International Mathematical Forum, Vol. 6, 2011, no. 58, 2849-2856 Algebraic Properties of Slant Hankel Operators M. R. Singh And M. P. Singh Department of Mathematics, Manipur University Imphal 795003, Manipur, India mranjitmu@rediffmail.com, mpremjit@yahoo.com Abstract Notion of Slant Hankel Operator S φ with symbol φ in L ( D) is introduced and studied. Many algebraic properties of the operator are obtained. It is shown that the only Hyponormal Slant-Hankel operator is zero operator. Mathematics Subject Classification: Primary 47B35, Secondry 47B20 Keywords: Slant Toeplitz operator, Slant Hankel operator, infinite matrix, projection, isometry, co-isometry, bounded function, multiplication operator, compact, hyponormal 1. Introduction M.C. Ho [2] defined the notion of slant-toeplitz operator A φ using the doubly infinite Toeplitz matrix. Motivated by this we introduce Slant-hankel operator as follows: Let D be the unit disc {z : z < 1} in the complex plane and {z : z = 1}, the unit circle, D,the boundary of D. Let φ(z) = a i z i be a bounded measurable function on the unit circle. Thus φ L ( D) and a i = φ, z i denotes the i th fourier co-efficient of φ. The Slant-Hankel operator S φ is defined to be an operator on L 2 ( D) by the following matrix w.r.t. the usual basis {z i : i Z} of L 2 ( D)... a 2 a 1 a 0......... a 0 a 1 a 2............ a 3 a 4 a 5............ a 6 a 7... (1.1)
2850 M. R. Singh And M. P. Singh The construction of this matrix has been done keeping the construction of the matrix of Slant-Toeplitz operator[2] in mind. Let V be the operator on L 2 ( D) such that Vz 2n = z n and Vz 2n 1 =0 for each n in Z. It is easy to see that V is a bounded operator on L 2 ( D) with norm 1. The matrix of V is given by... 0 0 1......... 1 0 0............ 0 0 0............ 0 0... (1.2) It is easy to see that V = S 1. Further, the adjoint of V is given by, for each n in Z, V z n = z 2n. Then VV = I and V V = P e,where P e is the projection on the closed span of {z 2n : n Z} in L 2 ( D). Thus V is a co-isometry, also S φ = VM φ. This can be seen by the following argument: From the definition of S φ, S φ (z k ) = a 2i k z i. But, M φ (z k ) = a i k z i. So for each k Z, Then VM φ (z k )=V ( a i k z i )= a 2i k z i = S φ (z k ). S φ = VM φ V M φ = M φ = φ, Therefore, S φ φ. We now proceed to study the properties of Slant-Hankel operators. It is easy to check that the mapping φ S φ is 1 1, if φ = I, then S φ = V. If φ = ακ + βψ, for some κ, ψ L then, S φ = αs κ + βs ψ. Let P be the projection from L 2 ( D) ontoh 2 ( D). Also let K φ be the compression of S φ to H 2 ( D). Then for each k 0, K φ (z k )=VM φ (z k )+a k. It is so because VM φ does not contain the constant term. The matrix of K φ is given by the lower right-half portion of the Slant-Hankel matrix w.r.t. the decomposition L 2 =(H 2 ) H 2 a 0 a 1 a 2...... a 2 a 3 a 4 a 5... a 4 a 5 a 6 a 7.................. (1.3)
Algebraic properties of Slant Hankel operators 2851 2. Properties of V and V We have seen that V is a co-isometry on L 2 ( D) and isometry on P e (L 2 ). Range of V is L 2 ( D). V has similar properties as that of operator W defined by M.C. Ho in [2]. Let f be a function on L 2 ( D). Now V f P e (L 2 ). Thus zv f is in the closed span of {z 2k 1 : k Z}. Therefore, VM z V = 0 (2.1) Let f,g L 2 ( D). also assume that fg L 2 ( D). Then V (fg)=(fg)(z 2 )=f(z 2 )g(z 2 )=(V f)(v g) (2.2) So, V {(V f)(v g)} = V (V (fg)) = fg (2.3) Using this result, we obtain that if one of f and g is in L, then V (fg)=(vf)(vg)+z(vzf)(vzg). (2.4) We derive the above result as follows: Let f be any function in L 2. Then f can be written as f 1 (z 2 )+zf 2 (z 2 ) where f 1 (z 2 )=P e f and zf 2 (z 2 )=f f 1 (z 2 ). Also suppose g is written in