z ) } . (2) If r , ( z ) < r , ( z ) , then ϕ z, is the unique (up to rotation about the origin) extremalfunction for the squeezing problem; if r , ( z ) > r , ( z ) , then ϕ z, is the uniqueextremal function; and if r , ( z ) = r , ( z ) , then both ϕ z, and ϕ z, are extremal. Mathematics Subject Classification.
Primary 30C75; Secondary 30C35.
Key words and phrases.
Squeezing function, circularly slit disk, doubly connected domain,Jenkins’s module problem. a r X i v : . [ m a t h . C V ] J a n A. YU. SOLYNIN
This theorem, except for the uniqueness part, was proved in [8] using the Loewnerdifferential equation and tricky calculations with special functions. A simpler proofbased on the potential theory, which also includes the proof of the uniquenessstatement, was presented in [3].In this note, we first show that Theorem 1 is immediate from Theorem 1.2 inour 1993 paper [10]. Then we discuss how the method used in [10] can be appliedto study the squeezing problem in a more general setting.Let 0 < r ≤ r < E be a compact subset in the closure ¯ A ( r , r ) ofthe annulus A ( r , r ) = { z : r < | z | < r } . Let D ( E ) = { ( D , D ) } be the set ofpairs ( D , D ) of non-overlapping domains in D \ E , where D is a simply connecteddomain containing the origin and D is a doubly connected domain separating E from T . In what follows, m ( D, z ) stands for the reduced module of a simplyconnected domain D with respect to the point z ∈ D and m ( D ) stands for themodule of a doubly connected domain D . For the definitions and properties ofthese moduli, we refer to Jenkins’s monograph [4] as the primary source and alsoto [6], [2], and [11] . Figure 1 illustrates our notations introduced above.Consider the following module problem . Problem M.
Given nonnegative numbers α and α , at least one of which ispositive, identify all pairs ( D ∗ , D ∗ ) ∈ D ( E ) , which maximize the weighted sum ofmoduli α m ( D ,
0) + α m ( D ) (3) over the set D ( E ) . Theorem 2 ([10, Theorem 1.2]) . (1) There is a unique pair ( D ∗ , D ∗ ) ∈ D ( E ) maximizing the sum (3) over the class D ( E ) . The domains D ∗ and D ∗ are, re-spectively, a circle domain and a ring domain of a quadratic differential Q ( z ) dz defined on D \ E , which is positive on T and has a second order pole with circularlocal structure of trajectories at z = 0 .(2) Let L ( E ) denote the free boundary of the module problem; i.e. L ( E ) =( ∂D ∗ ∪ ∂D ∗ ) ∩ ( D \ E ) . Then L ( E ) consists of arcs of critical trajectories of Q ( z ) dz and their endpoints in D \ E .(3) The following holds: (a) if α ≤ α , then L ( E ) ⊂ ¯ D r , where D r = { z : | z | Proof of Theorem 1. Using an auxiliary conformal mapping, we may assume withoutloss of generality that Ω is the unit disk D slit along a proper arc C , of the circle { z : | z | = r , } , that z = 0, and that r , ≥ r , . Consider Problem M with α = α = 1 for the set E = C , . We call it Problem P . It follows from Gr¨otzsch’slemma [4, Theorem 2.6] (or from Theorem 2(3) above) that ( D r , , A ( r , , P and the complementary arc C (cid:48) , = { z : | z | = r , } \ C , is the free boundary of Problem P .Now, suppose, by contradiction, that S Ω (0) > r , . The latter implies that thereis a function f ∈ U ( D \ C , ) such that f (0) = 0 and dist(0 , f ( C , )) > r , . Let ρ = inf {| z | : z ∈ f ( C , ) } , ρ = sup {| z | : z ∈ f ( C , ) } . Consider Problem M,again with α = α = 1, for the set f ( C , ) ⊂ ¯ A ( ρ , ρ ) } . We call it Problem P .Since Jenkins’s problem on extremal partitioning is conformally invariant (in the NOTE ON THE SQUEEZING FUNCTION 3 Figure 1. Extremal partitioning into two domains.sense that conformal mappings preserve extremal configurations), it follows that thepair ( f ( D r , ) , f ( A ( r , , P . Also, conformal mappingspreserve the module of a doubly connected domain. Hence, m ( f ( A ( r , , m ( A ( r , , − π log r , . (4)Now, it follows from Theorem 2(3) that the free boundary of Problem P liesin the annulus ¯ A ( ρ , ρ ). This implies that f ( A ( r , , ⊂ A ( ρ , m ( f ( A ( r , , ≤ m ( A ( ρ , − π log ρ . Since r , < ρ , the latter inequalitycontradicts (4). Thus, our assumption leads to a contradiction and therefore wemust have S Ω (0) = r , .It remains to show that rotations about the origin are the only extremal functionsfor the problem under consideration. If f is not a rotation, then our argumentabove implies that f ( A ( r , , A ( r , , m ( f ( A ( r , , < m ( A ( r , , − π log r , , which again contradicts equation(4). The proof is complete. (cid:3) More results and questions. Let Ω be a domain (of any connectivity) and let γ be a nondegenerate boundary continuum of Ω separated from the rest of ∂ Ω, i.e.such that γ ∩ ( ∂ Ω \ γ ) = ∅ . If Ω is finitely connected, the latter separation propertyalways holds. Then, for each z ∈ Ω, we can consider a non-empty class U γ (Ω) offunctions f ∈ U (Ω) such that f ( γ ) = T , in the sense of boundary correspondence.Then the function S Ω ,γ : Ω → R defined by S Ω ,γ ( z ) = sup { dist(0 , ∂f (Ω)) : f ∈ U γ (Ω) , f ( z ) = 0 } (5)can be thought as the squeezing function toward the boundary continuum γ . Forthis function we have the following result. Lemma 1. Let Ω and γ be as above and let a ∈ Ω . Suppose that there is a function f a,γ ∈ U γ (Ω) such that f a,γ ( a ) = 0 and D \ f a,γ (Ω) ⊂ { z : | z | = r } , < r < .Then S Ω ,γ ( a ) = r .Furthermore, f a,γ is a unique (up to rotation about ) function in U γ (Ω) that isextremal for the problem on the squeezing toward γ for the point a . A. YU. SOLYNIN Proof. Since the squeezing problem is conformally invariant, we may assume thatΩ is the disk D slit along a proper compact subset of the circle T r = { z : | z | = r } ,0 < r < 1. Then our argument used in the proof of Theorem 3 shows that if f ∈ U T (Ω) such that f (0) = 0 is extremal for the squeezing toward T problem,then D \ f (Ω) ⊂ T r . Furthermore, same argument shows that f ( D r ) = D r . Since f (0) = 0, the latter implies that f is a rotation about the origin. (cid:3) Below, we assume that Ω is a finitely-connected domain with n ≥ γ , . . . , γ n . In this case, the following properties are eitherexist in the literature or easy to prove:(1) Existence of extremal functions. For each z ∈ Ω and 1 ≤ k ≤ n , thereis an extremal function for the problem on squeezing toward γ k ; i.e. a func-tion f z,k ∈ U γ k (Ω) such that f z,k ( z ) = 0 and S Ω ,γ k ( z ) = dist(0 , ∂f z,k (Ω)).Therefore, for each z ∈ Ω, there is a function f Ω ∈ U (Ω) extremal for thesqueezing problem on Ω.(2) Continuity. The functions S Ω ( z ) and S Ω ,γ k ( z ), k = 1 , . . . , n , are continu-ous on Ω.