Boundedness and compactness of weighted composition operators and monomial operators

This paper characterises the boundedness and compactness of Agler–McCarthy monomial operators by reducing them to weighted composition operators and deriving explicit Carleson measure criteria on the half-plane. The results are illustrated by examples.


Introduction
Motivated by questions of polynomial approximation, Agler and McCarthy [1] defined a monomial operator as a bounded linear operator T : L 2 (0, 1) → L 2 (0, 1) mapping monomials to monomials; that is, such that for each n ∈ N we have (with a slight abuse of notation) T (x n ) = c n x pn for some p n ∈ C with Re p n > −1/2.Such an operator is flat if there is a constant b with p n = n + b for each n.We shall also say that it has affine index if p n = an + b for real constants a, b ≥ 0.
Well-known examples of monomial operators include the Volterra operator V with and the Hardy operator H (also known as the continuous Cesàro operator [2]) with which are clearly flat operators.Agler and McCarthy gave conditions for flat operators to be bounded, and showed that they were never compact except in trivial cases, but the general case was left open.Their work was based on a transformation to the Hardy space of the half-plane S = {s ∈ C : Re s > −1/2} and showed that the adjoint of a bounded monomial operator can be expressed as an operator on H 2 (S).By means of a closely-related transformation we shall work on the Hardy space H 2 (C + ) on the standard right half-plane, and derive results in the theory of weighted composition operators, which will be applied to characterise bounded and compact operators.The results are particularly simple for flat and affine-index operators, as will be illustrated by examples.

Boundedness and Compactness
The approach here is very similar to Agler-McCarthy [1], but adapted for own purposes.
We transfer the operator T to L 2 (0, ∞).Writing x = e −t we have Then by taking Laplace transforms and using the Paley-Wiener theorem we have a unitarily equivalent operator T 0 on H 2 (C + ) which maps 1/(s + n + 1/2) to c n /(s + p n + 1/2).
Since the reproducing kernel for H 2 (C + ) is k w (s) = 1/(s + w) we have The Blaschke sequences (zero sequences) (z n ) for H 2 (C + ) are those with , and so the sequence (n + 1/2) is not a Blaschke sequence for the Hardy space of the right half-plane.Thus we have a unique formula for T * 0 , namely the weighted composition operator W h,φ with Remark 2.1.The classical Müntz-Szász theorem says that the functions {x n : n ∈ S} span a dense subspace of L 2 (0, 1) if and only if Thus under this condition a bounded monomial operator is completely defined by {T x n : n ∈ S}.We see this from the above discussion, since such an S does not constitute a Blaschke sequence, and the interpolation problem has a unique solution.
Boundedness of the weighted composition operator W h,φ : f → h(f •φ) can be characterised using a Carleson measure criterion by proving a result similar to those in [3,6].Namely, for Borel subsets E ⊆ C + define a measure by Theorem 2.2.The weighted composition operator W h,φ is bounded if and only if µ is a Carleson measure; that is, based on an interval I ⊂ iR.Hence the monomial operator T is bounded if and only if the same condition holds for the associated weighted composition operator.
Proof.In [3] the analogous result is proved for weighted composition operators on the disc and in [6] a similar result is derived for the Bergman space.For the half-plane an identical proof works with obvious modifications, as follows: We claim that for g ≥ 0 Borel measurable we have (2.3) As in [3] we can easily verify this result for simple functions, and then for arbitrary g take an increasing sequence of positive simple functions (g n ) with . Now, the Carleson measure condition that for some fixed C > 0 is equivalent to the condition that W h,φ is bounded.
In the context of a monomial operator T the adjoint operator is unitarily equivalent to W h,φ and the result follows.
Remark 2.3.It is interesting to consider the case h = 1, an unweighted composition operator C φ on the right half-plane.We see that C φ is bounded if and only if there is a constant K > 0 such that µ(Q I ) = m(φ −1 (E) ∩ iR) ≤ K|I| for all Carleson squares (here m is Lebesgue measure).This may be compared with the alternative condition given by Elliott and Jury [5]; namely, that there is an angular derivative at ∞, the non-tangential limit Example 2.5.We can have c n → 0 and the monomial operator not even bounded.For example, take c n = (n + 1/2)e −(n+1/2) , for which the associated function h can only be se −s , which does not satisfy the Carleson criterion.This fact may also be proved using the fact that the monomials do not form an unconditional basis, but this is more explicit.
We know from [1] that there are no compact monomial operators of the form x n → c n x n , except the zero operator, and we also know that there are no compact composition operators on the half-plane [7].
If T is compact, then observe first that the normalized functions e n : x → √ 2n + 1 x n tend weakly to zero.For they form a bounded sequence tending weakly to zero on the dense subspace consisting of polynomials, since Hence compactness implies that T e n → 0 and this is given by (2.5).That is, for compact affine-index operators, c n → 0.
Using the completely continuous characterization of compactness in reflexive Banach spaces, namely, that if f n → 0 weakly then W h,φ f n → 0, we see that a necessary and sufficient condition on compactness will be compactness of the mapping J : H 2 (C + ) → L 2 (C + , µ).This is characterised for the disc in [3].
On the disc this is the familiar vanishing Carleson measure criterion for µ.On the half-plane the condition that where Q I = [0, |I|] × I is a Carleson square based on an interval I ⊂ iR, is not sufficient, as we now show.
Consider the mapping x n → x n+1 , which is bounded but not compact.We have h(s) = 1 and φ(s) = s + 1.Now according to (2.2) we have µ(E) = m(E ∩ (1 + iR)), for E ⊂ C + , and thus The measure µ of a square Q I is 0 if |I| < 1 and so this µ satisfies (2.6), but nonetheless we do not have compactness.
In fact the correct test for compactness for embeddings from H 2 (C + ) into L 2 (C + , µ) is given in [8] in the more general context of Zen spaces, namely, it can be written (2.7) sup where C(I) is the centre of a Carleson square Q I and for 0 < r < 1 we define which is the union of the complement of a large semi-circle and a narrow strip near the axis.Equivalently, for each ǫ > 0 there is a compact subset K ⊂ C + such that µ(Q I )/|I| < ǫ for every Q I with centre lying outside K.
We therefore have the following result.
Theorem 2.6.The monomial operator T is compact if and only if the associated measure µ satisfies the condition (2.7).
This can be made more explicit for operators with affine index.Proof.Equation (2.8) is a direct rewriting of the vanishing condition (2.7), using the fact that a square [0, 2L] × [t − L, t + L] with centre L + it meets the line {s ∈ C : Re s = b} if and only if L ≥ b/2.Now if h ∈ L 2 (iR), the inequality (2.8) holds for all t if L > L 0 := h 2 2 /ǫ.For b/2 ≤ L ≤ L 0 , we may make the integral less than ǫb/2 by choosing |t| sufficiently large.

