Strict singularity of Volterra type operators on Hardy spaces

TL;DR

This paper characterizes the boundedness, compactness, and spectrum of Volterra type operators on Hardy spaces, showing that non-compactness implies the operator fixes copies of ^p and ^2, thus linking strict singularity with compactness.

Contribution

It provides a complete characterization of the spectral and compactness properties of Volterra type operators on Hardy spaces, establishing their strict singularity equivalence with compactness.

Findings

The spectrum of the operator is explicitly determined.

Non-compact operators fix isomorphic copies of ^p and ^2.

Strict singularity coincides with compactness for these operators.

Abstract

In this paper, we first characterize the boundedness and compactness of Volterra type operator $S_gf(z) = \int_0^z f'(\zeta)g(\zeta)d\zeta, \ z \in \mathbb{D},$ defined on Hardy spaces $H^p, \, 0< p <\infty$. The spectrum of $S_g$ is also obtained. Then we prove that $S_g$ fixes an isomorphic copy of $\ell^p$ and an isomorphic copy of $\ell^2$ if the operator $S_g$ is not compact on $H^p (1\leq p<\infty)$. In particular, this implies that the strict singularity of the operator $S_g$ coincides with the compactness of the operator $S_g$ on $H^p$. At last, we post an open question for further study.

Equations

$$H^p:=\{f\in H(\mathbb{D}):\Vert f{\Vert }_{H^p}={\left(\underset{r\to 1^{-}}{\lim }{\int }_{\partial \mathbb{D}}|f(r\xi ){|}^{p}dm(\xi )\right)}^{1/p}<\infty \},$$

$$H^p:=\{f\in H(\mathbb{D}):\Vert f{\Vert }_{H^p}={\left({\int }_{\partial \mathbb{D}}|f(\xi ){|}^{p}dm(\xi )\right)}^{1/p}<\infty \},$$

$$H^{\infty }=\{f\in H(\mathbb{D}):\Vert f{\Vert }_{\infty }:=\underset{z\in \mathbb{D}}{\sup }\{|f(z)|\}<\infty \}.$$

$$(T_{g}f)(z)={\int }_0^{z}f(\omega )g^{\prime }(\omega )d\omega ,z\in \mathbb{D},f\in H(\mathbb{D}),$$

$$(S_{g}f)(z)={\int }_0^z{f}^{\prime }(\omega )g(\omega )d\omega ,z\in \mathbb{D},f\in H(\mathbb{D}).$$

$$\underset{a\in \mathbb{D}}{\sup }{\int }_{\mathbb{D}}|f^{\prime }(z){|}^p(1-|z{|}^2{)}^{p-1}(1-|{\phi }_a(z){|}^2)dA(z)<\infty$$

$$\underset{a\in \mathbb{D}}{\sup }{\int }_{\mathbb{D}}|(S_{g}f{)}^{\prime }(z){|}^p(1-|z{|}^2{)}^{p-1}(1-|{\phi }_a(z){|}^2)dA(z)\le C\Vert f{\Vert }_{H^p}^p.$$

$$f_a(z)=\frac{(1-|a{|}^2{)}^{2-1/p}}{(1-\bar{a}z{)}^2}$$

$$\underset{a\in \mathbb{D}}{\sup }{\int }_{\mathbb{D}}|(S_g{f}_a{)}^{\prime }(z){|}^p(1-|z{|}^2{)}^{p-1}(1-|{\phi }_a(z){|}^2)dA(z)\le C\Vert f_a{\Vert }_{H^p}^p,$$

$$\underset{a\in \mathbb{D}}{\sup }{\int }_{\mathbb{D}}|\bar{a}{|}^p|g(z){|}^p\frac{(1-|a{|}^2{)}^{2p-1}}{|1-\bar{a}z{|}^{3p}}(1-|z{|}^2{)}^{p-1}(1-|{\phi }_a(z){|}^2)dA(z)\le C.$$

