# Characterizations of some function spaces associated with Bloch type spaces on the unit ball of $$C^{n}$$

## Abstract

Motivated by characterizations of Bloch type spaces, we define some related function spaces and characterize them in this paper. Our results reveal the reason in theory that some equivalent characterizations of the Bloch type space $$B^{\alpha}$$ require extra conditions for α.

## Introduction

Let $$B_{n}$$ be the unit ball of the complex Euclidean space $$C^{n}$$ and $$\partial B_{n}$$ be the unit sphere. The class of all holomorphic functions on $$B_{n}$$ is denoted by $$H(B_{n})$$. For $$0\leq\alpha<\infty$$, let $$H_{\alpha}^{\infty}$$ be the space of holomorphic functions f on $$B_{n}$$ satisfying $$\sup _{z\in B_{n}}(1-|z|^{2})^{\alpha}|f(z)|<\infty$$. When $$\alpha=0$$, we write $$H^{\infty}$$ for $$H_{0}^{\infty}$$.

For $$f\in H(B_{n})$$, its complex gradient is defined by

$$\nabla f(z)= \biggl(\frac{\partial f}{\partial z_{1}}(z),\ldots, \frac{\partial f}{\partial z_{n}}(z) \biggr).$$

As introduced by Timoney in , the Bloch space on the unit ball $$B_{n}$$ is the space of all $$f\in H(B_{n})$$ such that $$\sup _{z\in B_{n}}Q_{f}(z)<\infty$$, where

$$Q_{f}(z)=\sup \biggl\{ \frac{(1-|z|^{2})|\langle\nabla f(z),\overline{w}\rangle |}{\sqrt{(1-|z|^{2})|w|^{2}+|\langle z,w\rangle|^{2}}}:0\neq w\in C^{n} \biggr\} .$$

Let $$0<\alpha<\infty$$. The Bloch type space $$B^{\alpha}$$ consists of functions $$f\in H(B_{n})$$ satisfying

$$\sup_{z\in B_{n}}\bigl(1-|z|^{2}\bigr)^{\alpha}\bigl| \nabla f(z)\bigr|< \infty.$$

It is well known that $$B^{\alpha}=H_{\alpha-1}^{\infty}$$ for $$\alpha>1$$. Moreover, when $$\alpha>1/2$$, a function $$f\in B^{\alpha}$$ if and only if

$$\sup_{z\in B_{n}}\bigl(1-|z|^{2} \bigr)^{\alpha-1}Q_{f}(z)< \infty.$$
(1)

But it is not true for the case $$0<\alpha\leq1/2$$ in the setting of several complex variables. For example (see ), when $$0<\alpha<1/2$$, $$f\in B^{\alpha}$$ if and only if $$\sup _{z\in B_{n}}(1-|z|^{2})^{\alpha-1}G_{f}(z)<\infty$$, where

$$G_{f}(z)=\sup \biggl\{ \frac{(1-|z|^{2})|\langle\nabla f(z),\overline{w}\rangle|}{ \sqrt{(1-|z|^{2})^{2\alpha}|w|^{2}+|\langle z,w\rangle|^{2}}}:0\neq w\in C^{n} \biggr\} .$$

Thus an interesting question arises naturally: what is the property for a holomorphic function f on $$B_{n}$$ satisfying (1) when $$0<\alpha\leq1/2$$. Below we define the space

$$T_{\alpha}= \Bigl\{ f\in H(B_{n}):\sup_{z\in B_{n}} \bigl(1-|z|^{2}\bigr)^{\alpha -1}Q_{f}(z)< \infty \Bigr\} .$$

In Section 2 of this paper, we prove that $$f\in T_{1/2}$$ if and only if the directional derivatives of f in the directions perpendicular to the radial direction are uniformly bounded. In particular, $$T_{\alpha}$$ is a trivial space consisting of constants for $$\alpha <1/2$$.

In 1986 Holland and Walsh  gave another characterization for the Bloch space on the unit disc D, namely, f belongs to the Bloch space if and only if

$$\mathop{\sup_{z,w\in D }}_{z\neq w}\bigl(1-|z|^{2}\bigr)^{{1}/{2}} \bigl(1-|w|^{2}\bigr)^{{1}/{2}} \frac{|f(z)-f(w)|}{|z-w|}< \infty.$$

Ren and Tu  extended the above form to the unit ball $$B_{n}$$. Zhao  generalized these results as follows.

