1 Introduction

Motivated by questions of polynomial approximation, Agler and McCarthy [1] defined a monomial operator as a bounded linear operator \(T: L^2(0,1) \rightarrow L^2(0,1)\) mapping monomials to monomials; that is, such that, for each \(n \in {\mathbb {N}}\), we have (with a slight abuse of notation) \(T(x^n)=c_n x^{p_n}\) for some \(p_n \in {\mathbb {C}}\) with \(\mathop {\textrm{Re}}\nolimits p_n > -1/2\). 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 \(a,b \ge 0\).

Well-known examples of monomial operators include the Volterra operator V with

$$\begin{aligned} Vf(x)=\mathop {\int }\limits _{0}^{x} f(t) \, dt, \quad Vx^n=\frac{x^{n+1}}{n+1}, \end{aligned}$$

and the Hardy operator H (also known as the continuous Cesàro operator [3]) with

$$\begin{aligned} Hf(x)=\frac{1}{x} \mathop {\int }\limits _{0}^{x} f(t) \, dt, \quad Hx^n=\frac{x^{n}}{n+1}, \end{aligned}$$

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 \({\mathbb {S}}=\{s \in {\mathbb {C}}: \mathop {\textrm{Re}}\nolimits s > -1/2 \}\) and showed that the adjoint of a bounded monomial operator can be expressed as an operator on \(H^2({\mathbb {S}})\). By means of a closely-related transformation, we shall work on the Hardy space \(H^2({\mathbb {C}}_+)\) 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.

2 Boundedness and compactness

The approach here is very similar to Agler–McCarthy [1], but adapted for own purposes.

We transfer the operator T to \(L^2(0,\infty )\). Writing \(x=e^{-t}\), we have

$$\begin{aligned} \mathop {\int }\limits _{0}^{1} |f(x)|^2 \, dx = \mathop {\int }\limits _{0}^{\infty } |f(e^{-t})|^2 e^{-t} \, dt, \end{aligned}$$

giving an isometry by \(Jf(t)= e^{-t/2} f(e^{-t})\). So \(J^{-1}g(x) = \frac{1}{\sqrt{x}} g(-\log x)\). The unitarily equivalent operator \({{\tilde{T}}}\) on \(L^2(0,\infty )\) maps \(e^{-nt-t/2}\) to \(c_n e^{-p_nt-t/2}\).

Then by taking Laplace transforms and using the Paley–Wiener theorem, we have a unitarily equivalent operator \(T_0\) on \(H^2({\mathbb {C}}_+)\) which maps \(1/(s+n+1/2)\) to \(c_n /(s+p_n+1/2)\).

Since the reproducing kernel for \(H^2({\mathbb {C}}_+)\) is \(k_w(s)=1/(s+{{\overline{w}}}) \), we have

$$\begin{aligned} T_0^* G(n+1/2) = \overline{c_n} G(p_n+1/2). \end{aligned}$$

The Blaschke sequences (zero sequences) \((z_n)\) for \(H^2({\mathbb {C}}_+)\) are those with \( \sum _n \frac{\mathop {\textrm{Re}}\nolimits z_n}{1+|z_n|^2}<\infty \) [5], 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 \(W_{h,\phi }: f \mapsto h (f \circ \phi )\) with

$$\begin{aligned} \phi (n+1/2)=p_n+1/2 \quad \hbox {and} \quad h(n+1/2)=\overline{c_n} \end{aligned}$$
(2.1)

since \((T_0^*-W_{h,\phi })(G)=0\) for all \(G \in H^2({\mathbb {C}}_+)\).

For affine-index operators with \(p_n=an+b\), we have \(T_0^* G=h( G\circ \phi )\) with \(h(n+1/2)={{\overline{c}}}_n\) and \(\phi (s)=as+b\). This is consistent with Agler–McCarthy [1] except that we use the “usual” right half-plane not the shifted version.

