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A type of Volterra operator

Complex Analysis and its Synergies20162:3

  • Received: 27 May 2016
  • Accepted: 7 September 2016
  • Published:


In Diamantopoulos and Siskakis (Studia Math 140:191–198, 2000), the authors study the action of the classical Cesaro matrix C on the Taylor coefficients of analytic functions on the Hardy spaces \(H^p(\mathbb {D}), \;\; 1<p < \infty .\) They convert the matricial action of C on sequences into a Volterra type integral operator \(\mathbb {H}\) on \(H^P.\) They show that it is bounded for \(1<p<\infty \) and derive estimates on the operator norm of \(\mathbb {H}.\) We continue this study and show that \(\mathbb {H}\) maps boundedly from \(H^1(\mathbb {D})\) into the space of Cauchy transforms of finite Borel measures on unit circle. We show that \(\mathbb {H}\) is one to one on \(H^2(\mathbb {D}).\)


  • Hardy spaces
  • Volterra type operators

1 Introduction and definitions

We use \(\mathbb {D}\) as the notation for the unit disk in the complex plane and \(\mathbb {T}\) as its boundary. For \(\zeta = \exp (\imath \theta )\), we denote the normalized Lebesgue measure on \(\mathbb {T}\) by \(\text {d} \mu (\theta ) = \text {d} \theta / 2 \pi .\) The classical Hardy space on \(\mathbb {D}\) is written as \(H^p(\mathbb {D})\) and consists of those analytic functions f for which
$$\begin{aligned} \mathop{\text{sup}}\limits_{r<1}^{} \left( \int _{\mathbb {T}} |f(r \exp (\imath \theta ))|^p \text {d}\mu (\theta )\right) <\infty . \end{aligned}$$
For \(p \ge 1\), taking the p-th root of this sup yields a norm and with this norm \(H^p(\mathbb {D})\) is a Banach space. When \(p=2\), this is Hilbert space and identifying the Taylor coefficients of such an f with a sequence we obtain an isometry from \(H^2(\mathbb {D})\) onto the classical sequence space \(l^2.\) In addition, there is a natural identification of functions in \(H^p(\mathbb {D})\) with functions in a closed subspace of \(L^p(\mathbb {T})\) by way of the non-tangential limits of functions f in \(H^p(\mathbb {D})\) at points of \(\mathbb {T}\), and it shall be clear in our discussion whether we are dealing with points in \(\mathbb {D}\) or with points in \(\mathbb {T}.\,\,\) See [1] for more details.

2 The main result

In [2], the authors study the integral operator
$$\begin{aligned} (\mathcal {H}f)(z) = \int _{0}^{1} \frac{f(t)}{1 - z t} \text {d}t \end{aligned}$$
on the classical Hardy spaces \(H^p\) of the open unit disk \(\mathbb {D}\) [1]. Making this operator well defined is the integrability of \(f \in H^p\) on the interval [0, 1] which follows from the well-known Fejer–Riesz inequality [1, p. 46]
$$\begin{aligned} \int _{-1}^{1} |f(t)|^p \text {d}t \le \frac{1}{2} \int _{0}^{2 \pi } |f(e^{i \theta })|^p \text {d} \theta , \quad f \in H^p. \end{aligned}$$
In [2], they prove that when \(p \in (1, \infty )\) the operator \(\mathcal {H}\) maps \(H^p\) to itself. This is no longer the case when \(p = 1\) or \(p = \infty \).

The purpose of this paper is to discuss the range of \(\mathcal {H}\) on \(H^1\). The main result is the following.

Theorem 1

The operator \(\mathcal {H}\) maps \(H^1\) to the space of Cauchy transforms of measures on the unit circle \(\partial \mathbb {D}\). Furthermore, \(\mathcal {H}\) is injective