### Theorem A

Let $$0<\alpha\leq2$$. Let λ be any real number satisfying the following properties:

1. (1)

$$0\leq\lambda\leq\alpha$$ if $$0<\alpha<1$$;

2. (2)

$$0<\lambda<1$$ if $$\alpha=1$$;

3. (3)

$$\alpha-1\leq\lambda\leq1$$ if $$1<\alpha\leq2$$.

Then a holomorphic function f on $$B_{n}$$ is in $$B^{\alpha}$$ if and only if

$$\mathop{\sup_{z,w\in B_{n} }}_{z\neq w}\bigl(1-|z|^{2} \bigr)^{\lambda}\bigl(1-|w|^{2}\bigr)^{\alpha-\lambda} \frac{|f(z)-f(w)|}{|z-w|}< \infty.$$
(2)

Zhao gave some examples showing that the conditions on α and λ in Theorem A cannot be improved.

Motivated by Theorem A, we denote the following function space:

$$S_{\alpha,\lambda}= \biggl\{ f\in H(B_{n}):\mathop{\sup_{z,w\in B_{n} }}_{ z\neq w} \bigl(1-|z|^{2}\bigr)^{\lambda}\bigl(1-|w|^{2} \bigr)^{\alpha-\lambda} \frac{|f(z)-f(w)|}{|z-w|}< \infty \biggr\} ,$$

where α and λ are real numbers. Our purpose is to characterize the space $$S_{\alpha,\lambda}$$ for α and λ without satisfying those conditions in Theorem A. If $$\lambda<0$$ or $$\lambda>\alpha$$, by the maximum modulus principle, it is easy to see that $$S_{\alpha,\lambda}$$ consists of constants. So we always assume $$0\leq\lambda\leq\alpha$$. In Section 3 we show $$S_{\alpha,\lambda}\subset B^{\alpha}$$ when α and λ do not satisfy the conditions of Theorem A. More explicitly, $$S_{\alpha,\lambda}$$ coincides with the bounded space $$H^{\infty}$$, or the Bloch type space $$B^{\lambda+1}$$ or $$B^{\alpha-\lambda+1}$$ in terms of different numbers α and λ. Our results reveal in theory instead of examples that the conditions on α and λ in Theorem A cannot be improved.

Throughout this paper, constants are denoted by C, and they are positive finite quantities and not necessarily the same in each occurrence.

## Characterizations of $$T_{\alpha}$$

### Theorem 2.1

The following statements are equivalent:

1. (i)

$$f\in T_{1/2}$$;

2. (ii)

There exists a constant $$C>0$$ such that

\begin{aligned} \bigl|\bigl\langle \nabla f(z),\overline{\zeta}\bigr\rangle \bigr|\leq C \end{aligned}
(3)

for all $$z\in B_{n}$$ and $$\zeta\in\partial B_{n}$$ with $$\langle z,\zeta\rangle=0$$;

3. (iii)

For all $$1\leq i,j\leq n$$,

\begin{aligned} \sup_{z\in B_{n}}\biggl\vert \overline{z_{i}} \frac{\partial f}{\partial z_{j}}(z)-\overline{z_{j}}\frac{\partial f}{\partial z_{i}}(z)\biggr\vert < \infty. \end{aligned}
(4)

### Proof

(i) (ii): For $$z\in B_{n}$$ and $$\zeta\in\partial B_{n}$$ with $$\langle z,\zeta\rangle=0$$, we have

\begin{aligned} \bigl(1-|z|^{2}\bigr)^{-1/2}Q_{f}(z) \geq\frac{(1-|z|^{2})^{1/2}|\langle\nabla f(z),\overline{\zeta}\rangle|}{\sqrt{(1-|z|^{2})|\zeta|^{2}+|\langle z,\zeta \rangle|^{2}}}= \bigl|\bigl\langle \nabla f(z),\overline{\zeta}\bigr\rangle \bigr|. \end{aligned}
(5)

This shows that (i) (ii).

