Throughout this paper, the unit disc is denoted by $\mathbb{D}$. The Möbius transformation of $\mathbb{D}$ is defined by

$ \varphi_{a}(z)=\frac{a-z}{1-\bar{a}z}, \quad a\in\mathbb{D}. $

Definition 1.1 Let $K : [0, \infty) \rightarrow [0, \infty) $ be a right-continuous and nondecreasing function. An analytic function $f$ on the unit disc is said to belongs to the space $Q_K$ if

$ \|f\|^2_{Q_K}=\mathop {\sup }\limits_{a \in \mathbb{D}} \int_{\mathbb{D}} |f'(z)|^2K\left(\log\frac{1}{|\varphi_a(z)|}\right)dA(z)<\infty, $

(1.1)

where $dA(z)$ is the Euclidean area element on $\mathbb{D}$ so that $A(\mathbb{D})=1$.

Equipped with the norm $|f(0)|+\|f\|_{Q_K}$, the space $Q_K$ is Banach. We know that the space $Q_K$ is Möbius invariant in the sense that $\|f\circ\varphi_a\|_{Q_K}=\|f\|_{Q_K}$ for any $a\in\mathbb{D}$. See [5] and [6] for a general theory of $Q_K$ spaces. By [5] we know that $Q_K$ spaces are contained in the Bloch space. Recall that an analytic function $f$ on the unit disc is said to belong to the Bloch space, denoted by $\mathcal{B}$, if

$ \|f\|_{\mathcal{B}}=\mathop {\sup }\limits_{z \in \mathbb{D}} (1-|z|^{2})|f'(z)|<\infty. $

The Hardy space $H^\infty$ consists of all analytic $f$ on the unit disk $\mathbb{D}$ for which $\sup\limits_{z\in\mathbb{D}}|f(z)|<\infty.$ A sequence $\{z_n\}\subset\mathbb{D}$ is called an interpolating sequence for $Q_K\cap H^\infty$ if each bounded sequence $\{w_n\}$ of complex numbers there exists $f\in Q_K\cap H^\infty$ such that $f(z_n)=w_n$ for all $n$.

In this paper the weight function $K$ satisfies

$ \int_0^{\frac{1}{e}}K\left(\log\frac{1}{r}\right)dr<\infty. $

Otherwise, the space $Q_K$ only contains constant functions. By Theorem 2.1 in [5] we may assume that $K$ is defined on [0, 1] and extend its domain to $[0, \infty)$ by setting $K(t)=K(1)$ for $t>1$.

We need the following two conditions on $K$:

$ \int^\infty_1\frac{\varphi_K(s)}{s^2}ds<\infty$

(1.2)

and

$ \int^1_0\frac{\varphi_K(s)}{s}ds<\infty, $

(1.3)

where

$ \varphi_K(s)=\mathop {\sup }\limits_{0<t\leq1}K(st)/K(t). $

For a subarc $I\subset\partial\mathbb{D}$, let $\theta$ be the midpoint of $I$ and denote the Carleson box

$ S(I)=\left\{z\in\mathbb{D}: 1-|I|<|z|<1, |\theta-\arg z|<\frac{|I|}{2}\right\} $

for $|I|\leq 1$ and $S(I)=\mathbb{D}$ for $|I|>1$.

Definition 1.2 A positive measure $d\mu$ is said to be a $K$-Carleson measure on $\mathbb{D}$ provided

$ ||\mu||_K=\mathop {\sup }\limits_{I\subset\mathbb{D}}\int_{S(I)} K\left(\frac{1-|z|}{|I|}\right)d\mu(z)<\infty.$

(1.4)

The interpolation problems for different analytic function spaces were discussed in many paper (see [2, 3, 10, 13, 14]). A well-known result of Carleson (see [3]) asserts that $\{z_n\}$ is an interpolating sequence for $H^\infty$ if and only if $d\mu=\sum\delta_{z_n}$ is a Carleson measure and $\{z_n\}$ is separated, that is

$\mathop {\inf }\limits_{m \ne n} \left| {\frac{{{z_n} - {z_m}}}{{1 - \overline {{z_n}} {z_m}}}} \right| = \delta ＞ 0.$

(1.5)

In this paper, we study the interpolation problem in the analytic function space $Q_K\cap H^\infty$. The main result is the following:

Theorem 1.1 Let $K$ satisfy conditions (1.2) and (1.3). Then $\{z_n\}$ is an interpolating sequence for $Q_K\bigcap H^\infty$ if and only if $\{z_n\}$ is separted and $d\mu=\sum\delta_{z_n}$ is a $K$-Carleson measure.