(3) Monotonicity of S Ω ,γ n ( z ) with respect to γ n . For a fixed z ∈ Ω, S Ω ,γ n ( z ) is monotone with respect to γ n in the following sense. Let Ω (cid:48) (cid:54) = Ωbe a finitely connected domain with boundary continua γ (cid:48) , . . . , γ (cid:48) n (cid:48) , 2 ≤ n (cid:48) ≤ n , such that a ∈ Ω (cid:48) ⊂ Ω and γ (cid:48) k = γ k for k = 1 , . . . , n (cid:48) − 1. Then S Ω (cid:48) ,γ (cid:48) n (cid:48) ( a ) > S Ω ,γ n ( a ) . (4) Non-monotonicity of S Ω ( z ) with respect to Ω . For a fixed z ∈ Ω, S Ω ( z ) is not monotone as a function of the domain.(5) Boundary values of S Ω ( z ) . S Ω ,γ k ( z ) → z → γ k and therefore S Ω ( z ) → z → ∂ Ω.(6) Boundary values of S Ω ,γ k ( z ) on γ j , j (cid:54) = k . For each k and j (cid:54) = k , thereis 0 < c j,k < z → γ j S Ω ,γ k ( z ) ≤ c j,k . (7) Separation property 1. If c , 0 < c < 1, is sufficiently close to 1, thenthe level set { z ∈ Ω : S Ω ,γ k ( z ) = c } separates γ k from ∂ Ω \ γ k .(8) Separation property 2. Let j (cid:54) = k . The set { z ∈ Ω : S Ω ,γ j ( z ) = S Ω ,γ k ( z ) } separates γ j from γ k inside Ω.(1) The existence of extremal functions and therefore existence of extremal do-mains as well was established in [1, Theorem 2.1].(2) To prove the continuity property, one can use extremal functions f z, Ω ( ζ )composed with the Moebius automorphisms of the unit disk D . The details are leftto the interested reader.(3) To prove the monotonicity property of the squeezing function S Ω ,γ n ( z ) definedby (5), we consider the composition ϕ ◦ f Ω ,γ n of the extremal function f Ω ,γ n withthe function ϕ that is the Riemann mapping function from a simply connecteddomain with the boundary continuum γ (cid:48) n (cid:48) such that ϕ (0) = 0. Then the desiredresult follows from the Schwarz’s lemma.(4) To show that the monotonicity with respect to Ω is absent, consider twosimple examples. Let Ω be a circularly slit disk D \ { re iθ : | θ | ≤ α } with 0 < r < < α < π . For 0 < r < r , r < r < 1, let Ω = Ω \ [ r , r ], Ω = Ω \ [ r, r ]. Thenthe argument involving Theorem 2, as it was used in the proof of Theorem 1, showsthat S Ω (0) < S Ω (0) < S Ω (0). Therefore, the monotonicity property of S Ω ( z ) asa function of the domain does not hold, in general. NOTE ON THE SQUEEZING FUNCTION 5 Figure 2. Mapping onto a circularly slit disk.(5) To show that S Ω ,γ k ( z ) → z → γ k , we may assume that Ω ⊂ D and γ k = T . Then the Moebius mapping ϕ ( ζ ) = ( ζ − z ) / (1 − ¯ zζ ) is in U γ k (Ω) and it iseasy to see that that dist(0 , ∂ϕ (Ω)) → | z | → l be a Jordan curve in Ω separating γ j from ∂ Ω \ γ j and let Ω l be adoubly connected domain with boundary components γ j and l . It follows from theproperty (3) that S Ω ( z ) < S Ω l ( z ) for all z ∈ Ω l .Fix z ∈ Ω l . Let m = m ( D l ) be the module of the doubly connected domain D l .Then there is a function ϕ that maps Ω l conformally onto the annulus A ( s, 1) with s = e − πm . In addition, we may assume that ϕ ( z ) = a , s < a < 1. Furthermore,there exists a function ψ , such that ψ ( a ) = 0, which maps A ( s, 1) conformally ontothe disk D slit along and arc C on a circle T ρ , with 0 < ρ < 1, where ρ = ρ ( a )depends on a . The mapping ψ is one of the well-studied canonical mappings. A par-ticular property of ψ , we need here, was proved by E. Reich and S. E. Warschawskiin 1960 [9, Lemma 3]. These authors showed that ρ ( a ) = a . This result impliesthat a → s as z → γ j . It follows from Theorem 1 that S Ω l ( z ) = ρ ( a ) = a . This,being combined with property (3), implies thatlim sup z → γ j ≤ lim a → s a = s = e − πm < , as required.