1 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 [1], 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 [2]), 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 [3] 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 [4] extended the above form to the unit ball \(B_{n}\). Zhao [5] 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.

2 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 [6]). 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 [6], 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. □

3 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 [7])

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