In addition, we say that $f\lesssim g$ (for two functions $f$ and $g$) if there exists a constant $C$ (independent of $f$ and $g$) such that $f(t)\leq Cg(t), t>0$. We say $f\thickapprox g$ (that is, $f$ is comparable with $g$) whenever $f\lesssim g\lesssim f$.

We give some auxiliary lemmas that will be needed in the proof of the main result.

Lemma 2.1 If conditions (1.2) and (1.3) hold, then the weight $K$ has the following proposition:

(1) $K(t)/t^c$ is non-decreasing for some $c>0$.

(2) $K(t)/t^{p}$ is non-increasing for some $0<p<1$.

(3) $K(2t)\approx K(t)$ for any $0<t<\infty$.

The above result can be found in [6].

Remark 2.2 If conditions (1.2) and (1.3) hold, then the space $Q_K$ is a subset of $D_p$ for some $0<p<1$ by Lemma 2.1, where the space $D_p$ consists of all analytic function $f$ on $\mathbb{D}$ such that

$ \int_\mathbb{D}|f'(z)|^2(1-|z|^2)^pdA(z)<\infty. $

Lemma 2.3 Let $K$ satisfy conditions (1.2) and (1.3). There is a constant C (independent of $w$) such that

$ \int_{\mathbb{D}} \frac{(1-|z|^2)^{q}}{K(1-|z|^2)|1-\overline{w}z|^{b+q}} dA(z) \leq\frac{\mbox{C}}{(1-|w|^2)^{b-2}K(1-|w|^2)}, \quad b\geq 2$

(2.1)

for any given $w\in\mathbb{D}$ and $q\geq 0$.

Proof In our proof we borrowed a technique from [15]. Without loss of generality, we may assume Im$(w)=0$. Set $\alpha=1-w$, then $\alpha$ belongs to $(0, 1)$. We divide the unit disk into $S_1\bigcup S_2\bigcup S_3$, where

$\begin{eqnarray*} &&S_1=\left\{z: 0<1-|z|\leq\alpha, |\arg z|\leq\alpha/2\right\}, \\ &&S_2=\left\{z: \alpha<1-|z|\leq 1, |\arg z|\leq\alpha/2\right\}, \end{eqnarray*}$

and

$S_3=\left\{z: 0<1-|z|\leq 1, |\arg z|>\alpha/2\right\}.$

Then

$\begin{eqnarray*} \int_{S_1} \frac{(1-|z|^2)^{q}}{K(1-|z|^2)|1-\overline{w}z|^{b+q}} dA(z) &\lesssim&\alpha\int^\alpha_0\frac{t^qdt}{K(t)(\alpha+t(1-\alpha))^{b+q}}\\ &\leq&\int^\alpha_0\frac{t^qdt}{K(t)(\alpha+t(1-\alpha))^{b-1+q}} \end{eqnarray*}$

and

$\begin{eqnarray*} \int_{S_2} \frac{(1-|z|)^q}{K(1-|z|^2)|1-\overline{w}z|^{b+q}} dA(z) &\lesssim&\alpha\int^1_\alpha\frac{t^qdt}{K(t)(\alpha+t(1-\alpha))^{b+q}}\\ &\leq&\int^1_\alpha\frac{t^qdt}{K(t)(\alpha+t(1-\alpha))^{b-1+q}}.\\ \end{eqnarray*}$