$$\underset{a\in \mathbb{D}}{\sup }{\int }_{\mathbb{D}}|\bar{a}{|}^p|g({\phi }_a(\omega )){|}^p\frac{(1-|a{|}^2{)}^{2p-1}}{|1-\bar{a}{\phi }_a(\omega ){|}^{3p}}(1-|{\phi }_a(\omega ){|}^2{)}^{p-1}(1-|\omega {|}^2)|{\phi }_a^{\prime }(\omega ){|}^{2}dA(z)\le C.$$

$$\underset{a\in \mathbb{D}}{\sup }{\int }_{\mathbb{D}}|\bar{a}{|}^p|g({\phi }_a(\omega )){|}^p|1-\bar{a}\omega {|}^{p-2}(1-|\omega {|}^2{)}^{p}dA(z)\le C.$$

$$\underset{a\in \mathbb{D}}{\sup }|G_a(0)|=\underset{a\in \mathbb{D}}{\sup }|\bar{a}{|}^p|g(a){|}^p\le \underset{a\in \mathbb{D}}{\sup }{\int }_{\mathbb{D}}|\bar{a}{|}^p|g({\phi }_a(\omega )){|}^p|1-\bar{a}\omega {|}^{p-2}(1-|\omega {|}^2{)}^{p}dA(z)\le C$$

$$\underset{|a|\to 1^{-}}{\sup }|g(a){|}^p\le C,$$

$$\Vert f{\Vert }_{H^p}^p\asymp {\int }_{\partial \mathbb{D}}{\left({\int }_{S(\xi )}|f^{(n)}(z){|}^2(1-|z{|}^2{)}^{2n-2}dA(z)\right)}^{p/2}dm(\xi )+\sum _{j=0}^{n-1}|f^{(j)}(0){|}^p.$$

$$\underset{n\to \infty }{\lim }{\int }_{\mathbb{D}}|(S_g{f}_{a_n}{)}^{\prime }(z){|}^p(1-|z{|}^2{)}^{p-1}(1-|{\phi }_{a_n}(z){|}^2)dA(z)=0,$$

$$\underset{n\to \infty }{\lim }|g(a_n){|}^p=0.$$

$$f-\frac{1}{\lambda }{S}_{g}f=h$$

$$f(z)=R_{\lambda ,g}h(z):={\int }_0^z\frac{h^{\prime }(\zeta )}{1-\frac{1}{\lambda }g(\zeta )}d\zeta +h(0).$$

$$f=S_{(1-\frac{1}{\lambda }g)}h+h(0)\in H^p$$

$$g(\mathbb{D})\cup \{0\}\subset \sigma (S_g)\subset \overline{g(\mathbb{D})}\cup \{0\}.$$

$$\Vert S_g(f_{a_n}){\Vert }_{H^p}\ge h>0$$

$$\underset{\epsilon \to 0}{\lim }{\int }_{A_{\epsilon }}|S_g{f}_{a_n}{|}^{p}dm=0$$

$$\underset{n\to \infty }{\lim }{\int }_{\partial \mathbb{D}\setminus A_{\epsilon }}|S_g{f}_{a_n}{|}^{p}dm=0$$

$$|f_{a_n}^{\prime }(r{e}^{i\vartheta }){|}^p=\frac{|{\bar{a}}_n{|}^p(1-|a{|}^2{)}^{2p-1}}{|1-{\bar{a}}_{n}r{e}^{i\vartheta }{|}^{3p}}\le \frac{|{\bar{a}}_n{|}^p(1-|a_n{|}^2{)}^{2p-1}}{{\gamma }^{3p}},$$

$$\begin{array}{rl}|(S_g{f}_{a_n})(\xi ){|}^p& ={\left|{\int }_0^1{f}_{a_n}^{\prime }(r\xi )g(r\xi )\xi dr\right|}^p\le {\left({\int }_0^1|f_{a_n}^{\prime }(r\xi )g(r\xi )|dr\right)}^p\\ & \le \Vert g{\Vert }_{\infty }^p{\left({\int }_0^1|f_{a_n}^{\prime }(r\xi )|dr\right)}^p\le \Vert g{\Vert }_{\infty }^p\frac{|{\bar{a}}_n{|}^p(1-|a_n{|}^2{)}^{2p-1}}{{\gamma }^{3p}}.\end{array}$$

