On planar Beurling and Fourier transforms

a r X i v :m a t h /0701696v 1 [m a t h .C V ] 24 J a n 2007On planar Beurling and Fourier transformsH ˚akan HedenmalmAbstract.We study the Beurling and Fourier transforms on subspaces of L 2(C )defined by aninvariance property with respect to the root-of-unity group.This leads to generalizations of these transformations acting unitarily on weighted L 2-spaces over C .1.IntroductionBeurling and Fourier transforms.In this note,we shall study certain extensions of the Beurling and Fourier transforms on the complex plane C .The Fourier transform of an appropriately area-integrable function f isF [f ](ξ)=Ce −2i R e [z ¯ξ]f (z )d A (z ),ξ∈C ,while the Beurling transform is the singular integral operatorB C [f ](z )=pvCf (w )π,z =x +i y,is normalized area measure.The two transforms are connected viaFB C [f ](ξ)=ξ1991Mathematics Subject Classification.Primary 32A25,32A36;Secondary 46E15,47A15.Key words and phrases.Beurling transform.Research supported by the G¨o ran Gustafsson Foundation.2HedenmalmIt is easy to see that f∈L2n,N(C)if and only f∈L2(C)is of the form(1.2)f(z)=z n g(z N),z∈C,where g some other complex-valued function.We shall study the Beurling and Fourier transforms on the subspaces L2n,N(C).This will be shown to lead to interesting generalizations of these transforms to the weighted spaces L2θ(C),with normf 2L2θ(C)= C|f(z)|2|z|2θd A(z).Here,θis a real parameter.We apply the results obtained regarding the Beurling transform to conformal mapping,and obtain Grunsky-type identities in the spirit of[3].The Grunsky-type identity obtained here implies the Prawitz inequality(see,e.g.,[5],[6])as a special case.The Grunsky-type inequality that follows from the Grunsky-type identity is more or less equivalent to the general Grunsky inequality that forms the backdrop to Louis de Branges’work leading up to the solution of the Bieberbach conjecture(see[1]).2.The Beurling transformThe Cauchy transform.The Cauchy transform C C is the integral transformC C[f](z)= C f(w)2 ∂∂y,¯∂z=1∂x+i∂(w−z)2d A(w)=pv CαnNpv C α∈A NαnN α∈A NαnOn planar Beurling and Fourier transforms3 The function G then has a simple pole at1,and is analytic everywhere else in the complex plane. Moreover,F vanishes at infinity,so G vanishes there,too.This leaves us but one possibility,that G has the formG(z)=CN α∈A Nαn1−z N,z∈C.As a consequence,we getH(z):=F(z)+zF′(z)=[zF(z)]′=1(1−αz)2=z N−n N1−z N .This allows us to compute the sum we need:1(αw−z)2=1z =z n−2w N−nNz N z N−w N .For f∈L2n,N(C),we thus get the representationB C[f](z)=z n−2pvC Nz N z N−w N w N−n f(w)d A(w),z∈C.Let f and g be connected via(1.2),and implement this relationship into the above formula:(2.2)B C[f](z)=z n−2pv C Nz N z N−w N w N g(w N)d A(w),z∈C.A similar expression may be found for the Cauchy transform as well:(2.3)C C[f](z)=z n−N−1 C w NzC C[h](z),and introduce,for0≤θ≤1,the modified Beurling transform(2.4)BθC[h](z)=B C[h](z)+T C[h](z),z∈C;here,h is assumed to be a nice enough function so that the above Beurling and Cauchy transforms make sense.It is easy to check that withh(z)=z g(z)4Hedenmalmwhere we suppose thatθ=(n−1)/N.However,fractions of this type are dense in the interval [0,1],so that(2.5)extends to allθwith0≤θ≤1.In other words,for0≤θ≤1,the operator BθC is unitary on the space L2θ(C),which was defined