A genuine analogue of Wiener Tauberian theorem for $\mathrm {SL}(2, \R)$
aa r X i v : . [ m a t h . F A ] O c t A GENUINE ANALOGUE OF WIENER TAUBERIAN THEOREM FOR
SL(2 , R ) TAPENDU RANA
Abstract.
We prove a genuine analogue of Wiener Tauberian theorem for integrable functions onSL(2 , R ) . Introduction
Let f ∈ L ( R ) and b f be its Fourier transform. The celebrated Wiener-Tauberian (W-T) theoremsays that the ideal generated by f in L ( R ) is dense in L ( R ) if and only if b f is nowhere vanishing on R . This theorem has been extended to abelian groups. In 1955, Ehrenpreis and Mautner observedthat the exact analogue of the theorem above fails for the commutative algebra of the integrable K -biinvariant functions on the group SL(2 , R ), where K = SO(2) is a maximal compact subgroup.Nonetheless the authors proved that if a K -biinvariant integrable function f on G satisfies a “not-to-rapid decay” condition and nonvanishing condition on a extended strip S ,δ = { λ ∈ C | |ℜ λ | ≤ δ } for δ >
0, etc, that is, b f ( λ ) = 0 for all λ ∈ S ,δ , and “not-to-rapid decay” conditionlim sup | t |→∞ | b f ( it ) | e Ke | t | > K > f in L ( G//K ) is dense in L ( G//K ) (see [EM] for the precise statements). Using the extended strip condition the resultshas been generalised to the full group SL(2 , R ) (see [Rs1]) and to the real rank one semi simpleLie groups (see [BBH], [BWH], [Rs2], [As]). We also refer [En], and [NS] for an analogue of W-Ttheorem for semisimple Lie groups of arbitrary real rank.Y. Ben Natan, Y. Benyamini, H. Hedenmalm and Y. Weit (in [BBH, BWH]) proved a genuineanalogue of the W-T theorem without the extended strip condition for L (SL(2 , R ) // SO(2)). In[PS] the authors extended this result to real rank one semisimple Lie group in the K -biinvariantsetting. In this article we generalize the result to the full group SL(2 , R ) and therefore this improvesthe corresponding result of [Rs1]..Let G be the group SL(2 , R ) and K be its maximal compact subgroup SO(2). A complex valuedfunction f on G is said to be of left (resp. right) K -type n if f ( kx ) = e n ( k ) f ( x ) (resp. f ( xk ) = e n ( k ) f ( x )) for all k ∈ K and x ∈ G, (1.1)where e n ( k θ ) = e inθ . For a class of functions F on G (e.g. L ( G )), F n denotes the correspondingsubclass of functions of right n type and F m,n will denote the subclass of F n which are also of lefttype m . We denote the subclass of F consisting of functions with integral zero by F .The main result (Theorem 1.1) of this article is an analogue of W-T theorem to the the full groupwithout the redundant extended strip condition. We first prove the W-T theorem for L ( G ) n,n (Theorem 1.3) for all n ∈ Z . This is the most crucial step in the direction of proving the W-Ttheorem to the full group. Before stating our main result we introduce some notation. For a Mathematics Subject Classification.
Primary 43A85; Secondary 22E30.
Key words and phrases.
Wiener Tauberian theorem, estimate of hypergeometric functions, resolvent transform. unction f ∈ L ( G ) its principal and discrete parts of the Fourier transform will be denoted by b f H and b f B respectively. Let M = {± I } and c M = { σ + , σ − } which consists the trivial ( σ + ) andthe non-trivial ( σ − ) irreducible representations of M . The representation π σ − , has two irreduciblesubrepresentations, so called mock series. We will denote them by D + and D − . The representationspaces of D + and D − contain e n ∈ L ( K ) respectively for positive odd n ’s and negative odd n ’s.For each σ ∈ c M , Z σ stands for the set of even integers for σ = σ + , and the set of odd integers for σ = σ − . Moreover, we express − σ by − σ + = σ − and − σ − = σ + . We define, S = { λ ∈ C | |ℜ λ | ≤ } and Γ n = ( { k | < k < n and k ∈ Z − σ } if n > { k | n < k < k ∈ Z − σ } if n < . For any function F on i R , we let δ ±∞ ( F ) = − lim sup t →∞ e − π t log | F ( ± it ) | . (1.2)We now state our main theorem. Theorem 1.1.
Let { f α | α ∈ Λ } be a collection of functions in L ( G ) such that the collections { c f αH | α ∈ Λ } and { c f αB | α ∈ Λ } have no common zero in c M × S ∪ { D + , D − } and Z ∗ respectively. If inf α ∈ Λ m,n ∈ Z δ ±∞ ( c f αH ) m,n = 0 , then the L ( G ) -bimodule generated by { f α | α ∈ Λ } is dense in L ( G ) . Moreover, if the integral of f α is zero for all α , then the ideal is dense in L ( G ) . For f ∈ L ( G ) n , the natural domain of the principal part b f H and the discrete part b f B of theFourier transform is S and Γ n respectively. To prove Theorem 1.1 we will prove the followingtheorem. Theorem 1.2.
Let { f α | α ∈ Λ } be a collections of functions in L ( G ) n such that the collection { c f αH | α ∈ Λ } and { c f αB | α ∈ Λ } have no common zero in S and Γ n respectively . Moreover, if inf α ∈ Λ ,m ∈ Z δ ±∞ ( c f αH ) m,n = 0 , then the left L ( G ) module generated by { f α | α ∈ Λ } is dense in L ( G ) n . Theorem 1.2 will follow from the theorem below.
Theorem 1.3.
Let { f α | α ∈ Λ } be a collection of functions in L ( G ) n,n and I be the smallestclosed ideal in L ( G ) n,n containing { f α | α ∈ Λ } such that the collection { ( c f αH ) n,n | α ∈ Λ } and { ( c f αB ) n,n | α ∈ Λ } have no common zero in S and Γ n respectively . Moreover, if inf α ∈ Λ δ ±∞ ( c f αH ) n,n = 0 ,then I = L ( G ) n,n . Proof of the theorem above borrows heavily from the ideas and methods of [BBH], [PS] whichuses the method of the resolvent transform. In the following we give a sketch of our proof.1. We will begin by showing that for all λ in C + = { λ ∈ C | ℜ λ > } except for a finite set B there is a family b λ such that b b λH ( iξ ) = λ + ξ for all ξ ∈ R and b b λB ( k ) = λ − k for all k ∈ Γ n . For ℜ λ > b λ ∈ L ( G ) n,n and { b λ | ℜ λ > λ / ∈ B } spans a dense subset of L ( G ) n,n . We will show || b λ || → λ → ∞ along the positive real axis.2. By the Banach algebra theory (using the fact that principal part and discrete part of theFourier transforms of the elements of I have no common zero), we define λ B λ as a L ( G ) n,n /I valued even entire function.3. Let g ∈ L ∞ ( G ) n,n such that g annihilates I . We define the resolvent transfrom R [ g ] by R [ g ]( λ ) = h B λ , g i , Considering g as a bounded linear functional on L ( G ) n,n /I , we write R [ g ]( λ ) = h B λ , g i , here B λ = b λ + I ∈ L ( G ) n,n /I , for all λ with ℜ λ > λ / ∈ B .4. We need an explicit formula of the function R [ g ]( λ ). For this we find a representative T λ f in L ( G ) n,n of the cosets B λ for 0 < ℜ λ < f ∈ I such that ( b f H ( σ, λ )) n,n = 0.5. By the estimates of || b λ || , || T λ f || and using a continuity argument we get the necessaryestimate of R [ g ]( λ ). Then using a log-log type theorem [PS, Theorem 6.3] we show R [ g ] = 0 .