$$\underset{n\to \infty }{\lim }{\int }_{\partial \mathbb{D}\setminus A_{\epsilon }}|S_g{f}_{a_n}{|}^{p}dm\le \underset{n\to \infty }{\lim }\Vert g{\Vert }_{\infty }^p\frac{|{\bar{a}}_n{|}^p(1-|a_n{|}^2{)}^{2p-1}}{{\gamma }^{3p}}=0.$$

$${\left({\int }_{A_n}|S_g{f}_{b_k}{|}^{p}dm\right)}^{1/p}<4^{-n}\delta h,k=1,\dots ,n-1;$$

$$\begin{array}{rl}& \Vert \sum _{j=1}^{\infty }{c}_j{S}_g(f_{b_j}){\Vert }_{H^p}^p=\sum _{n=1}^{\infty }{\int }_{A_n\setminus A_{n+1}}{\left|\sum _{j=1}^{\infty }{c}_j{S}_g(f_{b_j})\right|}^{p}dm\\ & \ge \sum _{n=1}^{\infty }{\left(|c_n|{\left({\int }_{A_n\setminus A_{n+1}}|S_g{f}_{b_n}{|}^{p}dm\right)}^{1/p}-\sum _{j\ne n}|c_j|{\left({\int }_{A_n\setminus A_{n+1}}|S_g{f}_{b_j}{|}^{p}dm\right)}^{1/p}\right)}^p.\end{array}$$

$$\begin{array}{rl}{\left({\int }_{A_n\setminus A_{n+1}}|S_g{f}_{b_n}{|}^{p}dm\right)}^{1/p}& ={\left({\int }_{A_n}|S_g{f}_{b_n}{|}^{p}dm-{\int }_{A_{n+1}}|S_g{f}_{b_n}{|}^{p}dm\right)}^{1/p}\\ & \ge {\left({\left(\frac{h}{2}\right)}^p-{\left(4^{-n-1}\delta h\right)}^p\right)}^{1/p}\ge \frac{h}{2}-4^{-n-1}\delta h\end{array}$$

$${\left({\int }_{A_n\setminus A_{n+1}}|S_g(f_{b_j}){|}^{p}dm\right)}^{1/p}\le {\left({\int }_{A_n}|S_g(f_{b_j}){|}^{p}dm\right)}^{1/p}<4^{-n}\delta h$$

Full text

*Corresponding author

This work was supported by NNSF of China (Grant No. 11801094).

1. Introduction

Let $\mathbb{D}$ be the unit disk of the complex plane $\mathbb{C}$ and $H(\mathbb{D})$ the space consisting of all analytic functions on $\mathbb{D}$. Then for $0<p<\infty$, the Hardy space $H^{p}$ on $\mathbb{D}$ consists of all analytic functions $f\in H(\mathbb{D})$ satisfying

[TABLE]

where $m$ is the normalized Lebesgue measure on $\partial\mathbb{D}$. By [13, Theorem 2.6], this norm is equal to the following norm:

[TABLE]

where for any $\xi\in\partial\mathbb{D}$, $f(\xi)$ is the radial limit which exists almost everywhere (see [34, Theorem 9.4]).

When $p=\infty$, the space $H^{\infty}$ is defined by

[TABLE]

For any analytic function $g\in H(\mathbb{D})$, there are two kinds of Volterra type operators defined, respectively, by

[TABLE]

and

[TABLE]

The boundedness and compactness of these two operators on some spaces of analytic functions were extensively studied. Pommerenke [30] firstly studied the boundedness of $T_{g}$ on Hardy-Hilbert space $H^{2}$. After his work, Aleman, Siskakis and Cima [2, 3] systematically studied the boundedness and compactness of $T_{g}$ on Hardy space $H^{p}$, in which they showed that $T_{g}$ is bounded (or compact) on $H^{p},\ 0<p<\infty$, if and only if $g\in BMOA$ (or $g\in VMOA$). What’s more, Aleman and Siskakis [4] studied the boundedness and compactness of $T_{g}$ on Bergman spaces while Galanopoulos, Girela and Peláez [14, 15] investigated the boundedness of $T_{g}$ and $S_{g}$ on Dirichlet type spaces and Xiao [32] studied $T_{g}$ and $S_{g}$ on $Q_{p}$ spaces.