For the converse, if (ii) holds, we have $$(1-|z|^{2})^{1/2}|\langle\nabla f(z),\overline{z}\rangle|\leq C$$ for all $$z\in B_{n}$$ (see ). When $$1/2\leq|z|<1$$ and $$0\neq w\in C^{n}$$, by the projection theorem, there exists $$\zeta\in\partial B^{n}$$ such that $$\langle\zeta,z\rangle=0$$ and

$$w={w_{1}z}+w_{2}\zeta,$$

where $$w_{1}=\langle w,z\rangle/|z|^{2}$$ and $$|w|^{2}=|w_{1}|^{2}|z|^{2}+|w_{2}|^{2}$$. Thus,

\begin{aligned} \bigl|\bigl\langle \nabla f(z),\overline{w}\bigr\rangle \bigr|&\leq\bigl|w_{1}\bigl\langle \nabla f(z),\overline{z}\bigr\rangle \bigr|+\bigl|w_{2}\bigl\langle \nabla f(z),\overline{\zeta}\bigr\rangle \bigr| \\ &\leq C\bigl|\langle z,w\rangle\bigr|\bigl(1-|z|^{2}\bigr)^{-1/2}+C|w|. \end{aligned}

It follows that

\begin{aligned} \frac{(1-|z|^{2})^{1/2}|\langle\nabla f(z),\overline{w}\rangle|}{\sqrt{(1-|z|^{2})|w|^{2}+|\langle z,w\rangle |^{2}}}\leq C \end{aligned}

for all $$1/2\leq|z|<1$$ and $$0\neq w\in C^{n}$$. On the other hand, for $$0\leq|z|<1/2$$, we have

\begin{aligned} \frac{(1-|z|^{2})^{1/2}|\langle\nabla f(z),\overline{w}\rangle|}{\sqrt{(1-|z|^{2})|w|^{2}+|\langle z,w\rangle|^{2}}} \leq \frac{(1-|z|^{2})^{1/2}|\langle\nabla f(z),\overline{w}\rangle|}{\sqrt{(1-|z|^{2})|w|^{2}}}=\bigl|\nabla f(z)\bigr|\leq C. \end{aligned}

This proves (i).

(ii) (iii): Without loss of generality, we only need to show that

\begin{aligned} \sup_{z\in B_{N}}\biggl\vert \overline{z_{2}} \frac{\partial f}{\partial z_{1}}(z)-\overline{z_{1}}\frac{\partial f}{\partial z_{2}}(z)\biggr\vert < \infty. \end{aligned}

If $$z_{1}=z_{2}=0$$, there is nothing to prove. If $$|z_{1}|^{2}+|z_{2}|^{2}\neq0$$, put

$$\zeta= \biggl(\frac{\overline{z_{2}}}{\sqrt{|z_{1}|^{2}+|z_{2}|^{2}}}, \frac{-\overline{z_{1}}}{\sqrt{|z_{1}|^{2}+|z_{2}|^{2}}},0,\ldots,0 \biggr).$$

Obviously, $$\zeta\in\partial B_{n}$$ and $$\langle z,\zeta\rangle=0$$. Therefore,

\begin{aligned} \biggl\vert \overline{z_{2}}\frac{\partial f}{\partial z_{1}}(z)- \overline{z_{1}}\frac{\partial f}{\partial z_{2}}(z)\biggr\vert = \sqrt{|z_{1}|^{2}+|z_{2}|^{2}}\bigl|\bigl\langle \nabla f(z),\overline{\zeta}\bigr\rangle \bigr|\leq C. \end{aligned}