6. By denseness of { b λ | ℜ λ > λ / ∈ B } , it follows that g = 0.Let Ω be the Casimir element of G . In [PS] the solutions φ , σ,λ and Φ , σ,λ ofΩ f = (cid:18) λ − (cid:19) f (1.3)played a crucial role and they are given in terms of hypergeometric functions in [Er]. From [AKS,p.31 ] observing a formula of φ n,nσ,λ we found a way to derive the second solution of (1.3) in terms ofhypergeometric functions. We are also able to write φ n,nσ,λ as a linear combination of Φ n,nσ,λ and Φ n,nσ, − λ that is, φ n,nσ,λ = c n,nσ ( λ )Φ n,nσ,λ + c n,nσ ( − λ )Φ n,n − λ . (1.4)It is an essential tool to find the Fourier transforms of b λ ’s.As in [PS, Lemma 8.1], using asymptotic behaviour Φ n,nσ,λ ( a t ) for λ ∈ C near t = ∞ we show,lim t →∞ φ n,nσ,iξ Φ n,nσ,λ ( a t ) e λt = 0 . (1.5)This directly gives b b λH ( iξ ) = λ + ξ , for all ξ ∈ R . But for a general n , we also need to find b b λB ( k )for all k ∈ Γ n and for that we need to showlim t →∞ φ n,nσ, | k | Φ n,nσ,λ ( a t ) e λt = 0 . (1.6)Here asymptotic behaviour of Φ n,nσ,λ is not enough. We have to use the full potential of decay ofdiscrete series matrix coefficient φ n,nσ, | k | . From [Ba, Theorem 8.1] we get φ n,nσ, | k | has sufficient decay as t → ∞ . Using this we proved (1.6) and consequently d b λB ( k ) = λ − k for all k ∈ Γ n .By inverse Fourier transform we find the representative of B λ . For all but finitely many λ with ℜ λ > B λ = b λ + I . In general there are some zeros of c n,nσ ( − λ ) in C + for n ∈ Z . Forthis reason we have to remove a neighbourhood B to find estimate of || b λ || on |ℜ λ | >
1. Usingestimate of || b λ || we find the estimate of R [ g ]( λ ) on { λ ∈ C | |ℜ λ | > } \ B . We use continuityto get estimate of R [ g ]( λ ) on B . Similarly using the estimate of || T λ f || we find the estimate of R [ g ]( λ ) on 0 < ℜ λ < R [ g ]( λ ). Finally using alog-log type theorem it will follow that R [ g ] = 0.Next we extend W-T theorem for L ( G ) n,n to L ( G ) n . From the given collection { f α | α ∈ Λ } ⊂ L ( G ) n and using the isomorphism between L -Schwartz space and its image under Fouriertransform we construct a new collection of L ( G ) n,n functions { g m ∗ f α | m ∈ Z σ } . We show thenew collection satisfy the hypothesis of Theorem 1.3. Therefore the collection { g m ∗ f α | m ∈ Z σ } is dense in L ( G ) n,n and consequently Theorem 1.2 will follow. Following similar idea as above weprove W-T theorem for L ( G ). . Preliminaries
In this article most of our notations are standard can be found in [Ba], [AKS], [PS] and [Rs1].We will denote C as constant and its value can change from one line to another. For any twopositive expressions f and f , f ≍ f stands for that there are positive constants C , C suchthat C f ≤ f ≤ C f . For z ∈ C we will use ℜ z and ℑ z to denote real and imaginary parts of z respectively.For k ∈ Z ∗ and σ ∈ c M be determined by k ∈ Z − σ and we define, Z ( k ) = ( { m ∈ Z σ : m ≥ k + 1 } if k ≥ { m ∈ Z σ : m ≤ k − } if k ≤ − . (2.1)The Iwasawa decomposition for G gives a gives a diffeomorphism of K × A × N onto G where A = { a t | t ∈ R } and N = { n ξ | ξ ∈ R } . That is, by Iwasawa decomposition x ∈ G can be uniquelywritten as x = k θ a t n ξ and using this we define K ( x ) = k θ and H ( x ) = t . Let A + = { a t = | t > } .The Cartan decomposition for G gives G = KA + K . Let dg , dn , dk and dm be the Haar meausresof G , N , K and M respectively where R K dk = 1 and R M dm = 1. We have the following integralformulae corresponding to the Cartan decomposition, which holds for any integrable function: Z G f ( x ) dx = Z K Z R + Z K f ( k a t k )∆( t ) dk dtdk , (2.2)where ∆( t ) = 2 sinh 2 t .For all λ ∈ C let us define, φ n,nσ,λ ( x ) = Z K e ( λ − H ( xk ) e n ( k − ) e n ( K ( xk ) − ) dk, for all x ∈ G. Then we have for all λ ∈ C , φ n,nσ,λ is a smooth eigenfunctions of the Casimir element Ω that is,Ω φ n,nσ,λ = λ − φ n,nσ,λ . Let Π n,n (Ω) be the differential operator on A \ { } defined byΠ n,n (Ω) f = d dt f ( a t ) + 2 coth 2 t ddt f ( a t ) + 14 n cosh t f ( a t ) , t > . Then from [Ba, p.62, eqn. (13.2)]) we get that φ n,nσ,λ is a solution of the following equation,Π n,n (Ω) f = ( λ − f. (2.3)We also have the following properties of φ n,nσ,λ :(1) φ n,nσ,λ is a ( n, n ) type function.(2) φ n,nσ,λ = φ n,nσ, − λ , φ n,nσ,λ ( a t ) = φ n,nσ,λ ( a − t ) .(3) For any fixed x ∈ G , λ φ n,nσ,λ ( x ) is an entire function.(4) | φ n,nσ,λ ( x ) | ≤ x ∈ G if λ ∈ S . For f ∈ L ( G ) n,n the principal and discrete parts of the Fourier transform are defined by, c f H ( σ, λ ) n,n = Z G f ( x ) φ n,nσ,λ ( x − ) dx for all λ ∈ S , (2.4) c f B ( k ) n,n = Z G f ( x ) ψ n,nk ( x − ) dx for all k ∈ Γ n . (2.5) t follows from Riemann-Lebesgue lemma that if f ∈ L ( G ) n,n then | c f H ( σ, λ ) n,n | → |ℑ λ | → ∞ in S . We also have from [Ba, p.30 propn 7.3] ψ n,nk = φ n,nσ, | k | for all k ∈ Γ n . (2.6)We denote C ( G ) the L - Schwartz space of G . Suppose σ ∈ c M , m, n ∈ Z σ then the space C H ( b G ) m,n denotes the collection of functions F : S → C such that(1) F is continuous on S and homomorphic on Int S ,(2) F ( λ ) = ϕ m,nλ F ( − λ ) for all λ ∈ S , where ϕ m,nλ = P m,n ( λ ) /P m,n ( − λ ) (2.7)is the rational function defined in [Ba, Prop. 7.1],(3) b ρ H,l,r ( F ) < ∞ for all l ∈ N , r ∈ R + , where b ρ H,l,r ( F ) = sup λ ∈ S (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:18) ddλ (cid:19) l F ( λ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (1 + | λ | ) r , (4) F ( k ) = 0 if nm < k ∈ Z − σ and | k | ≤ min {| m | , | n | , } . We note that for the particular case m = n , P n,n = 1 so the property 2 in the definition of C H ( b G ) n,n reduces to F ( λ ) = F ( − λ ) and property 4 becomes irrelevant. Let, Z m,n = { k | < k < min { m, n } and k ∈ Z − σ } if mn > , m > { k | max { m, n } < k < k ∈ Z − σ } if mn > , m < φ if mn ≤ C B ( b G ) m,n is the set of all functions F : Z m,n → C . Then from [Ba, Theorem 18.2] we have thefollowing result: Lemma 2.1.
