Composition operators and Gleason’s problem on weighted Fock spaces

Abstract

Let $F_\alpha ^p$ be the weighted Fock space in $\mathbb {C}^{n}$ with $\alpha$ real and $0<p\le \infty$. In this paper, we completely characterize the topological components and isolated elements of the set of bounded composition operators acting on $F_\alpha ^p$ in the uniform operator topology. Meanwhile, we show that Gleason’s problem on $F_\alpha ^p$ in $\mathbb {C}^{n}$ is solvable.

1 Introduction

Let $\mathbb {C}^{n}$ be the Euclidean space of complex dimension n. For $z= (z_1,\dots ,z_n)$ and $w=(w_1,\dots ,w_n)$ in $\mathbb {C}^n$, we write $\langle z,w\rangle =\sum _{j=1}^nz_j\overline{w_j}$ and $|z|=\sqrt{\langle z,z\rangle }$. Let $\mathrm{d}v$ be the volume measure on $\mathbb {C}^n$ normalized so that $\int _{\mathbb {C}^n}e^{-|z|^2}\mathrm{d}v(z)=1$. For $\alpha$ real, we let

$$\begin{aligned} \mathrm{d}v_\alpha (z)=\frac{\mathrm{d}v(z)}{(1+|z|)^\alpha }. \end{aligned}$$

For $0<p<\infty$, let $L_\alpha ^p$ denote the space of Lebesgue measurable functions f on $\mathbb {C}^n$ such that

$$\begin{aligned} \Vert f\Vert _{p,\alpha }^p=\int _{\mathbb {C}^n}\left| f(z)e^{-\frac{1}{2}|z|^2}\right| ^p\mathrm{d}v_\alpha (z) <\infty , \end{aligned}$$

where we use the term norm $\Vert f\Vert _{p,\alpha }$ for $0<p<1$ only for convenience. For $p=\infty$, we use the notation $L_\alpha ^\infty$ to denote the space of Lebesgue measurable functions f on $\mathbb {C}^n$, such that

$$\begin{aligned} \Vert f\Vert _{\infty ,\alpha }=\sup \left\{ \frac{|f(z)|e^{-\frac{1}{2}|z|^2}}{(1+|z|)^\alpha }: z\in \mathbb {C}^n\right\} <\infty . \end{aligned}$$

Let $F^p_\alpha$ denote the class of holomorphic functions in $L_\alpha ^p$. Clearly, $F_\alpha ^p\subset F_\beta ^p$ when $\alpha <\beta$. For $1\le p\le \infty$, $F_\alpha ^p$ is a Banach space. When $0<p<1$, $F_\alpha ^p$ is a complete metric space with the distance $d(f,g)=\Vert f-g\Vert _{p,\alpha }^p$. In particular, when $\alpha =0$, $F_0^p$ is called the Fock space (see [25]). Therefore, we call $F^p_\alpha$ the weighted Fock space, which is identified with the Fock–Sobolev space of fractional order. For more detailed information about weighted Fock spaces $F^p_\alpha$, see [6].

For a holomorphic mapping $\varphi :\mathbb {C}^n\rightarrow \mathbb {C}^n$, the composition operator $C_{\varphi }$ on the weighted Fock space is defined by $C_{\varphi }f=f\circ \varphi$ for $f\in F_\alpha ^p$. The main subject in the study of composition operators is to describe operator theoretic properties of $C_{\varphi }$ in terms of function theoretic properties of $\varphi$, see [7]. Let ${\mathcal {C}}(X)$ denote the space of bounded composition operators with the operator norm topology on a space X of holomorphic functions. An area of considerable interest is the topological structure of ${\mathcal {C}}(X)$. In 1981, Berkson [2] initiated the topological structure with his isolation results on composition operators acting on the Hardy space $H^2$. After that, many experts studied the same questions on various holomorphic function spaces (see [3, 9, 11, 13, 17,18,19, 22]). However, a complete description of topological components of ${\mathcal {C}}(H^2)$ is still unknown so far. In [8], the second author of this paper completely characterized the topological structure of composition operators acting on the Fock space $F^2_0$ in $\mathbb {C}^n$. Our main purpose here is to extend the result of [8] to weighted Fock spaces $F^p_\alpha$ with $\alpha$ real and $0<p\le \infty$. Because the weighted Fock space $F^p_\alpha$ is not a Hilbert space when $p\ne 2$, we cannot use the approach in [8] to prove our main result (see Theorem 3.7). Our work with the full range of p requires some new techniques.

Let X be a space of holomorphic functions on a domain $\Omega$ in $\mathbb {C}^n$. For $a=(a_1,\dots , a_n)$ in $\Omega$ and f in X, if there exists functions $f_1, \dots , f_n$ in X, such that

$$\begin{aligned} f(z)-f(a)=\sum _{k=1}^n (z_k-a_k)f_k(z) \end{aligned}$$

for all z in $\Omega$, we say that Gleason’s problem on X is solvable. Note that

$$\begin{aligned} f(z)-f(a)=\int _0^1\frac{df(a+t(z-a))}{\mathrm{d}t}\mathrm{d}t=\sum _{k=1}^n(z_k-a_k) \int _0^1\frac{\partial f}{\partial z_k}(a+t(z-a))\mathrm{d}t. \end{aligned}$$

Thus, if

$$\begin{aligned} A_kf(z)=\int _0^1\frac{\partial f}{\partial z_k}(a+t(z-a))\mathrm{d}t \end{aligned}$$

(1.1)

is in X for each $k(1\le k\le n)$, then Gleason’s problem on X is solvable. The difficulty of the problem depends on $\Omega$ and X. Gleason’s problem was investigated on various function spaces and domains in $\mathbb {C}^n$ (see [1, 12, 15, 20, 21, 23]). In [10], the second author of the present paper and Zhou solved Gleason’s problem on the Fock–Sobolev space, or equivalently, a weighted Fock space $F_\alpha ^p$ for $\alpha$ real satisfying certain conditions. In this paper, we prove that Gleason’s problem is solvable on any wighted Fock space $F_\alpha ^p$ with $\alpha$ real and $0<p\le \infty$ (see Theorem 4.3), which extends the corresponding result in [10].

Throughout this paper, the notation $X\preceq Y$ or $Y\succeq X$ for two nonnegative quantities X and Y means that $X\le CY$ for some inessential positive constant C. Similarly, we write $X\approx Y$ if $X\preceq Y$ and $Y\preceq X$.

2 Preliminary results

In this section, we will collect some preliminary results. The following proposition characterizes completely the boundedness and compactness of a composition operator on the weighted Fock space, and it is an extension of the corresponding result on the Fock space in [4].

Proposition 2.1

([16]) Let $\varphi :\mathbb {C}^n\rightarrow \mathbb {C}^n$ be a holomorphic mapping. Then, the following statements hold.

(i) $C_{\varphi }$ is bounded on $F_{\alpha }^p$ if and only if $\varphi (z)=Az+B$, where A is an $n\times n$ matrix with $\Vert A\Vert \le 1$ and B is an $n\times 1$ vector satisfying $\langle A\zeta , B\rangle =0$ whenever $|A\zeta |=|\zeta |$ for $\zeta \in \mathbb {C}^n$.

(ii) $C_{\varphi }$ is compact on $F_{\alpha }^p$ if and only if $\varphi (z)=Az+B$, where A is an $n\times n$ matrix with $\Vert A\Vert <1$ and B is an $n\times 1$ vector.

The proof of Proposition 2.1 need to use the notion of the singular value decomposition of a matrix. Suppose that A is an $n\times n$ matrix of rank r. Let $\sigma _j(j=1, \dots ,n)$ be the nonnegative square roots of the eigenvalues of $AA^*$ listing in decreasing order so that $\sigma _1\ge \sigma _2\ge \cdots \ge \sigma _r>\sigma _{r+1}=\cdots =\sigma _n=0$. By the singular value decomposition of a matrix, there are unitary matrices U and V such that $A=U\Lambda V$, where $\Lambda$ is the diagonal matrix whose jth diagonal entry is $\sigma _j$. If $\Vert A\Vert \le 1$, then $0\le \sigma _j\le 1$ for all j and at least one will equal 1 if $\Vert A\Vert =1$. Let

$$\begin{aligned} k=\max \{j:\sigma _j=1\}. \end{aligned}$$

(2.1)

If $\sigma _j<1$ for all $j=1,\ldots , n$, that is $\Vert A\Vert <1$, we write $k=0$. Let $\varphi (z)=Az+B$ and $\psi (z)=\Lambda z+B'$, where the singular value decomposition of A is $U\Lambda V$ and $B'=U^*B$. Then we call $\psi$ a normalization of $\varphi$.

The following lemmas are very useful in the proof of our main results.

Lemma 2.2

([4]) Let $\varphi (z)=Az+B$, where A is an $n\times n$ matrix with $\Vert A\Vert \le 1$ and B is an $n\times 1$ vector satisfying $\langle A\zeta , B\rangle =0$ whenever $|A\zeta |=|\zeta |$ for $\zeta \in \mathbb {C}^n$. Let $\psi (z)=\Lambda z+B'$ be a normalization of $\varphi$. Then $C_\varphi =C_{Vz}C_\psi C_{Uz}$ and the first k coordinates of $B'$ are 0, where k is defined by (2.1).

Lemma 2.3

([8]) Let A and $A_1$ be $n\times n$ matrices with $\Vert A\Vert =\Vert A_1\Vert =1$. Let $E_k$ be a $k\times k$ unit matrix, where k is defined by (2.1). Suppose that $A\zeta = A_1\zeta$ for all $\zeta$ with $|A\zeta |=|\zeta |$ or $|A_1\zeta |=|\zeta |$. Then there exist $n\times n$ unitary matrices U and V, such that

$$\begin{aligned} A=U\left( \begin{array}{cc}E_k &{} O \\ O &{}D \end{array}\right) V, \qquad A_1=U\left( \begin{array}{cc}E_k &{} O \\ O &{}D_1 \end{array}\right) V, \end{aligned}$$

where D and $D_1$ are $(n-k)\times (n-k)$ matrices with $\Vert D\Vert <1$ and $\Vert D_1\Vert <1$.