Conversely, suppose (iii) holds. When $$|z|\leq1/2$$, it is clear that (3) holds. For $$|z|= \sqrt{|z_{1}|^{2}+\cdots+|z_{n}|^{2}}>1/2$$, there exists $$z_{i}$$ ($$1\leq i\leq n$$) such that $$|z_{i}|>1/(2\sqrt{n})$$. We may assume that $$|z_{1}|>1/(2\sqrt{n})$$. Let $$V=\{w\in C^{n}:\langle z,w\rangle=0\}$$. Then V is a subspace of $$C^{n}$$ and a basis of V is $$\{v_{1},\ldots,v_{n-1}\}$$, where

\begin{aligned} v_{i}=(-\overline{z_{i+1}},\underbrace{0,\ldots 0}_{i-1{\scriptsize}},\overline{z_{1}},\underbrace{0,\ldots ,0}_{n-(i+1){\scriptsize}}),\quad i=1,\ldots,n-1. \end{aligned}
(6)

Therefore, for $$\zeta=(\zeta_{1},\ldots,\zeta_{n})\in\partial B_{n}$$ with $$\langle z,\zeta\rangle=0$$, there exist scalars $$k_{1},\ldots,k_{n-1}$$ such that ζ is expressed as a linear combination of $$v_{1},\ldots,v_{n-1}$$ in only one way. That is,

\begin{aligned} \zeta=k_{1}v_{1}+k_{2}v_{2}+ \cdots+k_{n-1}v_{n-1}. \end{aligned}
(7)

By (6) and (7), we get

$$\zeta_{1}=-k_{1}\overline{z_{2}}-k_{2} \overline{z_{3}}-\cdots-k_{n-1}\overline{z_{n}}$$

and

\begin{aligned} \zeta_{i+1}=k_{i}\overline{z_{1}},\quad i=1,\ldots,n-1. \end{aligned}
(8)

Note that $$|\zeta|=1$$ and $$|z_{1}|>1/(2\sqrt{n})$$. Hence, it follows from (8) that

\begin{aligned} |k_{i}|=\frac{|\zeta_{i+1}|}{|z_{1}|}< 2\sqrt{n},\quad i=1,\ldots, n-1. \end{aligned}
(9)

Thus we have

\begin{aligned} \bigl\langle \nabla f(z),\overline{\zeta}\bigr\rangle =\sum _{i=1}^{n-1}k_{i}\bigl\langle \nabla f(z), \overline{v_{i}}\bigr\rangle =\sum_{i=1}^{n-1}k_{i} \biggl(-\overline{z_{i+1}} \frac{\partial f}{\partial z_{1}}(z)+\overline{z_{1}} \frac{\partial f}{\partial z_{i+1}}(z) \biggr). \end{aligned}

The desired result follows from (4) and (9). This finishes the proof of Theorem 2.1. □

From Theorem 2.1 and the result of , we see that $$T_{1/2} \subset B^{1/2}$$. Meanwhile, it is evident that $$B^{\alpha}\subset T_{1/2}$$ for $$0<\alpha<1/2$$. Below we give two examples to show that these inclusions are strict.

### Example 1

Let

$$f(z_{1},z_{2})=(1-z_{1})^{1/2}.$$

Note that $$|z_{1}|^{2}+|z_{2}|^{2}<1$$ for $$z=(z_{1},z_{2})\in B_{2}$$. Then

$$\biggl\vert \overline{z_{2}}\frac{\partial f}{\partial z_{1}}(z)- \overline{z_{1}}\frac{\partial f}{\partial z_{2}}(z)\biggr\vert =\frac{|z_{2}|}{2|1-z_{1}|^{1/2}}< \frac{(1-|z_{1}|^{2})^{1/2}}{ 2(1-|z_{1}|)^{1/2}}< \frac{\sqrt{2}}{2}.$$

On the other hand, when $$0<\alpha<1/2$$ and $$z=(r,0)\rightarrow(1,0)$$ ($$0< r<1$$),

$$\bigl(1-|z|^{2}\bigr)^{\alpha}\bigl|\nabla f(z)\bigr| =\frac{(1-|z|^{2})^{\alpha}}{2|1-z_{1}|^{1/2}}= \frac{(1-r^{2})^{\alpha}}{ 2(1-r)^{1/2}}\rightarrow\infty.$$

Therefore $$f\in T_{1/2}$$ by Theorem 2.1, but $$f\,\overline{\in}\, B^{\alpha}$$ ($$0<\alpha<1/2$$).