Remark 2.1

The classical Müntz–Szász theorem [2, 9, 11] says that the functions \(\{x^n: n \in S \}\) span a dense subspace of \(L^2(0,1)\) if and only if \( \sum _{n\in S} \frac{1}{n+1} = \infty \). Thus under this condition a bounded monomial operator is completely defined by \(\{T x^n: n \in S\}\). 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 \(W_{h,\phi }\) can be characterised using a Carleson measure criterion by proving a result similar to those in [4, 7]. Namely, for Borel subsets \(E \subseteq \overline{{\mathbb {C}}_+}\), define a measure by

$$\begin{aligned} \mu (E)= \mathop {\int }\limits _{\phi ^{-1}(E) \cap i{\mathbb {R}}} |h(s)|^2 \, |ds|. \end{aligned}$$
(2.2)

Theorem 2.2

The weighted composition operator \(W_{h,\phi }\) is bounded if and only if \(\mu \) is a Carleson measure; that is, \(\sup _I \mu (Q_I)/|I| < \infty \), where \(Q_I=[0, |I|] \times I\) is a Carleson square based on an interval \(I \subset i{\mathbb {R}}\). Hence the monomial operator T is bounded if and only if the same condition holds for the associated weighted composition operator.

Proof

In [4], the analogous result is proved for weighted composition operators on the disc and, in [7], 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 \(g \ge 0\) Borel measurable, we have

$$\begin{aligned} \mathop {\int }\limits _{\overline{{\mathbb {C}}_+}} g \, d\mu = \mathop {\int }\limits _{i{\mathbb {R}}} |h|^2 (g \circ \phi ) |ds|. \end{aligned}$$
(2.3)

As in [4], 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 \(g_n \rightarrow g\) pointwise in \(\overline{{\mathbb {C}}_+}\).

From (2.3), we deduce that \(\Vert W_{h,\phi }f\Vert = \Vert f\Vert _{L^2(\overline{{\mathbb {C}}_+},\mu )}\). Now, the Carleson measure condition that, for some fixed \(C>0\),

$$\begin{aligned} \mathop {\int }\limits _{\overline{{\mathbb {C}}_+}} |f|^2 \, d\mu \le C \Vert f\Vert ^2 \quad (f \in H^2({\mathbb {C}}_+)) \end{aligned}$$

is equivalent to the condition that \(W_{h,\phi }\) is bounded.

In the context of a monomial operator T, the adjoint operator is unitarily equivalent to \(W_{h,\phi }\) and the result follows. \(\square \)

Remark 2.3

It is interesting to consider the case \(h=1\), an unweighted composition operator \(C_\phi \) on the right half-plane. We see that \(C_\phi \) is bounded if and only if there is a constant \(K>0\) such that \(\mu (Q_I)= m(\phi ^{-1}(E) \cap i{\mathbb {R}}) \le K|I|\) for all Carleson squares (here m is the Lebesgue measure). This may be compared with the alternative condition given by Elliott and Jury [6]; namely, that there is an angular derivative at \(\infty \), the non-tangential limit \(\phi '(\infty )=\lim _{z \rightarrow \infty \ \mathrm{n.t.}}z/\phi (z)\) satisfying \(0< \phi '(\infty )< \infty \).

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

$$\begin{aligned} \sup _{t \in {\mathbb {R}}} \mathop {\int }\limits _{t}^{t+1} |h(iy)|^2 \, dy < \infty . \end{aligned}$$
(2.4)

Proof

By the simple structure of \(\phi \), we have \(\mu \) concentrated on \(b+i{\mathbb {R}}\) and the Carleson condition is

$$\begin{aligned} \mathop {\int }\limits _{u/a}^{v/a} |h(iy)|^2 \, dy \le C (v-u) \end{aligned}$$

but only for \(v-u \ge b\); that is, equivalently, (2.4). \(\square \)

Example 2.5

We can have \(c_n \rightarrow 0\) with 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 \(x_n \mapsto c_n x^n\), except the zero operator, and we also know that there are no compact composition operators on the half-plane [8].

