Sampling and interpolation in the Paley-Wiener spaces Lp π, 0 < p ≤ 1

Publicacions Matem`atiques, Vol 42 (1998), 103–118. SAMPLING AND INTERPOLATION IN THE PALEY-WIENER SPACES Lp π,0<p≤1 Kristin M. Flornes Abstract Following Beurling’s ideas concerning sampling and interpolation in the Paley-Wiener space L∞ τ, we find necessary and sufficient density conditions for sets of sampling and interpolation in the Paley-Wiener spaces Lp τfor 0 <p≤1. 1. Introduction This work is inspired by Beurling’s lectures on balayage of FourierStieltjes transforms and interpolation for an interval on R. In our terms, his problem concerned the so called Paley-Wiener space Lp τwith p=∞. This space consists of entire functions of exponential type at most τ, bounded on the real axis. Beurling proved that a discrete set of real numbers is a set of sampling for this space if and only if its lower uniform density is bounded by τ/π, and the set is a set of interpolation if and only if its upper uniform density does not exceed τ/π. We prove that the same density results are valid for sampling and interpolation for functions which belong to Lp τ,0<p≤1. The Paley-Wiener spaces with 1 <p<∞have different properties. The density restrictions turn out to be sufficient but not necessary conditions for sampling and interpolation. In [4], Lyubarskii and Seip describe complete interpolating sequences in the Paley-Wiener spaces for 1 <p<∞. Their work shows that the difference between 0 <p≤1 and p=∞on the one hand and 1<p<∞on the other is related to the problem of boundedness of the Hilbert transform. To see how to proceed in our case, we have been guided by Seip’s analysis of corresponding problems for the Bargmann-Fock space [5]. Keywords. Sampling, interpolation. The research of the author has been supported by the Research Council of Norway under grant 100552/410. 104 K. M. Flornes 2. Main results For 0 <p<∞the space Lp τis defined to be the collection of entire functions fof exponential type at most τfor which kfkp=µZ|f(x)|pdx¶1/p <∞. Unlike the regular Lpspaces, these spaces are nested, i.e., Lp τ⊆Lq τ for 0 <p≤q. According to classical results |f(x+iy)|≤Ceτ|y|kfkp so kfk∞≤Ckfkpwhere Cdepends only on τand p. Basic facts about entire functions can be found in e.g. [8]. Lp τis a Banach space for 1≤p≤∞. For 0 <p<1 the norm k·k pis a quasinorm and Lp τ is complete with respect to this quasinorm [3]. We fix the type of the functions to be πfor the rest of the paper. The other cases are handled by a change of variables. For f∈Lp πand Λ = {λj}a discrete set of real numbers, we write kf|Λkp p=Pλj∈Λ|f(λj)|p. The set Λ is said to be a set of sampling if there exist positive numbers Aand Bsuch that Akfkp p≤kf|Λk p p≤Bkfk p p for all f∈Lp π. Λ is said to be a set of interpolation if to every sequence w={wj}∈l pwe can find a function f∈Lp πsuch that f(λj)=w jfor all j. We will consider sampling and interpolation for functions in Lp π for 0 <p≤1. For the description of the density of a set of real numbers, we use the following concept introduced by Beurling. Let Λ = {λj}be a uniformly discrete set, i.e. there exists δ>0 such that |λi−λj|≥δ,i6=j. Let n(r) and n(r) denote respectively the largest and smallest number of points in any interval [x, x +r] for r>0. We define the upper uniform density ofΛ(u.u.d.