The Fourier transform, f ( b f H , b f B ) is a topological isomorphism between C ( G ) m,n and C ( b G ) m,n = C H ( b G ) m,n × C B ( b G ) m,n . Moreover, the restriction of that isomorphism gives, (i) C H ( G ) m,n ismorphic to C H ( b G ) m,n , (ii) C B ( G ) m,n ismorphic to C B ( b G ) m,n . Hypergeometric function:
We are going to use the following properties of hypergeometric func-tion,(a) The hypergeometric function has the following integral representation for ℜ c > ℜ b > F ( a, b ; c ; z ) = Γ( c )Γ( b )Γ( c − b ) Z s b − (1 − s ) c − b − (1 − sz ) − a ds, | z | < . (2.8)(see [Ln, p, 239, eqn 9.1.4])(b) c ( c + 1) F ( a, b ; c ; z ) = c ( c − a + 1) F ( a, b + 1; c + 2; z )+ a [ c − ( c − b ) z ] F ( a + 1 , b + 1; c + 2; z ) , z ∈ C \ [1 , ∞ ) . (2.9)(see [Ln, p. 240, eqn. (9.1.7)])(c) Z x d − (1 − x ) b − d − F ( a, b ; c ; x ) dx = Γ( c )Γ( d )Γ( b − d )Γ( c − a − d )Γ( b )Γ( c − a )Γ( c − d ) , if ℜ d > , ℜ ( b − d ) > , ℜ ( c − a − d ) > . (2.10) see [Gr, p. 813, 7.512 (3)]) The functions b λ : Representatives of B λ , ℜ λ > φ n,nσ,λ and the second solutions of (2.3) in terms ofhypergeometric functions. We substitute f ( t ) = g ( t ) cosh n t, in the equation (2.3). Then we get the following ODE, d gdt + ((2 n + 1) tanh t + coth t ) dgdt + (( n + 1) − λ ) g = 0 , t > . By the change of variable z := − sinh t the equation reduces to the following hypergeometricdifferential equation z (1 − z ) d gdz + ( c − (1 + a + b ) z ) dgdz − abg = 0 , (2.11)with a = n +12 + λ , b = n +12 − λ , c = 1. Therefore g ( t ) = F (cid:18) n + 12 + λ , n + 12 − λ − sinh t (cid:19) , is a solution of (2.11) which is regular at origin, so by the uniqueness of regular solution, φ n,nσ,λ ( a t ) = (cosh t ) n F (cid:18) n + 12 + λ , n + 12 − λ − sinh t (cid:19) , t > . Also from [Er, p.105, 2.9 (11)] second solutions of (2.11) on (0 , ∞ ) are, g ( t ) = (cosh t ) − (1+ λ + n ) 2 F (cid:18) λ n , λ − n λ ; cosh − t (cid:19) ,g ( t ) = (cosh t ) − (1 − λ + n ) 2 F (cid:18) − λ n , − λ − n − λ ; cosh − t (cid:19) . We now define,Φ n,nσ,λ ( a t ) = (2 cosh t ) − (1+ λ ) 2 F (cid:18) λ | n | , λ − | n | λ ; cosh − t (cid:19) . (2.12)Then Φ n,nσ,λ and Φ n,nσ, − λ both are solutions of (2.3), both has singularity at t = 0 and they are linearlyindependent. For λ ∈ C \ Z , from [Er, p.110, 2.10(2,3 and 5)] we have, φ n,nσ,λ = c n,nσ ( λ )Φ n,nσ,λ + c n,nσ ( − λ )Φ n,nσ, − λ , (2.13)where σ is determined by n ∈ Z σ and c n,nσ ( λ ) = 2 λ Γ( − λ )Γ( − λ −| n | )Γ( − λ + | n | ) . (2.14)We have for t → ∞ , Φ n,nσ,λ ( a t ) = e − ( λ +1) t (1 + O (1)) . (2.15)Hence for ℜ λ < t → ∞ ,φ n,nσ,λ ( a t ) = c n,nσ ( λ ) e − ( λ +1) t (1 + O (1)) . (2.16)For simplicity if f ∈ L ( G ) n,n then we denote the principal and discrete parts of the Fouriertransform by c f H and c f B respectively. Since n ∈ Z σ determines σ we will use c n,n ( λ ) instead of c n,nσ ( λ ). et C + = { z ∈ C | ℜ ( z ) > } be the open right half plane and B = { λ ∈ Z − σ : 0 < λ < n } arethe zeros of c n,n ( − λ ) in the open right half plane.For λ ∈ C + \ B , we define b λ ( a t ) := 12 λc n,n ( − λ ) Φ n,nσ,λ ( a t ) , for t > . (2.17)Then for ξ ∈ R and ξ > n + 1, b ξ ( a t ) is positive. Now by Cartan decomposition we extend b λ as a( n, n ) type function on G \ K and so b λ is a solution of (1.3) on G \ K .Later on we shall need estimates of b λ ( a t ) near t = 0 and t = ∞ . For this purpose, we need thefollowing lemma. Lemma 2.2.
Let B = S k − i =0 B ( | n | − i −
1; 1) where B ( z ; 1) denotes a Euclidean ball of radius centered at z and k = [ | n | ] + 1 . Then for λ ∈ C + \ B , we have (i) There is a positive constant C independent of λ such that for all t ∈ (0 , / , | b λ ( a t ) | ≤ C log 1 t . (ii) There is a positive constant C independent of λ such that for all t ∈ [1 / , ∞ ) , | b λ ( a t ) | ≤ Ce − ( ℜ λ +1) t . Proof. (i) SinceΦ n,nσ,λ ( a t ) = (2 cosh t ) − (1+ λ )2 F (cid:18) λ | n | , λ − | n | λ ; cosh − t (cid:19) , we first find the estimate of the hypergeometric function near t = 0 and then by polynomialapproximation of gamma functions we will finally find the estimate of b λ . Now by (2.9), F (cid:18) λ + | n | , λ − | n | λ ; cosh − t (cid:19) (2.18)= 1(1 + λ )(2 + λ ) " (1 + λ ) (cid:18) λ − | n | (cid:19) F (cid:18) λ + | n | , λ − | n | λ + 2; cosh − t (cid:19) + 1 + λ + | n | (cid:18) (1 + λ ) − λ + | n | − t (cid:19) F (cid:18) λ + | n | , λ − | n | λ + 2; cosh − t (cid:19) Since (1 + λ )(2 + λ ) has no zero in C + and cosh t ≍ t = 0, so for all t ∈ (0 , ] we have(1 + λ )( λ −| n | + 1)(1 + λ )(2 + λ ) ≍ C and (cid:18) ( λ + | n | ) (cid:18) (1 + λ ) − λ + | n | − t (cid:19)(cid:19) (1 + λ )(2 + λ ) ≍ C for all λ ∈ C + .By the same argument and applying the formula (2.9) k = [ | n | ] + 1 times we can write , F (cid:18) λ + | n | , λ − | n | , ; 1 + λ ; cosh − t (cid:19) = k X i =0 P i ( λ, cosh − t ) Q i ( λ ) F (cid:18) λ + | n | i, λ − | n | k ; 1 + λ + 2 k ; cosh − t (cid:19) where P i ’s are polynomials in λ and cosh − t and Q i ’s are polynomials in λ which has no zero in C + such that for all t ∈ (0 , ], P i ( λ, cosh − t ) Q i ( λ ) ≍ C or all λ ∈ C + . Now since ℜ (1 + λ + 2 k ) > ℜ ( λ −| n | + k ) from (2.8), F (cid:18) λ + | n | i, λ − | n | k ; 1 + λ + 2 k ; cosh − t (cid:19) = C ( λ ; n ) Z s λ −| n | + k − (1 − s ) λ + | n | + k − (1 − s cosh − t ) − λ + | n | − i ds where C ( λ ; n ) = Γ(1 + λ + 2 k )Γ (cid:16) λ −| n | + k (cid:17) Γ (cid:16) λ + | n | + k (cid:17) . = C ( λ ; n )(cosh t ) λ + | n | +2 i Z s λ −| n | + k − (1 − s ) λ + | n | + k − (cosh t − s ) − λ + | n | − i ds ( writing cosh t = 1 + x and making the change of variable s → − s we get)= C ( λ ; n )(cosh t ) λ + | n | +2 i Z (1 − s ) λ −| n | + k − s λ + | n | + k − ( x + s ) − λ + | n | − i ds. (2.19)Let I be the integration above. Then, | I | ≤ Z (1 − s ) ℜ λ −| n | + k − s ℜ λ + | n | + k − ( x + s ) − ℜ λ +2 − i ds. Now we let I be the integration on (0 , ] and I on ( , I ≤ C Z (1 − s ) − −| n | + k −ℜ λ ℜ λ + | n | + i ds ≤ C (independent of λ ) , and I ≤ C Z s ℜ λ + | n | + k − ( x + s ) − ℜ λ + | n | − i ds ≤ C − ℜ λ + | n | − k ( x + ) − ℜ λ + | n | − i ℜ λ + | n | + k + C ℜ λ + | n | + i ℜ λ + | n | + k Z s ℜ λ + | n | + k ( x + s ) − ( ℜ λ + | n | + i ) − ds ≤ C + C Z (cid:18) sx + s (cid:19) ℜ λ + | n | + i x + s ds ≤ C + C log (cid:18) x (cid:19) . Since x = sinh t and log is an increasing function we have, I ≤ C log t , hence it follows | I | ≤ C log t for all t ∈ (0 , ]. e now turn to the estimate of b λ . Using the inequality above of | I | and applying the expressionof c n,n ( λ ) in (2.17) we get, | b λ ( a t ) | ≤ C (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Γ (cid:18) λ − | n | (cid:19) Γ( λ + | n | )Γ(1 + λ + 2 k )Γ(1 + λ )Γ( λ −| n | + k )Γ( λ + | n | + k ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) log 1 t ≤ C (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Γ( λ −| n | + k )Γ( λ + | n | )Γ(1 + λ + 2 k )( λ −| n | )( λ −| n | + 1) · · · ( λ −| n | + k − λ )Γ( λ −| n | + k )Γ( λ + | n | + k ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) log 1 t ≤ C (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (1 + | λ | ) k ( λ −| n | )( λ −| n | + 1) · · · ( λ −| n | + k − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) log 1 t The last line of the inequality above follows from [PS, Appendix, Lemma A.3 ] and the fact that k − | n | ≥ . Therefore for all λ ∈ C + \ B , (where B = S k − i =0 B ( | n | − i −
1; 1) ) | b λ ( a t ) | ≤ C log 1 t , (2.20)for all t ∈ (0 , ] and C is independent of λ .(ii) Since Φ n,nσ,λ ( a t ) ≍ e − ( ℜ λ +1) t near ∞ and by the definition of b λ and c n,n ( λ ) we get for all t ∈ [ , ∞ ) , | b λ ( a t ) | ≤ C (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Γ( λ −| n | )Γ( λ + | n | )Γ(1 + λ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) | Φ n,nσ,λ ( a t ) |≤ C (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Γ( λ −| n | + k )Γ( λ + | n | )( λ −| n | )( λ −| n | + 1) · · · ( λ −| n | + k − λ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) e − ( ℜ λ +1) t ≤ C (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (1 + | λ | ) k − ( λ −| n | )( λ −| n | + 1) · · · ( λ −| n | + k − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) e − ( ℜ λ +1) t (using [PS, Appendix, Lemma A.3 ]) ≤ C e − ( ℜ λ +1) t (Since λ / ∈ B ).Therefore for all t ∈ [ , ∞ ) , | b λ ( a t ) | ≤ Ce − ( ℜ λ +1) t where C is independent of λ . (cid:3) Remark 2.3.
The proof of the lemma above shows that to get the estimate of b λ near 0 and ∞ we only need to remove a neighbourhood of the zeros of c n,n ( − λ ) and origin (when n is odd). If weonly remove the zeros of c n,n ( − λ ) and origin but not the neighbourhoods then on both cases theconstant on the right hand side will depend on λ , for example for t ∈ (0 , ] , | b λ ( a t ) | ≤ C λ log t . Lemma 2.4. (a)
For all λ ∈ C + \ B , b λ is locally integrable at e . (b) For ℜ λ > and λ B , b λ ∈ L ( G ) n,n . (c) For all λ ∈ C + and λ B , b λ is in L outside neighbourhood of e . (d) For each λ ∈ C + \ B , there exists p < (depending on λ ) such that b λ is in L p outsideneighbourhood of e .Proof. Proof of this lemma follows directly from previous Lemma 2.2 and the asymptotic behaviourof ∆( t ) near 0 and ∞ . (cid:3) emark 2.5. By the lemma above b λ can be written as a sum of L and L p ( p <
2) functions on G . Therefore its principal part of the Fourier transform is a continuous function on C , vanishingat infinity in C . In fact in the next lemma we are going to find the Fourier transforms of b λ . Lemma 2.6.
Let λ ∈ C + \ B . Then, b b λH ( iξ ) = 1 λ + ξ , for all ξ ∈ R and b b λB ( k ) = 1 λ − k , for all k ∈ Γ n . Proof.
For two smooth functions f and g on (0 , ∞ ), we define[ f, g ]( t ) = ∆( t ) (cid:2) f ′ ( t ) g ( t ) − f ( t ) g ′ ( t ) (cid:3) , t > . An easy calculation shows that [ f, g ] ′ ( t ) = [Π n,n (Ω) f · g − f · Π n,n (Ω) g ]( t )∆( t ). Therefore, for any b > a >
0, we have Z ba (Π n,n (Ω) f · g − f · Π n,n (Ω) g )( t )∆( t ) = [ f, g ]( b ) − [ f, g ]( a ) . (2.21)Then by similar calculations in [PS, Lemma 8.1] we have the following two results,[ φ n,nσ,λ , Φ n,nσ,λ ]( · ) = 2 λc n,n ( − λ )and if f is an even smooth function on R thenlim t → + [ f, Φ n,nσ,λ ]( t ) = 2 λc n,n ( − λ ) f (0) . (2.22) CASE 1 : b b λH ( iξ ) = λ + ξ , for all ξ ∈ R .For ξ ∈ R , we put f = φ n,nσ,iξ , g = Φ n,nσ,λ in equation (2.21) we get, Z ba Φ n,nσ,λ ( t ) φ n,nσ,iξ ( t )∆( t ) dt = 1 − λ − ξ (cid:16) [ φ n,nσ,iξ , Φ n,nσ,λ ]( b ) − [ φ n,nσ,iξ , Φ n,nσ,λ ]( a ) (cid:17) . Taking a → + , we get from (2.22) Z b Φ n,nσ,λ ( t ) φ n,nσ,iξ ( t )∆( t ) dt = 2 λc n,n ( − λ ) λ + ξ − [ φ n,nσ,iξ , Φ n,nσ,λ ]( b ) λ + ξ . Therefore if we could show [ φ n,nσ,iξ , Φ n,nσ,λ ]( b ) → b → ∞ then we will be done.We note that the existence of limit is guaranteed by the equation above. As like before we canwrite, lim b →∞ [ φ n,nσ,iξ , Φ n,nσ,λ ]( b ) = lim b →∞ e − λb φ n,nσ,iξ Φ n,nσ,λ ! ′ ( b ) . By the asymptotic behavior of φ n,nσ,iξ and Φ n,nσ,λ ,lim b →∞ φ n,nσ,iξ Φ n,nσ,λ ( b ) e λb = 0 . Finally for all λ ∈ C + \ B , b b λH ( iξ ) = 1 λ − ( iξ ) , for all ξ ∈ R . ASE 2 : b b λB ( k ) = 1 λ − k for all k ∈ Γ n .We note that from [Ba, p.30 propn 7.3] we have ψ n,nk = φ n,nσ, | k | for all k ∈ Γ n .Let k ∈ Γ n , we put f = ψ n,nk , g = Φ n,nσ,λ in equation (2.21) to get, Z ba Φ n,nσ,λ ( t ) ψ n,nk ( t )∆( t ) dt = 1 − λ + k (cid:0) [ ψ n,nk , Φ n,nσ,λ ]( b ) − [ ψ n,nk , Φ n,nσ,λ ]( a ) (cid:1) . Taking a → + , we get from (2.22) Z b Φ n,nσ,λ ( t ) ψ n,nk ( t )∆( t ) dt = 2 λc n,n ( − λ ) λ − k − [ ψ n,nk , Φ n,nσ,λ ]( b ) λ − k . Therefore if we could show [ ψ n,nk , Φ n,nσ,λ ]( b ) → b → ∞ then we will be done.From [Ba, p.33 Theorem 8.1] we get that there exist constants C, r , r , r ≥ | ψ n,nk ( t ) | ≤ C (1 + | n | ) r (1 + | k | ) r (1 + t ) r e − t for all k ∈ Z ∗ for which | k | ≥ n ∈ Z ( k ) . Now by the asymptotic behaviour of Φ n,nσ,λ we get, (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ψ n,nk ( b ) e λb Φ n,nσ,λ ( b ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C (1 + | k | ) r (1 + b ) r e − b e λb e − ( λ +1) b ≤ C (1 + | k | ) r (1 + b ) r e ( λ +1) b for a fixed n . Therefore lim b →∞ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ψ n,nk ( b ) e λb Φ n,nσ,λ ( b ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 0. This completes the proof. (cid:3) Remark 2.7.
