1 Introduction

Let \(g \in L^{2}({\mathbb{R}})\). The Gabor system generated by g with respect to the lattice \({\mathbb{Z}}\times{\mathbb{Z}}\) is the set

$$ G(g) := \bigl\{ \mathrm{e}^{2\pi \mathrm {i}n t} g(t - m)\bigr\}_{(m,n) \in{\mathbb{Z}}^2}. $$

Lattice Gabor systems play an important role in time-frequency analysis and its applications [6, 11, 16]. A central problem in this area is to describe the optimal time-frequency localization of generators of “good” Gabor systems, e.g., orthonormal bases and Riesz bases.

The classical Balian-Low theorem [1, 5, 13] is an uncertainty principle for generators of Gabor Riesz bases. It states that if G(g) is such a system, then

$$ \int_{\mathbb{R}}\bigl|g(t)\bigr|^2 t^2 \,\mathrm {d}t =\infty\quad \text{or}\quad \int_{\mathbb{R}}\bigl|\hat{g}(\xi)\bigr|^2 \xi^2\,\mathrm {d}\xi=\infty. $$
(1)

In the last 20 years, additional versions of this theorem were established. In particular we mention the (p,q) Balian-Low theorem and the Amalgam Balian-Low theorem (see discussion in Sect. 5). However, these versions do not include any quantitative estimate for the decay of the generator.

A quantitative version of the classical uncertainty principle states that there exist constants A,C>0 such that for \(g\in L^{2}({\mathbb{R}})\)

$$ \int_{|t| \geq R} \bigl|g(t)\bigr|^2 \,\mathrm {d}t + \int_{|\xi| \geq L} \bigl|\hat{g}(\xi)\bigr|^2 \,\mathrm {d}\xi\geq C \mathrm{e}^{-ARL}\|g\|^2_{L^2({\mathbb{R}})},\quad \forall R,L>0. $$
(2)

(Although not explicitly stated, this result follows from [8, Theorem 1], and is a special case of the uncertainty principle due to Nazarov [14]. See also the discussion in Sect. 6.) In light of the analogy between the classical uncertainty principle and the classical Balian-Low theorem, the following question was asked by K. Gröchenig: Is it possible to obtain a version of (2) for generators of Gabor Riesz bases?

Our main result is as follows.

Theorem 1

Let \(g \in L^{2}({\mathbb{R}})\). If G(g) is a Riesz basis, then for R,L≥1 we have

$$ \int_{|x| \geq R} \bigl|g(t)\bigr|^2 \,\mathrm {d}t + \int_{|\xi| \geq L} \bigl|\hat{g}(\xi)\bigr|^2 \,\mathrm {d}\xi\geq \frac{C}{RL}, $$
(3)

where C is a positive constant depending only on the Riesz basis bounds.

This result is sharp in the sense that, for functions with Riesz bounds A=B=1, the lower bound cannot be replaced by log(RL)/(RL).

Theorem 1 is strictly stronger than the classical Balian-Low theorem as well as its (p,q) version (see Sects. 5.1 and 5.2). The Amalgam Balian-Low theorem is independent of Theorem 1 (see Sect. 5.3). However, in Sect. 5.3, we find a scale of uncertainty estimates that interpolate Theorem 1 and the Amalgam Balian-Low theorem as (2,2) and (1,∞) endpoints.

The structure of the paper is as follows. In Sect. 2, we give some needed background. In Sect. 3 we prove Lemma 1, which is the main step in the proof of Theorem 1. We prove Theorem 1 in Sect. 4. Finally, as mentioned above, in Sect. 5, we discuss the connection between Theorem 1 and other Balian-Low type estimates.

2 Preliminaries

In this section, we recall the definitions of Riesz bases, the Zak transform, and quasi-periodic functions.

2.1 The Zak Transform

The following definition provides the main tool in the study of lattice Gabor systems (see, for example, [11, Chap. 8]).