On the other hand,

$\begin{eqnarray*} &&\int_{S_3} \frac{(1-|z|)^{q}}{K(1-|z|^2)|1-\overline{w}z|^{b+q}}dA(z)\\ &\leq&2\int^1_0\frac{t^q}{K(t)}\left( \int^{\pi}_\alpha\frac{d\theta}{[(\alpha+t(1-\alpha))^2 +\sin^2(\theta/2)]^{\frac{b+q}{2}}}\right)dt\\ &\lesssim&\int^1_0\frac{t^qdt}{K(t)(\alpha+t(1-\alpha))^{b-1+q}}\\ &\leq&\int^\alpha_0\frac{t^qdt}{K(t)(\alpha+t(1-\alpha))^{b-1+q}}+ \int^1_\alpha\frac{t^qdt}{K(t)(\alpha+t(1-\alpha))^{b-1+q}}. \end{eqnarray*}$

By Lemma 2.1, $K(t)/t^p$ is non-increasing for some $0<p<1$. This gives

$\begin{eqnarray*} \int^\alpha_0\frac{t^qdt}{K(t)(\alpha+t(1-\alpha))^{b-1+q}} &\leq&\frac{\alpha^p}{K(\alpha)}\int^{\alpha}_0\frac{t^{q-p}}{(\alpha+t(1-\alpha))^{b-1+q}}dt\\ &\leq& \frac{1}{K(\alpha)\alpha^{b-1+q-p}}\int^{\alpha}_0t^{q-p}dt\\ &\lesssim&\frac{1}{K(\alpha)\alpha^{b-2}}. \end{eqnarray*}$

Applying Lemma 2.1 again we obtain that $K(t)/t^c$ is non-decreasing for some small enough $c>0$. Note that $c<p$ and $b\geq 2$. Then we have

$\begin{eqnarray*} \int^1_\alpha\frac{t^qdt}{K(t)(\alpha+t(1-\alpha))^{b-1+q}} &\leq& \frac{\alpha^c}{K(\alpha)}\int^1_\alpha\frac{t^{q-c}dt}{(\alpha+t(1-\alpha))^{b-1+q}}\\ &\leq&\frac{\alpha^c}{K(\alpha)}\int^1_\alpha\frac{dt}{t^{b-1+c}}\\ &\lesssim&\frac{1}{K(\alpha)\alpha^{b-2}}. \end{eqnarray*}$

The above estimates give

$\begin{eqnarray*} &&\int_\mathbb{D} \frac{(1-|z|^2)^{q}}{K(1-|z|^2)|1-\overline{w}z|^{b+q}} dA(z)=\sum^3_{j=1}\int_{S_j} \frac{(1-|z|^2)^{q}}{K(1-|z|^2)|1-\overline{w}z|^{b+q}} dA(z) \\ &\lesssim&\int^{\alpha}_0\frac{t^q}{ K(t)(\alpha+t(1-\alpha))^{b-1+q}}dt+ \int^1_\alpha\frac{t^{q}dt}{K(t)(\alpha+t(1-\alpha))^{b-1+q}}\\ &\lesssim&\frac{1}{K(\alpha)\alpha^{b-2}}. \end{eqnarray*}$

Hence (2.1) holds.

Remark 2.4 If $K(t)=t^s, 0<s<1$ and $p\geq 0$, Lemma 2.3 gives the result

$ \int_{\mathbb{D}} \frac{(1-|z|^2)^{q-s}}{|1-\overline{w}z|^{b+q}} dA(z) \leq\frac{C}{(1-|w|^2)^{b-2+s}}, \quad b\geq 2. $

This case wsa proved in [18].

Given finitely many complex numbers $w_1, \cdots, w_n$, consider the $2^n$ possible sumes $\sum\limits^n_{j=1}\pm w_j$ obtained as the plus-mius signs vary in the $2^n$ possible ways. For $p>0$, denote by $\mathbb{E}\left(\left|\sum\limits^n_{j=1}\pm w_j\right|^p\right)$ the average value of $\left|\sum\limits^n_{j=1}\pm w_j\right|^p $ over the $2^n$ choices of sign. The following Lemma is called Khinchin's inequality (see [7]).

Lemma 2.5 If $0<q<\infty, $ then

$ \mathbb{E}\left( {{{\left| {\sum\limits_{j = 1}^n \pm {w_j}} \right|}^q}} \right) \le {C_q}{\left( {\sum\limits_{j = 1}^n | {w_j}{|^2}} \right)^{q/2}},$

(2.2)

where $C_q$ is a constant that does not depend on $n$.