Lemma 2.4

([24]) Suppose that $(X,\mu )$ is a measure space, H is a non-negative measurable function on the product space $X\times X$ and T is the integral operator induced by H as follows:

$$\begin{aligned} Tf(x)=\int _X H(x,y)f(y)\mathrm{d}\mu (y). \end{aligned}$$

Let $1<p<\infty$ with $1/p+1/q$=1. If there exists a positive measurable function h on X satisfying

$$\begin{aligned} \int _X H(x,y)h^q(y)\mathrm{d}\mu (y)\preceq h^q(x) \end{aligned}$$

for almost all $x\in X$, and

$$\begin{aligned} \int _X H(x,y)h^p(x)\mathrm{d}\mu (x)\preceq h^p(y) \end{aligned}$$

for almost all $y\in Y$, then the operator T is bounded on the space $L^p(X,\mathrm{d}\mu )$.

Lemma 2.5

([6]) For any real number $\alpha$ and $0<p\le \infty$, then the reproducing kernel $e^{\langle z,w\rangle }$ of $F_0^2$ can reproduce functions in any weighted Fock space $F_\alpha ^p$, that is

$$\begin{aligned} f(z)=\int _{\mathbb {C}^n}f(w)e^{\langle z,w\rangle }e^{-|w|^2}\mathrm{d}v(w), \qquad f\in F_\alpha ^p, \quad z\in \mathbb {C}^n. \end{aligned}$$

(2.2)

Lemma 2.6

([16]) For $0<p\le \infty$, $\alpha$ real and $f\in F_\alpha ^p(\mathbb {C}^n)$, let $g(z)=f(z_1,\ldots ,z_{n-1},0)$. Then $g\in F_\alpha ^p$ and $\Vert g\Vert _{p,\alpha }\preceq \Vert f\Vert _{p,\alpha }$.

Lemma 2.7

Let $\alpha$ be a real number and $0<p, r<\infty$. Then

$$\begin{aligned} \frac{|f(z)|^pe^{-\frac{p}{2}|z|^2}}{(1+|z_j|)^p(1+|z|)^\alpha } \preceq \int _{|w-z|<r}\frac{|f(w)|^pe^{-\frac{p}{2}|w|^2}}{(1+|w_j|)^p}\mathrm{d}v_\alpha (w) \end{aligned}$$

for all holomorphic functions f and $z\in \mathbb {C}^n$, where $z_j(1\le j\le n)$ is the jth component of z.

Proof

For any z, w in $\mathbb {C}^n$, we see that

$$\begin{aligned} 1+|z|\le 1+|w|+|z-w||\le (1+|w|)(1+|z-w|). \end{aligned}$$

It follows that

$$\begin{aligned} (1+|z-w|)^{-|\alpha |}\le \left( \frac{1+|z|}{1+|w|}\right) ^\alpha \le (1+|z-w|)^{|\alpha |} \end{aligned}$$

(2.3)

for any real number $\alpha$. Since

$$\begin{aligned} 1+|z_j|\le (1+|w_j|)(1+|z_j-w_j|)\le (1+|w_j|)(1+|z-w|), \end{aligned}$$

we also have

$$\begin{aligned} (1+|z-w|)^{-|\alpha |}\le \left( \frac{1+|z_j|}{1+|w_j|}\right) ^\alpha \le (1+|z-w|)^{|\alpha |}. \end{aligned}$$

(2.4)

Using (2.3) and (2.4), we obtain

$$\begin{aligned}&(1+|z_j|)^{p}(1+|z|)^{\alpha } \int _{|w-z|<r}\frac{|f(w)|^pe^{-\frac{p}{2}|w|^2}}{(1+|w_j|)^p}\mathrm{d}v_\alpha (w)\\&=\int _{|w-z|<r}|f(w)|^pe^{-\frac{p}{2}|w|^2}\frac{(1+|z_j|)^{p}(1+|z|)^{\alpha }}{(1+|w_j|)^p(1+|w|)^\alpha }\mathrm{d}v(w) \\&\ge \int _{|w-z|<r}\frac{|f(w)|^pe^{-\frac{p}{2}|w|^2}}{(1+|w-z|)^{p+|\alpha |}}\mathrm{d}v(w)\\&=\int _{|w|<r}\frac{|f(w+z)|^pe^{-\frac{p}{2}|w+z|^2}}{(1+|w|)^{p+|\alpha |}}\mathrm{d}v(w)\\&=e^{-\frac{p}{2}|z|^2}\int _{|w|<r}\frac{|f(w+z)e^{-\langle w,z\rangle }|^pe^{-\frac{p}{2}|w|^2}}{(1+|w|)^{p+|\alpha |}}\mathrm{d}v(w). \end{aligned}$$

Let $d\sigma$ denote the surface measure on the unit sphere $\partial \mathbb {B}_n=\{\zeta \in \mathbb {C}^n: |\zeta |=1\}$. By integrating in polar coordinates (see [21]) and using the subharmonicity of the function $|f(w+z)e^{-\langle w,z\rangle }|^p$ with respect to w, we have

$$\begin{aligned}&\int _{|w|<r}\frac{|f(w+z)e^{-\langle w,z\rangle }|^pe^{-\frac{p}{2}|w|^2}}{(1+|w|)^{p+|\alpha |}}\mathrm{d}v(w) \\&\approx \int _0^r \frac{t^{2n-1}e^{-\frac{p}{2} t^2}}{(1+t)^{p+|\alpha |}}\mathrm{d}t\int _{\partial \mathbb {B}_n} |f(t\zeta +z)e^{-t\langle \zeta ,z\rangle }|^pd\sigma (\zeta )\succeq |f(z)|^p. \end{aligned}$$

So the desired inequality follows. $\square$

It is known that the (weighted) Fock space can be characterized in terms of partial derivatives. For example, if f belongs to the Fock space $F^p_0(0<p<\infty )$, Hu proved in [14] that $\partial _j f/(1+|z|)\in L^p_0$, where $\partial _jf$ is denoted by the partial derivative of f with respect to $z_j(1\le j\le n)$. We will improve Hu’s result and generalize it to the weighted Fock space, that is $\partial _j f/(1+|z_j|)\in L^p_\alpha$ if $f\in F^p_\alpha$ for any $\alpha$ real.

Lemma 2.8

Let $\alpha$ be a real number, $0<p\le \infty$ and $f\in F^p_\alpha$. Then

$$\begin{aligned}&\int _{\mathbb {C}^n}\frac{|\partial _jf(z)|^pe^{-\frac{p}{2}|z|^2}}{(1+|z_j|)^{p}}\mathrm{d}v_\alpha (z)\preceq \Vert f\Vert _{p,\alpha }^p, \qquad 0<p<\infty ; \\&\sup _{z\in \mathbb {C}^n}\frac{|\partial _jf(z)|e^{-\frac{|z|^2}{2}}}{(1+|z_j|)(1+|z|)^\alpha } \preceq \Vert f\Vert _{\infty ,\alpha }, \qquad p=\infty . \end{aligned}$$

Proof

For any $f\in F^p_\alpha$, by (2.2) we have

$$\begin{aligned} \partial _jf(z)=\int _{\mathbb {C}^n}f(w)e^{\langle z,w\rangle }\overline{w_j}e^{-|w|^2}\mathrm{d}v(w), \end{aligned}$$

(2.5)

which, together with (2.4), yields

$$\begin{aligned} \frac{|\partial _jf(z)|}{1+|z_j|}&\le \int _{\mathbb {C}^n}\left| f(w)e^{\langle w,z\rangle }\right| \frac{1+|w_j|}{1+|z_j|}e^{-|w|^2}\mathrm{d}v(w) \\&\le \int _{\mathbb {C}^n}|f(w)|(1+|w-z|)e^{-|w|^2+R e\langle w,z\rangle }\mathrm{d}v(w). \end{aligned}$$

When $1<p<\infty$, let

$$\begin{aligned} H(z,w)=(1+|w-z|)(1+|w|)^\alpha e^{(\frac{p}{2}-1)|w|^2+R e\langle w,z\rangle }, \quad \mathrm{d}\mu (w)=e^{-\frac{p}{2}|w|^2}\mathrm{d}v_\alpha (w). \end{aligned}$$

We define an integral operator T on the space $L^p(\mathbb {C}^n,\mathrm{d}\mu )$ as follows:

$$\begin{aligned} Tf(z)=\int _{\mathbb {C}^n}f(w)H(z,w)\mathrm{d}\mu (w). \end{aligned}$$

Let $1<q<\infty$ with $\frac{1}{p}+\frac{1}{q}=1$ and consider the positive function $h(z)=e^{\frac{|z|^2}{2q}}$. Then

$$\begin{aligned} \int _{\mathbb {C}^n}H(z,w)h^q(w)\mathrm{d}\mu (w)&=\int _{\mathbb {C}^n}e^{-\frac{|w|^2}{2}+R e\langle w,z\rangle } (1+|w-z|)\mathrm{d}v(w)\\&=e^{\frac{|z|^2}{2}}\int _{\mathbb {C}^n}e^{-\frac{|w-z|^2}{2}} (1+|w-z|)\mathrm{d}v(w)\\&=e^{\frac{|z|^2}{2}}\int _{\mathbb {C}^n}e^{-\frac{|w|^2}{2}}(1+|w|)\mathrm{d}v(w)\\&\preceq e^{\frac{|z|^2}{2}}=h^q(z). \end{aligned}$$

Meanwhile, by (2.3), we have

$$\begin{aligned}&\int _{\mathbb {C}^n}H(z,w)h^p(z)\mathrm{d}\mu (z)\\&= e^{(\frac{p}{2}-1)|w|^2} \int _{\mathbb {C}^n}e^{-\frac{|z|^2}{2}+R e\langle w,z\rangle }(1+|w-z|) \frac{(1+|w|)^\alpha }{(1+|z|)^\alpha }\mathrm{d}v(z)\\&\le e^{\frac{p-1}{2}|w|^2}\int _{\mathbb {C}^n}e^{-\frac{|w-z|^2}{2}} (1+|w-z|)^{1+|\alpha |}\mathrm{d}v(z)\\&=e^{\frac{p}{2q}|w|^2}\int _{\mathbb {C}^n}e^{-\frac{|w|^2}{2}}(1+|w|)^{1+|\alpha |}\mathrm{d}v(z))\preceq h^p(w). \end{aligned}$$

It follows from Lemma 2.4 that the operator T is bounded on $L^p(\mathbb {C}^n,\mathrm{d}\mu )$, and so

$$\begin{aligned} \int _{\mathbb {C}^n}\frac{|\partial _jf(z)|^pe^{-\frac{p}{2}|z|^2}}{(1+|z_j|)^{p}}\mathrm{d}v_\alpha (z)&=\int _{\mathbb {C}^n}\frac{|\partial _jf(z)|^p}{(1+|z_j|)^{p}}\mathrm{d}\mu (z)\le \int _{\mathbb {C}^n}[T|f|(z)]^{p}\mathrm{d}\mu (z)\\&\preceq \int _{\mathbb {C}^n}|f(z)|^{p}\mathrm{d}\mu (z)=\Vert f\Vert _{p,\alpha }^p. \end{aligned}$$

