The Newman--Shapiro problem

We give a negative answer to the Newman--Shapiro problem on weighted approximation for entire functions formulated in 1966 and motivated by the theory of operators on the Fock space. There exists a function in the Fock space such that its exponential multiples do not approximate some entire multiples in the space. Furthermore, we establish several positive results under different restrictions on the function in question.


Introduction and the main results
Let F = F (1) be the classical Bargmann-Segal-Fock space, where and m stands for the area Lebesgue measure. This space serves as a model of the phase space of a particle in quantum mechanics and so plays an important role in theoretical physics. Moreover, this space appears in time-frequency analysis, as a spectral model of L 2 (R) via the Bargmann transform (see, e.g., [15,Section 3.4]). Note also that the complex exponentials e λ , e λ (z) = e λz are the reproducing kernels of F , i.e., F, k λ F = F (λ), F ∈ F , where k λ = πe πλ .
In 1966, D. J. Newman and H. S. Shapiro posed in [20] the following problem about the structure of the operator adjoint to the multiplication operator in Fock space. Let F be an entire function such that, for every A > 0, This condition is equivalent to the following one: e λ · F ∈ F for every λ ∈ C. Now we can define the multiplication operator M F : G → F G This work was supported by Russian Science Foundation grant 17-11-01064. 1 on the linear span of the exponentials L = Span{e λ : λ ∈ C}.
The natural domain of the operator M F is given by Thus, we can consider the adjoint operator M * F as well as the operator adjoint to the restriction M F L , which we (following [20]) denote by F * d dz . This notation is motivated by the fact that when F = P is a polynomial, we have P (λ)G(λ) = M P G, πe πλ = G, P * (d/dz)(πe πλ ) , where P * (z) = P (z) and P * ( d dz ) is understood in the usual sense as a differential operator. In this case it is easy to see that M * P = P * d dz . The Newman-Shapiro problem (related to a much earlier work of E. Fischer [13]) is whether M * F = F * d dz for all F satisfying (1.1). In [20] (see also [21] and an extended unpublished manuscript [22]) Newman and Shapiro proved that this is the case when F is an exponential polynomial (i.e., F = n k=1 P k e λ k , where P k are polynomials and λ k ∈ C) and for some other special cases (i.e. F has no zeros or F (z) = sin z/z). Moreover, they revealed some connections of this problem with weighted polynomial approximation in F . More precisely, they proved the following result (to avoid inessential technicalities we assume that F has simple zeros only). Denote by E the space of all entire functions. Theorem 1.1 ([20, Theorem 1], [21]). For every F satisfying estimates (1.1) the following statements are equivalent: (1) Span{z n F, n ≥ 0} = EF ∩ F ; The Newman-Shapiro problem remained open since 1966. Several similar questions were studied, e.g., in [19] (see also [12,Chapter X.8]). For related questions on Toeplitz operators on the Fock space see [11] and the references therein.
It should be mentioned that the Newman-Shapiro problem is closely related to the spectral synthesis (hereditary completeness) problem for systems of reproducing kernels in the Fock space (or of Gabor-type expansions with respect to time-frequency shifts of the Gaussian). In the Paley-Wiener space setting, the spectral synthesis problem was solved in [2], whereas for the reproducing kernels of the Fock space the solution (in general, also negative) was recently given in [3].
In this article we prove that the answer to the Newman-Shapiro problem is in general negative and establish several positive results under different restrictions on the growth and regularity of the function F .
The original Newman-Shapiro problem is formulated for the Fock spaces on C n , n ≥ 1. Here, we restrict ourselves to the case n = 1. The negative answer to the Newman-Shapiro problem in the case n = 1 means the negative answer for every n ≥ 1. It seems plausible that one should use different techniques to obtain positive results in the case n > 1.
Theorem 1.2. For any α ∈ (1, 2), there exist two entire functions F and G such that G, GF ∈ F and for every entire function h of order at most α we have hF ∈ F , but GF / ∈ Span pF : p ∈ P = Span e λ F : λ ∈ C .
Thus, the equivalent conditions of Theorem 1.1 do not hold for F . Here and later on, P is the space of the polynomials.
Next we establish that under more restrictive growth and regularity conditions on the function F the answer to the Newman-Shapiro problem becomes positive.
Given α ≥ 0, denote by E 2,α the class of all entire functions of type at most α for order 2, that is Given F ∈ E 2 , consider its indicator function for order 2, We say that F ∈ E 2 is of completely regular growth if log |F (re iθ )|/r 2 converges uniformly in θ ∈ [0, 2π] to h F (θ) as r → ∞ and r ∈ E F for some set E F ⊂ [0, ∞) of zero relative measure, that is Suppose that there exist G ∈ E 2 of completely regular growth and α < 1 such that Span e λ F : λ ∈ C = EF ∩ F .