With \(T(x^n)=c_n x^{p_n}\), if T is bounded, then the sequence

$$\begin{aligned} \Vert Tx^n\Vert /\Vert x^n\Vert =|c_n| \sqrt{\frac{2n+1}{2\mathop {\textrm{Re}}\nolimits p_n+1}} \end{aligned}$$
(2.5)

is bounded. So for affine-index operators, if T is bounded, we clearly have \((c_n)\) bounded.

If T is compact, then observe first that the normalized functions \(e_n : x \mapsto \sqrt{2n+1} \, x^n\) tend weakly to zero. For they form a bounded sequence tending weakly to zero on the dense subspace consisting of polynomials since

$$\begin{aligned} \langle e_m, e_n \rangle = \frac{\sqrt{2m+1}\sqrt{2n+1}}{m+n+1} \rightarrow 0 \quad \hbox {as} \ n \rightarrow \infty . \end{aligned}$$

Hence compactness implies that \(\Vert Te_n\Vert \rightarrow 0\) and this is given by (2.5). That is, for compact affine-index operators, \(c_n \rightarrow 0\).

Using the completely continuous characterization of compactness in reflexive Banach spaces, namely, that if \(f_n \rightarrow 0\) weakly, then \(\Vert W_{h,\phi }f_n \Vert \rightarrow 0\), we see that a necessary and sufficient condition on compactness will be compactness of the mapping \(J: H^2({\mathbb {C}}_+) \rightarrow L^2(\overline{{\mathbb {C}}_+},\mu )\). This is characterised for the disc in [4].

On the disc this is the familiar vanishing Carleson measure criterion for \(\mu \). On the half-plane, the condition that

$$\begin{aligned} \mu (Q_I)/|I| \rightarrow 0 \quad \hbox {as} \ |I| \rightarrow 0, \end{aligned}$$
(2.6)

where \(Q_I=[0, |I|] \times I\) is a Carleson square based on an interval \(I \subset i{\mathbb {R}}\), is not sufficient, as we now show.

Consider the mapping \(x^n \mapsto x^{n+1}\), which is bounded but not compact. We have \(h(s)=1\) and \(\phi (s)=s+1\). Now according to (2.2), we have \(\mu (E)=m(E \cap (1+i{\mathbb {R}}))\) for \(E \subset {\mathbb {C}}_+\), and thus

$$\begin{aligned} \Vert C_\phi f\Vert ^2 = \mathop {\int }\limits _{\overline{{\mathbb {C}}_+}} |f|^2 \, d\mu = \mathop {\int }\limits _{1+i{\mathbb {R}}} |f|^2 \, |ds|. \end{aligned}$$

The measure \(\mu \) of a square \(Q_I\) is 0 if \(|I|<1\) and so this \(\mu \) satisfies (2.6), but nonetheless, we do not have compactness.

In fact, the correct test for compactness for embeddings from \(H^2({\mathbb {C}}_+)\) into \(L^2(\overline{{\mathbb {C}}_+},\mu )\) is given in [10] in the more general context of Zen spaces, namely, it can be written as

$$\begin{aligned} \sup _{I: C(I) \in S_r} \mu (Q_I)/|I| \rightarrow 0 \quad \hbox {as} \ r \rightarrow 0, \end{aligned}$$
(2.7)

where C(I) is the centre of a Carleson square \(Q_I\) and for \(0<r<1\), we define

$$\begin{aligned} S_r = \{z \in {\mathbb {C}}_+: 0<\mathop {\textrm{Re}}\nolimits z<r \hbox { or } |z|>1/r\}, \end{aligned}$$

which is the union of the complement of a large semi-circle and a narrow strip near the axis.

Equivalently, for each \(\epsilon >0\), there is a compact subset \(K \subset {\mathbb {C}}_+\) such that \(\mu (Q_I)/|I| < \epsilon \) 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 \(\mu \) satisfies the condition (2.7).