When $0<p\le 1$, by (2.5), we have

$$\begin{aligned} |\partial _jf(z)|\le \int _{\mathbb {C}^n}|f(w)e^{\langle w,z\rangle }w_j|e^{-|w|^2}\mathrm{d}v(w). \end{aligned}$$

Noting that $f(w)e^{\langle w,z\rangle }w_j$ is holomorphic with respect to w on $\mathbb {C}^n$, we have by Lemma 4 in [5]

$$\begin{aligned} |\partial _jf(z)|^p\preceq \int _{\mathbb {C}^n}|f(w)e^{\langle w,z\rangle }w_j|^pe^{-p|w|^2}\mathrm{d}v(w). \end{aligned}$$

This yields together with (2.4)

$$\begin{aligned} \frac{|\partial _jf(z)|^p}{(1+|z_j|)^p}&\preceq \int _{\mathbb {C}^n}|f(w)e^{\langle w,z\rangle }|^p \left( \frac{1+|w_j|}{1+|z_j|}\right) ^pe^{-p|w|^2}\mathrm{d}v(w)\\&\le \int _{\mathbb {C}^n}|f(w)|^p(1+|w-z|)^p e^{p(R e\langle w,z\rangle -|w|^2)}\mathrm{d}v(w). \end{aligned}$$

It follows from Fubini’s theorem and (2.3) that

$$\begin{aligned}&\int _{\mathbb {C}^n}\frac{|\partial _jf(z)|^pe^{-\frac{p}{2}|z|^2}}{(1+|z_j|)^{p}}\mathrm{d}v_\alpha (z)\\&\preceq \int _{\mathbb {C}^n}\int _{\mathbb {C}^n} \frac{e^{p(R e\langle w,z\rangle -\frac{|z|^2}{2})}(1+|z-w|)^p}{(1+|z|)^{\alpha }}\mathrm{d}v(z) |f(w)|^pe^{-p|w|^2}\mathrm{d}v(w)\\&=\int _{\mathbb {C}^n}\left[ \int _{C^n} e^{-\frac{p}{2}|z-w|^2}(1+|z-w|)^p\frac{(1+|w|)^\alpha }{(1+|z|)^\alpha } \mathrm{d}v(z)\right] |f(w)|^pe^{-\frac{p}{2}|w|^2}\mathrm{d}v_\alpha (w)\\&\le \int _{C^n}\left[ \int _{\mathbb {C}^n} e^{-\frac{p}{2}|z-w|^2}(1+|z-w|)^{p+|\alpha |}\mathrm{d}v(z)\right] |f(w)|^pe^{-\frac{p}{2}|w|^2}\mathrm{d}v_\alpha (w)\\&\preceq \int _{\mathbb {C}^n}|f(w)|^pe^{-\frac{p}{2}|w|^2}\mathrm{d}v_\alpha (w)=\Vert f\Vert _{p,\alpha }^p. \end{aligned}$$

For $p=\infty$ and $f\in F^\infty _\alpha$, by (2.5), (2.3) and (2.4) we obtain

$$\begin{aligned}&\frac{|\partial _jf(z)|e^{-\frac{|z|^2}{2}}}{(1+|z_j|)(1+|z|)^\alpha }\\&\le e^{-\frac{|z|^2}{2}}\Vert f\Vert _{\infty ,\alpha }\int _{\mathbb {C}^n}\frac{(1+|w_j|)(1+|w|)^\alpha }{(1+|z_j|)(1+|z|)^\alpha } e^{-\frac{|w|^2}{2}+R e\langle z,w\rangle }\mathrm{d}v(w)\\&\le \Vert f\Vert _{\infty ,\alpha }\int _{\mathbb {C}^n}(1+|w-z|)^{1+|\alpha |} e^{-\frac{|w-z|^2}{2}}\mathrm{d}v(w)\preceq \Vert f\Vert _{\infty ,\alpha }. \end{aligned}$$

The proof is complete. $\square$

3 Topological structures of ${\mathcal {C}}(F_\alpha ^p)$

Proposition 3.1

For $0<p\le \infty$ and $\alpha$ real, let A be an $n\times n$ matrix and B be an $n\times 1$ vector such that $C_{Az+B}$ is bounded on $F^p_\alpha$. Then $C_{Az}$ and $C_{Az+B}$ are in the same path component in ${\mathcal {C}}(F^p_\alpha )$.

Proof

The boundedness of $C_{Az+B}$ on $F^p_\alpha$ implies the boundedness of $C_{Az}$ by Proposition 2.1. Let $\varphi _t(z)=t(Az+B)+(1-t)Az=A z+tB~(0\le t\le 1)$. It suffices to prove that $t\mapsto C_{\varphi _t}$ is a continuous path joining $C_{Az}$ to $C_{Az+B}$. Let $\psi (z)=\Lambda z+B'$ be the normalization of $Az+B$ as described previously. For a unitary matrix U, the composition operator $C_{Uz}$ is unitary on $F_\alpha ^p$. By Lemma 2.2 we only need to show that the map $t\mapsto C_{\psi _t}$ is continuous, where $\psi _t(z)=\Lambda z+tB'$. Let t and s be distinct points in [0, 1]. We may take $t>s$ without loss of generality. Let $b_j'(j=k+1,\ldots , n)$ be the j-th entry of $B'$. It suffices to prove $\Vert C_{\psi _t}-C_{\psi _s}\Vert _{p,\alpha }\preceq (t-s)$.

If $\Lambda$ is invertible, for $0<p<\infty$ and $f\in F_\alpha ^p$, we have by a change of variables:

$$\begin{aligned}&\Vert (C_{\psi _t}-C_{\psi _s})f\Vert _{p,\alpha }^p= \int _{\mathbb {C}^n}|f(\Lambda w+tB')-f(\Lambda w+sB')|^pe^{-\frac{p}{2}|w|^2}\mathrm{d}v_\alpha (w)\\&=(\det \Lambda )^{-2}\int _{\mathbb {C}^n}|f(z+tB')-f(z+sB')|^pe^{-\frac{p}{2}|\Lambda ^{-1}z|^2} (1+|\Lambda ^{-1}z|)^{-\alpha }\mathrm{d}v(z)\\&=(\det \Lambda )^{-2}\int _{\mathbb {C}^n}\left| \int _s^t\sum _{j=k+1}^nb_j'\partial _jf( z+\tau B')\mathrm{d}\tau \right| ^p e^{-\frac{p}{2}|\Lambda ^{-1}z|^2}(1+|\Lambda ^{-1}z|)^{-\alpha }\mathrm{d}v(z)\\&\preceq \int _{\mathbb {C}^n}\left( \sum _{j=k+1}^n\int _s^t|\partial _jf(z+\tau B')|\mathrm{d}\tau \right) ^p e^{-\frac{p}{2}|\Lambda ^{-1}z|^2}(1+|z|)^{-\alpha }\mathrm{d}v(z). \end{aligned}$$

Below we assume $B'\ne 0$, since there is nothing to prove when $B'=0$. By Lemma 2.7, we obtain

$$\begin{aligned} |\partial _jf(z+\tau B')|^p&\preceq (1+|z_j+\tau b_j'|)^{p}(1+|z+\tau B'|)^\alpha e^{\frac{p}{2}|z+\tau B'|^2}\\&\times \int _{|w-z-\tau B'|<|B'|} \frac{|\partial _jf(w)|^pe^{-\frac{p}{2}|w|^2}}{(1+|w_j|)^p}\mathrm{d}v_\alpha (w)\\&\preceq (1+|z_j|)^p(1+|z|)^\alpha e^{\frac{p}{2}|z+\tau B'|^2}F(z) \end{aligned}$$

for all $0\le \tau \le 1$ and $z\in \mathbb {C}^n$, where

$$\begin{aligned} F(z)&=\int _{|w-z|<2|B'|}\frac{|\partial _jf(w)|^pe^{-\frac{p}{2}|w|^2}}{(1+|w_j|)^p}\mathrm{d}v_\alpha (w)\\&=\int _{|w|<2|B'|}\frac{|\partial _jf(w+z)|^pe^{-\frac{p}{2}|w+z|^2}}{(1+|w_j+z_j|)^p(1+|w+z|)^\alpha }\mathrm{d}v(w). \end{aligned}$$

Hence

$$\begin{aligned}&\Vert (C_{\psi _t}-C_{\psi _s})f\Vert _{p,\alpha }^p\\&\preceq \int _{\mathbb {C}^n}\left( \sum _{j=k+1}^n\int _s^t(1+|z_j|) e^{\frac{1}{2}(|z+\tau B'|^2-|\Lambda ^{-1}z|^2)} \mathrm{d}\tau \right) ^pF(z)\mathrm{d}v(z). \end{aligned}$$

Let $\sigma _j$ be the j-th diagonal of $\Lambda$. Then, $0<\sigma _j<1$ for $k+1\le j\le n$. Noting that

$$\begin{aligned} |z+\tau B'|^2-|\Lambda ^{-1}z|^2=\sum _{j=k+1}^n\left( \frac{|\tau b_j'|^2}{1-\sigma _j^2} -\frac{1-\sigma _j^2}{\sigma _j^2}\left| z_j-\frac{\tau b_j'\sigma _j^2}{1-\sigma _j^2}\right| ^2\right) , \end{aligned}$$

we have

$$\begin{aligned} e^{\frac{1}{2}(|z+\tau B'|^2-|\Lambda ^{-1}z|^2)}(1+|z_j|)\preceq 1 \end{aligned}$$

for all $z\in \mathbb {C}^n$ and $\tau \in [0,1]$ and $k+1\le j\le n$. Thus, by Fubini’s theorem and Lemma 2.8, we have

$$\begin{aligned}&\Vert (C_{\psi _t}-C_{\psi _s})f\Vert _{p,\alpha }^p\preceq (t-s)^p\int _{\mathbb {C}^n}F(z)\mathrm{d}v(z)\\&=(t-s)^p\int _{|w|<2|B'|} \left[ \int _{\mathbb {C}^n}\frac{|\partial _jf(w+z)|^pe^{-\frac{p}{2}|w+z|^2}}{(1+|w_j+z_j|)^p(1+|w+z|)^\alpha }\mathrm{d}v(z)\right] \mathrm{d}v(w)\\&=(t-s)^p\int _{|w|<2|B'|} \left[ \int _{\mathbb {C}^n}\frac{|\partial _jf(z)|^pe^{-\frac{p}{2}|z|^2}}{(1+|z_j|)^p}\mathrm{d}v_\alpha (z)\right] \mathrm{d}v(w)\\&\preceq (t-s)^p\Vert f\Vert _{p,\alpha }^p \end{aligned}$$

as desired.

