\input amstex \documentstyle{amsppt} \magnification = 1200 \hcorrection{.25in} \def\({\left(} \def\){\right)} \def\[{\left[} \def\]{\right]} \def\lvert{\left\vert} \def\rvert{\right\vert} \def\grad{\text{grad}} \def\ABFP{1} \def\BFP{2} \def\Ha{3} \def\Hb{4} \def\Hc{5} \def\HKZ{6} \def\Ho{7} \def\Ko{8} \def\R{9} \def\Sa{10} \def\Sb{11} \def\SzNF{12} \NoBlackBoxes \NoRunningHeads \TagsOnRight %\refstyle{C} \topmatter \title Complete interpolating sequences for Paley-Wiener spaces and Muckenhoupt's $(A_p)$ condition \endtitle \thanks{Research supported by NATO linkage grant LG 930329. While at MSRI, Berkeley, during the fall 1995, the second author was supported in part by NSF grant DMS-9022140 and by the Norwegian Research Council.} \endthanks \author Yurii I. Lyubarskii and Kristian Seip \endauthor \address \hskip-\parindent Yurii I. Lyubarskii, Institute for Low Temperature Physics and Engineering, 47 Lenin Prospect, 310164 Kharkov, Ukraine \endaddress \email lyubarskii\@ilt.kharkov.ua \endemail \address \hskip-\parindent Kristian Seip, Department of Mathematical Sciences, The Norwegian Institute of Technology, N--7034 Trondheim, Norway \endaddress \email seip\@imf.unit.no \endemail %\dedicatory %\enddedicatory %\date %\enddate %\translator %\endtranslator %\keywords %\endkeywords \subjclass 30E05, 42A50 \endsubjclass \abstract We describe the complete interpolating sequences for the Paley-Wiener spaces $L^p_\pi$ ($1-\infty$. Finally, Minkin \cite{6} solved the problem without any a priori assumption on $\L$. The methods of \cite{3,6,7,8} are of a geometric nature and make crucial use of the Hilbert space structure of $L^2_\pi$. In this paper, we shall give a simpler proof, which works equally well for all $p$, $1a\}$ for each $\ a\in \R$. It follows that the sequence $\L\cap\Cpl_a$ is $H^p$-interpolating in $\Cpl_a$ (see \cite{4}, Chapter 9). Similarly, $\L\cap\Cmi_a$ is $H^p$-interpolating in the half-plane $\Cmi_a:=\{z\in \C; \Im za}\prod_{\Im \l_k>a, k\neq j}\left|\frac{\l_j-\l_k} {\l_j-\overline{\l_k}-i2a}\right|>0, \quad \sup_{\Im \l_j0. \tag 3 $$ We note that, by standard manipulations (which we omit) with the Carleson condition, this is equivalent to the following condition: $$\sup_j \sum_{k,k\neq j}\frac{(1+|\eta_j|)(1+|\eta_k|)}{|\l_j-\l_k|^2}<\infty. \tag 4$$ In particular, for some $\eps > 0$, the disks $$ K(\l_k):=\{z: |z-\l_k|<10\eps (1+|\eta_k|)\} $$ are pairwise disjoint. (We fix this value of $\eps$ until the end of the paper.) Moreover, the measure $$ \mu^+_\L:=\sum_{\eta_k>0} \eta_k\delta_{\l_k} $$ ($\delta_\l$ is the unit point measure at $\l$) is a Carleson measure, i.e., $$ \int_{\Cpl} |f|^s d\mu^+_\L \leq C \|f\|_{H^s}^s $$ for each function $f$ in the Hardy space $H^s(\Cpl)$, $s\geq 1 $. Similarly, $\L$ generates a Carleson measure in the lower half-plane as well as in each of the half-planes $\C^\pm_a$. If $\L$ is a complete interpolating sequence