earlier.It is known[7]that B C is a bounded operator on L2θ(C)for−1<θ<1(but not forθ=±1).This means that for−1<θ<1,both terms in(2.4)are bounded operators on Lθ(C).We suspect that the second term in(2.4),the operator T C,is compact with small spectrum.Extension to realθ.Wefirst note that M z,multiplication by the independent variable,is an isometric isomorphism L2θ+1(C)→L2θ(C)for all realθ.Therefore,for integers k and0≤θ≤1, the operatorBθ+kC:=M−k z BθC M k zis unitary on L2θ+k(C).It supplies an extension of Bθto all realθ.Note that since1w−z =w z (z).3.Fourier transformsThe Fourier transform and root-of-unity invariance.The Fourier transform of a function in L2(C) is given byF[f](ξ)= C e−2i R e[z¯ξ]f(z)d A(z),ξ∈C.By the Plancherel identity,we haveF[f] L2(C)= f L2(C),f∈L2(C).Now,suppose f∈L2n,N(C),so that f has the invariance property(1.1).Then,by the change of variables formula,F[f](ξ)= C e−2iR e[z¯ξ]f(z)d A(z)=αn C e−2iR e[αz¯ξ]f(z)d A(z)=αn F[f](¯αξ),ξ∈C, forα∈A N.Taking the average over A N,we get the identityF[f](ξ)= C E n,N(z¯ξ)f(z)d A(z),ξ∈C,whereE n,N(z)=1On planar Beurling and Fourier transforms5The extended Fourier transform.We now introduce the generalized Fourier transformF n,N[h](ξ)=|ξ|−2(n−1)/N|z|2−2/N,we get thatC |f(z)|2d A(z)=1N C F n,N[h](ξ) 2|ξ|2(n−1)/N d A(ξ).The Plancherel identity thus states that F n,N is a unitary transformation on L2(n−1)/N (C).Theinverse transformation is quite similar:F−1n,N[h](z)=(−1)N−n F n,N[h] (−1)N z . We need to express the function D n,N in a different manner.Sincee−2i R e[αz]=e−iαz e−i¯α¯z=+∞j,k=0(−i)j+kN α∈A Nαm=+∞ l=−∞δm,Nl,where delta stands for the Kronecker delta,we haveE n,N(z)=1N+∞j,k=0(−i)j+kj!k!z j¯z kδj−k+n,Nl=+∞l=−∞+∞ k=0(−1)k(−i)−n+Nl(k−n+Nl)!k!|z|2k z Nl,with the understanding that1(k−n+Nl)!k!|z|2k/N z l−1.6HedenmalmWe should mention that as E n,N (z )is bounded by 1in modulus,we have D n,N (z )≤|z |n/N −1,z ∈C .The formula for D n,N allows us to express the modified Fourier transform accordingly (provided f (z )=O (|z |−m )as |z |→+∞for every positive integer m ):F n,N [f ](ξ)=|ξ|−2(n −1)/N(k −n +Nl )!k !|ξ|2k/N ξl −1C |z |2k/N z l −1f (z )d A (z ).An application involving the confluent hypergeometric function.Next,we consider a function f ofthe formf (z )=¯z m |z |2αe −β|z |2/N,where m is an integer,α,βare real with β>0,andm +2α+n −1N,where r is an integer,we obtain by calculation that F n,N [f ](ξ)=(−i)−n +N (m +1)|ξ|−2(n −1)/N ξm (k −n +N (m +1))!k !|ξ|2k/N×C|z |2(k −n +r +1)/N +2m e −β|z |2/Nd A (z )=(−i)−n +N (m +1)ξm |ξ|−2(n −1)/N(k −n +N (m +1))!k !−|ξ|2/Nβr −n +1+N (m +1)(N (m +1)+r −n )!β,where 1F 1stands for the standard confluent hypergeometric function.We have from one of the classical identities that1F 1 N (m +1)−n +r +1;N (m +1)−n +1;−|ξ|2/Nβ,where the right hand side is easy to compute for positive r ,as the sum is then finite.As a result of the unitarity of F n,N ,we find that (with β=1and M =N (m +1)−n +1)+∞01F 1 −r ;M ;t2e −2t t M −1d t =(M +2r −1)![(M −1)!]2On planar Beurling and Fourier transforms 74.Applications of Beurling transforms to conformal mappingTransfer to the unit disk.We need to introduce some general notation.Let M F denote the operator of multiplication by the function F .We also need the Hilbert space L 2θ(X )with the normh 2L 2θ(X )=X|h (z )|2|z |2θd A (z ),where X is some Borel