When $p=\infty$ and $f\in F_\alpha ^\infty$,

$$\begin{aligned} |f(\Lambda z+tB')-f(\Lambda z+sB')|&= \left| \int _s^t\sum _{j=k+1}^nb_j'\partial _jf(\Lambda z+\tau B')\mathrm{d}\tau \right| \\&\preceq \sum _{j=k+1}^n\int _s^t|\partial _jf(\Lambda z+\tau B')|\mathrm{d}\tau . \end{aligned}$$

It follows from Lemma 2.8 that

$$\begin{aligned}&\frac{|\partial _jf(\Lambda z+\tau B')|e^{-\frac{|z|^2}{2}}}{(1+|z|)^\alpha }\\&\preceq \Vert f\Vert _{\infty ,\alpha } \frac{(1+|\Lambda z+\tau B'|)^\alpha }{(1+|z|)^\alpha }(1+|\sigma _j z_j+\tau b_j'|) e^{\frac{1}{2}(|\Lambda z+\tau B'|^2-|z|^2)}\\&\preceq \Vert f\Vert _{\infty ,\alpha } (1+|z_j|)e^{\frac{1}{2}(|\Lambda z+\tau B'|^2-|z|^2)}. \end{aligned}$$

Since

$$\begin{aligned} |\Lambda z+\tau B'|^2-|z|^2= \sum _{j=k+1}^n\left( \frac{|\tau b_j'|^2}{1-\sigma _j^2} -(1-\sigma _j^2)\left| z_j-\frac{\tau b_j'\sigma _j}{1-\sigma _j^2}\right| ^2\right) , \end{aligned}$$

we conclude that $(1+|z_j|)e^{\frac{1}{2}(|\Lambda z+\tau B'|^2-|z|^2)}$ is uniformly bounded for $\tau \in [0,1]$ and $k+1\le j\le n$, and so

$$\begin{aligned} |f(\Lambda z+tB')-f(\Lambda z+sB')|e^{-\frac{|z|^2}{2}}(1+|z|)^{-\alpha }\preceq (t-s)\Vert f\Vert _{\infty ,\alpha }. \end{aligned}$$

This means $\Vert (C_{\psi _t}-C_{\psi _s})f\Vert _{\infty ,\alpha }\preceq (t-s)\Vert f\Vert _{\infty ,\alpha }$ and $\Vert C_{\psi _t}-C_{\psi _s}\Vert _{\infty ,\alpha }\preceq (t-s)$.

If $\Lambda$ is not invertible, then we see that $\sigma _m=0$ for $r<m\le n$, where r=rank $\Lambda$. Let $\Sigma$ be the diagonal matrix with $\Sigma =\Lambda$ except in the last $n-r$ positions along the diagonal, where $\Lambda$ has entries equal to 0 and $\Sigma$ has entries equal to 1/2. Let $\phi _t(z)=\Sigma z+tB'$. Using Lemma 2.6 we obtain $\Vert (C_{\psi _t}-C_{\psi _s})f\Vert _{p,\alpha }\preceq \Vert (C_{\phi _t}-C_{\phi _s}) f\Vert _{p,\alpha }$ for any $f\in F_\alpha ^p$. By the argument just finished we see that $\Vert (C_{\phi _t}-C_{\phi _s}) f\Vert _{p,\alpha }\preceq (t-s)\Vert f\Vert _{p,\alpha }$. Therefore, $\Vert (C_{\psi _t}-C_{\psi _s})f\Vert _{p,\alpha }\preceq (t-s)\Vert f\Vert _{p,\alpha }$ This proves $\Vert C_{\psi _t}-C_{\psi _s}\Vert _{p,\alpha }\preceq (t-s)$. $\square$

Lemma 3.2

Let $a>0$ and $b>0$. Then, for all $x\ge 0$,

$$\begin{aligned} F(x)=\int _0^\infty (1+r)^ae^{-b(r-x)^2}dr\preceq (1+x)^{a+1}. \end{aligned}$$

Proof

We write

$$\begin{aligned} F(x)&=\int _{-x}^\infty (1+r+x)^ae^{-br^2}\mathrm{d}r\\&=\int _{-x}^0 (1+r+x)^ae^{-br^2}\mathrm{d}r+ \int _{0}^\infty (1+r+x)^ae^{-br^2}\mathrm{d}r. \end{aligned}$$

For $x\ge 0$, we have

$$\begin{aligned} \int _{-x}^0 (1+r+x)^ae^{-br^2}\mathrm{d}r\le \int _{-x}^0 (1+r+x)^a\mathrm{d}r\le \frac{(1+x)^{a+1}}{a+1} \end{aligned}$$

and

$$\begin{aligned} \int _{0}^\infty (1+r+x)^ae^{-br^2}\mathrm{d}r\le 2^a\int _{0}^\infty [r^a+(1+x)^a]e^{-br^2}\mathrm{d}r\preceq (1+x)^a. \end{aligned}$$

The desired estimate follows. $\square$

For simplicity, we use the notation $z=(z_k',z')$ with $1\le k<n$, where $z_k'=(z_1,\ldots ,z_k)$ and $z'=(z_{k+1},\cdots z_n)$.

Lemma 3.3

Let $0<p\le \infty$ and $\alpha$ real. Suppose that k is a positive integer with $1\le k<n$ and $\Lambda$ is an arbitrary $(n-k)\times (n-k)$ diagonal matrix whose jth diagonal entry is $\sigma _j$ with $0\le \sigma _j<1$. Then

$$\begin{aligned} \Vert f(z_k',t\Lambda z')-f(z_k',s\Lambda z')\Vert _{p,\alpha }\preceq (t-s)\Vert f\Vert _{p,\alpha } \end{aligned}$$

for any $f\in F_\alpha ^p$ and $0\le s<t\le 1$.

Proof

By Lemma 2.6 it is sufficient to prove for the case that $\Lambda$ is invertible. We split the proof into three cases for p.

Case 1: $0<p\le 1$. For any $f\in F_\alpha ^p$, it follows from (2.2) and Lemma 4 in [5] that

$$\begin{aligned}&|f(z_k',t\Lambda z')-f(z_k',s\Lambda z')|^p\\&=\left| \int _{\mathbb {C}^n}f(w)e^{\langle z_k',w_k'\rangle } \left( e^{t\langle \Lambda z',w'\rangle }-e^{s\langle \Lambda z',w'\rangle }\right) e^{-|w|^2}\mathrm{d}v(w)\right| ^p \\&\le \left( \int _{\mathbb {C}^n}\left| f(w)e^{\langle w_k',z_k'\rangle } \left( e^{t\langle w', \Lambda z'\rangle }-e^{s\langle w',\Lambda z'\rangle }\right) \right| e^{-|w|^2}\mathrm{d}v(w)\right) ^p\\&\preceq \int _{\mathbb {C}^n}|f(w)|^p|e^{p\langle w_k',z_k'\rangle }| \left| e^{t\langle \ w',\Lambda z'\rangle }-e^{s\langle w',\Lambda z'\rangle }\right| ^pe^{-p|w|^2}\mathrm{d}v(w). \end{aligned}$$

Integrating both sides and using Fubini’s theorem, we obtain

$$\begin{aligned} \int _{\mathbb {C}^n}|f(z_k',t\Lambda z')-f(z_k',s\Lambda z')|^pe^{-\frac{p}{2}|z|^2}\mathrm{d}v_\alpha (z) \preceq \int _{\mathbb {C}^n}|f(w)|^pe^{-p|w|^2}g(w)\mathrm{d}v(w), \end{aligned}$$

where

$$\begin{aligned} g(w)=\int _{\mathbb {C}^n}|e^{p\langle w_k',z_k'\rangle }|e^{-\frac{p}{2}|z|^2} \left| e^{t\langle w',\Lambda z'\rangle }-e^{s\langle w',\Lambda z'\rangle }\right| ^p\mathrm{d}v_\alpha (z). \end{aligned}$$

We easily see that

$$\begin{aligned} g(w)\preceq \int _{\mathbb {C}^{n-k}}\left| e^{t\langle w', \Lambda z'\rangle }-e^{s\langle w',\Lambda z'\rangle }\right| ^p e^{-\frac{p}{2}|z'|^2}h(w_k')\mathrm{d}v(z'), \end{aligned}$$

where

$$\begin{aligned} h(w_k')=\int _{\mathbb {C}^k}\frac{|e^{p\langle w_k',z_k'\rangle }|e^{-\frac{p}{2}|z_k'|^2}}{(1+|z_k'|+|z'|)^\alpha }\mathrm{d}v(z_k'). \end{aligned}$$

Meanwhile, for any $c\ge 1$ and z, w in $\mathbb {C}^n$

$$\begin{aligned} c+|z|\le c+|w|+|z-w|\le (c+|w|)(1+|z-w|). \end{aligned}$$

(3.1)

Then

$$\begin{aligned} h(w_k')&=e^{\frac{p}{2}|w_k'|^2}\int _{\mathbb {C}^k}\frac{e^{-\frac{p}{2}|z_k'-w_k'|^2}}{(1+|z_k'|+|z'|)^\alpha }\mathrm{d}v(z_k')\\&\le \frac{e^{\frac{p}{2}|w_k'|^2}}{(1+|w_k'|+|z'|)^\alpha } \int _{\mathbb {C}^k}e^{-\frac{p}{2}|z_k'-w_k'|^2}(1+|z_k'-w_k'|)^{|\alpha |}\mathrm{d}v(z_k')\\&\preceq \frac{e^{\frac{p}{2}|w_k'|^2}}{(1+|w_k'|+|z'|)^\alpha }\le \frac{e^{\frac{p}{2}|w_k'|^2}(1+|z'-w'|)^{|\alpha |}}{(1+|w_k'|+|w'|)^\alpha }\\&\preceq \frac{e^{\frac{p}{2}|w_k'|^2}(1+|z'-w'|)^{|\alpha |}}{(1+|w|)^\alpha } \le \frac{e^{\frac{p}{2}|w_k'|^2}(1+|w'|)^{|\alpha |}(1+|z'|)^{|\alpha |}}{(1+|w|)^\alpha }. \end{aligned}$$