Examples
For the Volterra operator, we have from (2.1) that h(s) = (s + 1/2) −1 and φ(s) = s + 1.Now we can apply Corollary 2.7 and since we can deduce that the operator is compact, as is well known.
For the bounded but non-compact Hardy operator we have h(s) = (s+1/2) −1 and φ(s) = s.Corollary 2.7 does not apply since a = 0, and we must consider all Carleson squares, since they all meet the imaginary axis.We have as v − u → 0, and so the operator is bounded but not compact, as is again well known.
For some non-flat examples consider the following operators defined on L 2 (0, 1): . These operators are not flat, but they are affine-index.
It is elementary to see that T 1 is unbounded, since it sends x −1/3 to x −2/3 .We shall see that T 2 is unbounded, but T 3 is bounded (even compact).

Corollary 2 . 4 .
Let T be an affine-index monomial operator with p n = an + b (a, b > 0).Then T is bounded if and only if the associated function h satisfies )| 2 dy < ∞.Proof.By the simple structure of φ, we have µ concentrated on b + iR and the Carleson condition is v/a u/a |h(iy)| 2 dy ≤ C(v − u) but only for v − u ≥ b; that is, equivalently, (2.4).

Corollary 2 . 7 .
Let T be an affine-index monomial operator with p n = an + b (a, b > 0).Then T is compact if and only if the associated function h satisfies the condition that for each ǫ > 0 there is an M > 0 such that(2.8)(t+L)/a (t−L)/a |h(iy)| 2 dy < ǫLfor all |t| ≥ M and L ≥ b/2.This holds whenever h ∈ L 2 (iR).