In this section we will prove Theorem 1.1.

Proof of Theorem 1.1 By Lemma 2.1, we know that $Q_K$ is a subset of $D_p$ for some $0<p<1$. By the reproducing formula of Rochberg and Wu (see [12] or p.75, [18]), we have

$ f(z)=f(0)+\int_\mathbb{D} f'(w)B(z, w)(1-|w|^2)^pdA(w), \quad z\in\mathbb{D}, $

(3.1)

where

$ B(z, w)=\frac{1-(1-z\overline{w})^{1+p}}{\overline{w}(1-z\overline{w})^{1+p}}. $

We now assume that $\{z_n\}$ is an interpolating sequence for $Q_K\cap H^\infty$. Then for $\epsilon_k^{(j)}=\pm 1, j=1, \cdots, 2^n, k=1, \cdots, n$, there are $f_j\in Q_K\cap H^\infty$ such that $f_j(z_k)=\epsilon_k^{(j)}, k=1, \cdots, n$ and

$ \|f_j\|_{H^\infty}+\|f_j\|_{Q_K}\lesssim 1 . $

Replaced $f$ by $f_j\circ\varphi_{a}$ in (3.1) we have

$ f_j\circ\varphi_{a}(z)=f_j\circ\varphi_{a}(0)+\int_\mathbb{D} (f_j\circ\varphi_{a})'(w)B(z, w)(1-|w|^2)^pdA(w). $

Let $z=\varphi_a(z_k), k=1, 2, \cdots, n$. Then

$ f_j(z_k)=f_j\circ\varphi_{a}(0)+\int_\mathbb{D} (f_j\circ\varphi_{a})'(w)B(\varphi_a(z_k), w)(1-|w|^2)^pdA(w), $

and we have

$\begin{eqnarray*} &&\sum^n_{k=1}K(1-|\varphi_a(z_k)|^2)=\sum^n_{k=1}\epsilon^{(j)}_kf_j(z_k)K(1-|\varphi_a(z_k)|^2)\\ &=&f_j(a)\sum^n_{k=1}\epsilon^{(j)}_kK(1-|\varphi_a(z_k)|^2)\\ &&+\sum^n_{k=1}\epsilon^{(j)}_kK(1-|\varphi_a(z_k)|^2)\int_\mathbb{D} (f_j\circ\varphi_{a})'(w)B(\varphi_a(z_k), w)(1-|w|^2)^pdA(w)\\ &=&A+B. \end{eqnarray*}$

We now compute the expectation of both side of the equality. Observe that by (2.2) with $q=1$, we have

$ \mathbb{E}(A) \mathbin{\lower.3ex\hbox{$\buildrel＜\over {\smash{\scriptstyle\sim}\vphantom{_x}}$}} {\left( {\sum\limits_{k = 1}^n {(K(} 1 - |{\varphi _a}({z_k}){|^2}){)^2}} \right)^{1/2}} \mathbin{\lower.3ex\hbox{$\buildrel＜\over {\smash{\scriptstyle\sim}\vphantom{_x}}$}} {\left( {\sum\limits_{k = 1}^n K (1 - |{\varphi _a}({z_k}){|^2})} \right)^{1/2}}.$

There exists a constant C such that

$\mathop {\sup }\limits_{a,{z_k},w \in \mathbb{D}} \left| {\frac{{1 - {{(1 - {\varphi _a}({z_k})\bar w)}^{1 + p}}}}{{\bar w}}} \right| \le {\rm{C}}.$

(3.2)

Applying the Cauthy-Schwarz inequality, (2.1), (2.2) with $q=2$ and (3.2) we have