measurable subset of C with positive area.In the sequel,we fix θto theinterval 0≤θ≤1.Fix a bounded simply connected domain Ωin C ,which contain the origin,and let ϕ:D →Ωdenote the conformal mapping with ϕ(0)=0and ϕ′(0)>0.Let f ∈L 2(Ω),and extend it to the whole complex plane so that it vanishes on C \Ω.Let B Ω[f ]denote therestriction to Ωof B C [f ],and do likewise to define the operators C Ω,T Ω,T ′Ω,B θΩ,as well as B −θΩ.We introduce transferred operators on spaces over the unit disk in the following fashion.First,we suppose f ∈L 2θ(Ω).Then the associated function(4.1)g (z )=¯ϕ′(z )ϕ(z )zθC Ω[f ](z )=Dw ϕ(z )ϕ(w )−ϕ(z )g (w )d A (w ).The transferred modified Beurling transform is defined analogously:B θϕ[g ](z )=ϕ′(z )ϕ(z )zθB Ω[f ]◦ϕ(z )+θϕ(z )C θϕ[g ](z ),whereB θ,0ϕ[g ](z )=pvDw ϕ(z )(ϕ(w )−ϕ(z ))2g (w )d A (w ).It is clear thatB θϕis a norm contraction on L 2θ(D ).Let P θbe the integral operatorP θ[f ](z )= D11−z ¯w f (w )|w |2θd A (w );it is the orthogonal projection to the subspace of analytic functions in L 2θ(D ).As both B θϕand P θare contractions on L 2θ(D ),so is their product P θB θϕ.It remains to represent the operator P θB θϕin a reasonable fashion.The main observation is thatw ϕ(z )(ϕ(w )−ϕ(z ))2=1ϕ(z )−1w −z+O (1)near the diagonal z =w ,so that(4.3) w ϕ(z )(ϕ(w )−ϕ(z ))2+θϕ′(z )z ϕ(w )θϕ′(w )(w −z )2+θ8Hedenmalmagain near the diagonal.We observe that in view of(4.3),weget the Grunsky-type identity(4.4)PθBθϕ=Bθϕ−B D+PθB D+θPθT D−θT D.To make the involved operators PθB D and PθT D appearing in the right hand side of(4.4)more concrete,it is helpful to know that forλ∈D,Pθ[fλ](z)=¯λ|λ|2θ 10 11−t¯λz tθd t,fλ(z)=1(1−t¯λz)2+θ(λ−z)2.In view of these relations,we quickly verify thatPθB D+θPθT D=0.The Grunsky-type identity(4.4)thus simplifies a bit:(4.5)PθBθϕ=Bθϕ−B D−θT D.The corresponding Grunsky-type inequality reads(4.6) Bθϕ−B D−θT D [f] L2θ(D)≤ f L2θ(D),f∈L2θ(D).To get a concrete example of how the Grunsky-type inequality works,we pickfλ(z)=|z|−2θ 11−¯zλ ,z∈D,and computeBθϕ−B D−θT D [f](z)= λϕ(z)(ϕ(λ)−ϕ(z))2−1ϕ(z) λϕ(z)ϕ(λ)−ϕ(z)−θzϕ(λ) θϕ′(z)ϕ′(λ)(λ−z)2+θϕ′(z)zϕ(λ)θϕ′(λ)z(λ−z)2|z|2θd A(z)≤ D|fλ(z)|2|z|2θd A(z)= D 11−¯zλ 2|z|−2θd A(z)=11−|λ|2. The special caseλ=0gives us the inequality of Prawitz(see[5]and[6];we assumeϕ′(0)=1):Dϕ′(z)ϕ(z)1−θ.On planar Beurling and Fourier transforms9A dual version.We carry out the corresponding calculations on the basis of the fact that B−θCis unitary on L2−θ(C)for0≤θ≤1.In analogy with the above treatment,we connect two functions f,g via(4.8)g(z)=¯ϕ′(z) ϕ(z)z −θB−θΩ[f]◦ϕ(z)=ϕ′(z) ϕ(z)z ◦ϕ(z) =B−θ,0ϕ[g](z)−θϕ′(z)C−θϕ g(1−w¯z)2−θ10Hedenmalm[4]H.Hedenmalm,H.,B.Korenblum,K.Zhu,Theory of Bergman spaces.Graduate Texts in Mathematics199,Springer-Verlag,New York,2000.[5]H.Hedenmalm,S.Shimorin,Weighted Bergman spaces and the integral means spectrum of conformalmappings,Duke Math.J.,vol.127(2005),341-393.[6]in,Univalent functions and orthonormal systems.Translated from the Russian.Translationsof Mathematical Monographs,Vol.49.American Mathematical Society,Providence,R.I.,1977. [7]S.Petermichl,A.L.Volberg,Heating of the Ahlfors-Beurling operator:weakly quasiregular maps on theplane are quasiregular.Duke Math.J.112(2002),281–305.Hedenmalm:Department of Mathematics,The Royal Institute of Technology,S–10044Stock-holm,SWEDENE-mail address:haakanh@math.kth.se。