(Λ)) and the lower uniform density of Λ (l.u.d.(Λ)) to be u.u.d.(Λ) = lim r→∞ n(r) rand l.u.d.(Λ) = lim r→∞ n(r) r, where the limits exist because of the subadditivity of the function r7→ n(r) and the superadditivity of the function r7→ n(r). The following two theorems are our main results. Theorem 2.1. A discrete set Λis a set of sampling for Lp πif and only if it can be expressed as a finite union of uniformly discrete sets and contains a uniformly discrete subset Λ0for which l.u.d.(Λ0)>1. Theorem 2.2. A discrete set Λis a set of interpolation for Lp πif and only if it is uniformly discrete and u.u.d.(Λ) <1. Sampling and Interpolation 105 3. Auxiliary results The following classical result of Plancherel-P´olya [8, p. 97] states that the upper sampling inequality holds for any uniformly discrete set. Lemma 3.1. Let f∈Lp πand {λk}be an increasing sequence of real numbers such that |λk−λj|≥²>0,k6=j, then kf|Λkp≤Bkfkp, where Bis a constant depending only on pand ². Moreover we have the following lemma. Lemma 3.2. There exists a positive constant Bsuch that kf|Λkp≤Bkfkp for all f∈Lp πif and only if Λcan be expressed as a finite union of uniformly discrete sets. Proof: The “only if” follows from Lemma 3.1. For the converse suppose that such a Bexists and that there is no bound on the number of points from Λ to be found in translates of a unit interval, In={x:n≤ x<n+1},n∈Z. This means that we can find a sequence {nj}such that #(Λ∩Inj)→∞. Pick xj∈Injand let hxj(z)=hsin(π(xj−z)/m) π(xj−z)/m im. Choose ma positive integer such that hxj(z)∈Lp π. Then khxj|Λkp→∞, which is a contradiction. We conclude that there has to be a bound, say N, on the number of points found in In. The set Λ can be divided into 2Nuniformly discrete sets by letting Λkfor k=1,2,... ,N consist of point number kin Infor neven, and Λkfor k=N+1,N+2,... ,2N consist of point number k−Nin Infor nodd. We conclude that every set of sampling is a finite union of uniformly discrete sets and we do not have to consider the upper sampling inequality which always holds for such sets. For a discrete set Λ, let K(Λ) = K(Λ,p) denote the smallest number Ksuch that (1) kfkp≤Kkf|Λkp for all f∈Lp π. We shall refer to K(Λ) as the sampling constant. From now on, we assume every set of sampling to be uniformly discrete. The following lemma shows that this assumption can be made without loss of generality. The lemma was proved by Seip in [6, p. 141], for the space L2 π. 106 K. M. Flornes Lemma 3.3. If Λis a set of sampling for Lp π, then it contains a uniformly discrete subset Λ0⊂Λwhich is also a set of sampling for Lp π. Proof: The Paley-Wiener spaces Lp π,p>0 are closed under differentiation, i.e. kf0kp≤Ckfkp, (see e.g. [8, p. 99]). Seip’s proof for L2 πis thus valid for any p>0. Suppose that Λ= N [ n=1 Λ(n)where inf λ∈Λ(n) i6=j |λj−λi|≥δ>0 for all n. Let 0 <²<δ/4 and construct a uniformly discrete subset ˜ Λ⊂Λ s.t. |λi−˜ λ|<²for every λi∈Λ. Then for arbitrary λ(n) i∈Λ(n), we can find a point ˜ λ(n) i∈˜ Λ s.t. |λ(n) i−˜ λ(n) i|<². The points ˜ λ(n) iare distinct for fixed n. The mean value theorem gives |f(˜ λ(n) i)−f(λ(n) i)|=|f(µ(n) i)|| ˜ λ (n) i−λ (n) i|, where µ(n) iis a point between ˜ λ(n) iand λ(n) i. We use the inequality kf+gkp p≤2p(kfkp p+kgkp p) instead of the triangle inequality for p∈(0,1]. This yields kf|Λkp p≤2pNkf|˜ Λkp p+2 p² N X n=1 kf0|{µ(n) i}kp p ≤2pNkf|˜ Λkp p+C(δ, N, p)²kfkp p, where the last step follows from an application of the Plancherel-P´olya inequality and the fact that Lp πis closed under differentiation. The set ˜ Λ is a set of sampling provided ²<1/(C(δ, N, p)K) where Kis the sampling constant for the set Λ. For a given closed set Qand for t>0 let Q(t) denote the set of points which are a distance less than or equal to tfrom Q. The Fr´echet distance [R, Q] between two closed sets Rand Qis the smallest number tsuch that Q⊂R(t) and R⊂Q(t). Let Qibe a sequence of closed sets. Qiconverges weakly to Q, denoted by Qi*Q, if for every finite interval L=[−l,l]wehave[(Q n∩L)∪{−l, l},(Q∩L)∪{−l, l}]→0. If Qis a uniformly discrete set, then every sequence of translates Q+xn contains a subsequence converging weakly to another uniformly discrete set. Let W(Q) be the collection of weak limits of translates of Q. The next lemma implies that for a given set of sampling Λ, every set Λ0in W(Λ) will be a set of sampling for Lp π. Sampling and Interpolation 107 Lemma 3.4. Let Λ0be a uniformly discrete set. Λn*Λ0implies K(Λ0)≤lim K(Λn). Proof: By Lemma 3.1, given f∈Lp πand ²>0, we can find T>0 such that kf|˜ Λ∩{|x|>T}kp≤² for any uniformly discrete set ˜ Λ. We can of course assume that K(Λn) is finite for all n. The set Λnis a set of sampling, so we know that kfkp≤K(Λn)(kf|Λn∩[−T,T]kp+²). Since [−T,T] is compact, the Fr´echet distance [(Λn∩[−T,T])∪{−T,T},(Λ0∩[−T,T])∪{−T,T}]→0 and we have lim kf|Λn∩[−T,T]kp≤kf|Λ 0∩[−T,T]kp. This gives the inequality kfkp≤lim K(Λn)(kf|Λ0∩[−T,T]kp+²) for all n. Letting ²→0wegetK(Λ0)≤lim K(Λn). The sampling inequality (1) gives a bound for the density of the sampling set. Lemma 3.5. If Λis a uniformly discrete set of sampling for Lp π, then l.u.d.(Λ) >0. Proof: Let Λ be uniformly discrete with l.u.d.(Λ) = 0 and suppose that kfkp≤Akf|Λkp,∀f∈Lp π. Let Txbe the translation operator, Txf(y)=f(y−x). Since kTxfkp=kfkpand kTxf|Λkp=kf|Λ−xkp, it follows that for all x∈Rwe have (2) kfkp=kTxfkp≤AkTxf|Λkp=Akf|Λ−xkp,∀f∈Lp π. The fact that l.u.d.(Λ) = 0 implies that we can find an arbitrarily large interval I(xR,R) for which I(xR,R)∩Λ=∅. Choose nsuch that the function g(z)=hsin(πz)/n πz/n inis in Lp π. We can make the right-hand side of the above inequality arbitrarily small for this function by choosing R large, so (2) does not hold for g. We conclude that l.u.d.(Λ) >0. A set Λ is said to be a set of uniqueness if every function f∈Lp πthat vanishes on Λ vanishes identically. The sampling inequality (1) implies that a set of sampling is also a set of uniqueness. Beurling showed that for the sampling problem in L∞ πeven more is true (see [1, p. 345]): Theorem 3.6. The set Λis a set of sampling for L∞ πif and only if every set Λ0∈W(Λ) is a set of uniqueness. Beurling’s density result for L∞ π(Theorem 5 [1, p. 346]) is crucial in our analysis. 108 K. M. Flornes Theorem 3.7. The uniformly discrete set Λis a set of sampling for L∞ πif and only if d=l.u.d.