$\begin{eqnarray*} \mathbb{E}(B)&\leq&\left(\sum^n_{k=1}\left|\int_\mathbb{D} (f_j\circ\varphi_{a})'(w) \frac{B(\varphi_a(z_k), w)(1-|w|^2)^p}{\left(K(1-|\varphi_a(z_k)|^{2})\right)^{-1}}dA(w)\right|^2\right)^{1/2}\\ &\leq&\sup_j\|f\|_{Q_K} \left(\sum^n_{k=1}\int_\mathbb{D}\left(\frac{B(\varphi_a(z_k), w)}{K(1-|\varphi_a(z_k)|^2)^{-1}} \frac{(1-|w|^2)^{p}}{\sqrt{K(1-|w|^2)}}\right)^2dA(w)\right)^{1/2}\\ &\lesssim&\left(\sum^n_{k=1}\left(K(1-|\varphi_a(z_k)|^2)\right)^2 \int_\mathbb{D}\frac{(1-|w|^2)^{2p}dA(w)}{|1-\varphi_a(z_k)\overline{w}|^{2(1+p)}K(1-|w|^2)}\right)^{1/2}\\ &\lesssim&\left(\sum^n_{k=1}K(1-|\varphi_a(z_k)|^2)\right)^{1/2}. \end{eqnarray*}$

The estimates involving $\mathbb{E}(A)$ and $\mathbb{E}(B)$ indicate that $d\mu=\sum\delta_{z_n}$ is a $K$-Carleson measure by Corollary 3.1 in [6]. Since $\{z_n\}$ is an interpolating sequence sequence for $H^\infty$, then $\{z_n\}$ is separated.

We need the following result to prove sufficiency, which dues to Earl (see [7]).

Lemma 3.1 Let $\{z_n\}$ be a sequence in the unit disc such that (1.5) holds. Then there is a constant $C$ such that whenever $\{a_n\}\in l^\infty, $ there exists $f\in H^\infty$ such that

$ f(z_n)=a_n, \quad n=1, 2, \cdots$

(3.3)

and such that

$f(z) = C\left( {\mathop {\sup }\limits_n |{a_n}|} \right)B(z),$

where $B(z)$ is a Blaschke product. The zeros $\xi_n$ of $B(z)$ satisfy $ \left|\varphi_{\xi_n}(z_n)\right|\leq\delta/3 $ and $\{\xi_j\}$ is separated.

For any given $\{a_n\}\in l^\infty$, by Lemma 3.1 there exists a function $f\in H^\infty$ such that (3.3) holds. We now show that $d\mu(z)=\sum\delta_{\xi_n}$ is a $K$-Carleson measure. For any $a\in\mathbb{D}$, by Lemma 1.4 in [7], we have

$\begin{eqnarray*} 1-|\varphi_a(z_n)|^2&\geq& 1-|\varphi_a(z_n)|\geq 1-\frac{|\varphi_a(\xi_n)|+|\varphi_{z_n}(\xi_n)|}{1+|\varphi_a(\xi_n)|\varphi_{z_n}(\xi_n)|}\\ &\geq& 1-\frac{\delta/3+|\varphi_{z_n}(\xi_n)|}{1+\delta|\varphi_{z_n}(\xi_n)|/3}\geq\frac{1-\delta/3}{1+\delta/3}(1-|\varphi_{a}(\xi_n)|)\\ &\geq&\frac{1-\delta/3}{2(1+\delta/3)}(1-|\varphi_{a}(\xi_n)|^2). \end{eqnarray*}$

Note that $K(t)\approx K(2t)$ and $d\mu=\sum\delta_{z_n}$ is a $K$-Carleson measure. Then we have

$\sum\limits_{n = 1}^\infty K (1 - |{\varphi _a}({\xi _n}){|^2}) \mathbin{\lower.3ex\hbox{$\buildrel＜\over {\smash{\scriptstyle\sim}\vphantom{_x}}$}} \sum\limits_{n = 1}^\infty K (1 - |{\varphi _a}({z_n}){|^2})＜\infty .$

This shows that $d\mu(z)=\sum\delta_{z_n}$ is a $K$-Carleson measure by Corollary 3.1 in [6]. So we have $f\in Q_K$ by Theorem 5.1 in [6]. The proof is completed.

If we choose $K(t)=t^p, 0<p<1$, then the space $Q_K$ is just a $Q_p$ space. Note that the conditions (1.2) and (1.3) hold. This gives the following result which appeared in [10].

Corollary 3.2  The sequence $\{z_n\}$ is an interpolating sequence for $Q_p\bigcap H^\infty$ if and only if $\{z_n\}$ is separted and $d\mu=\sum\delta_{z_n}$ is a $p$-Carleson measure.