Definition 1

Let \(g \in L^{2}({\mathbb{R}})\). The Zak transform of g is given by

$$ Zg(x,y) = \sum_{k \in{\mathbb{Z}}} g(x-k) \mathrm{e}^{2\pi \mathrm {i}ky} \quad (x,y) \in {\mathbb{R}}^2. $$

For \(g \in L^{2}({\mathbb{R}})\), the function Zg is quasi-periodic on \({\mathbb{R}}^{2}\). Namely, it satisfies

$$ Zg(x,y+1) = Zg(x,y)\quad\text{and}\quad Zg(x+1,y) = \mathrm{e}^{2\pi \mathrm {i}y} Zg(x,y). $$

In particular, this implies that Zg is uniquely determined by its values on [0,1]2. It is well-known that the Zak transform induces a unitary operator from \(L^{2}({\mathbb{R}})\) onto L 2([0,1]2), i.e.,

$$ \|Zg\|_{L^2({[0,1]^2})}^2 = \|g\|_{L^2({\mathbb{R}})}^2. $$
(4)

Similarly, by the triangle inequality, it satisfies the estimate

$$ \|Zg\|_{L^\infty({[0,1]^2})} \leq\sum _k \|g\|_{L^\infty(k,k+1)}. $$
(5)

We will use some basic properties of the Zak transform (see e.g. [11, Chap. 8]): The Zak transform satisfies

$$ Z\hat{g}(x,y) = \mathrm{e}^{2\pi \mathrm {i}xy} Zg(-y,x), $$
(6)

where

$$ \hat{g}(\xi) = \int_{\mathbb{R}}g(t) \mathrm{e}^{-2\pi \mathrm {i}\xi t} \,\mathrm {d}t $$

is the Fourier transform for functions in \(L^{1}({\mathbb{R}})\), which has the usual extension to functions in \(L^{2}({\mathbb{R}})\). Furthermore,

$$ Z(g \ast\phi) = Zg \ast_x \phi, $$
(7)

for any Schwarz class function ϕ, where the convolution on the right-hand side is taken with respect to the first variable. This last property implies, in particular, that if |Zg|≤M on \({\mathbb{R}}^{2}\) then for all h>0 we have

$$ \bigl|Z(g \ast\phi) (x+h,y) - Z (g \ast\phi) (x,y) \bigr| \leq Mh \int _{\mathbb{R}}\bigl|\phi'(\tau)\bigr| \,\mathrm {d}\tau. $$
(8)

Indeed, by (7), this follows from the corresponding well-known fact for \(F \in L^{\infty}({\mathbb{R}})\):

$$\begin{aligned} \bigl|(F \ast\phi) (x+h) - (F\ast\phi) (x)\bigr| &\leq\|F\|_{L^\infty ({\mathbb{R}})} \int_{\mathbb{R}}\bigl|\phi(x+h - \tau) - \phi (x-\tau ) \bigr| \,\mathrm {d}\tau \\ &= \|F\|_{L^\infty({\mathbb{R}})} \int_{\mathbb{R}} \biggl|\int_0^h \phi'(x + u - \tau) \,\mathrm {d}u \biggr| \,\mathrm {d}\tau \\ &\leq\|F\|_{L^\infty({\mathbb{R}})} \int_{\mathbb{R}}\int_0^h \bigl|\phi '(x+u-\tau)\bigr| \,\mathrm {d}\tau \,\mathrm {d}u \\ &= h \|F\|_{L^\infty({\mathbb{R}})} \int_{\mathbb{R}}\bigl|\phi '( \tau )\bigr| \,\mathrm {d}\tau. \end{aligned}$$

2.2 Riesz Bases

Let H be a separable Hilbert space. A system {f n } in H is called a Riesz basis if it is the image of an orthonormal basis under a bounded and invertible linear operator. This means that {f n } is a Riesz basis if it is complete in H and

$$ A \sum_{n} |a_n|^2 \leq \biggl\|\sum_n a_nf_n \biggr\|^2 \leq B \sum _n |a_n|^2, \quad\forall \{a_n\}\in\ell^2, $$
(9)

where A and B are positive constants. The largest A and smallest B for which (9) holds are called the Riesz basis bounds. (The Riesz basis bounds are equal to the so called “frame bounds”, see e.g. [17, p. 189]).

It is well-known that if a Gabor system

$$ G(g,a,b):= \bigl\{ \mathrm{e}^{2\pi \mathrm {i}b n t} g(t - ma)\bigr\}_{(m,n) \in{\mathbb{Z}}^2} $$

is a Riesz basis, then ab=1 [15]. Although the results in this paper are stated for the case a=b=1, they also hold for all systems G(g,a,b) with ab=1. This can be seen by appropriately dilating the generator function g.

The Zak transform allows one to characterize generators of lattice Gabor Riesz bases. Indeed, G(g) is such a system if and only if

$$ A \leq|Zg(x,y)|^2 \leq B \quad\forall(x,y) \in [0,1]^2, $$
(10)

where A and B are the Riesz basis bounds (see, e.g., [11, Corollary 8.3.2]). In this case, if the argument of Zg is discontinuous at a point, then so is Zg (when the values of the argument are considered modulo 1, see the beginning of Sect. 3). In Sect. 4, we make use of this fact.