(7) Let c k = max j (cid:54) = k { c j,k } , where c j,k introduced in part (6) above. Take c , c k < c < 1, and consider the level set L ( c ) = { z ∈ Ω : S Ω ,γ k ( z ) = c } . Since S Ω ,γ k iscontinuous on Ω and has boundary value 1 on γ k and boundary values less than c on other boundary continua, it follows that L ( c ) separates γ k from the rest of ∂ Ω.(8) The same argument, which was used in part (7), can be used to prove theseparation property in question as well.Let Ω be an n -connected domain as above. Then, for each z ∈ Ω, there are n functions ϕ z,k , 1 ≤ k ≤ n , and for each k there are n − r j,k = r j,k ( z ) ofcircular arcs C j,k = C j,k ( z ), as described above. An example of the domain Ω andits image under one of the mappings ϕ z,k are shown in Figure 2. It was conjecturedin [8] that S Ω ( z ) = max k min j (cid:54) = k r j,k ( z ) . (6)This conjecture, though it sounds plausible, was quickly disproved by Gumenyukand Roth [3], whose engineering construction shows that for every n ≥ n -connected domain Ω, that is not a circularly slit disk, and there is a point z ∈ Ω such that S Ω ( z ) is strictly greater than the right-hand side of (6). This showsthat, in general, the family of domains extremal for the squeezing problem is notlimited to the set of circularly slit disks. In [8] and [3], the question whether or A. YU. SOLYNIN not there exist circularly slit disks of connectivity n ≥ Theorem 3. For each n ≥ , there is a domain Ω , that is a circularly slit disk ofconnectivity n , such that Ω is the extremal domain, unique up to rotation about theorigin, for the squeezing problem for the point z = 0 .Proof. Let n ≥ 3, 0 < r < 1, and 0 < α < π/ ( n − γ k = { z = re iθ : | θ − π ( k − / ( n − | ≤ α } for 1 ≤ k ≤ n − γ n = T . Let Ω = Ω( n, r, α )denote the disk D slit along the arcs Γ k , k = 1 , . . . , n − S Ω ,γ n (0) = r . Furthermore, since Ω possesses( n − ρ , 0 < ρ < 1, such that S Ω ,γ k (0) = ρ for k = 1 , . . . , n − n and r , ρ = ρ ( α ) → α → 0. To emphasizedependence on α , we use notations Ω = Ω( α ), γ k = γ k ( α ), etc. To prove the claim,we consider a circle Γ ε = { z : | z − r | = ε } with ε > α issmall, then Γ ε separates γ ( α ) from all other boundary components of Ω( α ). Let D ε ( α ) denote the doubly connected domain bounded by Γ ε and γ ( α ). It is clearthat m ( D ε ( α )) → ∞ as α → 0. Since the module of a doubly connected domainis conformally invariant, it follows that for every δ > α > f (Γ ε ) ⊂ D δ , whenever 0 < α < α and f ∈ U γ ( α ) (Ω( α )) is such that f (0) = 0.Since the images f ( T ) and f ( γ k ( α )), k = 2 , . . . , n − 1, lie in the bounded componentof C \ f (Γ ε ), the claim follows. Therefore, ρ ( α ) < r if α is small enough. Hence S Ω( α ) (0) = r for all such α and Ω( α ) is the only extremal domain up to rotationabout the origin. The proof is complete. (cid:3) Example. Given a > 1, let γ k = γ k ( a ) = { z = te πi ( k − / : 1 ≤ t ≤ a } , k = 1 , , 3. Then Ω = C \ ∪ k =1 γ k is a domain on C of connectivity 3. Let ϕ , mapΩ conformally onto a circularly slit disk D such that ϕ , ( γ ) = T , ϕ , (0) = 0,and ϕ (cid:48) , (0) > 0. Since Ω is invariant under rotation by angle 2 π/ ϕ , ( γ ) and ϕ , ( γ ) are circular arcs lying on thesame circle symmetrically to each other with respect to the real axis. Therefore, itfollows from Theorem 3 that ϕ , is