for $L^p_\pi$, then $$ \|f\|_{L^p(\R)} \leq C \(\sum_k |f(\l_k)|^p e^{-p\pi|\eta_k|}(1+|\eta_k|)\)^{1/p},\quad f\in L^p_\pi. \tag{5} $$ Indeed, since $\L$ is interpolating for $L^p_\pi$, the operator $$ T:f\mapsto \{f(\l_k)e^{-\pi|\eta_k|}(1+|\eta_k|)^{1/p}\} $$ is bounded from $L^p_\pi$ onto $l^p$. By the uniqueness of the solution of the interpolation problem, we have $\ker T = \{0\}$, and it suffices to apply the Banach theorem on inverse operators. Given $x\in\R,\;r>0$, let $Q(x,r)$ be the square with center at $x$, side length $2r$, and sides parallel to the coordinate axes. We say that a sequence $\L\subset \C$ is {\it relatively dense} if there exists $r_0>0$ such that $\L\cap Q(x,r_0)\neq \emptyset$ for each $x\in \R$. If $\L$ is a complete interpolating sequence for $L^p_\pi$, \thetag{5} forces $\L$ to be relatively dense: if this is not the case and there exist sequences $\{x_j\}\subset\R$ and $r_j\to \infty $ such that $Q(x_j,r_j)\cap\L=\emptyset$, then, setting $$ f_j(z)=\frac{\sin\frac{\pi}{2}(z-x_j)}{z-x_j}, $$ we get $$ \sum_k |f_j(\l_k)|^pe^{-p\pi|\eta_k|}(1+|\eta_k|) \;\;\to\;\;0,\quad j\to \infty, $$ while $\|f_j\|_{L^p}$ is independent of $j$. Suppose that $\L$ is a complete interpolating sequence for $L^p_\pi$. Take $r>r_0$, where $r_0$ is as above, define $$ Q_j=Q(4rj,r),\ \ \ \ j\in \Z, $$ and pick a sequence $\Gamma=\{\gamma_j\}\subset \Lambda$ such that $\gamma_j\in Q_j$. Let $\Sigma=\{\sigma_j\}$ be another sequence with $|\gamma_j-\sigma_j|=\eps$. Suppose $w=\{w_j\}$ is a positive weight sequence. Associate with it the weighted space $l^p_w$ consisting of all sequences $a=\{a_k\}$ satisfying $$ \|a\|_{w,p}^p:=\sum_k |a_k|^pw_k<\infty. $$ We are interested in the boundedness of the {\it discrete Hilbert operator} $\H_{\Gamma,\Sigma}$ defined by the relation $$ \H_{\Gamma,\Sigma}:a=\{a_j\}\mapsto \{(\H_{\Gamma,\Sigma}a)_j\};\ \ (\H_{\Gamma,\Sigma}a)_j=\sum_k \frac{a_k}{\sigma_j-\gamma_k}, $$ on $l^p_w$. The following definitions are needed. We say that $w$ satisfies the {\it discrete $(A_p)$ condition} if $$ \sup_{k,n}\left(\frac{1}{n}\sum_{j=k+1}^{k+n} w_j\right) \left(\frac{1}{n}\sum_{j=k+1}^{k+n} w_j^{-\frac{1}{p-1}}\right)^{p-1}<\infty. $$ This condition is analogous to the classical continuous $(A_p)$ condition for a positive weight $v(x)>0,\ x\in \R$: $$ \sup_I \left\{\(\frac1{|I|}\int_I vdx\) \(\frac1{|I|}\int_Iv^{-1/(p-1)}dx\)^{p-1}\right\}<\infty , \tag{6} $$ where $I$ ranges over all intervals in $\R$ (see \cite{2}). Recall that the latter condition is necessary and sufficient for boundedness of the classical Hilbert operator $$ \H:f\mapsto (\H f)(t)=\frac1{i\pi}\int \frac{f(\tau)}{t-\tau}d\tau $$ on the weighted space of functions $L^p(\R;v)$ consisting of all functions $f$ satisfying $$\|f\|_{v,p}^p:=\int |f(t)|^pv(t)dt <\infty. $$ We shall need the following lemma. \proclaim{Lemma 1} If $\H_{\Gamma,\Sigma}$ is bounded from $l^p_w$ to $l^p_w$, then $w$ satisfies the discrete $(A_p)$ condition. \endproclaim \demo{Proof} We adopt the proof for the continuous case (see \cite{2}.) Let $k$ and $n$ be given. For convenience, put $I_1=\{k+1,k+2,...,k+n\}$, $I_2=\{k+2n+1,k+2n+2,...,k+3n\}$. Suppose that a positive sequence $a$ is supported on $I_1$. Then, for $j\in I_2$, we have $$ |(\H_{\Gamma,\Sigma}a)_j| \geq\sum_l a_l\frac{\Re(\sigma_j-\gamma_l)}{|\sigma_j-\gamma_l|^2} \geq \frac{C}{n}\sum_l a_l, \tag{7} $$ where $C$ is independent of $k$ and $n$. Putting $a_l=1$, we get thus $$ \sum_{j\in I_2} w_j \leq C \sum_{l\in I_1} w_l, $$ and by symmetry $$ \sum_{j\in I_1} w_j \asymp \sum_{l\in I_2} w_l; \tag{8} $$ here and in what follows the sign $\asymp$ means that the ratio of the two sides lies between two positive constants. Now we put $a_l=w_k^{\alpha}$ and get by \thetag{7} $$ \left(\sum_{j\in I_2} w_j\right) \left(\frac{1}{n}\sum_{l\in I_1} w_l^{\alpha}\right)^p\leq C \sum_{m\in I_1}w_m^{1+\alpha p}. $$ Finally, we put $\alpha=-\frac{1}{p-1}$ and invoke \thetag{8}, and the lemma is proved. \qed \enddemo The converse of Lemma 1 is also true, but we will not need that fact. Note also that the boundedness of the operator $\H_{\Gamma,\Sigma}$ is independent of the choice of sequence $\Sigma$, provided the condition $|\gamma_j-\sigma_j|=\eps$ holds. Let $\L$ be a complete interpolating sequence for $L^p_\pi$. If the function $f_0\in L^p_\pi$ solves the interpolation problem $f_0(\l_k)=\delta_{0,k},\ k\in\Z$, then $f_0(\mu)\neq 0$ for $\mu \in \C\setminus\L$, since otherwise the function $(z-\l_0)(z-\mu)^{-1}f_0(z)$ belongs to $L^p_\pi$ and vanishes on $\L$. Since $f_0\in L^p_\pi$, $f_0$ belongs to the Cartwright class $\Ca$ (see \cite{5}, Lecture 15) and, in particular, the limit $$ S(z)=\lim_{R\to\infty}\prod_{|\l_k|0$. \item"{(ii)}" The limit $S(z)$ in \thetag{9} exists and represents an entire function of exponential type $\pi$. \item"{(iii)}" There exists a relatively dense subsequence $\Gamma=\{\gamma_j\}\subset\L$ such that the sequence $\{|S'(\gamma_j)|^p\}$ satisfies the discrete $(A_p)$ condition. \endroster Defining $F(x)=|S(x)|/\roman{dist}(x,\Lambda)$, we may replace statement $\roman{(iii)}$ by the following: \roster \item"{(iii')}" $F^p$ satisfies the (continuous) $(A_p)$ condition. \endroster \endproclaim Note that that condition (i) is equivalent to the statement that, for each $a\in \R$, the sequences $\L\cap\C^\pm_a$ satisfy the Carleson condition (3). Another, more compact way of expressing (i), is given by (4). \head 3. Proof of Theorem 1: Necessity \endhead We have already proved the necessity of (i) and (ii), and also the existence of a relatively dense sequence $\Gamma=\{\gamma_j\}\subset \L$. We prove now that (iii) is necessary as well. Let $\eps$ be as above. Then, for every $j$, we can find a point $\sigma_j$ with $|\sigma_j-\gamma_j|=\eps$ and $$ |S(\sigma_j)|=\eps |S'(\gamma_j)|. $$ This follows from the fact that $S(z)(z-\gamma_j)^{-1}\neq 0$ for $|z-\gamma_j|\leq\eps$, hence $$ \min_{|z-\gamma_j|=\eps}|S(z)(z-\gamma_j)^{-1}| \leq |S'(\gamma_j)| \leq \max_{|z-\gamma_j|=\eps}|S(z)(z-\gamma_j)^{-1}|. $$ Set $\Sigma=\{\sigma_j\}$. The Plancherel-P\'{o}lya inequality (see \cite{5}, Lecture 20) yields $$ \sum_j|f(\sigma_j)|^p\leq C \|f\|^p_{L^p},\ \ \ f\in L^p_\pi. \tag{11} $$ Now let $a=\{a_j\}$ be a finite sequence. By \thetag{10}, the unique solution of the interpolation problem $f(\gamma_j)=a_j$, $f(\lambda_k)=0$, $\lambda_j\not\in\Gamma$ has the form $$ f(z)=\sum_j \frac{a_j}{S'(\sigma_j)}\frac{S(z)}{(z-\gamma_j)}. $$ By \thetag{5} and \thetag{11}, we have $$ \sum_j |f(\sigma_j)|^p \leq C \sum_j |a_j|^p. $$ Now, by our particular choice of the sequence $\Sigma$, we obtain (iii) by observing that Lemma 1 applies with $w_j=|S'(\gamma_j)|^p$. To prove that (iii) implies (iii'), we need the following lemma. \proclaim{Lemma 2} Suppose $x\in \R$ and $\Re \gamma_j \leq x\leq \Re \gamma_{j+1}$. Then there exists an $\alpha =\alpha(x)\in [0,1]$ such that $$ |S'(\gamma_j)|^{\alpha} |S'(\gamma_{j+1})|^{1-\alpha} \asymp |S(x)|/\roman{dist}(x,\Lambda), $$ uniformly with respect to $x\in \R$. \endproclaim In fact, assuming this lemma to hold, we see that (9) with $v=F^p$ follows from (iii) and the inequality $t^\alpha s^{1-\alpha} \leq t+s, \ \ t,s>0,\ \ \alpha\in [0,1]$. \demo{Proof of Lemma 2} We assume that $x\in [\Re \gamma_j,\Re \gamma_{j+1}]$ and, for simplicity, $x\not\in\L$. Set $\L(x)=\{\l\in\L:|\l-x|<30r\}$. (Here $r$ is the number used for constructing $\Gamma$.) For $\alpha\in[0,1]$ we have $$ \thickmuskip=.5 \thickmuskip \medmuskip=.5 \medmuskip \multline \rho:= \frac{|S'(\gamma_j)|^\alpha |S'(\gamma_{j+1})|^{1-\alpha}} {|S(x)|\text{dist}(x,\Lambda)^{-1}} \\ = \left\{\frac{\left |\frac1{\gamma_j} \prod_{\l_k\in\L(x)\setminus\{\gamma_j\}} \(1-\frac{\gamma_j}{\l_k}\)\right |^\alpha \left |\frac1{\gamma_{j+1}} \prod_{\l_k\in\L(x)\setminus\{\gamma_{j+1}\}} \(1-\frac{\gamma_{j+1}}{\l_k}\)\right |^{1-\alpha}} {\left |\prod_{\l\in\L(x)} \(1-\frac x\l\)\right |} \text{dist}(x,\L)\right\} \\ \times\left\{\prod_{\l\in\L\setminus\L(x)} \frac{|\gamma_j-\l_k|^\alpha|\gamma_{j+1}-\l_k|^{1-\alpha}} {|x-\l_k|} \right \} =\Pi_1(x)\times\Pi_2(x). \endmultline $$ A simple estimation shows that $\Pi_1(x)\asymp 1$ uniformly with respect to $\alpha\in [0,1]$ so we need only estimate $\Pi_2(x)$. Let us put $$ \gamma_j=x-x_j+i y_j, \ \gamma_{j+1}=x+x_{j+1}+i y_{j+1}. $$ The values $x_j$ and $x_{j+1}$ depend upon $x$ and also satisfy the inequalities $0\leq x_j, x_{j+1} \leq 8r$. We may then write $$ \multline \rho^2\asymp \prod_{\l_k\not\in\L(x)} \frac{((x-x_j-\xi_k)^2+(y_j-\eta_k)^2)^{\alpha} ((x+x_{j+1}-\xi_k)^2+(y_{j+1}-\eta_k)^2)^{1-\alpha}} {(x-\xi_k)^2+\eta_k^2} \\ =\prod_{\l_k\not\in\L(x)} \left(1-\frac{2x_j(x-\xi_k)+2y_j\eta_k+O(1)} {(x-\xi_k)^2+\eta_k^2}\right)^{\alpha} \\ \times \left(1+\frac{2x_{j+1}(x-\xi_k)-2y_{j+1}\eta_k+O(1)} {(x-\xi_k)^2+\eta_k^2}\right)^{1-\alpha}. \endmultline $$ Choosing $\alpha=\alpha(x)$ so that $\alpha x_j-(1-\alpha)x_{j+1}=0$, i.e., $\alpha=x_{j+1}/(x_j+x_{j+1})$, we find that $$ \rho^2\asymp \exp\left(c\sum_{\l_k\not\in\L(x)} \frac{|\eta_k|}{(x-\xi_k)^2+\eta_k^2}\right). $$ By Carleson's condition \thetag{4}, the sum is uniformly bounded, and we are done. \qed \enddemo \head 4. Proof of Theorem 1: Sufficiency \endhead We will now prove that (i), (ii), (iii') imply that $\Lambda$ is a complete interpolating sequence. To begin with, note that $$ \int[ F(x)]^p\frac{dx}{1+|x|^p}<\infty \tag{12} $$ and $$ \int [F(x)]^pdx=\infty. \tag{13} $$ The first relation follows from the fact that $\int[F(x)]^p|\H f(x)|^pdx<\infty$ for each bounded finite function $f$; it suffices to take $f=\chi_{[0,1]}$. To obtain \thetag{12}, we may apply the operator $\H$ to a $\delta$-sequence $\{\delta_n(x)\}$. First, we check that $\Lambda$ is a uniqueness set. To this end, we need to estimate $|S(z)|$ from below. \proclaim{Lemma 3} Let $\eps$ be the number from relation \thetag{3}. Then $$ |S(z)|\geq C (1-|z|)^{-1/p} e^{\pi |\Im z|} \ \ \text{for} \ \ \roman{dist} (z,\L)>\eps(1+|\Im z|). \tag{14} $$ \endproclaim \demo{Proof of Lemma 3} Put $\L'=\L\cap\{z:|\Im z|<\eps\}$ and consider the auxiliary function $$ S_1(z)=S(z)\prod_{\l\in\L'}\frac{z-\l+2i\eps}{z-\l}. $$ It is plain that $$ |S_1(z)|\asymp |S(z)|, \ \ |\Im z|> 3\eps, \tag{15} $$ and, besides, $|S_1(x)|^p$ satisfies the $(A_p)$ condition. Consider the inner-outer factorization of $S_1$ in the upper half-plane, $$ S_1(z)=e^{-i\pi z} G(z) B_1(z), \ \ \Im z>0. \tag{16} $$ Here the Blaschke product $B_1$ corresponds to the Carleson sequence $(\L\cap\Cpl)\setminus\L'$ and, in particular, $$ |B_1(z)|>c>0 \ \ \text{for dist}(z,\L)>\eps|\Im z|. \tag{17} $$ Moreover, $G$ is an outer function and $|G(x)|^p$ satisfies the $(A_p)$ condition. Therefore, $|G(x)|^{-q}$ is an $(A_q)$ weight (here $1/p+1/q=1$), $G(x)^{-1}(1+|x|)^{-1}\in L^q(\R)$, and thus $$ \frac1{(z+i)G(z)}=\frac1{2\pi i} \int \frac1{(t+i)G(t)}\frac{dt}{t-z}, \ \ \Im z>0. $$ It follows that $$ \frac 1{|G(z)|} \leq C (1+|z|)^{1/p}. \tag{18} $$ Combining relations (15)--(18), we obtain (14) for $\Im z> 3\eps$. The estimate for $\Im z<-3\eps$ is similar, and to fill the gap $-3\eps <\Im z < 3\eps$, we may repeat the construction, taking another horisontal line instead of $\R$. \qed \enddemo Note that, since $|S_1(x)|^p$ is an $(A_p)$ weight, we have $$ \int|S_1(x)|^pdx=\infty. $$ The Phragm\'{e}n-Lindel\"{o}f theorem (see \cite{5}, Lecture 20) yields $$ \int|S_1(x+ia)|^pdx=\infty, \ \ a\in \R, $$ and, by \thetag{15}, $$ \int |S(x+i)|^pdx =\infty. $$ Again applying the Phragm\'{e}n-Lindel\"{o}f theorem, we get $$ \int |S(x)|^pdx=\infty. \tag{19} $$ We are now in position to prove the uniqueness. Indeed, if $f\in L^P_\pi$ and $f(\l)=0, \ \l\in\L$, then $\phi(z)=f(z)/S(z)$ is an entire function of exponential type $0$, and \thetag{14} yields that $|\phi(z)|$ is uniformly bounded for $z$ satisfying $\text{dist}(z,\L)>\eps(|\Im z|+1)$. Therefore $\phi(z)\equiv C$, which is incompatible with \thetag{19}, unless $C=0$. It remains only to check that we can actually solve the interpolation problem (1) for each sequence $a=\{a_k\}$ satisfying \thetag{2}. It suffices to consider a finite sequence $a$ and bound the norm of the solution by a constant times the left-hand side of \thetag{2}. After doing so, we can apply a limit procedure. If $a$ is a finite sequence, then, by (12), the unique solution of the interpolation problem (1) has the form $$ f(z)=\sum_j \frac{a_k}{S'(\lambda_k)}\frac{S(z)}{(z-\lambda_k)}. \tag{20} $$ We split the sum (20) into two parts, corresponding to points lying in $\Cpl\cup \R$ and in $\Cmi$, respectively. We may estimate the norm of each sum separately, so let us assume that all the $\lambda_k$ corresponding to $a_k\neq 0$ are in $\Cpl\cup\R$. Clearly, we may estimate the $L^p$ integral along $\Im(z)=-\ha$. Let us, however, for conventional reasons, estimate it along $\R$ and assume all the points $\lambda_k$ satisfy $\eta_k\geq \ha$. Now let $$ B(z)=\prod_k\frac{z+\frac{i}{2}-(\lambda_k+\frac{i}{2})} {z+\frac{i}{2}-(\ol{\lambda_k}-\frac{i}{2})}. $$ Writing $S(z)=B(z)e^{-i\pi z} G(z)$, where $G$ is an outer function in $\Cpl$, we observe that (iii') is equivalent to $|G(x)|^p$ satisfying the $(A_p)$ condition. We note that we have $$ |S'(\lambda_k)|\asymp |G(\lambda_k)|\frac{e^{\pi\eta_k}}{\eta_k}. $$ Thus it is enough to consider the $L^p$ boundedness of $$ \tilde{f}(x)=\sum_k\frac{a_k \eta_k e^{-\pi\eta_k}}{G(\lambda_k)} \frac{G(x)}{x-\lambda_k}. $$ By duality, $$ \multline \|\tilde{f}\|_p\asymp\sup_{\|h\|_q=1, h\in H^q} \left|\sum_k\frac{a_k \eta_k e^{-\pi\eta_k}} {G(\lambda_k)}\int_{\R} \frac{G(x)h(x)}{x-\lambda_k}dx\right| \\ \leq \sup_{\|h\|_q=1,h\in H^q}\left|\sum_k\frac{a_k \eta_k e^{-\pi\eta_k}} {G(\lambda_k)}{\Cal H}Gh(\lambda_k)\right| \\ \leq \sup_{\|h\|_q=1,h\in H^q} \left(\sum_k|a_k|^p \eta_k e^{-p\pi\eta_k}\right)^{1/p} \left(\sum_k \left|\frac{{\Cal H}Gh(\lambda_k)}{G(\lambda_k)}\right|^q\eta_k\right)^{1/q}. \endmultline $$ Since $|G(x)|^{-q}$ satisfies the $(A_q)$ condition, $G$ is an outer function in $\Cpl$, and $h\in H^q, \|h\|_q\leq 1$, we have ${\Cal H}Gh(z)/G(z)\in H^q$, and $\|{\Cal H}Gh(z)/G(z)\|_q\leq C$. Since $\sum_k\eta_k\delta_{\lambda_k}$ is a Carleson measure, we get the desired conclusion. The sum corresponding to points in $\Cmi$ is treated similarly. \head 5. A stability result \endhead We will now show how Theorem 1 can be used to obtain a result similar to the Kadets 1/4 theorem. The same technique implies more sophisticated stability results similar to the theorems of Avdonin and Katsnelson; for these results we refer to \cite{3}. For $1