(Λ) >1. If Λ = {λj}is a set of interpolation for Lp π, standard arguments based on the open mapping theorem for Fr´echet spaces [7, p. 75] shows that the interpolation is stable. This means that there exists a positive number K such that for every sequence {wj}∈l pwe can find f∈Lp πsuch that (3) kfkp≤Kkf|Λkp. The smallest such Kis denoted by K0(Λ). Lemma 3.8. Every set of interpolation for Lp πis uniformly discrete. Proof: Choose f(λk) = 1 for some arbitrary kand let f(λj)=0, ∀j6=k. Then kf|Λkp= 1. We know that we can find fsuch that kfkp≤K0. By Bernstein’s inequality [8, p. 84] for L∞ πand the fact that kfk∞≤Ckfkp,weget 1=kf(λ j)−f(λ k )k ∞≤kfk ∞ π|λ j−λ k |≤CK0π|λj−λk|. Lemma 3.9. Let Λbe a uniformly discrete set. Then Λn*Λimplies K0(Λ) ≤lim K0(Λn). Proof: Let Λ = {λk}and Λn={λ(n) k}. We may assume without loss of generality that K0(Λn)<∞for all n, and thus there exists a solution fn w∈Lp πto every interpolation problem fn w(λ(n) k)=w ksuch that kfn wkp≤K0(Λn)kwkp. Choose a subsequence Λnifor which K0(Λni)→lim K0(Λn). Then there exists a subsequence of ni,sayn i j, where fnij w→fwand fw(λk)=w k , kf w k p≤lim K0(Λn)kwkp. Assume that Λ is a set of interpolation and a set of uniqueness. Every function is then uniquely determined by its values on Λ. This implies that (3) holds for every f∈Lp πand thus Λ is a set of sampling. A set of interpolation can only be a set of uniqueness if it is also a set of sampling. The key lemma in the next section shows that this can not be the case, i.e., there are no discrete sets which are both sets of sampling and sets of interpolation for Lp π. Sampling and Interpolation 109 4. The key lemma The following lemma is our main auxiliary result. Lemma 4.1. For 0<p≤1there is no discrete subset of Rthat is both a set of sampling and a set of interpolation for Lp π. Proof: We argue by contradiction. Suppose that such a set Λ exists. Choose a λ0∈Λ and consider the unique function g0∈Lp πsatisfying g0(z)=½1ifz=λ 0 , 0ifz∈Λ\{λ 0 }. Let g(z)=(z−λ 0 )g 0 (z). It is clear that the function fλ(z)=g(z)/(z− λ) lies in Lp πfor arbitrary λ∈Λ. The fact that Λ is a set of sampling implies that kfλkp≤Kkfλ|Λkp=KÃX λk∈Λ |fλ(λk)|p!1/p =K|g0(λ)| because g(λk) = 0, for all λk∈Λ. The sampling constant Kis independent of the choice of λ. Using the subharmonicity of |fλ(z)|pwe get (with z=x+iy) |fλ(λ)|p=|g0(λ)|p≤C(²)ZZ R(λ,²,1) |g(z)|pdx dy where R(w,a, 1)={z:w−a≤x≤w+a, −1≤y≤1},w,a ∈R and ²is chosen so small that inf |λi−λk|>4²for λi6=λk. The fact that a set of interpolation is uniformly discrete implies that we have a finite number of points from Λ in any R(w,a,1). Collecting our results we obtain independently of w inf ξ∈R(w,T,1) X λ∈Λ∩R(w,T,1) |ξ−λ|−pZZ R(w,T,1) |g(z)|pdx dy(4) ≤X λ∈Λ∩R(w,T,1) ZB −BZ∞ −∞ |fλ(z)|pdx dy ≤X λ∈Λ∩R(w,T,1) C|g0(λ)|p (5) ≤CZZ R(w,T,1) |g(z)|pdx dy,(6) 110 K. M. Flornes where we have used a classical result by Plancherel-P´olya [8,p.94] ZB −BZ∞ −∞ |fλ(x+iy)|pdx dy ≤C1(B,p)Z∞ −∞ |fλ(x)|pdx. Inequality (6) implies that the sum is bounded even when Ttends to infinity, i.e., (7) inf |=z|≤1X λ∈Λ |λ−z|−p≤C. But in Lemma 3.5 we found that l.u.d.