Thus, the equivalent conditions of Theorem 1.1 hold for such F . The conditions of the theorem mean that the zero set of F can be complemented by a set of positive angular density to a set Λ such that the system {k λ } λ∈Λ is complete and minimal in F (α) .
When the zero set of F is sufficiently regular and not very dense, we get the following result. Corollary 1.4. Let F ∈ F be of completely regular growth. Suppose that the upper Beurling-Landau density D + Z(F ) of the zero set Z(F ) of F (with multiplicities taken into account) is less than 1/π: Then Span e λ F : λ ∈ C = EF ∩ F .
Here and later on D(z, r) stands for the open disc centered at z of radius r.
Condition (1.2) is indispensable here as demonstrates the example given in the proof of Theorem 1.2.
When we restrict the growth of F , there are no more regularity restrictions on the zeros: Span e λ F : λ ∈ C = EF ∩ F . Thus, the situation here could be compared to that of the cyclicity/invertibility problem in the Bergman space. Invertibility does not imply cyclicity there [10]; if we impose additional growth restrictions, then invertibility does imply cyclicity. (Stronger lower estimates also imply cyclicity [8]). The main difference is that the Bergman space cyclic/invertible functions f are zero free and one works with harmonic log |f | while in our situation the Fock space functions have a lot of zeros which makes the problem much more complicated.
The Fock space does not possess a Riesz basis of reproducing kernels. Instead, we have the system K = {k w } w∈Z 0 which is complete and minimal in F . Here and later on Z = Z + iZ ⊂ C, Z 0 = Z \ {0}. Let σ be the Weierstrass sigma function associated to Z, σ 0 (z) = σ(z)/z. For more information about these functions see Section 2. The system {g w } w∈Z 0 , g w = σ 0 /(σ ′ 0 (w)(· − w)), is biorthogonal to K. One of our main technical tools to get the completeness results is the following for some thin set Ω. Furthermore, we study related continuous Cauchy transforms corresponding to pairs of Fock space functions, whose asymptotics gives their scalar product. In particular, given F 1 , F 2 ∈ F , we have for some thin set Ω. Finally, we establish and use a number of uniqueness results on the Fock space functions outside thin sets (thin lattice sets). The plan of the paper is as follows. In Section 2 we introduce some notations and prove three uniqueness results for functions in the Fock space. Section 3 contains several auxiliary results on interpolation formulas and the scalar product in the Fock space. Theorem 1.3 together with some auxiliary lemmas is proved in Section 4. Theorem 1.5 and Corollary 1.4 are proved in Section 5. Finally, Theorem 1.2 is proved in Section 6 using techniques that are quite different from those in the previous part of the paper.
Notations. Throughout this paper the notation U(x) V (x) means that there is a constant C such that U(x) ≤ CV (x) holds for all x in the set in question,

Notations and some uniqueness results for the Fock space
In this section after introducing some notations, we establish three uniqueness results for the Fock space functions.
Given α ∈ C, the time-frequency shift operator T α given by is unitary on the Fock space F . Put dν(z) = e −π|z| 2 dm 2 (z). Given F ∈ E we denote by Z(F ) its zero set. It is known [17, Theorem 5, Chapter 3] that if F, G ∈ E 2 , F is of completely regular growth, then Together with F we consider its subspace Following [3] we say that a measurable subset of C is thin if it is the union of a measurable set Ω 1 of zero (area) density, and a measurable set Ω 2 such that The union of two thin sets is thin. If Ω is thin, then its lower density is zero. In particular, C is not thin. If Ω is thin, then its translations z + Ω are thin, z ∈ C.
We start with the following Liouville type result. Although we do not use it directly in the paper, it helps us to better understand how sparse thin sets are with respect to the small value sets of the Fock space functions. The lemma was originally formulated in [3,Lemma 4.2]. A corrected proof is given in [4]. Here we give an alternative proof.
Lemma 2.1. Let F be an entire function of finite order, bounded on C \ Ω for some thin set Ω. Then F is a constant.
Proof. Suppose that F is not a constant and that for some N < ∞. We can find w ∈ C and c ∈ R such that the subharmonic function u, for some a > 0.
For some R ≥ 1 to be chosen later on we set By the maximum principle, ψ increases on [0, R].
We say that a subset A of the lattice Z = Z + iZ is lattice thin if for some (every) c > 0, the set is thin.