### Example 2

Let

$$g(z_{1},z_{2})=z_{2}\log(1-z_{1}).$$

Then, for $$z=(z_{1},z_{2})\in B_{2}$$,

\begin{aligned} \bigl(1-|z|^{2}\bigr)^{1/2}\bigl|\nabla g(z)\bigr|&\leq \bigl(1-|z|^{2}\bigr)^{1/2} \biggl(\biggl\vert \frac{z_{2}}{1-z_{1}}\biggr\vert +\bigl|\log (1-z_{1})\bigr| \biggr)\\ &\leq\bigl(1-|z|^{2}\bigr)^{1/2} \biggl(\frac{(1-|z_{1}|^{2})^{1/2}}{1-|z_{1}|}+\bigl| \log (1-z_{1})\bigr| \biggr)\leq C . \end{aligned}

Meanwhile, let $$z=(r,0)\rightarrow(1,0)$$ ($$0< r<1$$), we have

$$\biggl\vert \overline{z_{2}}\frac{\partial g}{\partial z_{1}}(z)- \overline{z_{1}}\frac{\partial g}{\partial z_{2}}(z)\biggr\vert =-r\log(1-r) \rightarrow\infty.$$

Thus $$g\,\overline{\in}\, T_{1/2}$$ but $$g\in B^{1/2}$$.

### Lemma 2.1

Let $$n>1$$, $$1\leq i,j\leq n$$ and $$i\neq j$$. Suppose that $$f\in H(B_{n})$$, $$g\in H(B_{n})$$ and

$$\lim _{|z|\rightarrow1}\bigl|\overline{z_{i}}f(z)-\overline{z_{j}}g(z)\bigr|=0.$$

Then $$f\equiv0$$ and $$g\equiv0$$.

### Proof

We may assume $$i=1$$, $$j=2$$ without loss of generality. Let $$h(z)=\overline{z_{1}}f(z)-\overline{z_{2}}g(z)$$. For each fixed $$\zeta=(\zeta_{1},\ldots,\zeta_{n})\in \partial B_{n}$$, define the slice function $$h_{\zeta}(\lambda)=h(\lambda \zeta)$$ on the unit disk $$D=\{\lambda:|\lambda|<1\}$$. Then

$$\lim_{|\lambda|\rightarrow1}\bigl\vert \overline{\lambda \zeta_{1}}f( \lambda\zeta_{1},\ldots,\lambda\zeta_{n})- \overline{\lambda \zeta_{2}}g(\lambda\zeta_{1},\ldots,\lambda \zeta_{n})\bigr\vert =0.$$

That is,

$$\lim_{|\lambda|\rightarrow1}\bigl\vert \overline{ \zeta_{1}}f( \lambda\zeta_{1},\ldots,\lambda\zeta_{n})- \overline{ \zeta_{2}}g(\lambda\zeta_{1},\ldots,\lambda \zeta_{n})\bigr\vert =0.$$

Since the function $$\overline{ \zeta_{1}}f(\lambda\zeta_{1},\ldots,\lambda\zeta_{n})- \overline{ \zeta_{2}}g(\lambda\zeta_{1},\ldots,\lambda\zeta_{n})$$ is holomorphic on the unit disk $$|\lambda|<1$$, by the maximum modulus principle of one complex variable, it follows that

$$\overline{ \zeta_{1}}f(\lambda\zeta_{1},\ldots,\lambda \zeta_{n})- \overline{ \zeta_{2}}g(\lambda\zeta_{1}, \ldots,\lambda\zeta_{n})=0$$

for all $$\lambda\in D$$. Therefore, for any $$\zeta\in \partial B_{n}$$ and $$\lambda\in D$$,

$$h(\lambda\zeta)=\overline{\lambda \zeta_{1}}f(\lambda \zeta_{1},\ldots,\lambda\zeta_{n})- \overline{\lambda \zeta_{2}}g(\lambda\zeta_{1},\ldots,\lambda \zeta_{n})=0.$$

When $$0\neq z\in B_{n}$$, let $$\lambda=|z|$$ and $$\zeta=z/|z|$$. Then we get

$$h(z)=h(\lambda\zeta)=\overline{z_{1}}f(z)-\overline{z_{2}}g(z)=0.$$

Hence, for all $$z\in B_{n}$$,

\begin{aligned} \overline{z_{1}}f(z)=\overline{z_{2}}g(z). \end{aligned}

Since f and g are holomorphic on $$B_{n}$$, we can conclude $$f(z)=g(z)\equiv0$$. The proof is finished. □

An immediate consequence of Lemma 2.1 is the following theorem.