On the other hand

$$\begin{aligned} |e^{t\langle w',\Lambda z'\rangle }-e^{s\langle w',\Lambda z'\rangle }|=\left| \int _s^t \langle w',\Lambda z'\rangle e^{\tau \langle w',\Lambda z'\rangle }\mathrm{d}\tau \right| \le (t-s)|z'||w'|e^{|z'|\cdot |\Lambda w'|} \end{aligned}$$

for $0\le s<t\le 1$. Let

$$\begin{aligned} \sigma =\max \{\sigma _j: ~j=1,2,\ldots ,n-k\}. \end{aligned}$$

Then, $\sigma <1$. Combining above, we obtain by Lemma 3.2

$$\begin{aligned}&\int _{\mathbb {C}^n}|e^{p\langle w_k',z_k'\rangle }|e^{-\frac{p}{2}|z|^2} \left| e^{t\langle w',\Lambda z'\rangle }-e^{s\langle w',\Lambda z'\rangle }\right| ^p\mathrm{d}v_\alpha (z)\\&\preceq \frac{(t-s)^pe^{\frac{p}{2}|w_k'|^2}(1+|w'|)^{|\alpha |+p}}{(1+|w|)^\alpha } \int _{\mathbb {C}^{n-k}}(1+|z'|)^{|\alpha |+p} e^{p|z'|\cdot |\Lambda w'|}e^{-\frac{p}{2}|z'|^2}\mathrm{d}v(z')\\&=\frac{(t-s)^pe^{\frac{p}{2}(|w_k'|^2+|\Lambda w'|^2)}(1+|w'|)^{|\alpha |+p}}{(1+|w|)^\alpha } \int _{\mathbb {C}^{n-k}}\frac{(1+|z'|)^{|\alpha |+p}}{e^{\frac{p}{2}(|z'|-|\Lambda w'|)^2}} \mathrm{d}v(z')\\&\preceq \frac{(t-s)^pe^{\frac{p}{2}(|w_k'|^2+\sigma ^2|w'|^2)}(1+|w'|)^{|\alpha |+p}}{(1+|w|)^\alpha } \int _0^\infty \frac{(1+r)^{2(n-k)-1+|\alpha |+p}}{e^{\frac{p}{2}(r-|\Lambda w'|)^2}}\mathrm{d}r\\&\preceq \frac{(t-s)^pe^{\frac{p}{2}|w|^2}e^{-\frac{p}{2}(1-\sigma ^2)|w'|^2}(1+|w'|)^{2(n-k+|\alpha |+p)}}{(1+|w|)^\alpha }\\&\preceq \frac{(t-s)^pe^{\frac{p}{2}|w|^2}}{(1+|w|)^\alpha }. \end{aligned}$$

Therefore, we have

$$\begin{aligned} \Vert f(z_k',t\Lambda z')&-f(z_k',s\Lambda z')\Vert _{p,\alpha }^p =\int _{\mathbb {C}^n}|f(z_k',t\Lambda z')-f(z_k',s\Lambda z')|^pe^{-\frac{p}{2}|z|^2}\mathrm{d}v_\alpha (z)\\&\preceq (t-s)^p\int _{\mathbb {C}^n}|f(w)|^pe^{-\frac{p}{2}|w|^2}\mathrm{d}v_\alpha (w) =(t-s)^p\Vert f\Vert _{p,\alpha }^p. \end{aligned}$$

Case 2: $1<p<\infty$. Applying Hölder’s inequality and Fubini’s theorem, we obtain

$$\begin{aligned} \Vert f(z_k',t\Lambda z')&-f(z_k',s\Lambda z')\Vert _{p,\alpha }^p=\int _{\mathbb {C}^n}|f(z_k',t\Lambda z')-f(z_k',s\Lambda z')|^pe^{-\frac{p}{2}|z|^2}\mathrm{d}v_\alpha (z)\\&=\int _{\mathbb {C}^n}\left| \int _s^t\sum _{j=k+1}^n\sigma _jz_j\partial _jf(z_k',\tau \Lambda z')\mathrm{d}\tau \right| ^p e^{-\frac{p}{2}|z|^2}\mathrm{d}v_\alpha (z)\\&\preceq \int _{\mathbb {C}^n}\sum _{j=k+1}^n\left( \int _s^t|\partial _jf(z_k',\tau \Lambda z')|\mathrm{d}\tau \right) ^p |z'|^pe^{-\frac{p}{2}|z|^2}\mathrm{d}v_\alpha (z)\\&\preceq (t-s)^{p-1}\sum _{j=k+1}^n\left( \int _{\mathbb {C}^n}\int _s^t|\partial _jf(z_k',\tau \Lambda z')|^p\mathrm{d}\tau \right) |z'|^pe^{-\frac{p}{2}|z|^2}\mathrm{d}v_\alpha (z)\\&=(t-s)^{p-1}\sum _{j=k+1}^n\int _s^t \left( \int _{\mathbb {C}^n}|\partial _jf(z_k',\tau \Lambda z')|^p|z'|^pe^{-\frac{p}{2}|z|^2}\mathrm{d}v_\alpha (z)\right) \mathrm{d}\tau . \end{aligned}$$

For $0<\tau \le 1$, we have by a change of variables

$$\begin{aligned}&\int _{\mathbb {C}^n}|\partial _jf(z_k',\tau \Lambda z')|^p|z'|^pe^{-\frac{p}{2}|z|^2}\mathrm{d}v_\alpha (z)\\&\preceq \tau ^{-2(n-k)-p} \int _{\mathbb {C}^n}\frac{|\partial _jf(z_k',z')|^p|\Lambda ^{-1}z'|^pe^{-\frac{p}{2} (|z_k'|^2+|\tau ^{-1}\Lambda ^{-1}z'|^2)}}{(1+|z_k'|+|\tau ^{-1}\Lambda ^{-1}z'|)^{\alpha }}\mathrm{d}v(z)\\&\preceq \int _{\mathbb {C}^n}|\partial _jf(z_k',z')|^p(1+|z'|)^{-p}e^{-\frac{p}{2}|z|^2}F(z',\tau )\mathrm{d}v_\alpha (z), \end{aligned}$$

where

$$\begin{aligned} F(z',\tau )=\tau ^{-2(n-k)-p-|\alpha |}(1+|z'|)^p |\Lambda ^{-1}z'|^pe^{-\frac{p}{2}(|\tau ^{-1}\Lambda ^{-1}z'|^2-|z'|^2)}. \end{aligned}$$

Since

$$\begin{aligned} |\tau ^{-1}\Lambda ^{-1}z'|^2-|z'|^2\ge (\sigma ^{-2}-1)\tau ^{-2}|z'|^2, \end{aligned}$$

we conclude that $F(z',\tau )\preceq 1$ for all $z\in \mathbb {C}^n$ and $0<\tau \le 1$. This, together with Lemma 2.8, yields

$$\begin{aligned}&\int _{\mathbb {C}^n}|\partial _jf(z_k',\tau \Lambda z')|^p|z'|^pe^{-\frac{p}{2}|z|^2}\mathrm{d}v_\alpha (z)\\&\preceq \int _{\mathbb {C}^n}|\partial _jf(z_k',z')|^p(1+|z'|)^{-p}e^{-\frac{p}{2}|z|^2}\mathrm{d}v_\alpha (z) \preceq \Vert f\Vert _{p,\alpha }^p \end{aligned}$$

for $k+1\le j\le n$. Therefore, we have $\Vert f(z_k',t\Lambda z')-f(z_k',s\Lambda z')\Vert _{p,\alpha }^p\preceq (t-s)^{p}\Vert f\Vert _{p,\alpha }^p$ as desired.

Case 3: $p=\infty$. For any $z\in \mathbb {C}^n$, we have

$$\begin{aligned} |f(z_k',t\Lambda z')-f(z_k',s\Lambda z')|&=\left| \int _s^t\sum _{j =k+1}^n\sigma _jz_j\partial _jf(z_k',\tau \Lambda z')\mathrm{d}\tau \right| \\&\preceq |z'|\sum _{j=k+1}^n \int _s^t|\partial _jf(z_k',\tau \Lambda z')|\mathrm{d}\tau . \end{aligned}$$

When $k+1\le j\le n$ and $0\le \tau \le 1$, we obtain from Lemma 2.8

$$\begin{aligned} |\partial _jf(z_k',\tau \Lambda z')|&\preceq (1+|\tau \sigma _jz_j|)(1+|z_k'|+|\tau \Lambda z'|)^\alpha e^{\frac{1}{2}(|z_k'|^2+|\tau \Lambda z'|^2)}\Vert f\Vert _{\infty ,\alpha }\\&\preceq (1+|z'|)(1+|z_k'|+\tau |z'|)^\alpha e^{\frac{1}{2}(|z_k'|^2+|\Lambda z'|^2)}\Vert f\Vert _{\infty ,\alpha }. \end{aligned}$$

By (3.1), we have

$$\begin{aligned} (1+|z_k'|+\tau |z'|)^\alpha&\le (1+|z_k'|+|z'|)^\alpha [1+(1-\tau )|z'|]^{|\alpha |}\\&\preceq (1+|z|)^\alpha (1+|z'|)^{|\alpha |}. \end{aligned}$$

Hence,

$$\begin{aligned}&|f(z_k',t\Lambda z')-f(z_k',s\Lambda z')(1+|z|)^{-\alpha } |e^{-\frac{|z|^2}{2}}\\&\preceq (t-s) e^{-\frac{1}{2}(|z'|^2-|\Lambda z'|^2)}(1+|z'|)^{2+|\alpha |}\Vert f\Vert _{\infty ,\alpha }\\&\le (t-s)e^{-\frac{1-\sigma ^2}{2}|z'|^2}(1+|z'|)^{2+|\alpha |}\Vert f\Vert _{\infty ,\alpha } \preceq (t-s)\Vert f\Vert _{\infty ,\alpha }. \end{aligned}$$

This proves $\Vert f(z_k',t\Lambda z')-f(z_k',s\Lambda z')\Vert _{\infty ,\alpha }\preceq (t-s)\Vert f\Vert _{\infty ,\alpha }$. The proof is complete. $\square$

Remark 3.4

Lemma 3.3 still remains valid for the case $k=0$. That is, for an $n\times n$ diagonal matrix $\Lambda$ with $\Vert \Lambda \Vert <1$,

$$\begin{aligned} \Vert f(t\Lambda z)-f(s\Lambda z)\Vert _{p,\alpha }\preceq (t-s)\Vert f\Vert _{p,\alpha }, \qquad f\in F_\alpha ^p, \quad 0\le s<t\le 1. \end{aligned}$$

Corollary 3.5

Let $0<p\le \infty$ and $\alpha$ real. Suppose that k is a positive integer with $1\le k<n$ and D is an arbitrary $(n-k)\times (n-k)$ matrix with $\Vert D\Vert <1$. Let

$$\begin{aligned} P=\left( \begin{array}{cc}E_k &{} O \\ O &{}O \end{array}\right) , \qquad P_1=\left( \begin{array}{cc}E_k &{} O \\ O &{}D \end{array}\right) . \end{aligned}$$

Then $C_{Pz}$ and $C_{P_1z}$ are in the same path component in ${\mathcal {C}}(F_\alpha ^p)$.