Since for ℜ λ > λ B , b λ is in L ( G ) n,n and its principal Fourier transform isa well defined continuous function on the strip S , which is also holomorphic in S o . Therefore byanalytic continuation we can write for ℜ λ > λ B , b b λH ( z ) = 1 λ − z , for all z ∈ S . We now turn to the estimates of || b λ || which is essential in § Lemma 2.8. (i) If ℜ λ > and λ B , || b λ || ≤ C (1+ | λ | ) ℜ λ − for some C > .(ii) || b λ || → if λ → ∞ along the positive real axis.Proof. (i) Since ∆( t ) ≍ t near 0 and ∆( t ) ≍ e t near ∞ , from Lemma 2.2 we can write, || b λ || = Z / | b λ ( a t ) | ∆( t ) dt + Z ∞ | b λ ( a t ) | ∆( t ) dt ≤ Z t log 1 t + C Z ∞ e (1 −ℜ λ ) t ≤ C + C ℜ λ − ≤ C | λ |ℜ λ − . ii)If λ = ξ ∈ R and ξ > n + 1 then b ξ ( a t ) is nonnegative. Hence || b ξ || = Z R + b ξ ( a t )∆( t ) dt ≤ Z R + (cosh t ) | n | b ξ ( a t )∆( t ) dt = 1 ξ − ( n + 1) . The last line of the inequalities follows from similar calculation of [PS, Lemma 3.3], which uses(2.10). Hence the proof follows. (cid:3)
Lemma 2.9.
The functions { b λ | ℜ λ > and λ B } span a dense subset of L ( G ) n,n .Proof. We will show that span { b λ |ℜ λ > λ / ∈ B } contains C ∞ c ( G ) n,n and since C ∞ c ( G ) n,n isdense in L ( G ) n,n , the lemma will follow.Let f ∈ C ∞ c ( G ) n,n . Since b f H is entire and it has polynomial decay on any bounded vertical strip(by Paley-Wiener theorem) Cauchy’s formula implies that b f H ( w ) = 12 πi Z Γ b f H ( z ) z − w dz + 12 πi Z Γ b f H ( z ) z − w dz, for w ∈ C . where Γ = ( | n | + 2) + i R downward and Γ = − ( | n | + 2) + i R upward. Next by the change ofvariable z → − z in the second integral, b f H ( w ) = 12 πi Z Γ b f H ( z ) z − w dz + 12 πi Z Γ b f H ( − z ) − z − w ( − dz ) . We know b f H ( z ) is an even function, therefore for all w ∈ C b f H ( w ) = 12 πi Z Γ z b f H ( z ) z − w dz. (2.23)Since b f B ( k ) = b f H ( | k | ) for all k ∈ Γ n , so from (2.23) and together with Lemma 2.6 we get, b f H ( w ) = 12 πi Z Γ z b f H ( z ) b b zH ( w ) dz, for all w ∈ S , (2.24) b f B ( k ) = 12 πi Z Γ z b f H ( z ) b b zB ( k ) dz, for all k ∈ Γ n . (2.25)The decay condition on b f H and Lemma 2.8 imply that the L ( G ) n,n valued integral12 πi Z Γ z b f H ( z ) b z ( · ) dz converges and (2.24), (2.25) implies that it must converge to f . Thus the Riemann sums which arenothing but finite linear combinations of b λ ’s converge to f . So we can conclude that f is in theclosed subspace spanned by { b λ | ℜ λ > λ B } . The lemma follows. (cid:3) Resolvent transform
Let L δ ( G ) n,n be the unitization of L ( G ) n,n and δ , where δ is the ( n, n ) type distribution definedby δ ( φ ) = φ ( e ) for all φ ∈ C ∞ c ( G ) n,n . Maximal ideal space of L δ ( G ) n,n is (cid:8) L z : z ∈ S ∪ {∞} (cid:9) and (cid:8) L ′ k : k ∈ Γ n (cid:9) , where L z and L ′ k are the complex homomorphism on L δ ( G ) n,n defined by L z ( f ) = b f H ( z ) and L ′ k ( f ) = b f B ( k ) for all f ∈ L δ ( G ) n,n . rom now on we will denote I as a closed ideal of L ( G ) n,n such that { b f H : f ∈ I } and { b f B : f ∈ I } does not have common zero on S and Γ n respectively. Since δ ∗ f = f for all f ∈ L ( G ) n,n so I isalso an ideal of L δ ( G ) n,n and L δ ( G ) n,n /I makes sense.In Banach algebra theory if J is a closed ideal of a commutative Banach algebra A then themaximal ideal space of A /J is Σ( A /J ) = { h ∈ Σ( A ) : h = 0 on J } , where Σ( A ) denotes the maximal ideal space of A . From the theory above the maximal ideal space of L δ ( G ) n,n /I is the complex homomorphism ˜ L ∞ and it is defined by ˜ L ∞ ( f + I ) = b f H ( ∞ ) for all f ∈ L δ ( G ) n,n /I. It also follows that an element f + I in L δ ( G ) n,n /I is invertible if and only if b f H ( ∞ ) = 0.Let λ be a fixed complex number with ℜ λ > n + 1. Then by Lemma 2.4 b λ is in L ( G ) n,n .For λ ∈ C the function, λ b δ − ( λ − λ ) c b λ H does not vanish at ∞ and hence δ − ( λ − λ ) b λ + I is invertible in the quotient algebra L δ ( G ) n,n /I .We put B λ = (cid:0) δ − ( λ − λ ) b λ + I (cid:1) − ∗ ( b λ + I ) , for λ ∈ C . (3.1)Now let g ∈ L ∞ ( G ) n,n annihilates I , so we can take g as a bounded linear functional on L ( G ) n,n /I .We define the resolvent transform R [ g ] of g by R [ g ]( λ ) = h B λ , g i . (3.2)From (3.1), λ B λ is a Banach space valued even holomorphic function on C . So R [ g ] is an evenholomorphic function on C .We need an explicit formula of the function R [ g ] almost everywhere in C . We will show for ℜ λ > λ / ∈ B , B λ = b λ + I . Also, for 0 < ℜ λ < B λ inthe next section.4. Representatives of B λ , < ℜ λ < and properties of R [ g ]Let λ be such that 0 < ℜ λ <
1. For f ∈ L ( G ) n,n we define T λ f := b f H ( λ ) b λ − f ∗ b λ . (4.1)Since b λ is a sum of L and L p functions (by Lemma 2.4) T λ f is well defined and the principal anddiscrete part of Fourier transforms exist on i R and Γ n respectively. The proof follows directly fromLemma 2.6. Lemma 4.1.