### Theorem 2.2

Let $$n>1$$ and $$\alpha<1/2$$. If $$f\in T_{\alpha}$$, then f is constant.

### Proof

By (5) it follows that $$|\langle\nabla f(z),\overline{\zeta }\rangle| \leq C(1-|z|)^{1/2-\alpha}$$ for $$z\in B_{n}$$ and $$\zeta\in\partial B_{n}$$ with $$\langle z,\zeta\rangle=0$$. Since $$\alpha<1/2$$, we get $$\lim _{|z|\rightarrow1}|\langle\nabla f(z),\overline{\zeta}\rangle|=0$$. Using a similar argument as in the proof of Theorem 2.1, we have $$\lim _{|z|\rightarrow1}\vert \overline{z_{j}}\frac{\partial f}{\partial z_{i}}(z)-\overline{z_{i}}\frac{\partial f}{\partial z_{j}}(z)\vert =0$$ for all $$1\leq i,j\leq n$$. Thus the desired result follows from Lemma 2.1. □

## Characterizations of $$S_{\alpha,\lambda}$$

In the section we will characterize $$S_{\alpha,\lambda}$$ explicitly for real numbers α and λ in several cases. For this, we need the following lemma which plays an important role in the proof of Theorem 3.1.

### Lemma 3.1

Let $$\alpha>1$$. Let λ be any real number satisfying the following properties:

1. (1)

$$0<\lambda<\alpha-1$$ if $$1<\alpha\leq2$$;

2. (2)

$$0<\lambda\leq{\alpha}/{2}$$ if $$\alpha>2$$.

Let

$$H(x,y)=\frac{x^{\lambda}y^{\alpha-\lambda}}{y-x}\int_{x}^{y} \frac{d\tau}{\tau^{\lambda+1}}.$$

Then there exists a constant $$C>0$$ such that $$H(x,y)\leq C$$ for any x and y satisfying $$0< x,y\leq1$$ and $$x\neq y$$.

### Proof

Let $$t=x/y$$. Then $$t\in(0,1)\cup(1,\infty)$$, and

$$H(x,y)=H(ty,y)=\frac{t^{\lambda}y^{\alpha-1}}{(1-t)}\int_{ty}^{y} \frac{d\tau}{\tau^{\lambda+1}}.$$

Let $$s={\tau}/{y}$$, we have

$$H(x,y)=\frac{t^{\lambda}y^{\alpha-(\lambda+1)}}{(1-t)}\int_{t}^{1} \frac{ds}{s^{\lambda+1}}= y^{\alpha-(\lambda+1)}G(t),$$

where

$$G(t)=\frac{t^{\lambda}}{(1-t)}\int_{t}^{1} \frac{ds}{s^{\lambda+1}} =\frac{1-t^{\lambda}}{\lambda(1-t)}.$$

It is evident that $$G(t)$$ is continuous on $$(0,1)\cup(1,\infty)$$, and

$$\lim_{t\rightarrow0}G(t)=\frac{1}{\lambda},\qquad \lim_{t\rightarrow1}G(t)=1.$$

For the case (1), since $$0<\lambda<1$$, we get

$$\lim_{t\rightarrow\infty}G(t)=0.$$

Noticing that $$y^{\alpha-(\lambda+1)}\leq1$$ for $$y\in(0,1]$$, we conclude that $$H(x,y)$$ is bounded in this case. For the case (2), we easily see that $$0<\lambda<\alpha-1$$. Write

$$H(x,y)=H(x,x/t)=\frac{x^{\alpha-(\lambda+1)}}{t^{\alpha-(\lambda+1)}}G(t).$$

It is clear that

$$\lim_{t\rightarrow\infty}\frac{G(t)}{t^{\alpha-(\lambda+1)}}= \lim_{t\rightarrow\infty} \frac{t^{2\lambda-\alpha}}{\lambda}.$$