Proof

By the singular value decomposition of D, there exist $(n-k)\times (n-k)$ unitary matrices U and V such that $\Lambda =U\mathrm{d}v$, where $\Lambda$ is a diagonal matrix whose j-th diagonal entry is $\sigma _j$ with $0\le \sigma _j<1(1\le j\le n-k)$. Let

$$\begin{aligned} U_1=\left( \begin{array}{cc}E_k &{} O \\ O &{}U \end{array}\right) , \qquad V_1=\left( \begin{array}{cc}E_k &{} O \\ O&{}V \end{array}\right) . \end{aligned}$$

Then, $U_1$ and $V_1$ are $n\times n$ unitary matrices. Furthermore

$$\begin{aligned} Q=U_1PV_1=\left( \begin{array}{cc}E_k &{} O \\ O &{}O \end{array}\right) , \qquad Q_1=U_1P_1V_1=\left( \begin{array}{cc}E_k &{} O \\ O &{}\Lambda \end{array}\right) . \end{aligned}$$

We only need to show that $C_{Qz}$ and $C_{Q_1z}$ are in the same path component in ${\mathcal {C}}(F_\alpha ^p)$. Let

$$\begin{aligned} \varphi _t(z)=tQ_1z+(1-t)Qz=(z_k', t\Lambda z'). \end{aligned}$$

By Lemma 3.3, we conclude that $t\mapsto C_{\varphi _t}$ is a continuous path joining $C_{Qz}$ to $C_{Q_1z}$ as desired. $\square$

Lemma 3.6

Let $0<p\le \infty$ and $\alpha$ real. Suppose that $C_\varphi$ and $C_\psi$ are bounded on $F_\alpha ^p$ in $\mathbb {C}^n$ with $\varphi (z)=Az+B$ and $\psi (z)=A_1z+B_1$, where A and $A_1$ are $n\times n$ matrices, and B and $B_1$ are $n\times 1$ vectors. If there exists $\zeta \in \mathbb {C}^n$ such that $|A\zeta |=|\zeta |$ but $A_1\zeta \ne A\zeta$, then there is a positive constant $C_{p,\alpha }$ only depending on p and $\alpha$ such that $\Vert C_\varphi -C_\psi \Vert _{p,\alpha }\ge C_{p,\alpha }$.

Proof

For $0<p<\infty$, let $f_w(z)=e^{\langle z,w\rangle -\frac{|w|^2}{2}}(1+|w|)^{\frac{\alpha }{p}}$. Then, by Lemma 2.2 in [16] we have $\Vert f_w\Vert _{p,\alpha }\approx 1$ for all w in $\mathbb {C}^n$. Since $F_\alpha ^p\subset F^\infty _{\alpha /p}$ for $0<p<\infty$ and $\Vert f\Vert _{\infty ,\alpha /p}\preceq \Vert f\Vert _{p,\alpha }$ for any $f\in F_\alpha ^p$ (see [6]), it follows that

$$\begin{aligned}&\Vert C_\varphi -C_\psi \Vert _{p,\alpha }\ge \frac{\Vert (C_\varphi -C_\psi )f_{\varphi (w)}\Vert _{p,\alpha }}{\Vert f_{\varphi (w)}\Vert _{p,\alpha }} \approx \Vert (C_\varphi -C_\psi )f_{\varphi (w)}\Vert _{p,\alpha } \\&\succeq \Vert (C_\varphi -C_\psi )f_{\varphi (w)}\Vert _{\infty ,\alpha /p}\\&=e^{-\frac{|\varphi (w)|^2}{2}}(1+|\varphi (w)|)^{\frac{\alpha }{p}} \sup _{z\in \mathbb {C}^n}\left\{ e^{-\frac{|z|^2}{2}}(1+|z|)^{-\frac{\alpha }{p}} |e^{\langle \varphi (z),\varphi (w)\rangle }-e^{\langle \psi (z),\varphi (w)\rangle }|\right\} \\&\ge e^{-\frac{1}{2}(|\varphi (w)|^2+|w|^2)}\left( \frac{1+|\varphi (w)|}{1+|w|}\right) ^{\frac{\alpha }{p}} \left| e^{|\varphi (w)|^2}-e^{\langle \psi (w),\varphi (w)\rangle }\right| \\&=e^{\frac{1}{2}(|\varphi (w)|^2-|w|^2)}\left( \frac{1+|\varphi (w)|}{1+|w|}\right) ^{\frac{\alpha }{p}} \left| 1-e^{\langle \psi (w)-\varphi (w),\varphi (w)\rangle }\right| . \end{aligned}$$

For $\zeta \in \mathbb {C}^n$ with $|A\zeta |=|\zeta |$ and $A_1\zeta \ne A\zeta$, then $\zeta \ne 0$. Let $w=\lambda \zeta$, where $\lambda \in \mathbb {C}$. It follows from Proposition 3.1 that $\langle A\zeta ,B\rangle =0$. Thus we obtain

$$\begin{aligned} |\varphi (w)|^2-|w|^2=|\lambda A\zeta |^2+|B|^2+2Re \lambda \langle A\zeta ,B\rangle -|\lambda \zeta |^2 =|B|^2, \end{aligned}$$

and

$$\begin{aligned} \frac{1+|\varphi (w)|}{1+|w|}=\frac{1+\sqrt{|\lambda \zeta |^2+|B|^2}}{1+|\lambda \zeta |}\longrightarrow 1 \end{aligned}$$

as $|\lambda |\rightarrow \infty$. Let $\eta =A\zeta$ and $\eta _1=A_1\zeta$. We have

$$\begin{aligned}&\langle \psi (w)-\varphi (w),\varphi (w)\rangle = \langle \psi (\lambda \zeta )-\varphi (\lambda \zeta ),\varphi (\lambda \zeta )\rangle \\&=\langle \lambda (\eta _1-\eta )+B_1-B,\lambda \eta +B\rangle \\&=|\lambda |^2(\langle \eta _1,\eta \rangle -1)+\lambda \langle \eta _1-\eta ,B\rangle + \overline{\lambda }\langle B_1-B,\eta \rangle +\langle B_1-B,B\rangle . \end{aligned}$$

Noting that $\eta \ne \eta _1$ and $|\eta |=1$ and $|\eta _1|\le 1$, we have $Re \langle \eta _1,\eta \rangle -1<0$. By letting $w=\lambda \zeta$ and $|\lambda |\rightarrow \infty$, we obtain

$$\begin{aligned} \left| e^{\langle \psi (w)-\varphi (w),\varphi (w)\rangle }\right| = e^{Re \langle \psi (w)-\varphi (w),\varphi (w)\rangle }\longrightarrow 0. \end{aligned}$$

Combining these, we easily see that $\Vert C_\varphi -C_\psi \Vert _{p,\alpha }\succeq e^{\frac{1}{2}|B|^2}\ge 1$.

When $p=\infty$, put $g_w(z)=e^{\langle z,w\rangle -\frac{|w|^2}{2}}(1+|w|)^{\alpha }$. By (2.3) we have

$$\begin{aligned} e^{-\frac{|z-w|^2}{2}}(1+|z-w|)^{-|\alpha |}&\le \frac{|g_w(z)|e^{-\frac{|z|^2}{2}}}{(1+|z|)^{\alpha }} =e^{-\frac{|z-w|^2}{2}}\left( \frac{1+|w|}{1+|z|}\right) ^\alpha \\&\le e^{-\frac{|z-w|^2}{2}}(1+|z-w|)^{|\alpha |}. \end{aligned}$$

This yields

$$\begin{aligned} \sup _{z\in \mathbb {C}^n}\{e^{-\frac{|z|^2}{2}}(1+|z|)^{-|\alpha |}\} \le \Vert g_w\Vert _{\infty ,\alpha }\le \sup _{z\in \mathbb {C}^n}\{e^{-\frac{|z|^2}{2}}(1+|z|)^{|\alpha |}\}, \end{aligned}$$

which implies $\Vert g_w\Vert _{\infty ,\alpha }\approx 1$ for all w in $\mathbb {C}^n$. Thus

$$\begin{aligned}&\Vert C_\varphi -C_\psi \Vert _{\infty ,\alpha }\ge \frac{\Vert (C_\varphi -C_\psi )g_{\varphi (w)}\Vert _{\infty ,\alpha }}{\Vert g_{\varphi (w)}\Vert _{\infty ,\alpha }} \approx \Vert (C_\varphi -C_\psi )g_{\varphi (w)}\Vert _{\infty ,\alpha }\\&=e^{-\frac{|\varphi (w)|^2}{2}}(1+|\varphi (w)|)^{\alpha } \sup _{z\in \mathbb {C}^n}\left\{ e^{-\frac{|z|^2}{2}}(1+|z|)^{-\alpha } |e^{\langle \varphi (z),\varphi (w)\rangle }-e^{\langle \psi (z),\varphi (w)\rangle }|\right\} \\&\ge e^{\frac{1}{2}(|\varphi (w)|^2-|w|^2)}\left( \frac{1+|\varphi (w)|}{1+|w|}\right) ^{\alpha } \left| 1-e^{\langle \psi (w)-\varphi (w),\varphi (w)\rangle }\right| . \end{aligned}$$

Similar to the argument above for the case $0<p<\infty$, we can conclude $\Vert C_\varphi -C_\psi \Vert _{\infty ,\alpha }\succeq 1$. The proof is complete. $\square$

Now, we state our main theorem, which completely characterizes the topological structure of ${\mathcal {C}}(F^p_\alpha )$ with $\alpha$ real and $0<p\le \infty$.