3 Arguments of Quasi-Periodic Functions

It is known that an argument of a quasi-periodic function cannot be continuous (see, e.g., [11, Lemma 8.4.2] and the references therein). We now establish our main lemma, which is a stronger version of this statement.

First, we introduce some notation. Let G be a quasi-periodic function on \({\mathbb{R}}^{2}\) and H be a branch of its argument, i.e.,

$$ G(x,y) = \bigl|G(x,y)\bigr| \mathrm{e}^{2\pi \mathrm {i}H(x,y)}. $$

Fix two integers K,N≥8, and a point (x,y)∈[0,1/K)×[0,1/N). Set

$$ h_{i,j} = H \biggl(x + \frac{i}{K}, y + \frac{j}{N} \biggr) \quad \forall(i,j) \in{\mathbb{Z}}^2. $$

Note that for a given function G, the numbers h i,j are defined uniquely up to the addition of an integer. For \(h\in{\mathbb{R}}\) we denote by h (mod 1) its fractional part, i.e, the number \(0\leq\tilde{h}<1\) which satisfies that \(h-\tilde{h}\) is an integer. We say that

$$ |h| > \delta\quad(\mathrm{mod}\ 1) $$

if \(\inf_{n \in{\mathbb{Z}}} |h - n| > \delta\).

Lemma 1

There exist two integers 0≤i<K and 0≤j<N such that either

$$ |h_{i+1,j} - h_{i,j}| > \frac{1}{8} \quad(\mathrm{mod}\ 1) $$
(11)

or

$$ |h_{i,j+1} - h_{i,j}| > \frac{1}{8} \quad(\mathrm{mod}\ 1). $$
(12)

Proof

Assume that neither (11) nor (12) hold (observe that these conditions are independent of the choice of branch).

For any choice of the branch H, by the periodicity of G in the y-variable, there exist integers L i such that h i,N =h i,0+L i , 0≤i≤K. We will choose a specific branch of the argument H, for which the conditions imply that, on the one hand, L K =L 0 while, on the other hand, they also imply that L K =L 0+1. By this, we will arrive at a contradiction.

Choose the branch of the argument H as follows (we keep the notation h i,j for simplicity):

  • First, fix the number h 0,0 and choose \(\{h_{i,0}\} _{i=1}^{K}\) to satisfy |h i+1,0−h i,0|≤1/8 for 0≤i<K.

  • Next, for each fixed 0≤i<K, choose \(\{h_{i,j}\}_{j=1}^{N}\) so that |h i,j+1−h i,j |≤1/8 for 0≤j<N.

  • Finally, recall that the point (x,y) was fixed and use the quasi-periodicity of G to observe that h K,0=h 0,0+y+M for some integer M. Now, choose, for 0<j≤N,

    $$ h_{K,j}=h_{0,j}+\frac{j}{N}+y + M. $$
    (13)

    This implies that |h K,j+1−h K,j |≤1/4 for 0≤j<N.

All other values of the branch H may be chosen arbitrarily.

We claim that, with these choices, we also have

$$ |h_{i+1, N} - h_{i,N}| \leq\frac{1}{8}, \quad0 \leq i < K. $$
(14)

Indeed, to prove (14), first note that, since G is a quasi-periodic function, this inequality holds modulo 1. Now we may use a recursive argument. Indeed, for 0≤i<K−1, the above construction yields

$$ |h_{i+1,1} - h_{i,1} | \leq |h_{i+1,1} - h_{i+1,0} | + |h_{i+1,0} - h_{i,0} | + |h_{i,0} - h_{i,1} | \leq\frac{3}{8}, $$

while for i+1=K, we get

$$ |h_{K,1} - h_{K-1,1} | \leq |h_{K,1} - h_{K,0} | + |h_{K,0} - h_{K-1,0} | + |h_{K-1,0} - h_{K-1,1} | \leq\frac{1}{2}. $$

In both cases, since |h i+1,1−h i,1|≤1/8 modulo 1, it is clear that the inequality also holds without taking modulo 1. Repeating this argument N−1 times, one arrives at (14).

Recall that at the beginning of the proof we defined integers L i which satisfy h i,N =h i,0+L i . By (14), we have that |L i+1−L i |≤1/4. Hence, all L i are equal. In particular, L K =L 0.