the form of g 1 (z 2 )+zg 2 (z 2 ). Then Vfg= V (f 1 (z 2 )g 1 (z 2 )) + V (zf 2 (z 2 )zg 2 (z 2 )) as the remaining terms include only odd fourier co-efficients and V eliminates all of them. Using (6), we have Vfg=(Vf 1 (z 2 ))(Vg 1 (z 2 )) + V (z 2 )V (f 2 (z 2 ))V (g 2 (z 2 )) But, f 1 (z 2 ) = P e f and Vf 1 (z 2 ) = VP e f = VV Vf = Vf. Similarly Vg 1 (z 2 )=Vg. Again zf 2 (z 2 )=f f 1 (z 2 ). Then f 2 (z 2 )=zf zf 1 (z 2 ) as z = 1. Also zf 1 (z 2 ) has odd fourier co-efficients. Therefore Vf 2 (z 2 )= V (zf). Similarly,Vg 2 (z 2 )=V(zg). So, Vfg =(Vf)(Vg)+z(Vzf)(Vzg) as V (z 2 )=z. The proof will be complete if we show that (Vf)(Vg) and (Vzf)(Vzg) are functions of L 2. Later in this section, we will show that if φ is a L function, then Vφ is also a L function and this will complete the proof. Letf = h(z 2 ). Then (zh(z 2 )) has only odd fourier co-efficients. So,V (zh(z 2 )) = 0. Therefore Vfg=(Vf)(Vg)=(VV h)(vg)=hv g (2.5)
2852 M. R. Singh And M. P. Singh We have S φ (z k )= a 2i k z i.thuss φ (zk )= a i 2k z i. So the matrix of S φ is... a 1 a 1 a 3......... a 0 a 2 a 4............ a 3 a 5 a 7............ a 6 a 8... Also, the matrix of VS φ is the bounded doubly infinite toeplitz matrix... a 2 a 4 a 6......... a 0 a 2 a 4............ a 0 a 2 a 4............ a 0 a 2... So the constants on each diagonal are the even co-efficients of the function φ and so VSφ = M ψ, where ψ = V (φ) (2.6) Now M V (φ) = VS is bounded on φ L2 for any φ L. So,for any φ L, V (φ) is also a L function. Using (2.6), we have S φ Sφ = VM φ M φ V = VM φ 2V = V (VM φ 2) = V (S φ 2) = M ψ, where ψ = V ( φ 2 ). Using (2.6), we can obtain a characterization of Slant- Hankel operators as shown in the following Theorem 2.1. A bounded operator S on L 2 is a Slant- Hankel operator if and only if M z S = SM z 2. Proof: Let S be a Slant-Hankel operator. Then S = S φ for some φ L. Using (2.5), M z S φ = M z VM φ = VM z 2M φ = VM φ M z 2 = S φ M z 2 Conversely, suppose S satisfies M z S = SM z 2. We show that S = S φ for some L function φ. Consider f(z) =S1(z 2 ) and g(z) =z(sz)(z 2 ). For any h L 2, we have Then S(h(z 2 )) = S(1.h(z 2 )) = h.(s1) (S1)h 2 = S(h(z 2 )) S h(z 2 ) 2 = S h 2. So the multiplication on L 2 by the function S1 is bounded and so S1 isa bounded function and hence f =(S1(z 2 )) too. Similarly g is a bounded
Algebraic properties of Slant Hankel operators 2853 function as S(zh(z 2 )) = h.(sz). It gives that φ = f + g is a L function. Now we show that S = S φ. Let F be a function in L 2 ( D). Then F can be written as P (z 2 )+zq(z 2 ) for some P and Q in L 2. S φ F = VM φ F = V [(f + g)(p (z 2 )) + zq(z 2 )] = V [{(S1)(z 2 )+z(sz)(z 2 )}{P (z 2 )+zq(z 2 )}] = V [(S1)(z 2 )P (z 2 )] + V [(Sz)(z 2 )Q(z 2 )] = (S1)P +(Sz)Q On using (2.5) we obtain that for each F in L 2 ( D) S φ F = S(P (z 2 )) + S(zQ(z 2 )) = S(F ). To obtain another characterization of Slant-Hankel operator in terms of matrices, we introduce the following. Definition 2.2: A Slant-Hankel matrix is a two way infinite matrix (a ij ) such that a i+1,j 2 = a ij. Theorem 2.3: A necessary and sufficient condition that an operator S on L 2 ( D) be a Slant-Hankel operator is that its matrix w.r.t. the orthonormal basis {z n : n Z} is a Slant-Hankel matrix. Proof: Suppose that the operator S on L 2 ( D) is a Slant-Hankel operator. Then by the definition S = S φ for some φ in L ( D). If (α ij ) is the matrix of S φ with respect to the given orthonormal basis, then α ij = S φ z j,z i = a 2i j = a 2(i+1) (j 2) = α i+1,j 2. Therefore, (α ij ) is a Slant-Hankel matrix. Conversely, let the matrix (α ij )ofs be a Slant-Hankel matrix. Then for all i, j Z, Now, Sz j,z i = α ij = α i+1,j 2 = Sz j 2,z i+1. M z Sz j,z i = Sz j,m z z i = Sz j,z i 1 = Sz j 2,z i = SM z 2z j,z i. This implies that M z Sz j = SM z 2z j, j Z. Therefore, M z S = SM z 2 and hence by theorem1, S is a Slant-Hankel operator. 3. Relation between S φ and A φ M.C. Ho in [2] defined W on L 2 ( D) aswz 2n = z n and Wz 2n 1 = 0 for each n Z. Also W z n = z 2n for each n Z. Now,VW z n = Vz 2n =
