A harmonic self-homeomorphism of a disk does not increase the area of any
concentric disk.Comment: 7 page

Koh, Ngin-Tee

Kovalev, Leonid V.

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A harmonic self-homeomorphism of a disk does not increase the area of any concentric disk

Kovalev, Leonid V

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Syracuse University SURFACE Mathematics - Faculty Scholarship Mathematics 11-3-2009 Area Contraction for Harmonic Automorphisms of the Disk Ngin-Tee Koh Syracuse University Leonid V. Kovalev Syracuse University Follow this and additional works at: https://surface.syr.edu/mat  Part of the Mathematics Commons Recommended Citation Koh, Ngin-Tee and Kovalev, Leonid V., "Area Contraction for Harmonic Automorphisms of the Disk" (2009). Mathematics - Faculty Scholarship. 51. https://surface.syr.edu/mat/51 This Article is brought to you for free and open access by the Mathematics at SURFACE. It has been accepted for inclusion in Mathematics - Faculty Scholarship by an authorized administrator of SURFACE. For more information, please contact surface@syr.edu. arXiv:0911.0623v1  [math.CV]  3 Nov 2009AREA CONTRACTION FOR HARMONICAUTOMORPHISMS OF THE DISKNGIN-TEE KOH AND LEONID V. KOVALEVAbstract. A harmonic self-homeomorphism of a disk does not increasethe area of any concentric disk.1. IntroductionThe unit disk D = {z ∈ C : |z| < 1} can be endowed with the hyperbolicmetricdσ =|dz|1 − |z|2.The Schwarz-Pick lemma (e.g., [1]) implies that any holomorphic map f : D →D does not increase distances in the hyperbolic metric. This is no longertrue for harmonic maps, which verify the Laplace equation ∂∂̄f = 0 butnot necessarily the Cauchy-Riemann equation ∂̄f = 0. The harmonic ver-sion of the Schwarz lemma ([5], see also [2]) states that any harmonic mapf : D → D with normalization f(0) = 0 satisfies|f(z)| 64πarctan|z|, z ∈ D.This inequality is sharp [4, p. 77]. More precisely, for any r ∈ (0, 1) and anysmall ǫ > 0 there is a bijective harmonic map f : D → D such that f(0) = 0andf(r) = −f(−r) =4πarctan r − ǫ.This map is not a contraction in either Euclidean or hyperbolic metric. Withrespect to either metric, the diameter of the disk Dr = {z ∈ C : |z| < r} isstrictly less than the diameter of f(Dr).In this note we prove that a bijective harmonic map f : D → D does notincrease the area of Dr for any 0 < r < 1. We write |E| for the area (i.e.,planar Lebesgue measure) of a set E.Theorem 1.1. Let f : D → D be a bijective harmonic map. Then(1.1) |f(Dr)| 6 |Dr|, 0 < r < 1.If (1.1) turns into an equality for some r ∈ (0, 1), then f is an isometry.Date: November 2, 2009.2000 Mathematics Subject Classification. Primary 31A05; Secondary 30C62, 58E20.Key words and phrases. Harmonic map, homeomorphism, area.Kovalev was supported by the NSF grant DMS-0913474.12 NGIN-TEE KOH AND LEONID V. KOVALEVIt should be noted that the class of harmonic automorphisms of D ismuch wider than the class of holomorphic automorphisms, which consistsof Möbius maps only. Harmonic homeomorphisms of D form an interestingand much-studied class of planar maps, see [3, 7, 8] or the monograph [4].Theorem 1.1 is different from most known estimates for harmonic maps inthat it remains sharp when specialized to the holomorphic case.An immediate consequence of (1.1) is |f(D \ Dr)| > |D \ Dr|. If f issufficiently smooth, we can divide by 1 − r and let r → 1 to obtain thefollowing.Corollary 1.2. Let f : D → D be a bijective harmonic map that is contin-uously differentiable in the closed disk D. Then∫|z|=1|detDf | |dz| > 2π,where det Df = |∂f |2 − |∂̄f |2 is the Jacobian determinant of f .Corollary 1.2 was proved in a different way in [6] where it serves as an im-portant part of the proof of Nitsche’s conjecture on the existence of harmonichomeomorphisms between doubly-connected domains. In fact, Corollary 1.2is what led us to think that (1.1) might be true.If f : D → D is holomorphic, then (1.1) holds without the assumption of fbeing bijective. Indeed, in this case f(Dr) is contained in a hyperbolic diskD of the same hyperbolic radius as Dr. Since the density of the hyperbolicmetric increases toward the boundary, it follows that the Euclidean radiusof D is at most r, which implies (1.1).Question 1.3. Does the area comparison (1.1) hold for general harmonicmaps f : D → D? Does it hold in higher dimensions?We conclude the introduction by comparing the behavior of |f(Dr)| forholomorphic and harmonic maps. If f : D → C is holomorphic and injective,one can use the power series f(z) =∑cnzn to compute|f(Dr)| =∞∑n=1n|cn|2r2n.Since the right-hand side is a convex function of r2, it follows that(1.2) |f(Dr)| 6 r2|f(D)|,which includes (1.1) as a special case. However, (1.2) fails for harmonicmaps. Indeed, let f(z) = z + cz̄2 where 0 < |c| < 1/2. It is easy to see thatf : D → C is harmonic and one-to-one, but|f(Dr)| = r2 − 2|c|2r4is a strictly concave function of r2. Therefore, |f(Dr)| > r2|f(D)| for 0 <r < 1. This example does not contradict Theorem 1.1 since f(D) is not adisk.AREA CONTRACTION FOR HARMONIC AUTOMORPHISMS OF THE DISK 32. PreliminariesLet f be as in Theorem 1.1. We may assume that f is orientation-preserving; otherwise consider f(z̄) instead. In this section we derive anidentity that relates the area of f(Dr) with the boundary values of f , whichexist a.e. in the sense of nontangential limits.The Poisson kernel for D will be denoted Pr(t),Pr(t) =1 − r21 − 2r cos t + r2, 0 6 r < 1, t ∈ R.We represent f by the Poisson integral(2.1) f(reiθ) =ω2π∫ 2π0eiξ(t)Pr(θ − t) dt,where ξ : [0, 2π) → [0, 2π) is a nondecreasing function and ω is a unimodularconstant. By Green’s formula we have|f(Dr)| =12∫ 2π0Im(f(reiθ)fθ(reiθ))dθ,where fθ indicates the derivative with respect to θ. Sincefθ(reiθ) =ω2π∫ 2π0eiξ(t)P ′r(θ − t) dt,it follows that(2.2) f(reiθ)fθ(reiθ) =14π2∫ 2π0∫ 2π0e−iξ(t)eiξ(s)Pr(θ − t)P′r(θ − s) dt ds.Integrating (2.2) with respect to θ and reversing the order of integration, wefind(2.3) |f(Dr)| =14π∫ 2π0∫ 2π0Kr sin(ξ(s) − ξ(t)) dt dswhere Kr is a function of r, s, and t,Kr =12π∫ 2π0Pr(θ − t)P′r(θ − s)Recall that the Poisson kernel has the semigroup property [9, p.62],(2.4) Prσ(t) =12π∫ 2π0Pr(s)Pσ(t − s) ds, 0 6 r, σ < 1.We will only use (2.4) with σ = r. Differentiation with respect to t yields(2.5)12π∫ 2π0Pr(s)P′r(t − s) ds = P′r2(t) = −2r2(1 − r4) sin t(1 − 2r2 cos t + r4)2.Identity (2.5) provides an explicit formula for Kr,(2.6) Kr = Kr(s − t) =2ρ2(1 − ρ4) sin(s − t)(1 − 2ρ2 cos(s − t) + ρ4)2.4 NGIN-TEE KOH AND LEONID V. KOVALEVNow we can rewrite (2.3) as(2.7) |f(Dr)| =14π∫ 2π0∫ 2π0Kr(s − t) sin(ξ(s) − ξ(t)) dt ds.In the next