The proof of this theorem needs a few facts about Cauchy transforms which can be found in [3]. Let M be the Banach space of finite complex Borel measures on \(\partial \mathbb {D}\) endowed with the total variation norm \(\Vert \mu \Vert \) and define
$$\begin{aligned} \mathscr {C} := \left\{ (C \mu )(z) := \int _{\partial \mathbb {D}} \frac{\text {d} \mu (e^{i \theta })}{1 - e^{-i \theta } z}: \mu \in M\right\} , \end{aligned}$$
to be the space of Cauchy transforms. This is a Banach space of analytic functions on \(\mathbb {D}\) when given the norm
$$\begin{aligned} \Vert C \mu \Vert := \inf \{\Vert \nu \Vert : C \mu = C \nu \}. \end{aligned}$$
It is well known [3, Ch. 4] that \(\mathcal {A}\) (the disk algebra), the space of continuous functions on \(\overline{\mathbb {D}}\) which are analytic on \(\mathbb {D}\), endowed with the supremum norm \(\Vert g\Vert _{\infty } = \sup \{|g(z)|: z \in \overline{\mathbb {D}}\}\), can be identified with the pre-dual of \(\mathscr {C}\) via the (Cauchy) pairing
$$\begin{aligned} (g, C \mu ) := \lim _{r \rightarrow 1} \int _{0}^{2 \pi } g(r e^{i \theta }) \overline{C \mu (r e^{i \theta })} \frac{\text {d} \theta }{2 \pi }. \end{aligned}$$

Proof of Theorem 1

For \(f \in H^1\) and \(g \in \mathcal {A}\), observe the following: For \(r \in (0, 1)\),
$$\begin{aligned} \int _{0}^{2 \pi } g(r e^{i \theta }) \overline{(\mathcal {H}f)(r e^{i \theta })} \frac{\text {d} \theta }{2 \pi }&= \int _{0}^{2 \pi } g(r e^{i \theta }) \overline{\left( \int _{0}^{1} \frac{f(t)}{1 - t r e^{i \theta }} \text {d}t\right) } \frac{\text {d} \theta }{2 \pi } \\&= \int _{0}^{1} \overline{f(t)} \left( \int _{0}^{2 \pi } \frac{g(r e^{i \theta })}{1 - t r e^{- i \theta }} \frac{\text {d} \theta }{2 \pi }\right) \text {d}t \\&= \int _{0}^{1} \overline{f(t)} \left( \oint _{\partial \mathbb {D}} \frac{g(r \zeta )}{1 - t r \overline{\zeta }} \frac{\text {d} \zeta }{\zeta 2 \pi i}\right) \text {d}t \\&= \int _{0}^{1} \overline{f(t)}\left( \frac{1}{2 \pi i} \oint _{\partial \mathbb {D}} \frac{g(r \zeta )}{\zeta - t r} \text {d} \zeta \right) \text {d}t \\&= \int _{0}^{1} \overline{f(t)} g(r^2 t) \text {d}t. \end{aligned}$$
Thus, by (1) (making \(f \in L^1[0, 1]\)) and the fact that \(g(r t) \rightarrow g(t)\) uniformly on [0, 1], we have
$$\begin{aligned} \lim _{r \rightarrow 1} \int _{0}^{2 \pi } g(r e^{i \theta }) \overline{(\mathcal {H}f)(r e^{i \theta })} \frac{\text {d} \theta }{2 \pi } = \lim _{r \rightarrow 1} \int _{0}^{1} \overline{f(t)} g(r^2 t) \text {d} t = \int _{0}^{1} \overline{f(t)} g(t) \text {d}t \end{aligned}$$
exists. Using (1) again, we get
$$\begin{aligned} \left| \int _{0}^{1} \overline{f(t)} g(t) \text {d} t \right| \le \Vert p\Vert _{\infty } \cdot \frac{1}{2} \int _{0}^{2 \pi } |f(e^{i \theta })| \text {d} \theta , \end{aligned}$$
and so
$$\begin{aligned} g \mapsto \lim _{r \rightarrow 1} \int _{0}^{2 \pi } g(r e^{i \theta }) \overline{(\mathcal {H}f)(r e^{i \theta })} \frac{\text {d} \theta }{2 \pi } \end{aligned}$$
defines a continuous linear functional on the disk algebra \(\mathcal {A}\). The duality in (2) yields \(\mathcal {H}f = C \mu \) for some \(\mu \in M\).
To show that \(\mathcal {H}\) is injective, observe that for \(f \in H^1\), a geometric series computation with the definition of \(\mathcal {H}\) will show that
$$\begin{aligned} (\mathcal {H}f)(z) = \sum _{n = 0}^{\infty } z^n \left( \int _{0}^{1} t^n f(t) \text {d} t \right) . \end{aligned}$$
If \(\mathcal {H}f \equiv 0\), then
$$\begin{aligned} \int _{0}^{1} p(t) f(t) \text {d} t = 0 \end{aligned}$$
for all polynomials p. The Weisrstrass approximation theorem and the fact that \(f \in L^1[0, 1]\) yields
$$\begin{aligned} \int _{0}^{1} f(t) g(t) \text {d}t = 0 \end{aligned}$$
for all continuous functions g on [0, 1]. By the Riesz representation theorem for the space of finite Borel measures on [0, 1], we conclude that \(f(t) = 0\) almost everywhere on [0, 1]. The identity theorem for analytic functions implies that \(f \equiv 0\). \(\square \)