On the other hand, condition (13) implies that the sequence {b j }:={h K,j −h 0,j } satisfies

$$ b_{j+1}-b_j=\frac{1}{N}, $$

whence L K −L 0=b N −b 0=1. This gives the required contradiction. □

4 Proof of Theorem 1

In this section, we obtain some consequences of Lemma 1 which we then use to prove our main result.

4.1 Bounded Quasi-Periodic Functions

A basic measure theoretic argument yields the following consequence of Lemma 1.

Lemma 2

Fix D>0. There exists a positive constant δ=δ(D) such that, given any two integers K,N≥8 and any quasi-periodic function G with |G|≥D almost everywhere, one can find a set S⊆[0,1]2 of measure at least 1/NK such that all (x,y)∈S satisfy either

$$ \biggl|G \biggl( x+{\frac{1}{K}}, y \biggr)-G ( x, y ) \biggr| \geq \delta, $$
(15)

or

$$ \biggl|G \biggl( x, y+\frac{1}{N}\biggr)- G ( x, y) \biggr| \geq \delta. $$
(16)

Proof

Given any two integers 0≤i<K and 0≤j<N, let

$$ \tilde{S}_{i,j}:= \biggl\{(x,y)\in \biggl[0,\frac{1}{K}\biggr) \times \biggl[0,\frac{1}{N}\biggr): (x,y) \text{ satisfies (11) or (12) for } (i,j) \biggr\}. $$

Lemma 1 implies that

$$ \bigcup_{i,j}\tilde{S}_{i,j}=\biggl[0, \frac{1}{K}\biggr) \times\biggl[0,\frac{1}{N}\biggr). $$

Set

$$ S_{i,j}=\tilde{S}_{i,j}+(i,j), $$

and note the equality of the measures \(|S_{i,j}|=|\tilde{S}_{i,j}|\). Since the sets S i,j are pairwise disjoint, it follows that

$$ \biggl|\bigcup_{i,j}{S}_{i,j}\biggr|=\sum _{i,j}|S_{i,j}|\geq\biggl|\bigcup _{i,j}\tilde{S}_{i,j}\biggr|= \frac{1}{KN}. $$

Hence, the set S:=⋃ i,j S i,j satisfies the required measure condition. We turn to the conditions (15) and (16). Let (x,y)∈S. This implies that there exists 0≤i<K and 0≤j<N such that (x−i/K,y−j/N) satisfies (11) or (12). Assume, without loss of generality, that it satisfies (11). Then,

$$ \biggl|H\biggl( x+{\frac{1}{K}}, y \biggr)-H( x, y )\biggr|>\frac{1}{8} \quad (\mathrm{mod}\ 1), $$

whence

$$ \biggl|G \biggl( x+{\frac{1}{K}}, y \biggr)-G ( x, y ) \biggr| \geq D\bigl|e^{2\pi iH( x+{\frac{1}{K}}, y )}-e^{2\pi iH( x, y )}\bigr|\geq\delta $$

for some constant δ>0 depending only on D. □

Next, we use Lemma 2 to evaluate the difference between the Zak transform of a function and the Zak transform of its convolution with some smooth kernel.

Lemma 3

Fix m,M>0. There exist positive constants δ=δ(m) and η=η(m,M) such that, given any pair of Schwarz class functions ϕ,ψ on \({\mathbb{R}}\) and any \(g\in L^{2}({\mathbb{R}})\) with m<|Zg|<M almost everywhere, one can find a set S⊆[0,1]2 of measure at least η/(1+∫|ϕ′|)(1+∫|ψ′|) such that all (x,y)∈S satisfy either

$$ \bigl|Zg(x,y) - Z(g \ast\phi) (x,y)\bigr|\geq\delta, $$
(17)

or

$$ \bigl|Z\hat{g}(x,y) - Z(\hat{g} \ast\psi) (x,y)\bigr| \geq\delta. $$
(18)

Proof

Let δ 1>0 be the constant from Lemma 2, and set G=Zg and D=m. Choose the smallest integers N,K≥8 that satisfy

$$ K\geq\frac{2M}{\delta_1} \int \bigl|\phi'\bigr| \quad\text{and}\quad N \geq\frac{4\pi M}{\delta _1} \biggl( \int \bigl|\psi'\bigr| +1 \biggr), $$
(19)

and let S 1 be the set described in Lemma 2. Note that the measure of the set S 1 now satisfies

$$ |S_1| \geq\frac{1}{KN} \geq\frac{1}{1 + ({2M}/{\delta_1})\int |\phi'|} \frac{1}{1 + ({4\pi M}/{\delta_1}) (\int|\psi'| +1)}. $$

It follows that the measure of S 1 is at least c/(1+∫|ϕ′|)(1+∫|ψ′|) for some positive constant c=c(m,M).