section we will estimate (2.7) from above.3. Proof of Theorem 1.1We continue to use the Poisson representation (2.1). The function ξ,originally defined on [0, 2π), can be extended to R so that ξ(t+2π) = ξ(t)+2πfor all t ∈ R. By (2.7) we have(3.1) |f(Dr)| =14π∫ 2π0∫ 2π0Kr(s − t) sin(ξ(s) − ξ(t)) dt ds.When f is the identity map, (3.1) tells us that14π∫ 2π0∫ 2π0Kr(s − t) sin(s − t) dt ds = |Dr|.The desired inequality |f(Dr)| 6 |Dr| now takes the form(3.2)∫ 2π0∫ 2π0Kr(s − t){sin(s − t) − sin(ξ(s) − ξ(t))}dt ds > 0.Neither the kernel Kr, which is defined by (2.6), nor the other factor in theintegrand are nonnegative. We will have to transform the integral in (3.2)before effective pointwise estimates can be made. It will be convenient touse the notation(3.3) α = s − t, and γ = γ(α, t) = ξ(α + t) − ξ(t),so that the integral in (3.2) becomes∫ 2π0∫ π−t−π−tKr(α) (sin α − sin γ) dα dt.Since the integrand is 2π-periodic with respect to α, our goal can be equiv-alently stated as(3.4)∫ 2π0∫ 2π0Kr(α) (sin α − sin γ) dα dt > 0.Note that γ ∈ [0, 2π] for all α, t ∈ [0, 2π].Step 1. We claim that(3.5)∫ 2π0∫ 2π0Kr(α)(γ − α) cos α dα dt = 0.Indeed, the function ζ(t) := ξ(t) − t is 2π-periodic, which implies(3.6)∫ 2π0{ζ(α + t) − ζ(t)} dt = 0AREA CONTRACTION FOR HARMONIC AUTOMORPHISMS OF THE DISK 5for every α ∈ R. Multiplying (3.6) by Kr(α) cos α and integrating overα ∈ [0, 2π], we obtain∫ 2π0∫ 2π0Kr(α){ζ(α + t) − ζ(t)} cos αdα dt = 0It remains to note that ζ(α+t)−ζ(t) = γ−α, completing the proof of (3.5).We take advantage of (3.5) by adding it to (3.4), which reduces our taskto proving that(3.7)∫ 2π0∫ 2π0Kr(α) {sin α + (γ − α) cos α − sin γ} dα dt > 0.Step 2. Let us now consider the function(3.8) H(α, β) := sinα + (β − α) cos α − sin β, (α, β) ∈ [0, 2π] × [0, 2π]which appears in (3.7). It has a simple geometric interpretation in terms ofthe graph of the sine function y = sinx. Indeed, the tangent line to thisgraph at x = α has equation y = sinα+(x−α) cos α. The quantity H(α, β)represents the difference in the y-values of the tangent line and the graphat x = β. Since the sine curve is strictly concave on [0, π], it follows that(3.9) H(α, β) > 0, 0 6 α, β 6 π,with equality only when α = β. The upper bound on β in (3.9) can beweakened to β 6 2π − α thanks to the monotonicity with respect to β,∂H∂β= cos α − cos β > 0, 0 6 α 6 π, α 6 β 6 2π − α.Note that the product Kr(α)H(α, β) is invariant under the central symmetryof the square [0, 2π]×[0, 2π], i.e., the transformation (α, β) 7→ (2π−α, 2π−β).Hence(3.10) Kr(α)H(α, β) > 0, (α, β) ∈([0, 2π] × [0, 2π])\ (T1 ∪ T2)whereT1 = {(α, β) : 0 < α < π, 2π − α < β 6 2π};T2 = {(α, β) : π < α < 2π, 0 6 β < 2π − α}.Within the triangles T1 and T2 the product Kr(α)H(α, β) may be negative.However, for all (α, β) ∈ [0, 2π] × [0, 2π] the following holds.(3.11) Kr(α)H(α, β) + Kr(2π − α)H(2π − α, β) = 2Kr(α)H(α, π) > 0,where the last inequality follows from (3.10). We will use (3.11) to controlthe contribution of triangles T1 and T2 to the integral (3.7).Step 3. For each fixed t the function α 7→ γ(α, t) defined by (3.3)is nondecreasing and it maps the interval [0, 2π] onto itself. Thus, in-equality (3.7) will follow once we show that for any nondecreasing function6 NGIN-TEE KOH AND LEONID V. KOVALEVΓ: [0, 2π] → [0, 2π](3.12)∫ 2π0Kr(α)H(α,Γ(α)) dα > 0.The