Remark 1

It is well known that when \(p \in (1, \infty )\), the dual of \(H^p\) can be identified with \(H^q\), where q is the Holder conjugate index for p, via the pairing
$$\begin{aligned} (g, f)&= \lim _{r \rightarrow 1} \int _{0}^{2 \pi } g(r e^{i \theta }) \overline{f(r e^{i \theta })} \frac{\text {d} \theta }{2 \pi } \\&= \int _{0}^{2 \pi } g(e^{i \theta }) \overline{f(e^{i \theta })} \frac{\text {d} \theta }{2 \pi }, \quad g \in H^q,\, f \in H^p. \end{aligned}$$
With this in mind, one can, for \(f \in H^p\), \(p \in (1, \infty )\), repeat the proof of Theorem 1, replacing (3) with the estimate
$$\begin{aligned} \left| \int _{0}^{1} \overline{f(t)} g(t) \text {d}t\right| \lesssim \left( \int _{0}^{2 \pi } |g(e^{i \theta })|^q \text {d} \theta \right) ^{1/q} \left( \int _{0}^{2 \pi } |f(e^{i \theta })|^{p} \text {d} \theta \right) ^{1/p}. \end{aligned}$$
(note two uses of the Fejer–Riesz inequality along with Holder’s inequality), to see that \(\mathcal {H}f \in H^{p}\). This yields an alternate proof of this fact from [2].

Remark 2

One can prove a bit more here. Indeed, if \(f \in H^1\), then \(f \in L^1[0, 1]\), and thus given \(\epsilon > 0\), there is a \(\delta > 0\), so that
$$\begin{aligned} \int _{1 - \delta }^{1} |f(t)| \text {d}t < \epsilon . \end{aligned}$$
For any \(\theta \in [0, 2 \pi ]\) and \(r \in (0, 1)\)
$$\begin{aligned} (1 - r) |\mathcal {H}f(r e^{i \theta })|&\le (1 - r) \int _{0}^{1} \frac{|f(t)|}{|1 - t r e^{i \theta }|} \text {d}t \\&\le (1 - r) \int _{0}^{1 - \delta } \frac{|f(t)|}{1 - (1 - \delta ) r} \text {d}t + \int _{1 - \delta }^{1} |f(t)| \text {d}t \\&\le \frac{1 - r}{1 - (1 - \delta ) r} \int _{0}^{1} |f(t)| \text {d}t + \epsilon . \end{aligned}$$
We conclude that
$$\begin{aligned} \lim _{r \rightarrow 1} (1 - r) |(\mathcal {H}f)(r e^{i \theta })| = 0, \quad \theta \in [0, 2 \pi ]. \end{aligned}$$
Theorem 1 says that \(\mathcal {H}f = C \mu \) for some \(\mu \in M\) and a simple exercise with the dominated convergence theorem [3, p. 42] says that
$$\begin{aligned} \lim _{r \rightarrow 1} (1 - r) (C \mu )(r e^{i \theta }) = \mu (\{e^{i \theta }\}). \end{aligned}$$
This means that \(\mathcal {H}\) maps \(H^1\) to the space of Cauchy transforms of measures on the circle which have no point masses.



The author would like to thank the referee for a careful rewriting of the principal result of this paper.

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Authors’ Affiliations

University of North Carolina at Chapel Hill, Chapel Hill, USA


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