Observe that by (8) and (19), for all K>0, we have

$$ \biggl|Z(g \ast\phi) \biggl(x+\frac{1}{K},y\biggr) - Z (g \ast\phi) (x,y) \biggr| \leq \frac{ M}{K} \int_{\mathbb{R}}\bigl|\phi'(\tau)\bigr| \,\mathrm {d}\tau\leq \frac {\delta_1}{2}. $$
(20)

Hence, if (x,y)∈S 1 satisfies (15) with δ=δ 1, then, in light of (20) and the triangle inequality, we find that

$$\begin{aligned} &\bigl|Zg(x,y) - Z(g \ast\phi) (x,y)\bigr|+ \biggl|Zg\biggl(x+\frac{1}{K},y\biggr) - Z(g \ast\phi ) \biggl(x+\frac{1}{K},y\biggr)\biggr|\\ &\quad {}\geq \biggl|Zg (x,y) - Zg\biggl(x+\frac{1}{K},y\biggr)\biggr|- \biggl|Z(g \ast\phi) (x,y)- Z(g \ast\phi ) \biggl(x+\frac{1}{K},y\biggr)\biggr|\\ &\quad {}\geq \delta_1-\frac{\delta_1}{2}=\frac{\delta_1}{2}, \end{aligned}$$

which implies that either (x,y) or (x+1/K,y) satisfy (17) with δ=δ 1/4.

On the other hand, if (x,y)∈S 1 satisfies (16), then (6), (19) and the quasi-periodicity of Zg imply that

$$ \biggl|Z\hat{g}\biggl(1-y - \frac{1}{N}, x\biggr) - Z\hat{g}(1-y,x)\biggr| \geq\delta_1 - \frac{2\pi}{N}M \geq\frac{\delta_1}{2}. $$

Applying (8) in the same way as above, we find that either (1−y,x) or (1−y−1/N,x) satisfy (18) with δ=δ 1/8.

In conclusion, we found that for every (x,y)∈S 1 at least one of the four points (x,y), (x+1/K,y), (1−y,x) and (1−y−1/N,x) satisfies at least one of the inequalities (17) and (18). It follows that there exists a set S, (a translation of some subset of S 1), of measure at least |S 1|/4, with the required properties.

(It may happen that (x+1/K,y) or (1−y−1/N,x) fall outside of [0,1]2. However, as the functions discussed are quasi-periodic, the conclusion remains true if x+1/K and 1−y−1/N are taken modulo 1). □

4.2 Proof of Theorem 1

Proof

Let \(g \in L^{2}({\mathbb{R}})\) be such that G(g) is a Riesz basis with bounds A and B. By (10), we have m≤|Zg|≤M with \(m=\sqrt{A}\) and \(M=\sqrt{B}\).

Fix a Schwarz class function ρ such that \(\hat{\rho}\) is symmetric, satisfies \(|\hat{\rho}|\leq1\) and

$$ \hat{\rho}(\xi) = \begin{cases} 1 & \text{if } |\xi| \leq1, \\ 0 & \text{if } |\xi| \geq2. \end{cases} $$

Given R,L≥1, we apply Lemma 3 with ϕ(t)=Rρ(Rt) and ψ(t)=Lρ(Lt). We find that, for some constants C=C(A,B) and δ=δ(A,B), there exists a set S⊆[0,1]2 with measure at least C/RL such that all (x,y)∈S satisfy

$$ \delta^2 \leq\bigl|Z\hat{g} - Z(\hat{g} \ast\phi) \bigr|^2 + \bigl|Zg - Z(g \ast \psi)\bigr|^2. $$

From this and (6) it follows that

$$\begin{aligned} \frac{C\delta^2}{RL} &\leq\iint_{[0,1]^2} \bigl|Z\hat{g} - Z(\hat {g}\ast \phi)\bigr|^2 + \bigl|Zg - Z(g \ast\psi)\bigr|^2 \,\mathrm {d}x \,\mathrm {d}y \\ &= \iint_{[0,1]^2} \bigl|Z \bigl(g(1 - \hat{\phi}) \bigr)\bigr|^2 + \bigl|Z \bigl(\hat {g} (1 - \hat{\psi}) \bigr)\bigr|^2 \,\mathrm {d}x \,\mathrm {d}y. \end{aligned}$$