2854 M. R. Singh And M. P. Singh z n = Wz 2n = WV z n for each n Z. Thus VW = WV. Similarly, V W = W V. In other words,vw and V W are self adjoint operators. Theorem 3.1. For each φ, ψ L ( D) (i) S φa ψ = A φs ψ (ii) S φ A ψ = A φ( z)s ψ( z) and (iii) A ψ S φ = S ψ( z)a φ( z) Proof: By property[2] we know that A φ = WM φ S φa ψ = M φv WM ψ = M φw VM ψ = A φs ψ which is (i). S φ A ψ = VM φm ψw = M φ(z 2 )M ψ(z 2 )VW = M φ(z 2 )M ψ(z 2 )WV = WM φ( z) M ψ( z) V = A φ( z) S ψ( z) V which gives(ii). Similarly we can check (iii). Theorem 3.2. S φ S ψ = S ψ S φ iff φ(z 2 ).ψ = φ.ψ(z 2 ) Proof: First, let φ(z 2 ).ψ = φ.ψ(z 2 ). Then S φ S ψ = VM φ VM ψ = VM φ M ψ(z 2 )V = VM φψ(z 2 )V = VM ψ M φ(z 2 )V = VM ψ VM φ = S ψ S φ. Conversely, let S φ commute with S ψ. Then, VM φ VM ψ = VM ψ VM φ. i.e.v V M φ(z 2 )ψ = VVM ψ(z 2 )φ. On taking adjoints and taking the values of both sides at z 0, we get φ(z 2 ).ψ = φ.ψ(z 2 ). Theorem 3.3. If S φ S ψ is a Slant-Hankel operator, then S φ S ψ =0. Before proving this theorem, we first prove the following. Lemma 3.4: VS φ is a Slant-Hankel operator iff φ =0. Proof of the lemma: Suppose VS φ is a Slant-Hankel operator. Then for each i, j in Z, VS φ z j,z i = S φ z j 2,z i+1. Therefore, a 4i j = a 4i j+2 where φ =Σ a iz i. Putting i =0,a j = a j+2. It gives a 0 = a 2n and a 1 = a 2n 1 for each n Z. But as φ L ( D),a n 0. So a 0 = a 1 = 0. It gives φ = 0. The converse is obvious. Proof of the theorem: Suppose S φ S ψ is a Slant-Hankel operator. Then S φ S ψ = VM φ VM ψ = VVM φ(z 2 )M ψ = VS φ(z 2 )ψ. By the above lemma φ(z 2 )ψ = 0 which gives S φ S ψ =0. Corollary 3.5: If S 2 φ = S φ, then φ =0.
Algebraic properties of Slant Hankel operators 2855 Proof: S 2 φ = S φ shows that S φ S φ is a Slant-Hankel operator. So by theorem5,s φ S φ = 0 and so S φ =0. Hence, φ =0. Theorem 3.6: The only compact Slant- Hankel operator is 0. proof: Let S φ be compact. By(2.6), VS φ = M ψ, where ψ = V ( φ). As V is a bounded operator, VS φ = M ψ is also compact. This gives ψ = 0. Therefore for each n Z, ψ, z n = V ( φ),z n = φ, z 2n = a 2n =0. Again S φ M z is also compact. But V (S φ M z ) = M ψ, where ψ = V ( zφ). So M ψ is compact. It gives ψ = 0. Therefore for each n Z ψ, z n = zφ, z 2n = a 2n+1 =0. Hence, φ =0. Theorem 3.7. The only hyponormal Slant-hankel operator is 0. Proof: Let S φ be hyponormal. Then, for each j Z, Therefore,for each j Z (S φs φ S φ S φ)z j,z j 0. j= a j 2i 2 a 2j i 2. Taking j = 0, we get a 2i 1 = 0 for each i Z. Again, taking j = 1, we have a 2i+1 2 a j 2 2. This gives a i+2 =0. So, a i = 0. It gives φ = 0 and hence S φ =0. It is easy to check the following Cor 3.8 S φ is normal iff φ =0. Cor 3.9 S φ is self-adjoint iff φ =0. References [1] A. Brown and P.R. Halmos (1963/64) The Algebraic properties of Toeplitz operators, J.Reine Angew. Math. 213 89-102.
2856 M. R. Singh And M. P. Singh [2] Mark C. Ho. (1996) Properties of Slant-Toeplitz operators,indiana University mathematics journal(c) 45,3. [3] Mark C. Ho.(1997) Adjoint of Slant-Toeplitz operators, Integral equation and Operator theory 29, 301-312. [4] Young Joo Lee and Kehe Zhu (2002) Some Differential and Integral Equations with Applications to Toeplitz operators, Integral equation and Operator theory 44(4), 466-479. [5] Xiao Feng Wang(2002) Some Problems of Toeplitz operators on the Dirichlet s Space over a circular ring, Sichuan Daxue Xuebao 39, 1005-1010. Received: August, 2009