Recently, Lin, et al [21] studied the boundedness of $T_{g}$ and $S_{g}$ acting on the derivative Hardy spaces $S^{p}$. For these operators on other spaces like Fock spaces and weighted Banach spaces, see [5, 8, 10, 22, 24, 25, 31] and the references therein.

A bounded operator $T\colon X\to Y$ between Banach spaces is strictly singular if its restriction to any infinite-dimensional closed subspace is not an isomorphism onto its image. This notion was introduced by Kato [16]. The obvious example of strictly singular non-compact operators are inclusion mappings $i_{p,q}\colon\ell^{p}\hookrightarrow\ell^{q},$ when $1\leq p<q<\infty.$

A bounded operator $T\colon X\to Y$ between Banach spaces is said to fix a copy of the given Banach space $E$ if there is a closed subspace $M\subset X$, linearly isomorphic to $E$, such that the restriction $T_{|M}$ defines an isomorphism from $M$ onto $T(M)$. The bounded operator $T\colon X\to Y$ is called $\ell^{p}$-singular if it does not fix any copy of $\ell^{p}$ .

Miihkinen [26] studied the strict singularity of $T_{g}$ on Hardy space $H^{p}$ and showed that the strict singularity of $T_{g}$ coincides with its compactness on $H^{p},\,1\leq p<\infty.$ whose main ideas come from the recent paper [19] where the corresponding questions are investigated for composition operators.

Although the boundedness and compactness of the operator $T_{g}$ on $H^{p}$ had been studied, from the literature that we have looked at so far, the proofs of the boundedness and compactness for the operator $S_{g}$ on $H^{p}$ are still not been shown in detail, except for the case $p=2$ whose study seems to be elementary (see [20]). Thus, in this paper, We first characterize the boundedness and compactness of Volterra type operator $S_{g}f$ defined on Hardy spaces $H^{p}$ for $0<p<\infty$ . Base on the characterization of the boundedness for the operator $S_{g}$ on $H^{p}$, we are able to characterize the spectrum of $S_{g}$ on $H^{p}$, inspired by the idea in the papers [8, 9]. Then we prove that the bounded operator $S_{g}$ fixes an isomorphic copy of $\ell^{p}$ if the operator $S_{g}$ is not compact on $H^{p}$. In particular, this implies that the strict singularity of the operator $S_{g}$ coincides with the compactness of the operator $S_{g}$ on $H^{p}.$ Moreover, we show that $S_{g}$, when acting on $H^{p}(1\leq p<\infty)$, fixes an isomorphic copy of $\ell^{2}$.

In the last section, we post an open question for further study.

Our main results are as follows:

Proposition 1**.**

Let $g\in H(\mathbb{D})$ and $0<p<\infty.$ Then the operator $S_{g}\colon H^{p}\to H^{p}$ is bounded if and only if $g\in H^{\infty}$.

Proposition 2**.**

Let $g\in H(\mathbb{D})$ and $0<p<\infty.$ Then the operator $S_{g}\colon H^{p}\to H^{p}$ is compact if and only if $g=0$.

Proposition 3**.**

Let $g\in H(\mathbb{D})$ and $0<p<\infty.$ Then the spectrum of the bounded operator $S_{g}\colon H^{p}\to H^{p}$ is $\sigma(S_{g})=\{0\}\cup\overline{g(\mathbb{D})}$.