Let < ℜ λ < and f be a L ( G ) n,n function on G. Then d T λ f H ( iξ ) = b f H ( λ ) − b f H ( iξ ) λ + ξ , for all ξ ∈ R , d T λ f B ( k ) = b f H ( λ ) − b f B ( k ) λ − k , for all k ∈ Γ n . Lemma 4.2.
Let λ ∈ C + \ B . Then, Z K b λ ( a s ka t ) e n ( k − ) dk = ( b λ ( a s ) φ n,nσ,λ ( a t ) if s > t ≥ ,b λ ( a t ) φ n,nσ,λ ( a s ) if t > s ≥ . roof. Since b λ is smooth outside K and a s ka t K as s = t , the integral is well defined. Fix s > b λ is a ( n, n ) type eigenfunction of Ω on G \ K with eigenvalue λ − , the function g Z K b λ ( a s kg ) e n ( k − ) dk is smooth ( n, n ) type eigenfunction of Ω on the open ball B s = { k a r k ∈ KA + K | r < s } .Hence the function t R K b λ ( a s ka t ) e n ( k − ) dk is a solution of (2.3) on (0 , s ) which is regular at 0.Therefore, Z K b λ ( a s ka t ) e n ( k − ) dk = Cφ n,nσ,λ ( a t ) for all 0 ≤ t < s and for some constant C.Putting t = 0 in the equation above we get C = b λ ( a s ). Therefore for s > t ≥ Z K b λ ( a s ka t ) e n ( k − ) dk = b λ ( a s ) φ n,nσ,λ ( a t ) . Similarly the second case follows. (cid:3)
Next we will show T λ f is in L ( G ) n,n and to do that we will use the following representation of T λ f . Lemma 4.3.
Let < ℜ λ < and f ∈ L ( G ) n,n . Then for all t > , T λ f ( a t ) = b λ ( a t ) Z ∞ t f ( a s ) φ n,nσ,λ ( a s )∆( s ) ds − φ n,nσ,λ ( a t ) Z ∞ t f ( a s ) b λ ( a s )∆( s ) ds. Proof.
Here we are going to use the fact that there exists k ∈ K such that k a s k − = a − s for all s ≥ . Now f ∗ b λ ( a t ) = Z K Z ∞ Z K f ( k a s k ) b λ ( k − a − s k − a t )∆( s ) dk dsdk = Z ∞ f ( a s ) Z K b λ ( a − s k a t ) e n ( k − )∆( s ) dk ds ( change of variable k → k − )= Z ∞ f ( a s ) Z K b λ ( k a s k − k a t ) e n ( k − )∆( s ) dk ds = Z ∞ f ( a s ) Z K b λ ( a s k a t ) e n ( k − )∆( s ) dk ds ( change of variable k → k k )= Z t f ( a s ) b λ ( a t ) φ n,nσ,λ ( a s )∆( s ) ds + Z ∞ t f ( a s ) b λ ( a s ) φ n,nσ,λ ( a t )∆( s ) ds (4.2)The last line follows from Lemma 4.2. Next, b f H ( λ ) b λ ( a t ) = b λ ( a t ) Z K Z ∞ Z K f ( k a s k ) φ n,nσ,λ (( k − a − s k − )∆( s ) dk dsdk = b λ ( a t ) Z ∞ f ( a s ) φ n,nσ,λ ( a s )∆( s ) ds (Since φ n,nσ,λ ( a − s ) = φ n,nσ,λ ( a s )) . (4.3)Putting the expressions above (4.2) and (4.3) in the definition of T λ f the result follows. (cid:3) Next we show T λ f is in L ( G ) n,n for 0 < ℜ λ < || T λ f || . Lemma 4.4.
Let < ℜ λ < and f be a ( n, n ) type integrable function on G . Then T λ f ∈ L ( G ) n,n and moreover if λ / ∈ B (0; 1) ∪ B (1; 1) , its L norm satisfies, || T λ f || ≤ C || f || (1 + | λ | ) d ( λ, ∂S ) − , here d ( λ, ∂S ) denotes the Euclidean distance of λ from the boundary ∂S of the strip S .Proof. Proof of the Lemma above follows exactly in the same line as [PS, Lemma 4.4]. (cid:3)
Now we summarize the necessary properties of the resolvent transform.
Lemma 4.5.
Assume g ∈ L ∞ ( G ) n,n annihilates I and fix a function f ∈ I . Let Z ( b f H ) := { z ∈ S : b f H ( z ) = 0 } . Then(a) R [ g ]( λ ) is an even holomorphic function on C . It is given by the following formula : R [ g ]( λ ) = ( h b λ , g i , ℜ λ > , λ B h T λ f,g i b f H ( λ ) , < ℜ λ < , λ / ∈ Z ( b f H ) . (b) For |ℜ λ | > , |R [ g ]( λ ) | ≤ C || g || ∞ (1+ | λ | ) d ( λ,∂S ) , (c) For |ℜ λ | < , (cid:12)(cid:12)(cid:12) b f H ( λ ) R [ g ]( λ ) (cid:12)(cid:12)(cid:12) ≤ C || f || || g || ∞ (1+ | λ | ) d ( λ,∂S ) , where the constant C is independent of f ∈ I .Proof. (a) CASE-1 : Let ℜ λ > λ B then by (2.4) b λ is in L ( G ) n,n . For z ∈ S we havefrom Lemma 2.6 and 2.7, 1 b b λ H ( z ) − b b λH ( z ) = λ − λ so , (cid:16) − ( λ − λ ) b b λ H ( z ) (cid:17) b b λH ( z ) = b b λ H ( z ) . Similarly for k ∈ Γ n we have, (cid:16) − ( λ − λ ) b b λ B ( k ) (cid:17) b b λB ( k ) = b b λ B ( k ) . So (cid:0) δ − ( λ − λ ) b λ ( · ) (cid:1) b λ ( · ) = b λ ( · )as L δ ( G ) n,n functions. Hence in the quotient algebra L δ ( G ) n,n /I , (cid:0) δ − ( λ − λ ) b λ + I (cid:1) ∗ ( b λ + I ) = b λ + I, (4.4)Now ( δ − ( λ − λ ) b λ + I ) is invertible in L δ ( G ) n,n /I so from (3.1) and (4.4)we get B λ = b λ + I Therefore by the definition of R [ g ]( λ ), R [ g ]( λ ) = h b λ , g i . CASE-2 : Let 0 < ℜ λ < λ / ∈ Z ( b f H ). Then by Lemma 4.4 T λ f is in L ( G ) n,n . Similarly asin previous case we have from Lemma 4.1, (cid:16) − ( λ − λ ) b b λ H ( z ) (cid:17) d T λ f H ( z ) b f H ( λ ) = b b λ H ( z ) − b f H ( z ) c b λ H ( z ) b f H ( λ ) for all z ∈ S and (cid:16) − ( λ − λ ) b b λ B ( k ) (cid:17) d T λ f B ( k ) b f H ( λ ) = b b λ B ( k ) − b f B ( k ) c b λ B ( k ) b f H ( λ ) for all k ∈ Γ n . Therefore (cid:0) δ − ( λ − λ ) b λ ( · ) (cid:1) T λ f ( · ) b f H ( λ ) ! = b λ ( · ) − f ( · ) b λ ( · ) b f H ( λ n L δ ( G ) n,n . Since f ∈ I (cid:0) δ − ( λ − λ ) b λ + I (cid:1) ∗ T λ f b f ( λ ) + I ! = b λ + I. (4.5)Again from (3.1) and the equation above B λ = T λ f b f ( λ ) + I, which implies R [ g ]( λ ) = h T λ f , g i b f ( λ ) . (b) Since R [ g ]( λ ) is even we only need to consider the case ℜ λ >
1. For ℜ λ > λ B wehave from Lemma 2.8, || b λ || ≤ C (1 + | λ | ) d ( λ, ∂S ) for some C > . Now from (3.2) it follows that R [ g ]( λ ) is bounded on B . Hence |R [ g ]( λ ) | ≤ C || g || ∞ (1 + | λ | ) d ( λ, ∂S ) . (c) From Lemma 4.4 we get for 0 < ℜ λ < λ B (0; 1) ∪ B (1; 1), (cid:12)(cid:12)(cid:12) b f H ( λ ) R [ g ]( λ ) (cid:12)(cid:12)(cid:12) ≤ C || f || || g || ∞ (1 + | λ | ) d ( λ, ∂S ) . Since b f H ( λ ) R [ g ]( λ ) is an even continuous function on S , the same estimate is true for 0 < |ℜ λ | < , λ B (0; 1) ∪ B (1; 1). Now from (3.2) it follows that R [ g ]( λ ) is bounded on B (0; 1) ∪ B (1; 1) withbound independent of f , Therefore for 0 < |ℜ λ | < λ ∈ B (0; 1) ∪ B (1; 1) , (cid:12)(cid:12)(cid:12) b f ( λ ) R [ g ]( λ ) (cid:12)(cid:12)(cid:12) ≤ C || f || , where C is independent of f and λ . So we have for 0 < |ℜ λ | < (cid:12)(cid:12)(cid:12) b f H ( λ ) R [ g ]( λ ) (cid:12)(cid:12)(cid:12) ≤ C || f || || g || ∞ (1 + | λ | ) d ( λ, ∂S ) . Finally the constant in the inequality above is independent of f so by continuity of R [ g ] and b f thelemma follows. (cid:3) Results from complex analysis
For any function F on i R , we let δ + ∞ ( F ) = − lim sup t →∞ e − π t log | F ( it ) | and δ −∞ ( F ) = − lim sup t →∞ e − π t log | F ( − it ) | . Next from [PS, Theorem 6.3] we have the following theorem.