(Λ) >0, so the sum does not increase if we exchange the λ’s by points on a grid where the separation is large enough. Choose e.g. the grid {cn}n∈Z. Since Pn∈Z1/npdiverges for p≤1 we have a contradiction. This concludes the proof. From the key lemma we see that the spaces Lp π,0<p≤1 are fundamentally different from Lp π,1<p<∞for which there exist sets which are both sampling and interpolation sets. (See Lyubarskii and Seip [4].) The following three results are direct consequences of the key lemma. Lemma 4.2. We can remove a point from a set of sampling and still have a set of sampling. Proof: Removing a point from a set of sampling does not change the fact that the sampling operator has closed range. The operator is injective so the open mapping theorem yields the lower frame bound. We shall need the following notion of distance from a point xon the real axis to the set Λ. For x∈R, let ρ(x; Λ) = supf|f(x)|, where f ranges over all functions f(x)∈Lp πvanishing on the set Λ and for which kfkp≤1 (see [1, p. 352]). Lemma 4.3. If Λis a set of interpolation then ρ(x;Λ)>0,x/∈Λ. Proof: If Λ is a set of interpolation, it is not a set of uniqueness as remarked in the last paragraph of section 3. Given x0/∈Λ, pick f∈Lp π, where f|Λ= 0 and f6≡ 0. We can find an integer n,n≥0 such that the function g(x)= f(x) (x−x 0 ) n is analytic at x0and g(x0)6= 0. Hence ρ(x0;Λ)6=0. Sampling and Interpolation 111 Lemma 4.4. Adding a point to a set of interpolation yields another set of interpolation. Proof: Let w0be the value at x0and wnat λnwhere |w0|≤1 and |wn|≤1. Using the result in the above lemma we can find a function f0∈Lp π, such that kf0kp≤1, f0is vanishing on the set Λ and f0(x0)= ρ(x 0 )6=0. Iffsolves the interpolation on Λ, the function g(x)=f(x)+w 0−f(x 0 ) ρ(x 0 )f 0 (x) solves the interpolation on Λ ∪{x 0 }. 5. Sampling 5.1. The necessity part of Theorem 2.1. We assume that Λ is a set of sampling for Lp π. Let Λ0∈W(Λ). According to Lemma 3.4. every set in W(Λ) is a set of sampling, so Λ0is a set of sampling and thus a set of uniqueness. We want to show that Λ0is a set of uniqueness not only for Lp πbut also for L∞ π. If we can show this, we can use Beurling’s results for L∞ πas cited in Theorem 3.6 and Theorem 3.7. Suppose that Λ0is not a set of uniqueness for L∞ π, i.e., that there exits g∈L∞ πsuch that g6≡ 0 but g|Λ0= 0. Define the function f(z)= g(z) (z−λ 1 )(z−λ2)···(z−λ n)where λ1,... ,λ n∈Λ 0,np≥1+². This function is an entire function and f(λ) = 0 for every λ∈Λ0\ {λ1,... ,λ n}. Since g∈L∞ π, we have supx∈R|g(x)|p<∞so |f(x)|p∼O(|x|−1−²) which means that f∈Lp π. According to Lemma 4.2, Λ0\{λ 1 ,... ,λ n}is a set of sampling and thus a set of uniqueness. But f6≡ 0, so our original assumption about the set Λ0is false. The set Λ0is a set of uniqueness for L∞ π. We conclude that every Λ0∈W(Λ) is a set of uniqueness for L∞ π. By Beurling’s Theorem 3.6 this implies that Λ is a set of sampling for L∞ π, and applying Theorem 3.7 we find that l.u.d.(Λ) >1 . This completes the proof. 5.2. The sufficiency part of Theorem 2.1. We suppose now that a=l.u.d.(Λ) >1. By Theorem 3.7, Λ is a set of sampling for every space L∞ π+²,²<a−1. If, say, ²=1 2(a−1), we get kfk∞≤Ckf|Λk∞for all f∈L∞ π+². 118 K. M. Flornes Academic Press, London-New York, 1980. Department of Mathematical Sciences Norwegian University of Science and Technology N-7034 Trondheim NORWAY email: [email protected]tnu.no Primera versi´o rebuda el 8 de gener de 1997, darrera versi´o rebuda el 2 d’abril de 1997