The above limit is $$1/\lambda$$ for $$\lambda={\alpha}/{2}$$ and 0 for $$0<\lambda<{\alpha}/{2}$$. Therefore $$H(x,y)$$ is also bounded in the case (2) since $$x^{\alpha-(\lambda+1)}\leq1$$ for $$x\in(0,1]$$. The proof is complete. □

### Theorem 3.1

Let $$\alpha>1$$. Let λ be any real number satisfying the following properties:

1. (1)

$$0<\lambda<\alpha-1$$ if $$1<\alpha\leq2$$;

2. (2)

$$0<\lambda\leq{\alpha}/{2}$$ if $$\alpha>2$$.

Then $$S_{\alpha,\lambda}=B^{\lambda+1}$$.

### Proof

Let $$f\in B^{\lambda+1}$$. For any $$z,w\in B_{n}$$, since

\begin{aligned} f(z)-f(w)&=\int_{0}^{1}\frac{d[f(tz+(1-t)w)]}{dt}\,dt\\ &=\sum_{k=1}^{n}(z_{k}-w_{k}) \int_{0}^{1}\frac{\partial f}{\partial z_{k}}\bigl(tz+(1-t)w \bigr)\,dt, \end{aligned}

we get

\begin{aligned} \bigl|f(z)-f(w)\bigr|&\leq C|z-w|\int_{0}^{1}\bigl|(\nabla f) \bigl(tz+(1-t)w\bigr)\bigr|\,dt\\ &\leq C|z-w|\int_{0}^{1}\frac{dt}{(1-|tz+(1-t)w|)^{\lambda+1}}\\ &\leq C|z-w|\int_{0}^{1}\frac{dt}{(1-t|z|-(1-t)|w|)^{\lambda+1}}\\ &=C|z-w|\int_{0}^{1}\frac{dt}{[t(1-|z|)+(1-t)(1-|w|)]^{\lambda+1}}. \end{aligned}

If $$|z|=|w|$$, noting that $$\lambda+1<\alpha$$, we get

\begin{aligned} \int_{0}^{1}\frac{dt}{[t(1-|z|)+(1-t)(1-|w|)]^{\lambda+1}}&= \frac{1}{(1-|z|)^{\lambda+1}}\leq\frac{C}{(1-|z|^{2})^{\lambda+1}} \\ &\leq\frac{C}{(1-|z|^{2})^{\lambda}(1-|w|^{2})^{\alpha-\lambda}}. \end{aligned}

If $$|z|\neq|w|$$, let $$\tau=t(1-|z|)+(1-t)(1-|w|)$$. By Lemma 3.1, we have

\begin{aligned} \int_{0}^{1}\frac{dt}{[t(1-|z|)+(1-t)(1-|w|)]^{\lambda+1}}&= \frac{1}{(1-|z|)-(1-|w|)}\int_{1-|w|}^{1-|z|}\frac{dt}{\tau^{\lambda +1}} \\ &\leq\frac{C}{(1-|z|^{2})^{\lambda}(1-|w|^{2})^{\alpha-\lambda}}. \end{aligned}

Therefore,

$$\bigl|f(z)-f(w)\bigr|\leq \frac{C|z-w|}{{(1-|z|^{2})^{\lambda}(1-|w|^{2})^{\alpha-\lambda}}},$$

which shows that $$f\in S_{\alpha,\lambda}$$.

Conversely, if $$f\in S_{\alpha,\lambda}$$, it follows that

$$\mathop{\sup_{z\in B_{n} }}_{ z\neq 0}\bigl(1-|z|^{2}\bigr)^{\lambda}\frac{|f(z)-f(0)|}{|z|}< \infty.$$

Thus we get

$$\sup_{z\in B_{n}}\bigl(1-|z|^{2}\bigr)^{\lambda}\bigl|f(z)\bigr|< \infty,$$

namely, $$f\in H_{\lambda}^{\infty}$$ and so $$f\in B^{\lambda+1}$$. This finishes the proof of the theorem. □

### Theorem 3.2

Let $$\alpha>1$$. Let λ be any real number satisfying the following properties:

1. (1)

$$1<\lambda<\alpha$$ if $$1<\alpha\leq2$$;

2. (2)

$$\alpha/2<\lambda<\alpha$$ if $$\alpha>2$$.