Theorem 1**.**

Let $1\leq p<\infty$ and suppose that $S_{g}\colon H^{p}\to H^{p}$ is bounded but not compact. Then the operator $S_{g}\colon H^{p}\to H^{p}$ fixes an isomorphic copy of $\ell^{p}$ and an isomorphic copy of $\ell^{2}$ as well. In particular, the operator $S_{g}$ is not strictly singular, that is, strict singularity of bounded operator $S_{g}$ coincides with its compactness.

2. Boundedness and Compactness of $S_{g}$ on $H^{p}$

In this section, we provide a new proof for the conditions of boundedness and compactness of the operator $S_{g}$ on $H^{p}$ when $0<p<\infty$. Although the result for the boundedness can be deduced from [5, Lemma 2.1(i)], we give our new proof which is not only useful to the proof of compactness of $S_{g}$ on $H^{p}$ and the proof of Theorem 1, but also of interest itself.

Proof of Proposition 1.

Assume that $S_{g}\colon H^{p}\to H^{p}$ is bounded. From [33], we know that $H^{p}\subset BMOA^{1+1/p}_{p}$, where $BMOA^{1+1/p}_{p}$ is the space of analytic functions $f$ satisfying

[TABLE]

in which $\varphi_{a}(z)=\frac{a-z}{1-\bar{a}z}$ is the Möbius transformation on $\mathbb{D}$ and $A$ is the normalized Lebesgue measure on $\mathbb{D}$. Hence, if $S_{g}\colon H^{p}\to H^{p}$ is bounded, then $S_{g}\colon H^{p}\to BMOA^{1+1/p}_{p}$ is also bounded. Therefore, for any $f\in H^{p}$, we have

[TABLE]

It is easy to verify that for any $a\in\mathbb{D}$, the function

[TABLE]

is a unit vector in $H^{p}$. Thus,

[TABLE]

or equivalently,

[TABLE]

Then, let $z=\varphi_{a}(\omega)=\frac{a-\omega}{1-\bar{a}\omega}$ be a Möbius transformation on $\mathbb{D}$, we have

[TABLE]

Note that $|1-\bar{a}\varphi_{a}(\omega)|=(1-|a|^{2})/|1-\bar{a}\omega|$ and $(1-|\varphi_{a}(\omega)|^{2})=(1-|z|^{2})||\varphi^{\prime}_{a}(\omega)|$, we obtain

[TABLE]

Now, consider the analytic function $G_{a}(\omega):=\bar{a}^{p}g(\varphi_{a}(\omega))^{p}(1-\bar{a}\omega)^{p-2}$, we get that

[TABLE]

which implies that

[TABLE]

that is, $g\in H^{\infty}$.

Conversely, assume that $g\in H^{\infty}$, then by [3, 12], the operator $T_{g}\colon H^{p}\to H^{p}$ and the multiplication operator $M_{g}\colon H^{p}\to H^{p}$ are both bounded. Therefore, it follows from the obvious equality $(M_{g}f)(z)-(M_{g}f)(0)=(T_{g}f)(z)+(S_{g}f)(z)$ that $S_{g}\colon H^{p}\to H^{p}$ is also bounded. Accordingly, the proof is complete. ∎

Remark 1**.**

We note that the sufficiency of Proposition 1 can also be proven directly by using the following equivalent norms for $H^{p}$ (see [1, p. 125]):

[TABLE]

Proof of Proposition 2.

It is obvious that if $g=0$, then $S_{g}\colon H^{p}\to H^{p}$ is compact.

Conversely, if $S_{g}\colon H^{p}\to H^{p}$ is compact, $S_{g}\colon H^{p}\to BMOA^{1+1/p}_{p}$ is also compact. Since for any sequence $\{a_{n}\}_{n=1}^{\infty}$ such that $\lim_{n\rightarrow\infty}|a_{n}|=1$, $f_{a_{n}}$ converges to [math] uniformly on compact subsets of $\mathbb{D}$, it holds that

[TABLE]

then similar to the arguments in the proof of Proposition 1, we obtain

[TABLE]

That is, $g=0$. Accordingly, the proof is complete. ∎

3. The spectrum of $S_{g}$ on $H^{p}$

In this section, we characterize the the spectrum of the bounded operator $S_{g}$ on $H^{p}$ .