Theorem 3.7

Let $\alpha$ be a real number and $0<p\le \infty$. Then the following statements hold.

(a) Suppose that $C_\varphi$ and $C_\psi$ are noncompact bounded composition operators with $\varphi (z)=Az+B$ and $\psi (z)=A_1z+B_1$. Then $C_\varphi$ and $C_\psi$ are in the same path component in ${\mathcal {C}}(F^p_\alpha )$ if and only if $A\zeta = A_1\zeta$ for all $\zeta \in \mathbb {C}^n$ satisfying $|A\zeta |=|\zeta |$ or $|A_1\zeta |=|\zeta |$.

(b) All compact composition operators $\{C_\varphi :\varphi (z)=Az+B, \Vert A\Vert <1\}$ on $F^p_\alpha$ form a path component in ${\mathcal {C}}(F^p_\alpha )$.

(c) A composition operator $C_\varphi$ is isolated in ${\mathcal {C}}(F^p_\alpha )$ if and only if $\varphi (z)=Uz$, where U is a unitary matrix.

Proof

(a) Suppose that $C_\varphi$ and $C_\psi$ are in the same path component. Let $C_{p,\alpha }$ be a positive constant as in Lemma 3.6, Then there exists a finite sequence of bounded composition operators $\{C_{\varphi _i}\}_{i=1}^{m}(m\ge 2)$ with $C_{\varphi _1}=C_\psi$, $C_{\varphi _{m}}=C_\varphi$ and $\Vert C_{\varphi _{i+1}}-C_{\varphi _i}\Vert <C_{p,\alpha }/2$ for $i=1,\ldots ,m-1$. Let $\varphi _i(z)=A_iz+B_i$. It follows from Lemma 3.6 that $A_{i+1}\zeta = A_{i}\zeta$ for all $\zeta$ with $|A_i\zeta |=|\zeta |$ or $|A_{i+1}\zeta |=|\zeta |$. By transitivity we conclude the desired result.

Conversely, if $A\zeta = A_1\zeta$ for all $\zeta$ satisfying $|A\zeta |=|\zeta |$ or $|A_1\zeta |=|\zeta |$, by Proposition 3.1 it is sufficient to show that $C_{Az}$ and $C_{A_1z}$ with $\Vert A\Vert =\Vert A_1\Vert =1$ are in the same path component. By Lemma 2.3 there exist unitary matrices U and V such that $A=UHV$ and $A_1=UH_1V$ with

$$\begin{aligned} H=\left( \begin{array}{cc}E_k &{} O \\ O &{}D \end{array}\right) ,\qquad H_1=\left( \begin{array}{cc}E_k &{} O \\ O &{}D_1 \end{array}\right) , \end{aligned}$$

and $\Vert D\Vert<1, ~\Vert D_1\Vert <1$. We only need to show that $C_{Hz}$ and $C_{H_1z}$ are in the same path component in ${\mathcal {C}}(F_\alpha ^p)$. The desired result follows from Corollary 3.5.

(b) From Proposition 2.1 and the proof of Theorem 3.7(a), we see easily that a compact operator and a noncompact bounded operator are not path connected. Therefore, it is enough to show that compact composition operators on $F_\alpha ^p$ are in the same path component in ${\mathcal {C}}(F_\alpha ^p)$. Let A be an arbitrary $n\times n$ matrix with $\Vert A\Vert <1$. By Proposition 3.1 we only need to show that $C_{Az}$ and $C_{0}$ are in the same path component, where $C_{0}$ denotes a composition operator induced by $\psi (z)=0$. Let $\varphi (z)=\Lambda z$ be the normalization of Az. It suffices to prove that $C_{\Lambda z}$ and $C_{0}$ are in the same path component. From Remark 3.4 we conclude that $C_{t\Lambda z}$ is a continuous path joining $C_{0}$ to $C_{\Lambda z}$ as desired.

(c) Let U be a unitary matrix and $\varphi (z)=Az+B\ne Uz$. Then $A\ne U$ otherwise it follows that $B=0$ from Proposition 2.1. Thus there exists a vector $\zeta$ such that $A\zeta \ne U\zeta$. But $|U\zeta |=|\zeta |$. We conclude by Theorem 3.7(a) and (b) that $C_{Uz}$ and $C_\varphi$ are not in the same path component. On the other hand, if A is an $n\times n$ non-unitary matrix with $\Vert A\Vert \le 1$, we will show that there exists some $n\times 1$ non-zero vector B such that $C_{Az+B}$ is bounded on $F_\alpha ^p$. If $\Vert A\Vert <1$, then any $n\times 1$ non-zero vector B may satisfy the requirement. When $\Vert A\Vert =1$, we write $A=U\Lambda V$ as before. By hypothesis $\Lambda$ is not a unit matrix. Thus we may choose an $n\times 1$ non-zero vector $B'$ as described in Lemma 2.2 such that $C_{\Lambda z+B'}$ is bounded on $F_\alpha ^p$. Let $B={U}B'$. Then $B\ne 0$ and $C_{Az+B}$ is bounded on $F_\alpha ^p$. By Proposition 3.7, $C_{Az}$ and $C_{Az+B}$ are in the same path component. That is to say, $C_{Az+B}$ is not isolated with $Az+B\ne Uz$. The proof is complete. $\square$

Corollary 3.8

The components and the path components in ${\mathcal {C}}(F_\alpha ^p)$ are the same.

Proof

Let $S_1$ and $S_2$ be two distinct path components in ${{\mathcal {C}}(F_\alpha ^p)}$. For any $C_\varphi \in S_1$ with $\varphi (z)=Az+B$ and $C_\psi \in S_2$ with $\psi (z)=A_1z+B_1$, it is easy to see from Theorem 3.7 that there exists $\zeta \in \mathbb {C}^n$ with $|A\zeta |=|\zeta |$ or $|A_1\zeta |=|\zeta |$ such that $A\zeta \ne A_1\zeta$. Applying Lemma 3.6 we have $\Vert C_\varphi -C_\psi \Vert \ge C_{p,\alpha }$. Thus $\overline{S_1}\cap \overline{S_2}=\emptyset$. This means that $S_1$ and $S_2$ are distinct components. On the other hand, path connectedness implies connectedness. Therefore, this desired result follows. $\square$

4 Gleason’s problem on $F_\alpha ^p$

To solve Gleason’s problem on the weighted Fock space, we need two useful lemmas.

Lemma 4.1

Let $0<p<\infty$ and $\alpha$ real. For fixed a in $\mathbb {C}^n$, we have

$$\begin{aligned} \int _{\mathbb {C}^n}(1+|w|)^{\alpha }e^{\frac{-|w|^2}{2}+ Re \langle a, w\rangle } \left[ \int _0^1|e^{t\langle z-a, w\rangle }|\mathrm{d}t\right] \mathrm{d}v(w) \preceq (1+|z|)^{\alpha -2}e^{\frac{|z|^2}{2}} \end{aligned}$$

(4.1)

for all z in $\mathbb {C}^n$, and

$$\begin{aligned} \int _{\mathbb {C}^n}(1+|z|)^{\alpha }e^{-\frac{p}{2}|z|^2} \left[ \int _0^1|e^{t\langle z-a, w\rangle }|\mathrm{d}t\right] ^p\mathrm{d}v(z) \preceq (1+|w|)^{\alpha -2p}e^{\frac{p}{2}|w|^2-Re \langle pw, a\rangle } \end{aligned}$$

(4.2)

for all w in $\mathbb {C}^n$.

Proof

We first prove

$$\begin{aligned} (1+|z|)(1+|w|)\int _0^1|e^{e^{t\langle z-a, w\rangle }}|\mathrm{d}t\preceq e^{\frac{|z-a|^2+|w|^2}{2}-\frac{|z-a-w|^2}{8}}. \end{aligned}$$

(4.3)

An easy computation shows that

$$\begin{aligned} \max \left\{ Re( z\cdot \overline{w}), ~~\frac{|z|^2+|w|^2}{4}\right\} \le \frac{|z|^2+|w|^2}{2}-\frac{|z-w|^2}{8}. \end{aligned}$$

(4.4)

When $\min \{|z|,|w|\}\ge 1+2|a|$, then $1+|z|\le 2|z|\le 2(|z-a|+|a|)\le 4|z-a|$ and $1+|w|\le 2|w|$. Thus, we obtain by (4.4)

$$\begin{aligned}&(1+|z|)(1+|w|)\int _0^1|e^{t\langle z-a, w\rangle }|\mathrm{d}t =(1+|z|)(1+|w|)\int _0^1e^{tRe \langle z-a, w\rangle }\mathrm{d}t\\&\le 8|z-a|\cdot |w|\int _0^1e^{t(\frac{|z-a|^2+|w|^2}{2}-\frac{|z-a-w|^2}{8})}\mathrm{d}t\\&\le \frac{4(|z-a|^2+|w|^2)}{\frac{|z-a|^2+|w|^2}{2}-\frac{|z-a-w|^2}{8}} \left( e^{\frac{|z-a|^2+|w|^2}{2}-\frac{|z-a-w|^2}{8}}-1\right) \\&\le 16e^{\frac{|z-a|^2+|w|^2}{2}-\frac{|z-a-w|^2}{8}}. \end{aligned}$$

On the other hand, when $|z|<1+2|a|$ or $|w|<1+2|a|$, we easily see that

$$\begin{aligned} (1+|z|)(1+|w|)\int _0^1|e^{t\langle z-a, w\rangle }|\mathrm{d}t&\le (1+|z|)(1+|w|)e^{|z-a||w|} \preceq e^{\frac{|z-a|^2+|w|^2}{4}}\\&\le e^{\frac{|z-a|^2+|w|^2}{2}-\frac{|z-a-w|^2}{8}}. \end{aligned}$$

This completes the proof of (4.3). Next, we have by (4.3) and (2.3)

$$\begin{aligned}&\int _{\mathbb {C}^n}(1+|w|)^{\alpha } e^{-\frac{1}{2}|w|^2+Re \langle a, w\rangle } \left[ \int _0^1|e^{t\langle z-a, w\rangle }|\mathrm{d}t\right] \mathrm{d}v(w) \\&\preceq (1+|z|)^{-1}e^{\frac{|z|^2}{2}}\int _{\mathbb {C}^n}(1+|w|)^{\alpha -1} e^{-\frac{|w-z+a|^2}{8}+Re \langle a, w-z\rangle }\mathrm{d}v(w) \\&\preceq (1+|z|)^{\alpha -2} e^{\frac{|z|^2}{2}}\int _{\mathbb {C}^n} (1+|w-z|)^{|\alpha -1|} e^{-\frac{|w-z-3a|^2}{8}}\mathrm{d}v(w)\\&= (1+|z|)^{\alpha -2} e^{\frac{|z|^2}{2}}\int _{\mathbb {C}^n} (1+|w|)^{|\alpha -1|} e^{-\frac{|w-3a|^2}{8}}\mathrm{d}v(w)\\&\preceq (1+|z|)^{\alpha -2} e^{\frac{|z|^2}{2}}. \end{aligned}$$