Theorem 5.1.
Let M : (0 , ∞ ) → ( e, ∞ ) be a continuously differentiable decreasing function with lim t → + t log log M ( t ) < ∞ , Z ∞ log log M ( t ) dt < ∞ . Let Ω be a collection of bounded holomorphic functions on S such that inf F ∈ Ω δ + ∞ ( F ) = inf F ∈ Ω δ −∞ ( F ) = 0 . uppose H satisfies the following estimates for some nonnegative integer N: | H ( z ) | ≤ (1 + | z | ) N M ( d ( z, ∂S )) , z ∈ C \ S , | F ( z ) H ( z ) | ≤ (1 + | z | ) N M ( d ( z, ∂S )) , z ∈ S , for all F ∈ Ω . If in addition, H is a holomorphic function on S \{± } then H is dominated by a polynomialoutside a bounded neighbourhood of {± } . If H is an entire function, then it is a polynomial. Proof of W-T Theorem for L ( G ) n,n Proof of Theorem 1.3.
Since the ideal generated by { f α | α ∈ Λ } is same as the ideal generatedby the elements n f α || f α || | α ∈ Λ o and δ ±∞ ( b f H ) = δ ±∞ c f αH || f || ! , we can assume that the functions f α are of unit L norm. Let g ∈ L ∞ ( G ) n,n annihilates the closed ideal I generated by { f α | α ∈ Λ } .We will show that g = 0. Then by an application of Hahn Banach theorem it will follow that I = L ( G ) n,n . From the hypothesis we have,inf α ∈ Λ δ + ∞ ( c f αH ) = inf α ∈ Λ δ −∞ ( c f αH ) = 0 . By Lemma 4.5, the entire function R [ g ] satisfies the following estimates |R [ g ]( z ) | ≤ C (1 + | z | ) ( d ( z, ∂S )) − , z ∈ C \ S , | c f αH ( z ) R [ g ]( z ) | ≤ C (1 + | z | ) ( d ( z, ∂S )) − , z ∈ S , for all α ∈ Λ, where C is a constant and we choose it is greater than e . We can define M : (0 , ∞ ) → ( e, ∞ ) to be a continuously differentiable decreasing function such that M ( t ) = Ct for 0 < t < R ∞ log log M ( t ) dt < ∞ . With this definition of M , we have |R [ g ]( z ) | ≤ (1 + | z | ) M ( d ( z, ∂S )) z ∈ C \ S , | c f αH ( z ) R [ g ]( z ) | ≤ (1 + | z | ) M ( d ( z, ∂S )) z ∈ S , for all α ∈ Λ . Therefore, by Theorem 5.1, R [ g ]( z ) is a polynomial. From Lemma 4.5, R [ g ]( z ) ≤ || b z || || g || ∞ . Then Lemma 2.8 implies R [ g ]( z ) → z → ∞ along the positive real axis. Therefore R [ g ]must be the zero polynomial. Hence h b λ , g i = 0 whenever ℜ λ > λ B but the collection { b λ | ℜ λ > λ B } spans a dense subset of L ( G ) n,n by Lemma 2.9. So g = 0 and the prooffollows. (cid:3) Finally we like to mention here that we first started to prove a W-T theorem for L ( G ) m,n but our method fails in this general setting as L ( G ) m,n is not necessarily a commutative banachalgebra. 7. Final Results
Now we prove Wiener Tauberian theorem for L ( G ) n using Theorem 1.3. Here we will followsimilar technique as in [Rs1].For f ∈ L ( G ) we have from [Ba, p. 30, prop 7.3],( b f B ( k )) m,n = η m,n ( k ) ( b f H ( k )) m,n for all k ∈ {± } and m, n ∈ Z ( k ) , (7.1)where η m,n ( k ) is a positive number. Therefore( b f B ( k )) m,n = 0 ⇔ ( b f H ( k )) m,n = 0 . uppose b f B ( k ) = 0 for all k ∈ Γ n , then it implies the following:(a) If n is positive then for every m < n, ( b f B ( n − m,n = 0, so f has at least one non zerocomponent of left type m such that m ≥ n . Similarly when n is negative f has at least oneleft type m for some m ≤ n .(b) Let f ∈ L ( G ) n and n is even. If n > b f B (1) = 0 and sothere is an m such that m ∈ Z (1) and ( b f B (1)) m,n = 0. Therefore ( b f H (1)) m,n = 0 . For n < Proof of Theorem 1.2.
We first consider the case when the collection indexed by Ω contains exactlyone function, f ∈ L ( G ) n . Let f m ( x ) = R π e − imθ f ( k θ x ) dθ for all m ∈ Z . Then f m is an ( m, n )type function and ( m, n )-th matrix coefficient b f H , ( b f H ) m,n = c f mH .Now we will construct a family of functions in C , {G m ( · ) | m ∈ Z σ } such that G ∈ C H ( b G ) n,m .When mn ≥ G m ( λ ) = e − λ Q n,m ( λ ) where Q n,m = P n,m which is the numerator ofthe rational function ϕ n,mλ from (2.7). Hence e − λ Q n,m ( λ ) = ϕ n,mλ e − λ Q n,m ( − λ ) which shows that G m ( λ ) = e − λ Q n,m ( λ ) ∈ C H ( b G ) n,m (7.2)for the case mn ≥
0. Here we note that Q n,m (0) = 0.If mn < G m ( λ ) to satisfy all the properties of C H ( b G ) n,m . Case 1.
Let n be odd. Then we take the polynomial Q ′ n,m ( λ ) = P n,m ( λ ) · λ . Now Q ′ n,m (0) = 0 and e − λ Q ′ n,m ( λ ) = ϕ n,mλ e − λ Q ′ n,m ( − λ ). Therefore, in this case G m ( λ ) = e − λ Q ′ n,m ( λ ) ∈ C H ( b G ) n,m . (7.3) Case 2.