Let σ be the Weierstrass sigma function associated to Z, we have σ ∈ E 2,π/2 , h σ ≡ π/2 and Z 0 is a uniqueness set for F .
Then every ∆ z contains a finite family of intervals z + e iθ k,z J k z with disjoint J k z ⊂ R + of total length (1/2) log −2 (1 + |z|). Set Given z ∈ C and a > 0, set Now we are going to prove that, under condition (2.4), we have for some absolute constant γ > 0.
Given z ∈ C, set t = δ|z|, where m is one dimensional Lebesgue measure and c is a normalization constant such that µ is a probability measure. Then, under condition (2.4), we have c ≍ log 2 t/t for large t, and the logarithmic energy of µ is estimated below as follows: If δ is sufficiently small, 0 < δ ≤ (1 − γ) −1/2 − 1, then (2.6) and (2.7) imply together that ψ ≤ 0 and, hence, S is a constant. This completes the proof.
Proof. By the Lagrange interpolation formula, for every k ≥ 0, z ∈ C \ Z 0 we have and, hence, Therefore, Thus, It remains to use the Liouville theorem.

Interpolation formulas and duality in the Fock space
In this section we establish several results on relations between interpolation formulas, expansions with respect to some fixed complete and minimal systems of the reproducing kernels and their biorthogonal systems, and the scalar product in the Fock space.
Therefore, for every R > 0, we have The proof of the second inequality is analogous.
To prove the third inequality, we verify that Then .
The following result is contained in the proof of Lemma 4.3 of [3].
for some thin set Ω.
The system K = {k w } w∈Z 0 is a complete and minimal system in F , and the system {g w } w∈Z 0 , g w = σ 0 /(σ ′ 0 (w)(· − w)), is biorthogonal to K, see [6,3]. Lemmas 2.3 and 4.1 of [3] give us the following result: with the series converging absolutely in C \ Z 0 .
The following lemma establishes some relations between the orthogonality in the Fock space and the corresponding discrete Cauchy transform.
Here and later on, , then, by (3.5), we have

By (3.3) and (3.6), the set
Thus the total change of the argument of E/σ 0 along ∂D(w, r) is 0 (consider first the case P = 0, then the case P = 0), and, hence, E has one zero in D(w, r).
Finally, E has at least one zero in every disc D(w, γ) for w ∈ Z 0 outside a lattice thin set.

Proof of Theorem 1.3
We start this section with four lemmas dealing with the closed polynomial span [F ] F of F ∈ F . Then we pass to the proof of the theorem. Therefore, In the opposite direction, let H ∈ F be orthogonal to all e λ F , λ ∈ C. Set a n = C ζ n F (ζ)H(ζ) dν(ζ), n ≥ 0.
By (1.1), the series n≥0 a n z n /n! converges in the whole plane and equals to the zero function. Hence, a n = 0, n ≥ 0. By the Hahn-Banach theorem, we conclude that PF ⊂ Span e λ F : λ ∈ C .
Proof. If P ∈ P and H − P F < ε, Hence,    Proof. Without loss of generality, we can assume that F has infinitely many zeros. Shifting F and H, if necessary, by an operator T α , we can assume that Z(F H) ∩ Z 0 = ∅. Suppose that (4.2) does not hold and choose V ∈ E \ {0} such that F V ∈ F and
By (4.3), (4.5), (4.6), and the maximum principle, F HR belongs to P·F . Furthermore, F H(R − 1) ∈ P·F . Since the only entire multiples of F H in F are the constant ones, we conclude that R is a polynomial. By Lemma 3.3 (applied to F 1 = F 2 = F V ) and (4.4) we have By Lemma 3.5 (applied to F 1 = F 2 = F V ), we obtain that UV has at least one zero in every disc D(w, 1/10) for w ∈ Z 0 outside a lattice thin set. Since R is a polynomial, (4.4) and (4.5) show that UV ∈ P·F . Since U ∈ E 2,(π/2)−η , a subset of Z(V ) of positive lower density is contained in ∪ w∈Z 0 D(w, 1/10). Repeating the above argument for T 1/2 (F ) and T 1/2 (V ), we obtain U 1 ∈ E 2,(π/2)−η such that W = T 1/2 (V )U 1 has at least one zero in every disc D(w, 1/10) for w ∈ Z 0 outside a lattice thin set. Since W ∈ P · F , and a subset of Z(W ) of positive lower density is contained in C \ ∪ w∈Z 0 D(w, 1/10), we obtain a contradiction. Thus, relation (4.2) does hold.