This proves (4.1). Finally, we see from (4.3) and (2.3) that

$$\begin{aligned}&e^{Re \langle pw, a\rangle }\int _{\mathbb {C}^n}(1+|z|)^{\alpha }e^{-\frac{p}{2}|z|^2} \left[ \int _0^1|e^{t\langle z-a, w\rangle }|\mathrm{d}t\right] ^p\mathrm{d}v(z)\\&\preceq (1+|w|)^{-p}e^{\frac{p}{2}|w|^2}\int _{\mathbb {C}^n}(1+|z|)^{\alpha -p} e^{\frac{-p|z-a-w|^2}{8}-Re \langle p(z-w), a\rangle }\mathrm{d}v(z) \\&\preceq (1+|w|)^{\alpha -2p} e^{\frac{p}{2}|w|^2}\int _{\mathbb {C}^n} (1+|z-w|)^{|\alpha -p|} e^{-\frac{p|z-w+3a|^2}{8}}\mathrm{d}v(z)\\&\preceq (1+|w|)^{\alpha -2p} e^{\frac{p}{2}|w|^2}. \end{aligned}$$

The proof is complete. $\square$

Lemma 4.2

Let $1<p<\infty$ and $\alpha$ real. Fix $a\in \mathbb {C}^n$ and define an integral operator T by

$$\begin{aligned} Tf(z)=\int _{\mathbb {C}^n}(1+|w|)e^{-|w|^2+Re \langle a,w\rangle } \left[ \int _0^1|e^{t\langle z-a, w\rangle }|\mathrm{d}t\right] f(w)\mathrm{d}v(w). \end{aligned}$$

Then $T: L_\alpha ^p\rightarrow L^p_{\alpha -p}$ is bounded.

Proof

Note that $f(z)\in L^p_{\alpha -p}$ if and only if $(1+|z|)f(z)\in L_\alpha ^p$. We define an integral operator S associated with T by

$$\begin{aligned} Sf(z)&=(1+|z|)Tf(z)\\&=\int _{\mathbb {C}^n}(1+|z|)(1+|w|)e^{-|w|^2+Re \langle a,w\rangle } \left[ \int _0^1|e^{t\langle z-a, w\rangle }|\mathrm{d}t\right] f(w)\mathrm{d}v(w). \end{aligned}$$

It is sufficient to prove the operator S is bounded on $L_\alpha ^p$. Let

$$\begin{aligned} \mathrm{d}\lambda (w)=e^{-\frac{p}{2}|w|^2}\mathrm{d}v_\alpha (w). \end{aligned}$$

Then $L_\alpha ^p=L^p(\mathbb {C}^n, \mathrm{d}\lambda )$ and

$$\begin{aligned} Sf(z)=\int _{\mathbb {C}^n}H(z,w)f(w)\mathrm{d}\lambda (w), \end{aligned}$$

where

$$\begin{aligned} H(z,w)=(1+|z|)(1+|w|)^{\alpha +1}e^{(\frac{p}{2}-1)|w|^2+Re \langle a,w\rangle } \left[ \int _0^1|e^{t\langle z-a, w\rangle }|\mathrm{d}t\right] . \end{aligned}$$

For $1<q<\infty$ with $1/p+1/q$=1, let $h(z)=e^{\frac{|z|^2}{2q}}$. By (4.1) we have

$$\begin{aligned}&\int _{\mathbb {C}^n}H(z,w)h^q(w)\mathrm{d}\lambda (w)\\&=(1+|z|)\int _{\mathbb {C}^n}(1+|w|)e^{-\frac{|w|^2}{2}+Re \langle a,w\rangle } \left[ \int _0^1|e^{t\langle z-a, w\rangle }|\mathrm{d}t\right] \mathrm{d}v(w)\\&\preceq e^{\frac{|z|^2}{2}}=h^q(z). \end{aligned}$$

On the other hand, by (4.2) we obtain

$$\begin{aligned} \int _{\mathbb {C}^n}H(z,w)h^p(z)\mathrm{d}\lambda (z)&=(1+|w|)^{\alpha +1}e^{(\frac{p}{2}-1)|w|^2+Re \langle a,w\rangle }\\&\times \int _{\mathbb {C}^n}(1+|z|)^{1-\alpha }e^{-\frac{|z|^2}{2}} \left[ \int _0^1|e^{t\langle z-a, w\rangle }|\mathrm{d}t\right] \mathrm{d}v(z)\\&\preceq e^{\frac{p-1}{2}|w|^2}=h^p(w). \end{aligned}$$

By Lemma 2.4 we conclude that the operator S is bounded on $L^p(\mathbb {C}^n, \mathrm{d}\lambda )$ as desired. $\square$

Theorem 4.3

Let $\alpha$ be a real number and a be any fixed point in $\mathbb {C}^n$. For $0<p<\infty$, there exist bounded linear operators $A_1,\ldots ,A_n$ from $F^p_\alpha$ to $F^p_{\alpha -p}$ such that

$$\begin{aligned} f(z)-f(a)=\sum _{k=1}^n (z_k-a_k)A_kf(z) \end{aligned}$$

for all f in $F^p_\alpha$ and z in $\mathbb {C}^n$. In addition, there exist bounded linear operators $A_1,\ldots ,A_n$ from $F^\infty _\alpha$ to $F^\infty _{\alpha -1}$, such that

$$\begin{aligned} f(z)-f(a)=\sum _{k=1}^n (z_k-a_k)A_kf(z) \end{aligned}$$

for all f in $F^\infty _\alpha$ and z in $\mathbb {C}^n$. $\square$

Proof

Let $f\in F^p_\alpha$ and define $A_k$ by (1.1) $(1\le k\le n)$. Obviously $A_k$ is linear and $A_kf$ is holomorphic on $\mathbb {C}^n$. Furthermore, by (1.1), (2.2) and Fubini’s theorem we may write

$$\begin{aligned} A_k f(z)= \int _{\mathbb {C}^n}\overline{w_k}e^{\langle a, w\rangle }e^{-|w|^2} \left[ \int _0^1e^{t\langle z-a, w\rangle }\mathrm{d}t\right] f(w)\mathrm{d}v(w). \end{aligned}$$

To prove the boundedness of $A_k:F^p_\alpha \rightarrow F^p_{\alpha -p}$, we first consider the case $0<p\le 1$. It is easy to see that

$$\begin{aligned} |A_k f(z)|\le \int _{\mathbb {C}^n}|w|e^{-|w|^2} \left| e^{\langle a, w\rangle }\int _0^1e^{t\langle z-a, w\rangle }\mathrm{d}t\right| |f(w)|\mathrm{d}v(w). \end{aligned}$$

(4.5)

Since the following function

$$\begin{aligned} e^{\langle a, w\rangle }\int _0^1e^{t\langle z-a, w\rangle }\mathrm{d}t \end{aligned}$$

is anti-holomorphic with respect to w on $\mathbb {C}^n$, by Lemma 2.2 in [10] we have

$$\begin{aligned} |A_kf(z)|^p\preceq \int _{\mathbb {C}^n}|w|^p|e^{p\langle a, w\rangle }|e^{-p|w|^2} \left[ \int _0^1|e^{t\langle z-a, w\rangle }|\mathrm{d}t\right] ^p|f(w)|^p\mathrm{d}v(w). \end{aligned}$$

By Fubini’s theorem, we have

$$\begin{aligned} \int _{\mathbb {C}^n}\left| A_kf(z)e^{-\frac{1}{2}|z|^2}\right| ^p\mathrm{d}v_{\alpha -p}(z)\preceq \int _{\mathbb {C}^n}|f(w)|^pe^{-p|w|^2+Re \langle pa, w\rangle }|w|^p g(w)\mathrm{d}v(w), \end{aligned}$$

where

$$\begin{aligned} g(w)=\int _{\mathbb {C}^n}(1+|z|)^{p-\alpha }e^{-\frac{p}{2}|z|^2} \left[ \int _0^1|e^{t\langle z-a, w\rangle }|\mathrm{d}t\right] ^p\mathrm{d}v(z). \end{aligned}$$

Using (4.2), we deduce

$$\begin{aligned} \int _{\mathbb {C}^n}\left| A_kf(z)e^{-\frac{1}{2}|z|^2}\right| ^p\mathrm{d}v_{\alpha -p}(z) \preceq \int _{\mathbb {C}^n}|f(w)|^pe^{-\frac{p}{2}|w|^2}(1+|w|)^{-\alpha }\mathrm{d}v(w). \end{aligned}$$

This shows $\Vert A_kf\Vert _{p,\alpha -p}\preceq \Vert f\Vert _{p,\alpha }$.

Next, for $1<p<\infty$, it follows from Lemma 4.2 that $\Vert A_kf\Vert _{p,\alpha -p}\le \Vert T|f|\Vert _{p,\alpha -p}\preceq \Vert f\Vert _{p,\alpha }$.

Finally, when $p=\infty$ it remains to show the boundedness of $A_k: F^\infty _\alpha \rightarrow F^\infty _{\alpha -1}$. From the definition of $F^\infty _\alpha$, together with (4.5) and (4.1), we have

$$\begin{aligned}&\frac{e^{\frac{-|z|^2}{2}}|A_k f(z)|}{(1+|z|)^{\alpha -1}}\\&\preceq \frac{ e^{\frac{-|z|^2}{2}}\Vert f\Vert _{\infty ,\alpha }}{(1+|z|)^{\alpha -1}} \int _{\mathbb {C}^n}(1+|w|)^{\alpha +1}e^{\frac{-|w|^2}{2}+Re \langle a, w\rangle } \left[ \int _0^1|e^{t\langle z-a, w\rangle }|\mathrm{d}t\right] \mathrm{d}v(w)\\&\preceq \Vert f\Vert _{\infty ,\alpha }. \end{aligned}$$

This proves that $A_k: F^\infty _\alpha \rightarrow F^\infty _{\alpha -1}$ is bounded. $\square$