Proof of Proposition 3.

Since for any $f\in S^{p}\setminus\{0\}$, the function $S_{g}f$ has a zero at $z=0$, it holds that $0\in\sigma(S_{g})$.

Now, we assume that $\lambda\in\mathbb{C}\setminus\{0\}$. For any $h\in H(\mathbb{D})$, it is easy to show that the equation

[TABLE]

has the unique solution $f$ in $H(\mathbb{D})$ and the solution is

[TABLE]

Therefore, the resolvent set $\rho(S_{g})$ of the bounded operator $S_{g}$ consists precisely of all points $\lambda\in\mathbb{C}$ for which $R_{\lambda,g}$ is a bounded operator on $H^{p}$ .

If $\lambda\in\mathbb{C}\setminus(\{0\}\cup\overline{g(\mathbb{D})})$, then $1-\frac{1}{\lambda}g(\zeta)$ is bounded away from [math], that is, $\frac{1}{1-\frac{1}{\lambda}g(\zeta)}$ is bounded. Thus,

[TABLE]

by Proposition 1, which implies that the operator $R_{\lambda,g}$ is a bounded operator on $H^{p}$ . Accordingly, $\mathbb{C}\setminus(\{0\}\cup\overline{g(\mathbb{D})})\subset\rho(S_{g})$, that is, $\sigma(S_{g})\subset(\{0\}\cup\overline{g(\mathbb{D})})$ .

Conversely, if $\lambda\in g(\mathbb{D})$ and $\lambda\neq 0$, then $\frac{1}{1-\frac{1}{\lambda}g(\zeta)}$ is not bounded, which implies that the operator $R_{\lambda,g}$ is not bounded on $H^{p}$, hence we have $g(\mathbb{D})\setminus\{0\}\subset\sigma(S_{g})$ . Thus, in conjunction with the fact that $0\in\sigma(S_{g})$, it holds that

[TABLE]

Since the spectrum $\sigma(S_{g})$ is closed, we obtain that $\sigma(S_{g})=\overline{g(\mathbb{D})}\cup\{0\}$ . ∎

4. Proof of Theorem 1

First, we note that Theorem 1 holds for $p=2$ due to the fact that a bounded linear operator on $H^{2}$ is compact if and only if it is strict singular, if and only if it does not fix any copy of $\ell^{2}$ (see [29, 5.1-5.2]) . From now on, we suppose that $p\neq 2$ .

From the proof in Proposition 2, it can be easily checked that, if the bounded operator $S_{g}\colon H^{p}\to H^{p}$ is not compact, then there exists a sequence $(a_{n})\subset\mathbb{D}$ with $0<|a_{1}|<|a_{2}|<\ldots<1$ and $a_{n}\to\omega\in\partial\mathbb{D},$ such that there is a positive constant $h$ such that

[TABLE]

holds for all $n\in\mathbb{N}$ and $f_{a_{n}}$ defined in the previous section. We may assume without loss of generality that $a_{n}\rightarrow 1$ as $n\rightarrow\infty$ by utilizing a suitable rotation.

Lemma 1**.**

Let $(a_{n})\subset\mathbb{D}$ be a sequence as above. Let $A_{\varepsilon}=\{e^{i\theta}:|e^{i\theta}-1|<\varepsilon\}$ for each $\varepsilon>0$. Then for bounded operator $S_{g}\colon H^{p}\to H^{p}$, we have

[TABLE]

Proof.

(1) For each fixed $n$, this follows immediately from the absolute continuity of Lebesgue measure and the boundedness of operator $S_{g}\colon H^{p}\to H^{p}$ .