integral in (3.12) remains unchanged if we replace Γ(α) with Γ̃(α) =2π − Γ(2π − α). Thus we lose no generality in assuming that Γ(π) 6 π. Byvirtue of (3.10) the integrand in (3.12) is nonnegative outside of the interval[π, α0], whereα0 = sup{α ∈ [π, 2π] : α + Γ(α) 6 2π}We claim that(3.13) Kr(α)H(α,Γ(α)) > Kr(α)H(α,Γ(π)), 2π − α0 < α < α0.Indeed, the inequality∂H∂β= cos α − cos β 6 0, |α − π| 6 |β − π| 6 π,implies(3.14) H(α, β1) > H(α, β2), 0 6 β1 6 β2 6 min(α, 2π − α).To see that (3.14) applies in our situation, note that Γ(α) 6 2π − α0 forα < α0. Inequality (3.14) yieldsH(α,Γ(α)) 6 H(α,Γ(π)), π 6 α < α0;H(α,Γ(α)) > H(α,Γ(π)), 2π − α0 < α 6 π.(3.15)Multiplying (3.15) by Kr(α), we arrive at (3.13).Finally, we combine (3.10), (3.13), and (3.11) to obtain∫ 2π0Kr(α)H(α,Γ(α)) dα >∫ α02π−α0Kr(α)H(α,Γ(α)) dα>∫ α02π−α0Kr(α)H(α,Γ(π)) dα= 2∫ α0πKr(α)H(α, π) dα > 0,(3.16)completing the proof of (3.7).Step 4. It remains to prove the equality statement in Theorem 1.1. Sup-pose that Γ: [0, 2π] → [0, 2π] is a nondecreasing function such that Γ(π) 6 π,and equality holds everywhere in (3.16). Returning to the geometric inter-pretation of H(α, γ) in (3.8), we note thatKr(α)H(α, π) > 0, 0 < |α − π| < π.This forces α0 = π, which by definition of α0 implies(3.17) Kr(α)H(α,Γ(α)) > 0, 0 6 α 6 2π.Hence, (3.17) must turn into an equality for almost all α ∈ [0, 2π]. In viewof (3.9) and of the monotonicity of Γ this is only possible if Γ(α) = α for allα ∈ [0, 2π].AREA CONTRACTION FOR HARMONIC AUTOMORPHISMS OF THE DISK 7If |f(Dr)| = |Dr|, then equality holds in (3.7). Then for almost all t ∈[0, 2π] the function Γ(α) = ξ(α + t) − ξ(t), or its reflection Γ̃(α) = 2π −Γ(2π − α), turns (3.16) into an equality. Hence ξ(α + t) − ξ(t) = α foralmost all t ∈ [0, 2π] and all α ∈ [0, 2π]. Thus ξ is the identity function andf : D → D is an isometry. Theorem 1.1 is proved.AcknowledgementsWe thank Tadeusz Iwaniec and Jani Onninen for valuable discussions onthe subject of this paper.References1. L. V. Ahlfors, Conformal invariants: topics in geometric function theory, McGraw-HillBook Co., New York-Düsseldorf-Johannesburg, 1973.2. H. Chen, P. M. Gauthier, and W. Hengartner, Bloch constants for planar harmonicmappings, Proc. Amer. Math. Soc. 128 (2000), no. 11, 3231–3240.3. J. Clunie and T. Sheil-Small, Harmonic univalent functions, Ann. Acad. Sci. Fenn. Ser.A I Math. 9 (1984), 3–25.4. P. Duren, Harmonic mappings in the plane, Cambridge University Press, Cambridge,2004.5. E. Heinz, On one-to-one harmonic mappings, Pacific J. Math. 9 (1959), 101–105.6. T. Iwaniec, L. V. Kovalev, and J. Onninen, The Nitsche conjecture, preprint,arXiv:0908.1253.7. D. Kalaj, On harmonic diffeomorphisms of the unit disk onto a convex domain, Com-plex Var. Theory Appl. 48 (2003), no. 2, 175–187.8. M. Pavlović, Boundary correspondence under harmonic quasiconformal homeomor-phisms of the unit disk, Ann. Acad. Sci. Fenn. Math. 27 (2002), no. 2, 365–372.9. E. M. Stein, Singular integrals and differentiability properties of functions, PrincetonUniversity Press, Princeton, 1970.Department of Mathematics, Syracuse University, Syracuse, NY 13244, USAE-mail address: nkoh@syr.eduDepartment of Mathematics, Syracuse University, Syracuse, NY 13244, USAE-mail address: lvkovale@syr.edu

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