Toeplitz Operators on Abstract Hardy Spaces Built upon Banach Function Spaces

LetX be a Banach function space over the unit circle T and letH[X] be the abstractHardy space built uponX. If the Riesz projection P is bounded on X and a ∈ L∞, then the Toeplitz operator Taf = P(af) is bounded on H[X]. We extend well-known results by Brown andHalmos forX = L2 and show that, under certain assumptions on the spaceX, the Toeplitz operatorTa is bounded (resp., compact) if and only if a ∈ L∞ (resp., a = 0). Moreover, ‖a‖L∞ ≤ ‖Ta‖B(H[X]) ≤ ‖P‖B(X)‖a‖L∞ . These results are specified to the cases of abstract Hardy spaces built upon Lebesgue spaces with Muckenhoupt weights and Nakano spaces with radial oscillating weights.


Introduction
The Banach algebra of all bounded linear operators on a Banach space  will be denoted by B().Let T be the unit circle in the complex plane C. For  ∈ Z + fl {0, 1, 2, . ..}, a function of the form () = ∑  =−     , where   ∈ C for all  ∈ {−, . . ., } and  ∈ T, is called a trigonometric polynomial of order .The set of all trigonometric polynomials is denoted by P. The Riesz projection is the operator  which is defined on P by ||/(2).Denote by  0 the set of all measurable complexvalued functions on T, and let  0 + be the subset of functions in  0 whose values lie in [0, ∞].The characteristic function of a measurable set  ⊂ T is denoted by I  .A mapping  :  0 + → [0, ∞] is called a function norm if, for all functions , ,   ( ∈ N) in  0 + , for all constants  ≥ 0, and for all measurable subsets  of T, the following properties hold: with   ∈ (0, ∞) depending on  and  but independent of .When functions differing only on a set of measure zero are identified, the set  of all functions  ∈  0 for which (||) < ∞ is a Banach space under the norm ‖‖  fl (||).Such a space  is called a Banach function space.If  is a function norm, its associate norm   is defined on  0 + by fl sup {∫ T  ()  ()  () :  ∈  0 + ,  () ≤ 1} ,  ∈  0 + . ( The Banach function space   determined by the function norm   is called the associate space (or Köthe dual space) of .
This definition makes sense because  is continuously embedded in Then there is a function  ∈  ∞ such that  =   and â() =   for all  ∈ Z. Moreover We need the notion of a function with absolutely continuous norm to formulate the result on the noncompactness of nontrivial Toeplitz operators.Following [6,  The paper is organized as follows.Section 2 contains results on the density of the set of all trigonometric polynomials P (resp., the set of all analytic polynomials P  ) in a Banach function space  (resp., in the abstract Hardy space [] built upon ).We also show that the norm of a function  in  can be calculated in terms of ⟨, ⟩, where  ∈ P, under the assumption that   is separable.Further, we prove that every bounded linear operator on a separable Banach function space, whose matrix is of the form ( − ) ,∈Z , is an operator of multiplication by a function  ∈  ∞ and the sequence of its Fourier coefficients is exactly {  } ∈Z .Finally, we prove that if the characteristic functions of all measurable sets  ⊂ T have absolutely continuous norms in , then the sequence {  } ∈Z + converges weakly to zero on the abstract Hardy space [].In Section 3, we provide proofs of our main results, using auxiliary results from the previous section.In Section 4, we specify our main results to the case of Hardy spaces built upon weighted Lebesgue spaces   () with Muckenhoupt weights  and to the case of weighted Nakano spaces  (⋅) () with certain radial oscillating weights.In both cases, it is known that the Riesz projection is bounded.(b) the space  of all continuous functions on T is dense in the space ;

Density of Continuous Function and Trigonometric Polynomials in Banach Function
(c) the Banach function space  is separable.

Multiplication
It is easy see that The following lemma shows that every bounded operator with such a property is a multiplication operator.

Lemma 7.
Let  be a separable Banach functions space over the unit circle T. Suppose  ∈ B() and there exists a sequence {  } ∈N of complex numbers such that Then there exists a function  ∈  ∞ such that  =   and â() =   for all  ∈ Z.

Proof of the Main Results
Consider the following subset of the associate space: Combining inequality (40) with equalities (42) and (46), we arrive at the first inequality in (10).The second inequality in (10) is obvious.

Proof of
Passing in this inequality to the limit as  → ∞ and taking into account (47), we see that â() = 0 for all  ∈ Z.By the uniqueness theorem for Fourier series (see, e.g., [11, Chap.I, Theorem 2.7]), this implies that  = 0 a.e. on T.
1in view of axiom (d).It can be shown that[] is a closed subspace of .It is clear that if 1 ≤  ≤ ∞, then [  ] is the classical Hardy space   .It follows from axiom (d) that P ⊂  ∞ ⊂ .We will restrict ourselves to Banach function spaces  such that the Riesz projection defined initially on P by formula (1) extends to a bounded linear operator on the whole space .The extension will again be denoted by .If  ∈  Z and  ∈ T, put   () =   .Then the Fourier coefficients of a function  ∈  1 can be expressed by f() = ⟨,   ⟩ for  ∈ Z.
Chap. 1, Definition 3.1], a function  in a Banach function space  is said to have absolutely continuous norm in  if ‖I   ‖  → 0 for every sequence {  } ∈N of measurable sets satisfying I   → 0 almost everywhere as  → ∞.The set of all functions in  of absolutely continuous norm is denoted by   .It is known that a Banach function space  is reflexive if and only if  and   have absolutely continuous norm (see [6, Chap. 1, Corollary 4.4]).(main result 2).Let  be a Banach function space over the unit circle T such that I  ∈   for every measurable subset  ⊂ T. If the Riesz projection  is bounded on  and  ∈ ∞ , then the Toeplitz operator   ∈ B([]) is compact if and only if  = 0.
=0     , where   ∈ C for all  ∈ {0, . . ., } and  ∈ T, is said to be an analytic polynomial on T. The set of all analytic polynomials is denoted by P  .Let  be a separable Banach functions space over the unit circle T. If the Riesz projection  is bounded on , then the set P  is dense in [].Proof.If  ∈ [] ⊂ , then by Lemma 3, there exists a sequence   ∈ P such that ‖ −   ‖  → 0 as  → ∞.It is clear that  =  and   ∈ P  .Since  ∈ B(), we finally have      −  Let  be a Banach function space over the unit circle T. If the associate space   is separable, then for every Fix  ∈   such that 0 < ‖‖   ≤ 1.Since   is separable, it follows from Lemma 3 that there exists a sequence   ∈ P \ {0} such that ‖  − ‖   → 0 as  → ∞.For  ∈ N, put   fl (‖‖   /‖  ‖   )  ∈ P. Then for every  ∈ N            =           ≤ 1, (15)      −        ≤      −        +            (1 −                      ) .
for all  ∈ P. In view of Lemma 3, P is dense in .Then (39) implies that the linear mapping  defined in (34) extends to an operator  ∈ B() such that ‖‖ B() ≤ ‖‖ B([]) .