1 ON RIESZ TYPE INEQUALITIES FOR HARMONIC MAPPINGS ON THE UNIT DISK 7 DAVIDKALAJ 1 0 2 ABSTRACT. Weprovesomesharpinequalitiesforcomplexharmonicfunctions ontheunitdisk. TheresultsextendaM.Rieszconjugatefunctiontheoremand n somewell-knownestimatesforholomorphicfunctions.Weapplysomeofresults a totheisoperimetricinequalityforharmonicmappings. J 9 1 V] CONTENTS C 1. Introduction 1 . 2. Mainresults 3 h t 2.1. ApplicationtoHilberttransform 4 a 2.2. Applicationtotheisoperimetric inequality 8 m 3. Strategyoftheproofs 10 [ 4. ProofofTheorem2.1 11 2 5. ProofofTheorem2.3 27 v 6. Appendix 36 5 8 Acknowledgement 37 7 References 37 4 0 . 1 0 1. INTRODUCTION 7 1 LetUdenotetheunitdiskandTtheunitcircleinthecomplexplane. Forp > 1, v: wedefinetheHardyclasshp astheclassofharmonicmappingsf = g+h¯,where i g andhareholomorphic mappingsdefinedontheunitdiskU C,sothat X ⊂ ar kfkp = kfkhp = sup Mp(f,r)< ∞, 0 2, then we make use of inequality (2.4) below. We have by using Jensen inequality (asin[16]),thefollowing 1/2 π kgkHp = k u2+v2kLp(T) ≤ kuk2hp +kvk2hp 1/2 ≤ kuk2hp +sin2 2pkgk2Hp . (cid:18) (cid:19) p (cid:0) (cid:1) Therefore π g 2 (1 sin2 ) u 2 k kHp − 2p ≤ k khp andthisimpliesthecorollary. (cid:3) ONRIESZTYPEINEQUALITIESFORHARMONICMAPPINGSONTHEUNITDISK 4 Tomotivatethefollowingtheoremnoticethefollowingsimplesharpinequality z+w¯ √2(z 2+ w 2)1/2. | |≤ | | | | Thuswehave f hp √2 g 2+ h2 Lp(T). k k ≤ k | | | | k However the last inequality is not shaprp, and the sharp inequality has been given bythefollowingtheorem. Theorem 2.3. Let 1 < p < and assume that f = g +¯h hp is a harmonic ∞ ∈ mapping on the unit disk with (g(0) h(0)) 0. Then we have the following ℜ · ≤ sharpinequality 1/p π π (2.3) f hp √2max sin ,cos (g 2 + h2)p/2 . k k ≤ { 2p 2p} T | | | | (cid:18)Z (cid:19) Remark2.4. Ifp = 2, then inequalities (2.1)and (2.3)areopposite toeach other because 1 π π = 1 = √2max sin ,cos . 1/2 { 2p 2p} 1 cos π −| p| (cid:16) (cid:17) Thisisnotasurprisingfact,becausethegivenintegralscoincideif (h(0)g(0)) = 0. Inotherwordsforeveryf h2, h h2 = h h2. ℜ ∈ k k ||| ||| Corollary2.5. [16]Ifv isarealharmonicfunctionwithv(0) = 0andg = u+iv isananalyticfunction, thenforeveryp > 1wehavetheinequality π (2.4) v hp sin g Hp. k k ≤ 2p¯k k ProofofCorollary2.5. Byapplying Theorem2.3totherealharmonicfunction g g¯ 1 v = f = − = (ig+ig), 2i −2 inviewofthefact i2g(0)g(0) = g2(0) 0, − ≤ andbyusingthesimpleformula π π π max sin ,cos = sin { 2p 2p} 2p weobtain(2.4). (cid:3) 2.1. ApplicationtoHilberttransform. ONRIESZTYPEINEQUALITIESFORHARMONICMAPPINGSONTHEUNITDISK 5 2.1.1. Hilbert transform on the unit disk and unit circle. If f = u+iv is a har- monicfunction definedintheunitdiskU,thenaharmonicfunctionf˜= u˜+iv˜is calledtheharmonicconjugateoff ifu+iuãndv+ivãreanalyticfunctions. Notice thatf˜isuniquely determined uptoanadditive constant. Letf = g+h¯ = u+iv be a harmonic mapping, where h and g are holomorphic and h(0) = 0. Then f˜:= (ig+ih) = u˜+iv˜isaharmonicconjugateoff whichwedealwithinthis paper.