extremal for the squeezing toward γ problemfor the point 0. Since Ω is invariant under rotations by the angle 2 π/ 3, it follows thateach of the other two functions, ϕ , and ϕ , , where ϕ ,k maps γ k to the unit circle T such that ϕ ,k (0) = 0, k = 2 , 3, is also extremal for the squeezing problem toward,respectively, γ or γ , for the point 0. Moreover, S Ω ,γ (0) = S Ω ,γ (0) = S Ω ,γ (0).Thus, this example shows that there are domains Ω of connectivity n > z ∈ Ω, such that every function f extremal for the squeezing problem for Ωand z maps Ω onto a circularly slit disk. Questions. We finish this part with six questions for future study.(1) As Theorem 1 shows, if Ω is doubly connected, then every function f ex-tremal for the squeezing problem for some point z ∈ Ω maps Ω onto acircularly slit disk. We suspect that this property characterizes doublyconnected domains. So, the question: Is it true that if there is an opensubset G of Ω, such that any function f extremal for the squeezing prob-lem for some z ∈ G maps Ω onto a circularly slit disk, then Ω is doublyconnected?(2) A point z in a domain Ω with boundary continua γ , . . . , γ n , n ≥ 2, isan equilibrium point for the squeezing problem if S Ω ,γ ( z ) = S Ω ,γ ( z ) = . . . = S Ω ,γ n ( z ). An annulus A ( r, 1) has the circle { z : | z | = √ r } as its set ofequilibrium points. A domain D n ( a ) = C \ ∪ nk =1 { te πi ( k − /n : 1 ≤ t ≤ a } NOTE ON THE SQUEEZING FUNCTION 7 with a > n ≥ z = 0 and z = ∞ . Let D n ( a, R ) = D n ( a ) ∩ D R , where R > a . There is a unique value R > a suchthat D n ( a, R ) has exactly one equilibrium point z = 0, while, for R (cid:54) = R , D n ( a, R ) does not have equilibrium points.Is it true that every domain Ω of connectivity n ≥ γ n = T and γ k ⊂ D , k = 1 , . . . , n , n ≥ 3, that are extremal for the squeezing problem for z = 0,such that γ j and γ k lie on different circles when j (cid:54) = k ?(4) Are there domains Ω ⊂ D of connectivity n + m +1 with boundary continua γ n + m +1 = T and γ k ⊂ D , k = 1 , . . . , n + m , such that γ k is an arc ofa circle centered at 0 when 1 ≤ k ≤ n , γ k is not a circular arc when n + 1 ≤ k ≤ n + m , that are extremal for the squeezing problem for z = 0?(5) Is it possible to characterize domains Ω of connectivity n ≥ 3, that areextremal for the squeezing problem for a point z ∈ Ω, in terms of qua-dratic differentials similar to characterization of configurations of domainsextremal for Jenkins’s problem [5] on extremal partitioning?(6) Our proof of Theorem 2 on the extremal partitioning of the disk D with acompact barrier E ⊂ D can be extended to higher dimensions for the unitball B ⊂ R n , n ≥ 3, with a compact set E ⊂ B .It would be interesting to know whether such a generalization can beused to study the squeezing problem in C n . Proof of Theorem 2. Problem M is a particular case of Jenkins’s problemon extremal partitioning [5]. Therefore, parts (1) and (2) of Theorem 2 followfrom [5, Theorem 1]. Furthermore, same Jenkins’s theorem implies that the metric ρ ( z ) | dz | = α − | Q ( z ) | / | dz | is extremal for the module problem for the familyof closed curves γ ⊂ D ∗ separating z = 0 from ∂D ∗ and the metric ρ ( z ) | dz | = α − | Q ( z ) | / | dz | is extremal for the module problem for the family of closed curves γ ⊂ D ∗ separating T from ∂D ∗ \ T . Moreover, it implies that (cid:90) γ | Q ( z ) | , | dz | ≥ α (7)for