Let n be even( hence | n | , | m | ≥ nm < Q ′′ n,m ( λ ) = P n,m ( λ )(1 − λ ). So Q ′′ n,m ( ±
1) = 0 and e − λ Q ′′ n,m ( λ ) = ϕ n,mλ e − λ Q ′′ n,m ( − λ ). Therefore inthis case also, G m ( λ ) = e − λ Q ′′ n,m ( λ ) ∈ C H ( b G ) n,m . (7.4)Now for all n, m G m ( λ ) c f mH ( λ ) = e − λ Q n,m ( λ ) c f mH ( λ )= e − λ Q n,m ( − λ ) ϕ n,mλ ϕ m,nλ c f mH ( − λ )= G m ( − λ ) c f mH ( − λ ) . Since f m is an ( m, n ) type function on G so c f mH ( λ ) = ϕ m,nλ c f mH ( − λ ) and ϕ n,mλ = ( ϕ m,nλ ) − .This shows that for all m , G m ( λ ) c f mH ( λ ) is the Fourier transform of an ( n, n ) type function withrespect to principal series representation. Now we claim that λ ∈ S there is an m such that G m ( λ ) c f mH ( λ ) = 0. The only possible zeros of the polynomials Q n,m , Q ′ n,m and Q ′′ n,m in S are { , ± } and everywhere else it is non-zero. Given b f H ( λ ) = 0 for all λ ∈ S . If we could show thatfor each λ ∈ { , ± } there is an m such that G m ( λ ) c f mH ( λ ) = 0 then we will be done.Before proving our claim we find out exactly when { , ± } are zeros of the polynomials above.(i) P n,m ( −
1) = 0 if and only if n = 0 and m = 0, P n,m (+1) = 0 for all m = 0 and P n, (+1) = 0 when n = 0,therefore Q n,m ( ± = 0 when nm = 0 and Q ′ n,m ( ± = 0 for all m = 0 . (7.5) ii) Since Q n,m (0) = 0 so Q ′′ n,m (0) = 0 . (7.6) Q ′ n,m (0) = 0 and Q ′′ n,m ( ±
1) = 0 . (7.7)First we consider the case for λ = 0 . By hypothesis there is an m such that c f mH (0) = 0.If n is odd then mn >
0, otherwise φ m,nσ − , ≡ c f mH (0) = 0. Therefore G m (0) c f mH (0) =0 as Q n,m (0) = 0.Next suppose n is even. Now If nm ≥ G m (0) c f mH (0) = 0 as Q n,m (0) = 0. When nm < G m (0) c f mH (0) = 0 because from (7.6) Q ′′ n,m (0) = 0 .Now we prove our claim for λ = ±
1. Here we will consider several case for n . Case 1.
Let n = 0, then c f mH (1) = 0 for all m = 0 as φ m, σ + , ≡
0. Therefore b f H (1) = 0 and also b f H (1) = b f H ( − . Case 2.
Let n ( = 0) be an even no. If n > r ∈ Z (1) such that b f rH (1) = 0 and so G r (1) b f rH (1) = 0 (since Q n,m ( ± = 0 for nm > b f rH ( −
1) = ϕ n,r b f rH (1) and ϕ n,rλ has no zero at λ = 1 (see [Ba, prop 7.2].This shows that b f rH ( − = 0 and so G m ( − c f mH ( − = 0.When n < s ∈ Z ( −
1) such that b f sH ( − = 0 and so G s ( − b f sH ( − = 0. But b f sH ( −
1) = ϕ n,s b f sH (1)and ϕ n,sλ has no pole at λ = 1. This implies b f sH (1) = 0 hence G m ( − c f mH ( − = 0 (since Q n,m ( ± = 0 for nm > n is an even no. Case 3.
Let n be an odd no. Then by the hypothesis there exist m ∈ Z σ − such that c f mH (1) = 0.Then mn = 0 so from (7.5) it follows G m (1) c f mH (1) = 0. Proof for λ = − G ′ m ( k ) = e − k Q n,m ( k ) for all k ∈ Γ n where Q n,m is chosen in the same way as before .Now let for k ∈ Γ n , d f m B ( k ) = 0 then m ∈ Z ( k ) . Therefore Q n,m ( k ) = 0 as all the zerosof the polynomial P n,m are either between m and n or between − m and − n (see [Ba, prop.7.1]). Now from Lemma 2.1 isomorphism between C ( G ) n,m and C ( b G ) n,m for every m , there exists g m ∈ C ( G ) n,m such that c g mH ( λ ) = G m ( λ ) for all λ ∈ S and c g mB ( k ) = G m ( k ) for all k ∈ Γ n .Now we show the set of L ( G ) n,n functions { g m ∗ f m | m ∈ Z σ } satisfies all the conditions ofTheorem 1.3. Since Q n,m ’s are always polynomial in λ , by a simple argument of analysis showsthat lim t →∞ e − π t log | G m ( it ) | = 0 . (7.8)Hence,lim sup t →∞ e − π t log | G m ( it ) c f mH ( it ) | = lim t →∞ e − π t log | ( G m ( it )) | + lim sup t →∞ e − π t log | c f mH ( it ) | = lim sup t →∞ e − π t log | c f mH ( it ) | . (7.9)Therefore by the given hypothesis, inf m ∈ Z σ δ ± ( G m c f mH ) = 0 . (7.10)So we have established that the ideal generated by { g m ∗ f m | m ∈ Z σ } is dense in L ( G ) n,n . But g m ∗ f m = g m ∗ f ; so the result follows from the fact that the left L ( G ) module generated by L ( G ) n,n is all of L ( G ) n . Now suppose Λ is an arbitrary index set. Then out of each f α by projections we get f αj for all j ∈ Z which are functions of type ( j, n ). We apply previous arguments to the collection { d f αjH | α ∈ , j ∈ Z } of functions in L ( b G ) n and the theorem follows. (cid:3) Proof of Theorem 1.1.
As we have seen in the proof of previous theorem , it is enough to considerthe case when the collection contains a single function, namely f . Let f j be the projection of f to L ( G ) j , for every j ∈ Z . For each j, m ∈ Z , we choose a polynomial Q j,m in λ involving j and m so that e − λ Q j,m ( λ ) ∈ C H ( b G ) j,m .When jm ≥ Q j,m = P j,m is the numerator of the rational function ϕ j,mλ . Now suppose jm < j, m are odd integers we take Q ′ j,m = λ P j,m and if j, m are even integers then wechoose Q ′′ j,m = (1 − λ ) P j,m , where P j,m is as above. Then for m ∈ Z , e − λ Q j,m ( λ ) ∈ C H ( b G ) j,m .By the isomorphism of L Schwartz space C ( G ) j,m and C ( b G ) j,m (see Lemma 2.1) there exists g j,m ∈ C ( G ) j,m such that d g j,mH ( λ ) = e − λ Q j,m ( λ ) for all λ ∈ S and d g j,mB ( k ) = e − k Q j,m ( k ) forall k ∈ Γ j . Now for all m ∈ Z we consider the following collection of functions, F m = { f j ∗ g j,m | j ∈ Z } contained in L ( G ) m .As in (7.8) and (7.9) we have for each m ∈ Z ,lim sup t →∞ e − π t log | d g j,mH ( it ) c f i,jH ( it ) | = lim sup t →∞ e − π t log | c f i,jH ( it ) | (7.11)for all i, j ∈ Z . So, inf i,j ∈ Z δ ± ( d g j,mH c f i,jH ) = 0 . (7.12)Now for all m ∈ Z , Fourier transforms of the elements of F m does not have common zeros, followsfrom [Rs1, Theorem 1.2]. Therefore together with (7.12) it follows that for every m , elements of F m satisfies all the conditions of Theorem 1.2 and so F m generates L ( G ) m under left convolution. Now f j ∗ g j,m = f ∗ g j,m , for every m. So the two sided closed ideal generated by f contains L ( G ) m forall m . The smallest closed right G -invariant subspace of L ( G ) containing L ( G ) m for all m ∈ Z ,is L ( G ) itself. Hence the first part of the Theorem 1.1 follows. The second part of the theoremfollows similarly as in [Rs1, Theorem 1.2]. (cid:3) Acknowledgement.
The author would like to thank to his supervisor Prof. Sanjoy Pusti forintroducing him to the problem and for the many useful discussions during the course of this work.I am grateful to him for encouraging me in research and his guidance.
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