Then $$S_{\alpha,\lambda}=B^{\alpha-\lambda+1}$$.

### Proof

If $$1<\lambda<\alpha$$ for $$1<\alpha\leq2$$, then $$0<\alpha-\lambda<\alpha-1$$. If $$\alpha/2<\lambda<\alpha$$ for $$\alpha>2$$, then $$0<\alpha-\lambda <\alpha/2$$. Applying Theorem 3.1, we immediately conclude that $$S_{\alpha,\lambda}= B^{\alpha-\lambda+1}$$. □

For any point $$w\in B_{n}-\{0\}$$, we recall that the bi-holomorphic mapping $$\varphi_{w}$$ of $$B_{n}$$, which interchanges the points 0 and w, is defined by

$$\varphi_{w}(z)=\frac{w-P_{w}(z)-s_{w}Q_{w}(z)}{1-\langle z,w\rangle},\quad z\in B_{n},$$

where $$s_{w}=\sqrt{1-|w|^{2}}$$, $$P_{w}(z)=\frac{\langle z,w\rangle}{|w|^{2}}w$$ and $$Q_{w}(z)=z-P_{w}(z)$$. When $$w=0$$, let $$\varphi_{w}(z)=-z$$. The pseudo-hyperbolic distance between w and z is denoted by $$\rho(w,z)=|\varphi_{w}(z)|$$.

### Theorem 3.3

Let $$\alpha\geq1$$. Then $$S_{\alpha,\lambda}=H^{\infty}$$ for $$\lambda=0$$ or $$\lambda=\alpha$$.

### Proof

It suffices to prove for $$\alpha\geq1$$ and $$\lambda=0$$. If $$f\in S_{\alpha,0}$$, that is,

$$\mathop{\sup_{z,w\in B_{n} }}_{ z\neq w}\bigl(1-|w|^{2} \bigr)^{\alpha} \frac{|f(z)-f(w)|}{|z-w|}< \infty.$$
(10)

Then

$$\mathop{\sup_{z\in B_{n}}}_{ z\neq0} \frac{|f(z)-f(0)|}{|z|}< \infty,$$

which implies that $$f\in H^{\infty}$$.

Conversely, assume $$f\in H^{\infty}$$. Then we have (see Lemma 1 in )

$$\bigl|f(z)-f(w)\bigr|\leq C\rho(z,w)$$
(11)

for all $$z,w\in B_{n}$$. On the other hand,

\begin{aligned} \bigl|w-P_{w}(z)-s_{w}Q_{w}(z)\bigr|^{2} =|z-w|^{2}+\bigl|\langle z,w\rangle\bigr|^{2}-|z|^{2}|w|^{2} \leq|z-w|^{2}, \end{aligned}

and so

$$\rho(w,z)\leq\frac{|z-w|}{|1-\langle z,w\rangle|}\leq \frac{|z-w|}{1-|w|}.$$
(12)

By (11) and (12), we get (10) since $$\alpha\geq1$$. Thus $$f\in S_{\alpha,0}$$. The proof is complete. □

### Remark 3.1

We easily see that $$\lambda+1<\alpha$$ in Theorem 3.1, and $$\lambda>1$$ in Theorem 3.2. Combining with Theorem 3.3, we conclude that $$S_{\alpha,\lambda}\subset B^{\alpha}$$ and the inclusion is strict for real numbers α and λ which do not satisfy the conditions of Theorem A.

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## Acknowledgements

JD was supported by the Natural Science Foundation of China under 11301404 and the Fundamental Research Funds for the Central Universities (WUT: 2015IVA069). BW was supported by the Natural Science Foundation of China under 11201348, the Fund of China Scholarship Council and the special Fund of Basic Scientific Research of Central Colleges of South Central University for Nationalities under CZQ13015.

## Author information

Authors

### Corresponding author

Correspondence to Baobin Wang.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

BW proposed the problem. JD and BW together finished the proof. All authors read and approved the final manuscript.

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