every Jordan curve γ ⊂ D \ E , which separates T from E and 0 and also for γ = T .Let us we consider the case α ≤ α , assuming that E consists of a finite numberof connected components. In this case D ∗ (cid:54) = ∅ . Otherwise, T would be on theboundary of D ∗ , which implies that (cid:90) T | Q ( z ) | / | dz | < α ≤ α , contradicting equation (7). Furthermore, since E consists of a finite number ofcomponents, it follows that the free boundary L ( E ) consists of a finite number ofcritical trajectories of Q ( z ) dz and their endpoints in D \ E .Suppose by contradiction that r < ρ = max {| z | : z ∈ L ( E ) } < . (8)Since L ( E ) consists of a finite number of analytic arcs, it follows that the inter-section L ( E ) ∩ T ρ consists of a finite number of points z k , k = 1 , . . . , n . It followsfrom the local structure of trajectories of Q ( z ) dz near critical points, that each ofthe points z k is regular. This implies that, for all sufficiently small ε > 0, the set L ( E ) ∩ ¯ A ( ρ − ε, ρ ) consists of n disjoint analytic arcs s k , k = 1 , . . . , n , such that s k has its endpoints on the circle T ρ − ε and the point z k is an interior point of s k . A. YU. SOLYNIN Let ˜ s k denote the arc symmetric to s k with respect to the circle T ρ − ε . For every ε small enough and each k , the intersection ˜ s k ∩ L ( E ) is empty. In this case, there are n simply connected domains ∆ k , symmetric with respect to the circle T ρ − ε , suchthat ∂ ∆ k = s k ∪ ˜ s k , k = 1 , . . . , n .We claim that there is no k such that ∆ k ⊂ D ∗ . To prove this claim, suppose that∆ k j ⊂ D ∗ for j = 1 , . . . , n , n ≤ n . Let ∂ D ∗ = T and ∂ D ∗ denote the boundarycomponents of D ∗ . Let D p denote polarization of the doubly connected domain D ∗ (considered as a condenser whose plates are connected components of C \ D ∗ ) withrespect to the circle T ρ − ε . In the case under consideration, the polarized domain D p is a doubly connected domain having T as one of its boundary componentswhile the other boundary component of D p is obtained from ∂ D ∗ by replacingthe arcs s k j with the arcs ˜ s k j , j = 1 , . . . , n . For the definition and properties ofpolarization, we refer to [2, Chapter 3]. As well known, polarization increases themodule of a doubly connected domain. Thus, in our case, we have m ( D ∗ ) < m ( D p ) , with the sign of strict inequality because D p does not coincide with D ∗ up toreflection with respect to T ρ − ε . Therefore, α m ( D ∗ , 0) + α m ( D ∗ ) < α m ( D ∗ , 0) + α m ( D p ) . Since the pair ( D ∗ , D p ) is admissible for Problem M, i.e. ( D ∗ , D p ) ∈ D ( E ), inequal-ity (8) contradicts our assumption that the pair of domains ( D ∗ , D ∗ ) is extremalfor Problem M.Now we consider the case when ∆ k ⊂ D ∗ for all k = 1 , . . . , n . In this case,we consider a pair ( D , D ), where D = D ∗ \ ¯ A ( ρ − ε, ρ ) is a simply connecteddomain and D = D ∗ ∪ A ( ρ − ε, ρ ) ∪ T ρ is a doubly connected domain. One caneasily see that ( D , D ) ∈ D ( E ). Let ρ k ( z ) | dz | denote the extremal metric for thecorresponding module problem for the domain D k , k = 1 , 2; see [4, Chapter II].Since D ∗ ⊂ D , it follows that ρ ( z ) | dz | is admissible for the module problemfor D ∗ but it is not extremal for this problem. Therefore, m ( D ∗ ) < (cid:90) (cid:90) D ρ ( z ) dA − (cid:90) (cid:90) ∪ Nk =1 ∆ k ρ ( z ) dA (9)= m ( D ) − (cid:90) (cid:90) ∪ Nk =1 ∆ k ρ ( z ) dA. Next, we consider a metric ρ ( z ) | dz | on the domain D ∗ defined by ρ ( z ) = (cid:40) ρ ( z ) if z ∈ D , ρ ( z ) if z ∈ D ∗ \ D .We claim that ρ ( z ) | dz | is admissible for the problem on the reduced modulefor the domain D ∗ . To prove this, we consider an analytic Jordan curve γ ⊂ D ∗ separating 0 from ∂D ∗ . If γ ⊂ ¯ D , then (cid:90) γ ρ ( z ) | dz | = (cid:90) γ ρ ( z ) | dz | ≥ ρ ( z ) | dz | is admissible for the problem on the reduced module of D .Suppose now that γ (cid:54)⊂ ¯ D . Then the circle T ρ − ε divides γ into a finite number ofarcs. By τ k , k = 1 , . . . , m , we denote those of them, which lie in D . Let σ k ⊂ D ∗ denote the closed arc of T ρ − ε joining the endpoints of τ k , k = 1 , . . . , m . It followsfrom Lemma 2, presented below, that (cid:90) τ k ρ ( z ) | dz | ≥ (cid:90) σ k ρ ( z ) | dz | ≥ (cid:90) σ k ρ ( z ) | dz | . (11) NOTE ON THE SQUEEZING FUNCTION 9 Let ˜ γ denote the curve obtained from γ by replacing the arcs τ k with the arcs σ k , k = 1 , . . . , m . Then ˜ γ is the curve in ¯ D , that is not Jordan in general, thatseparates 0 from ∂D . Hence, (cid:90) ˜ γ ρ ( z ) | dz | ≥ . (12)Combining equations (11) and (12), we conclude that (cid:90) γ ρ ( z ) | dz | ≥ (cid:90) ˜ γ ρ ( z ) | dz | ≥ . (13)Equations (10) and (13) imply that ρ ( z ) | dz | is an admissible metric for the problemon the reduced module in D ∗ . Hence, m ( D ∗ , ≤ lim (cid:15) → (cid:40)(cid:90) (cid:90) D ∗ \ ¯ D (cid:15) ρ ( z ) dA + 12 π log (cid:15) (cid:41) (14)= lim (cid:15) → (cid:40)(cid:90) (cid:90) D \ ¯ D (cid:15) ρ ( z ) dA + (cid:90) (cid:90) ∪ Nk =1 ∆ k ρ ( z ) dA + 12 π log (cid:15) (cid:41) = m ( D , 0) + (cid:90) (cid:90) ∪ Nk =1 ∆ k ρ ( z ) dA. Combining (9) and (14), we obtain the following inequalities: α m ( D ∗ , 0) + α m ( D ∗ ) < α m ( D , 0) + α m ( D )+ ( α − α ) (cid:90) (cid:90) ∪ Nk =1 ∆ k ρ ( z ) dA ≤ α m ( D , 0) + α m ( D ) . The latter inequalities contradict to the assumption that the pair ( D ∗ , D ∗ ) is ex-tremal for Problem M. This contradiction shows that our assumption R > r iswrong and therefore, L ( E ) ⊂ ¯ D r , as required.The latter inclusion is proved in the case when E consists of a finite number ofcomponents. In the general case, we approximate E with a sequence of compact sets E j , j = 1 , , . . . , such that E j +1 ⊂ E j and such that E j is bounded by appropriatelevel curves of Green’s function g D \ E ( z, 0) of the domain D \ E with pole at 0.Then the required inclusion L ( E ) ⊂ ¯ D r will follow from the result already provedfor sets E with finite number of connected components and from Carath´eodory’sconvergence theorem for simply connected and doubly connected domains.The proof presented above, can be easily modified to show that if α ≥ α , then L ( E ) ⊂ ¯ A ( r , (cid:3) Lemma 2. Let σ be an open arc on T . Let D k , k = 1 , , be a doubly connecteddomain having the circle T r , < r < , as one of its boundary components.Suppose that the other boundary component of D , call it γ , is such that l ⊂ γ and D ⊂ A ( r, . Suppose further that the other boundary component of D , call it γ , is such that l ⊂ γ and A ( r, ⊂ D . Let ρ k ( z ) | dz | denote the extremal metricof the module problem in D k , k = 1 , . Then ρ ( e iθ ) ≤ / π ≤ ρ ( e iθ ) for all e iθ ∈ σ . (15) Proof. Let us prove the first inequality. Let f maps D conformally onto theannulus A ( r , 1) such that f ( γ ) = T . Since the module of a doubly connecteddomain increases under expansion, the following holds: r < r < 1. In terms of themapping function, the extremal metric can be expressed as follows (see [4, ChapterII]): ρ ( z ) | dz | = 12 π | f (cid:48) ( z ) || f ( z ) | | dz | . This shows that the desired result will follow if we prove thatmeas( f ( σ )) ≤ meas( σ ) (16)for every σ ⊂ T , if the assumptions of the lemma are satisfied.To prove (16), consider a family of curves Γ ( σ ) = { γ } consisting of all rectifiablearcs γ joining T r and σ inside the domain D . Let Γ( s, σ ) denote a similar familyof curves in the annulus A ( s, < s < 1. Since the module increases underexpansion of the family of curves, we havemod(Γ ( σ )) < mod(Γ( s, σ )) if D (cid:54) = A ( r, f ( σ )) > meas( σ ). Since the moduleof a family of curves is conformally invariant and since a family of shorter curves hasbigger module than a corresponding family of longer curves, we have the following:mod(Γ ( σ )) = mod(Γ( r , f ( σ ))) > mod(Γ( r, f ( σ ))) ≥ mod(Γ( s, σ )) , (18)contradicting (17). Thus, the inequality (16) is proved and therefore the first in-equality in (15) holds. The proof of the second inequality in (15) follows the samelines. (cid:3) Notice that (15) holds for doubly connected domains D k having the circle T r ofarbitrary small radius 0 < r < r → 0, we conclude that (15) remains valid if D and/or D isa simply connected domain containing 0. In the latter case, ρ k ( z ) | dz | will denotethe extremal metric for the problem on the reduced module m ( D k , Remarks. (1) As Gumenyuk and Roth showed in [3], Theorem 1 on the extremalityof circularly slit disks cannot be extended to domains of connectivity greater thantwo. This result sounds similar to the observation made in [10] about a possibilityto extend Theorem 2 stated above in this note to the case of several barriers E k contained in non-overlapping annuli ¯ A ( r k , r k ), 0 < r ≤ r < r ≤ r < · · · < r n ≤ r n < 1; i.e. such that E k ⊂ ¯ A ( r k , r k ). Precisely, if ( D ∗ , . . . , D ∗ n ) is the configurationof non-overlapping domains maximizing the weighted sum of moduli m ( D , 0) + n (cid:88) k =2 m ( D k ) , then the free boundary L ( E , . . . , E n ) = ∪ nk =1 ∂D ∗ k ∩ ( D \ ∪ nk =1 E k ) is not necessarilycontained in the union ∪ nk =1 ¯ A ( r k , r k ) of these annuli. Here, D ⊂ D \ ∪ nk =1 E k is asimply connected domain containing 0 and D k , k = 2 , . . . , n , is a doubly connecteddomain in D \ ∪ nk =1 E k separating the set ∪ k − j =1 E j from the unit circle T and the set ∪ nj = k E j .(2) Our proof of Theorem 2 relies on the technique, which uses weighted sums ofmoduli of free families of curves developed by J. A. Jenkins and others. As wellknown, a module of a doubly connected domain D with complementary components E and E is the reciprocal of the capacity, when D is considered as the field ofthe condenser with plates E and E . This observation suggests that an approachutilizing properties of capacities and potential functions of condensers also canbe used to study the squeezing problem. An example, demonstrating how thisapproach works, was given in the second proof of Theorem 1.3 in [10]. The proof ofTheorem 2 in [3] is a nice demonstration how this approach based on the potentialtheory can be used in the context of the squeezing problem. 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