We have V ∈ E 2,q for some q ≥ 1. Furthermore, F V ∈ F . Without loss of generality, we can assume that F V has simple zeros. Otherwise, we can shift a bit the zeros of F and G without changing our hypothesis and conclusions. By Lemmas 4.1 and 4.4, Next, let V s (z) = V (sz), 0 < s ≤ 1.
Let 0 < η < η 1 < β/q, 0 < t ≤ η. Let P n , n ≥ 0, be the n-th partial sum of the Taylor series of V t . Then Hence, Once again, without loss of generality, we can assume that F V s has simple zeros and By Lemma 4.2, Furthermore, set By (4.7) and by Lemma 3.4 (applied to Comparing the residues we see that S ∈ E. By Lemma 3.3 (applied to F 1 = F W , F 2 = F V s+δ ), the series defining C converges absolutely in C \ Z 0 , and As in the proof of Lemma 4.4, we obtain that for every ε > 0, By (4.8), Hence, By Lemma 4.3 applied with H = V s we have Therefore, S ∈ E 2,q(s+δ) 2 −qs 2 ⊂ E 2,2qδ . Hence, S = S (2qδ) −1/2 ∈ F . Next, by (4.11) and by Lemma 3.1, S is bounded on Z(V s+δ ) and, hence, S is bounded on a lattice of density at least η 2 (1 −α)/(2qδ) > 1. By Lemma 2.3, S = S is a constant. Thus, By (4.9) and (4.10), UV s+δ ∈ P · F . By Lemma 3.5 (applied to F W and F V s+δ ), we conclude that UV s+δ has at least one zero in every disc D(w, 1/10) for w ∈ Z 0 outside a lattice thin set. However, shifting F and V s+δ we obtain that a subset of Z(V s+δ ) of positive density belongs to C \ ∪ w∈Z 0 D(w, 1/10). We get a contradiction, which completes the proof.
Suppose that the claim of the theorem does not hold. Choose V ∈ E such that F V ∈ F and F V ⊥ [F ] and set U = σ 0 · I(F V, F ). By Lemma 3.4 (applied to F 1 = F V , F 2 = F , F 3 = H), we get U ∈ E 2,πε/2 . Next, arguing as in the proof of Lemma 4.4, we obtain that for some S ∈ E. Since V ∈ E 2,π/2+Kη , we have S ∈ E 2,πε . Replacing F and V by F α = T α (F ) and V α = T α (V ), correspondingly, we obtain If S α and S α 1 are polynomials for some α, α 1 ∈ C such that α − α 1 ∈ Z, then we choose γ > 0 such that D(α−α 1 , γ)∩Z = ∅. By Lemma 3.5 we obtain that both U α V α and U α 1 V α 1 have at least one zero in every disc D(w, γ) for w ∈ Z outside a lattice thin set. Then the lower density of the set {w ∈ Z : D(w + α, γ) ∩ Z(V ) = ∅} is at least 3/4. The same is true with α replaced by β. This contradicts the fact that the upper density of Z(V ) is at most 5/4. Thus, for some α 0 and for every α ∈ C \ (α 0 + Z), S α is not a polynomial. We can find α ∈ α 0 + Z and Z 1 ⊂ Z 0 of lower density at least 1 − ε such that V α has no zeros in ∪ z∈Z 1 D(z, log −2 (1 + |z|)). Correspondingly, for some Z 2 ⊂ Z of lower density at least 1 − 2ε we obtain that U α V α has no zeros in ∪ z∈Z 2 D(z, log −2 (1 + |z|)). By Lemma 3.1, (U α V α /σ 0 ) − S α has at most polynomial growth. By the Rouché theorem, for some C > 0, N < ∞ we obtain that inf ∂D(z,ρ log −2 (1+|z|)) |S α | < C|z| N .
Claim 6.2. Define by P 1 the family of the polynomials P such that (6.3) P F F ≤ 1.
Since the point evaluations are locally uniformly bounded in the Fock space, by the maximum principle we obtain that sup P ∈P 1 , |z|≤2 |P (z)| ≤ C 2 .
Step 4: End of the proof. Now, we argue as in [7]. Suppose that P n are polynomials such that Then for some C 1 > 0 we have {P n /C 1 } n≥1 ∈ P 1 , and by Claim 6.2 we get (6.6) |P n (x)| ≤ CC 1 exp(x γ ), n ≥ 1, x ≥ 0.
Since P n F tends to F G uniformly on compact subsets of the complex plane, (6.6) contradicts (6.4), and F G ∈ Clos F {PF }.