This can be made more explicit for operators with affine index.

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 \(\epsilon >0,\) there is an \(M>0\) such that

$$\begin{aligned} \mathop {\int }\limits _{(t-L)/a}^{(t+L)/a} |h(iy)|^2 \, dy < \epsilon L \end{aligned}$$
(2.8)

for all \(|t| \ge M\) and \(L \ge b/2\). This condition holds whenever \(h \in L^2(i{\mathbb {R}})\).

Proof

Equation (2.8) is a direct rewriting of the vanishing condition (2.7), using the fact that a square \([0,2L] \times [t-L,t+L]\) with centre \(L+it\) meets the line \(\{s \in {\mathbb {C}}: \mathop {\textrm{Re}}\nolimits s=b\}\) if and only if \(L \ge b/2\). Now if \(h \in L^2(i{\mathbb {R}})\), the inequality (2.8) holds for all t if \(L > L_0:= \Vert h\Vert _2^2/\epsilon \). For \(b/2 \le L \le L_0\), we may make the integral less than \(\epsilon b/2\) by choosing |t| sufficiently large. \(\square \)

3 Examples

For the Volterra operator, we have from (2.1) that \(h(s)=(s+1/2)^{-1}\) and \(\phi (s)=s+1\). Now we can apply Corollary 2.7 and since

$$\begin{aligned} \mathop {\int }\limits _{-\infty }^{\infty } \frac{1}{|1/2+iy|^2 } \, dy < \infty , \end{aligned}$$

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 \(\phi (s)=s\). 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

$$\begin{aligned} \mathop {\int }\limits _u^v |h(iy)|^2 \, dy = O(v-u), \quad \hbox {but not} \quad o(v-u) \end{aligned}$$

as \(v-u \rightarrow 0\), 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 \(L^2(0,1)\):

$$\begin{aligned} T_1 f(x)= & {} f(x^2), \\ T_2 f(x)= & {} \mathop {\int }\limits _0^x f(t^2) \, dt,\\ T_3 f(x)= & {} \mathop {\int }\limits _0^x t f(t^2) \, dt. \end{aligned}$$

We have \(T_1 (x^n)=x^{2n}\), \(T_2 (x^n)=\dfrac{x^{2n+1}}{2n+1}\), \(T_3 (x^n)= \dfrac{x^{2n+2}}{2n+2}\). 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).

For \(T_1\), we have \(\phi (n+1/2)=2n+1/2\), giving \(\phi (s)=2s-1/2\), which is not a self-map of \({\mathbb {C}}_+\).

For \(T_2\), take \(f(x)=1/(\sqrt{x} \log x)\) so that \(f \in L^2(0,1)\). Then

$$\begin{aligned} T_2 f(x)= \mathop {\int }\limits _0^x \frac{1}{t (2 \log t)} \, dt, \end{aligned}$$

and since this integral diverges, we see that \(T_2\) is unbounded. Now we have \(\phi (n+1/2)=2n+3/2\), giving \(\phi (s)=2s+1/2\). Also \(h(n+1/2)=1/(2n+1)\), giving \(h(s)=1/(2s)\).

We may also use the Carleson test so that \(\phi ^{-1}(E)\cap i{\mathbb {R}}=\phi ^{-1}(E \cap (1/2+i {\mathbb {R}}))\). However, since \(|h|^2\) is not locally integrable on the imaginary axis, we clearly do not get a Carleson measure.

For \(T_3\), we have \(\phi (n+1/2)=2n+5/2\), and \(\phi (s)=2s+3/2\). Then \(h(n+1/2)=1/(2n+2)\), giving \(h(s)=1/(2s+1)\). It is now easy to see that \(\mu \) is a Carleson measure, concentrated on the line \(\{s: \mathop {\textrm{Re}}\nolimits s=3/2\}\), and the operator is bounded. Indeed, since \(h \in L^2(i{\mathbb {R}})\), we see from Corollary 2.7 that \(T_3\) is indeed compact.