−Namelyu= (g+h),v = (g+h). Furtheru˜ = ( i(g h¯)) = (g h¯) andv˜= ( i(g h¯ℜ)) = (h¯ g)ℑ. Thusu+iu˜= (g+ℜh−)+i−(g h¯)ℑ= g−+h andv+iℑv˜=− (g−+h)+ℜi (h¯− g) =i(h g). ℜ ℑ − ℑ ℜ − − Further if f is real valued, i.e. if f(z) = h(z) + h(z), then f(z) = h(z) + h(0)+h(z) h(0) = g (z)+h (z). Hereh (0) = 0. Ifh(0) = 0,then f˜(z) = 1 1 1 − 6 i(g (z) h (z))isnotrealvaluedfunction,butfˆ(z) := f˜(z) f˜(0)isrealvalued 1 1 − − withfˆ(0) = 0. Thenfîstheharmonicconjugatewithrespecttostandardmeaning. Furthermoreforeveryz (2.5) fˆ(z)2 = f˜(z)2 f˜(0)2. | | | | −| | Letχbetheboundary valueoff andassumethatχ˜istheboundary valueoff˜. Thenχ˜iscalledtheHilberttransformofχandwedenoteitbyχ˜ = H[χ]. Assume thatχ˜ L1(T). The∈(periodic) Hilbert transform of a function χ L1(T) is also given by the ∈ formulas 1 π χ(τ +t) χ(τ t) (2.6) χ˜(τ) = − − dt. −π 2tan(t/2) Z0+ and (2.7) χ˜(τ) = i sign(k)χ(k)eikτ, − k∈Z X where b 1 χ(k) = χ(eit)e−iktdt 2π T Z π π and sign(0) = 1. Here Φ(t)dt := lim Φ(t)dt.Theintegral in (2.6)is 0+b ǫ→0+ ǫ improper andconverges fora.e. τ [0,2π]. IfP denote thePoisson extension on R ∈ R ^ theunitdisk,thenwehaveP[χ](z) = P[χ˜](z). 2.1.2. Hilbert transform on the real line and half-plan. Let p > 1 and let f Lp(R,C). Thenthe(nonperiodic) Hilberttransform off isdefinedby ∈ 1 ∞ φ(t) 1 ∞ φ(x+t) φ(x t) [φ](x) = φ˜(x) = dt = lim − − dt. H π Z−∞ x−t −π ǫ↓0 Zǫ t Further, themapping φinduces aharmonic mapping defined onthe upper half- plane H := z C : z > 0 , by the formula f(z) = P[φ](z), where P is the { ∈ ℑ } Poissonintegralontheupperhalf-plane. ONRIESZTYPEINEQUALITIESFORHARMONICMAPPINGSONTHEUNITDISK 6 Now let hp(H) be the Hardy space on the upper half-plane, i.e. the class of harmonicmappingsf definedonHsothat f hp(H) := sup f(x+iy) Lp(R) < . k k k k ∞ y>0 If f = P[χ](z), then we have f hp(H) = χ Lp(R) ([1, Theorem 7.17]). Fur- k k k k thermoretheharmonicmappingw˜ = P[φ˜]isharmonicconjugate ofw. The following theorem, in view of (2.5) extends the main result of S. K. Pi- chorides([14]) Theorem 2.6. Assume that p > 1 and f is a complex harmonic mapping so that f = g +h¯ hp(U) and h(0) = 0. Then f˜= i(g h¯) hp, and we have the ∈ − ∈ sharpinequality π (2.8) f˜ hp cot f hp. k k ≤ 2p¯k k Inotherwordsthenormoftheoperator :hp(U,C) hp(U,C) ( :Lp(T,C) Lp(T,C)) H → H → isequalto tan π , ifp 2; = 2p ≤ kHkp cot π , ifp > 2. (cid:26) 2p Proof. Letf = g+h¯ = u+iv beaharmonic mappingsthatbelongstohp. Then f˜ = i(g ¯h) = u˜ + iv˜. From Theorem 2.1 and Theorem 2.3, in view of the − − fact that f = f˜ and (( ig(0))( ih(0))) = (g(0)h(0)) = 0, we hp ||| ||| hp ℜ − − −ℜ obtainthat (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)f˜ hp Ap f hp ApBp f hp. k k ≤ ||| ||| ≤ k k Thetheoremfollowsfromtheequation √2cos π π 2p A B = = cot . p p 1/2 2p¯ 1 cos π −| p| (cid:16) (cid:17) (cid:3) Remark 2.7. The condition h(0) = 0 of Theorem 2.6 is not essential. Indeed, if h(0) = 0,then 6 f(z) =g(z)+h(0)+h(z) h(0) = g (z)+h (z), 1 1 − whereh (0) = 0. Inthiscase 1 f˜(z) = i(g (z) h (z))= i g(z)+h(0) h(z) h(0) . 1 1 − − − − − (cid:16) (cid:17) Corollary 2.8. Letp > 1andletχ Lp(R,C)andf = P[φ]. Thenwehavethe ∈ sharpinequalities π (2.9) χ˜ cot χ , Lp Lp k k ≤ 2p¯k k ONRIESZTYPEINEQUALITIESFORHARMONICMAPPINGSONTHEUNITDISK 7 and π (2.10) f˜ hp cot f hp. k k ≤ 2p¯k k Inotherwords π :Lp(R,C) Lp(R,C) = cot kH → k 2p¯ and π : hp(H,C) hp(H,C) = cot . kH → k 2p¯ ProofofCorollary2.8. The relation (2.10) follows from Theorem 2.6 and the approach in the proof of a result of Zygmund in [17, Chapter XVI, Theorem 3.8], whereitisprovedthattheconstantthatappearinthecaseofperiodicHilberttrans- form could be taken in non-periodic Hilbert transform case as well. Moreover, as itisshownbyPichoridesin[14,Theorem4.1],theconstantcot π issharpforreal 2p¯ valuedfunctions, andsoitissharpforourcomplexfunctions aswell. (cid:3) LetbpdenotetheBergmanclassofharmonicmappingsdefinedontheunitdisk, satisfying thecondition 1/p dxdy f bp := f(z)p < . k k U| | π ∞ (cid:18)Z (cid:19) Forcompressive studyofthisclasswerefertothebook[5]. By integrating the functions U (z) = r f(zr)p and V (z) = r(g(zr)2 + r r | | | | h(zr)2)p/2 overtheunitcircleT,usingtheinequalities (2.1)and(2.3),andinte- g|rating|forr [0,1]weobtainthefollowingresultfortheBergmanspacebp. ∈ Corollary 2.9. Let 1 < p < . Assume that f = g + h¯ bp is a harmonic ∞ ∈ mapping on the unit disk with (g(0)h(0)) = 0 . Then we have the following ℜ inequalities 1/p 1/p 1 (2.11) (g 2 + h2)p/2 g+h¯ p (cid:18)ZU | | | | (cid:19) ≤ 1 cos π 1/2 (cid:18)ZU| | (cid:19) −| p| (cid:16) (cid:17) and 1/p 1/p π (2.12) (g+h¯ )p √2sin (g 2 + h2)p/2 . U | | ≤ 2p¯ U | | | | (cid:18)Z (cid:19) (cid:18)Z (cid:19) The inequality (2.12) does hold under weaker condition (g(0)h(0)) 0, and ℜ ≤ (2.11)for (g(0)h(0)) 0andp < 3. ℜ ≥ Remark 2.10. We were not able to check if the inequalities (2.11)and (2.12)are sharpornotforp = 2. Wewanttoemphasisthefollowingfact. Somewell-known 6 extremal functions that works for Hardy space, are not suitable for the Bergman space. The following example suggested by A. Calderon (see [14]) shows that 2γ/π (2.2) and (2.4) are sharp. Namely if g(z) = 1+z , arg 1+z π, and 1−z | 1−z| ≤ 2 γ < π , then g = u + iv hp. Further u =(cid:16) tan(cid:17)γ v almost everywhere on 2p ∈ | | | | ONRIESZTYPEINEQUALITIESFORHARMONICMAPPINGSONTHEUNITDISK 8 T, but u tanγ v > 0 everywhere on U. This is why this example works for | |− | | HardyspacebutnotforBergmanspace. 2.2. Application to theisoperimetric inequality. Thestarting point of this sub- sectionisthewellknownisoperimetric inequality forJordandomainsandisoperi- metric inequality for minimal surfaces due to Carleman [2]. In that paper Carle- man,amongtheotherresultsprovedthatifuisharmonicandsmoothinUthen 1 2π e2udxdy ( eudt)2. ZU ≤ 4π Z0 By using a similar approach as Carleman, Strebel ([15]) proved the isoperimetric inequality forholomorphic functions; thatisiff H1(U)then ∈ 1 (2.13) f(z)2dxdy ( f(eit)dt)2. U| | ≤ 4π T| | Z Z ByusingthenormalizedmeasuresonTandU,respectively, thepreviousinequal- itycanbewrittenintheform 2 (2.14) f(z)2 f(z) . U| | ≤ T| | Z (cid:18)Z (cid:19) Thisinequality has been proved independently by Mateljevic´ and Pavlovic´ ([13]). In[4],F.Hang,X.Wang,X.Yanhavemadeacertaingeneralizations forthespace. Nowweprovethefollowingtheorem Theorem2.11. Letf beacomplexharmonicmappingdefinedontheunitdiskand assumethatn 2isapositiveinteger. Assumethatf hn,thenf b2n andwe ≥ ∈ ∈ havetheinequality 1 π (2.15) f b2n csc f hn. k k ≤ 2 4n k k h i Remark2.12. Theproofsofthesamestatementforn = 2andn = 4canbefound in [12] and in [10] respectively (where different approaches used, but applicable onlyforthosetwospecificcases). The proof here works only for positive integers n 2, but probably the same ≥ estimateistrueforeverypositivenumbern > 2. Ontheotherhand,wewherenot abletocheckiftheinequality (2.15)issharp. A positive real function u is called log-subharmonic, if logu is subharmonic. First we formulate a lemma whose proof can be also deduced from [8, Corol- lary1.6.8]. Lemma 2.13. The function a 2 + b 2 is log-subharmonic, provided that a and b | | | | areanalytic. Proof. Weneed toshowthat f(z) = log(a 2 + b 2)issubharmonic. Bycalcula- | | | | tionwefind a′a¯+b′¯b f = z a 2+ b 2 | | | | ONRIESZTYPEINEQUALITIESFORHARMONICMAPPINGSONTHEUNITDISK 9 andso a′a¯′+b′¯b′ a′a¯+b′¯b aa¯′+b¯b′ f = zz¯ a 2 + b 2 − a 2+ b 2 a 2+ b 2 | | | | | | | | | | | | (a′ 2+ b′ 2)(a 2 + b 2) a¯a′+¯bb′ 2 = | | | | | | | | −| | , (a 2+ b 2)2 | | | | whichisclearlypositive. (cid:3) Nowtheisoperimetricinequalityforlog-subharmonicfunctions(e.g. [11,Lemma2.2]), statesthat,ifuispositivelog-subharmonic function, then 2 u2 u . U ≤ T Z (cid:18)Z (cid:19) Hereasbefore, 1 f := f(z)dxdy, z = x+iy, U π U Z Z and 1 f := f(z)dz , z = x+iy. T 2π T | | Z Z Thusweinferthat Lemma2.14. Foreverypositive number pand analytic functions aand bdefined ontheunitdiskU wehavethat 2 (a 2 + b 2)2p (a 2 + b 2)p . U | | | | ≤ T | | | | Z (cid:18)Z (cid:19) ProofofTheorem2.11. Without loos of generality assume that f(z) = g(z) + h(z),whereh(0) = 0,andg andhareholomorphic ontheunitdisk. Let L = (g+h¯ 2)n = (g 2 + h2+2 (gh))n. U | | U | | | | ℜ Z Z Then n n L = (g 2 + h2)k(2 (gh))n−k k U | | | | ℜ k=0(cid:18) (cid:19)Z X n n ((g 2 + h2)n)k/n( 2 (gh)n)(n−k)/n. ≤ k U | | | | U| ℜ | k=0(cid:18) (cid:19)Z Z X Letp 2andletE = cos π . ≥ p 2p ONRIESZTYPEINEQUALITIESFORHARMONICMAPPINGSONTHEUNITDISK 10 FromLemma2.14andCorollary2.9andTheorem2.3(Corollary2.5)wehave n n L En−k( (g 2 + h2)n)k/n( (2gh)n)(n−k)/n ≤ k n U | | | | U | | k=0(cid:18) (cid:19) Z Z X n n En−k( (g 2 + h2)n/2)2k/n( (2gh)n/2)2(n−k)/n ≤ k n T | | | | T | | k=0(cid:18) (cid:19) Z Z X n n En−k( (g 2 + h2)n/2)2k/n( (g 2 + h2)n/2)2(n−k)/n ≤ k n T | | | | T | | | | k=0(cid:18) (cid:19) Z Z X n n = En−k( (g 2 + h2)n/2)2 k n T | | | | k=0(cid:18) (cid:19) Z X n n 1 = En−k ( g+h¯ n)2 Xk=0(cid:18)k(cid:19) n 1−|cos πp| n ZT| | (1+E )n (cid:16) (cid:17) = n ( (g+h¯ n))2 1 cos π n T | | − n Z (1+cos π )n = (cid:0) 2n(cid:1) ( g+h¯ n)2. 1 cos π n T| | − n Z Thus (cid:0) (cid:1) (1+cos π )n (g+h¯ 2)n 2n ( g+¯hn)2. U | | ≤ 1 cos π n T| | Z − n Z Further (cid:0) (cid:1) 1+cos π cos2 π 1 2n = 4n = . 1 cos π sin2 π 2sin2 π − n 2n 4n Thisfinishestheproof. (cid:3) 3. STRATEGY OF THE PROOFS Astheauthorsofthepaperdidin[7],weuse”pluri-subharmonic minorant”. Definition3.1. Aupper semi-continuous real function uiscalled subharmonic in anopensetΩofcomplexplane, ifforeverycompactsubset K ofΩandforevery harmonicfunctionf definedonK,theinequalityu(z) f(z)forz ∂K implies ≤ ∈ thatu(z) f(z)onK. ≤ A property which characterizes the subharmonic mappings is the sub-mean value property which states that. If u is a subharmonic function defined on a do- mainΩ,thenforeverycloseddiskD(z ,r) Ω,wehavetheinequality 0 ⊂ 1 u(z ) u(z)dz . 0 ≤ 2πr | | Z|z−z0|=r Definition 3.2. A function u defined in an open set Ω Cn with values in ⊂ [ ,+ )iscalledplurisubharmonic if −∞ ∞

On Riesz type inequalities for harmonic mappings on the unit disk

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