(2) For given $\varepsilon>0$, it is easy to see that there is a positive $\gamma>0$ such that $|1-\bar{a}_{n}re^{i\vartheta}|\geq\gamma$ for all $n\in\mathbb{N}$, $0\leq r<1$ and $\varepsilon\leq\vartheta\leq\pi$ . Therefore, for these $r$ and $\vartheta$, we get that

[TABLE]

for all $n\in\mathbb{N}$. Then, for any $\xi\in\partial\mathbb{D}\setminus A_{\varepsilon}$, we have

[TABLE]

Accordingly,

[TABLE]

The proof is complete. ∎

Now, we are prepared to give a proof of Theorem 1.

Proof of Theorem 1.

First, as noted above, there exists a sequence $(a_{n})\subset\mathbb{D}$ with $0<|a_{1}|<|a_{2}|<\ldots<1$ and $a_{n}\to 1,$ such that there is a positive constant $h$ such that $\|S_{g}(f_{a_{n}})\|_{H^{p}}\geq h>0$ holds for all $n\in\mathbb{N}$ .

Then by Lemma 1 and induction method, we can find a decreasing positive sequence $(\varepsilon_{n})$ such that $A_{\varepsilon_{1}}=\partial\mathbb{D}$ and $\lim_{n\rightarrow\infty}\varepsilon_{n}=0$, and a subsequence $(b_{n})\subset(a_{n})$ such that the following three conditions hold:

[TABLE]

for every $n\in\mathbb{N},$ where $A_{n}=A_{\varepsilon_{n}}$ and $\delta>0$ is a small constant whose value will be determined later.

Now we are ready to prove that $\|\sum_{j=1}^{\infty}c_{j}S_{g}(f_{b_{j}})\|_{H^{p}}\geq C\|(c_{j})\|_{\ell^{p}},$ where the constant $C>0$ may depend on $p.$

[TABLE]

Observe that for every $n\in\mathbb{N},$ we have

[TABLE]

according to conditions (1) and (3) above, where the last estimate holds for $1\leq p<\infty$.

Moreover, we have

[TABLE]

for $j<n$ by condition (1) and

[TABLE]

for $j>n$ by condition (2).

Thus it always holds that

[TABLE]

Consequently, by the triangle inequality in $L^{p}$, we obtain that

[TABLE]

where the last inequality holds when we choose $\delta$ small enough.

A straightforward variant of the above procedure also gives

[TABLE]

where the constant $C_{1}>0$ may depend on $p.$

By choosing $g=1$ and the fact that $\lim_{n\rightarrow\infty}f_{a_{n}}(0)=0$, we obtain that

[TABLE]

Thus, we have

[TABLE]

This implies that the operator $S_{g}\colon H^{p}\to H^{p}$ fixes an isomorphic copy of $\ell^{p}$ .

To prove that the operator $S_{g}\colon H^{p}\to H^{p}$ fixes an isomorphic copy of $\ell^{2}$, we consider the trivial equality

[TABLE]

where $M_{g}$ is the multiplication operator. It is proven that $M_{g}$ is not $l^{2}$-singular on $H^{p}$ while $T_{g}$ is $l^{2}$-singular on $H^{p}$ (see [23, 27]). Since the class of $l^{2}$-singular operators forms a linear subspace of the space of all bounded operators, it follows from the above equality that $S_{g}$ is not $l^{2}$-singular on $H^{p}$, that is, $S_{g}\colon H^{p}\to H^{p}$ fixes an isomorphic copy of $\ell^{2}$. The proof is complete. ∎

5. Open question

In Theorem 1, we exclude the case of $p=\infty$ since our method is not applicable in this case. Although Bourgain [6] has established that a bounded linear operator on $H^{\infty}$ is weakly compact if and only if it does not fix any copy of $\ell^{\infty}$, we do not know whether or not $S_{g}$ acting on $H^{\infty}$ is compact if and only if it does not fix any copy of $\ell^{\infty}$. So we post it as an open question as follows:

Open Question.* Suppose that $S_{g}\colon H^{\infty}\to H^{\infty}$ is bounded but not compact. Does the operator $S_{g}\colon H^{\infty}\to H^{\infty}$ fix an isomorphic copy of $\ell^{\infty}$? In particular, is the operator $S_{g}$ not strictly singular?*