Since the Zak transform is a unitary operator from \(L^{2}({\mathbb{R}})\) to L 2([0,1]2), this gives

$$\begin{aligned} \frac{C_1}{RL} &\leq\int_{{\mathbb{R}}} \bigl|g(t)\bigr|^2 \bigl|1 - \hat{\phi}(t)\bigr|^2 \,\mathrm {d}t + \int _{{\mathbb{R}}} \bigl|\hat{g}(\xi)\bigr|^2 \bigl|1 - \hat{\psi }( \xi)\bigr|^2 \,\mathrm {d}\xi \\ &\leq\int_{|t|>R} \bigl|g(t)\bigr|^2 \,\mathrm {d}t + \int _{|\xi|>L} \bigl|\hat{g}(\xi)\bigr|^2 \,\mathrm {d}\xi, \end{aligned}$$

where C 1=Cδ 2 is a positive constant depending only on A and B.

To see that the result is sharp, we use the function g defined in [2, Definition 2.3]. On the one hand, this function satisfies |Zg|=1 for all \((x,y) \in{\mathbb{R}}^{2}\), and therefore the system G(g) is an orthonormal basis. On the other hand, by following the proofs of Theorems 3.4 and 3.10 in the same reference, it follows that

$$ \int_{|t|>R} \bigl|g(t)\bigr|^2 \,\mathrm {d}t + \int _{|\xi| >L} \bigl|\hat{g}(\xi)\bigr|^2 \,\mathrm {d}\xi\leq \frac{1}{R^2} + \frac{\log L}{L^2}. $$

This ends the proof. □

5 Theorem 1 in Comparison to Related Results

In this section, we discuss the connection between Theorem 1, the classical Balian-Low theorem and other results in this field. In particular, in Sect. 5.2, we study the connection to the (p,q) Balian-Low theorem, and, in Sect. 5.3, the connection to the Amalgam Balian-Low theorem.

5.1 The Classical Balian-Low Theorem

We begin by observing that Theorem 1 implies the Balian-Low theorem. Indeed, it is enough to notice that if \(g\in L^{2}({\mathbb{R}})\) satisfies (3), then for every R>0 we have

$$ \begin{aligned} C &\leq R^2\int _{|t|>R}\bigl|g(t)\bigr|^2\,\mathrm {d}t+R^2\int _{|\xi|>R}\bigl|\hat{g}(\xi )\bigr|^2\,\mathrm {d}\xi \\ & \leq\int_{|t|>R}|t|^2\bigl|g(t)\bigr|^2\,\mathrm {d}t+ \int_{|\xi|>R}|\xi|^2\bigl|\hat{g}(\xi)\bigr|^2\,\mathrm {d}\xi, \end{aligned} $$
(21)

whence at least one of the integrals in (1) diverges.

This proves the first half of the following proposition.

Proposition 1

If a function in \(L^{2}({\mathbb{R}})\) satisfies condition (3), then it also satisfies condition (1). Moreover, there exists a function \(g\in L^{2}({\mathbb{R}})\) that satisfies condition (1) but does not satisfy condition (3).

Proof

Throughout the proof, we let C denote different constants that may change from line to line. Let

$$ g(t)=\sum_{k=1}^{\infty}\frac{\sin^2(t-2^k)}{k^{1/2}2^k(t-2^k)^2}, $$

and note that its Fourier transform is compactly supported. Similar constructions are known in the literature, see, e.g., [3, Examples 3.3 and 3.4]. The function g satisfies

$$ \int_{\mathbb{R}}|t|^2\bigl| g(t)\bigr|^2\,\mathrm {d}t\geq \sum_{k=1}^{\infty}\frac{1}{k}\int _{2^k}^{2^k+1}\frac{\sin ^4(t-2^k)}{(t-2^k)^4}\,\mathrm {d}t=+\infty. $$

Given a positive integer n, set R n =[2n+2(n+1)]/2. To estimate

$$ I:=\int_{|t|>R_n}\bigl| g(t)\bigr|^2\,\mathrm {d}t, $$

we write \(g=\sum_{k=1}^{n}+\sum_{k=n+1}^{\infty}\) and correspondingly I<2I 1+2I 2. By the Cauchy-Schwartz inequality, we have

$$\begin{aligned} I_1 &=\int_{|t|>R_n} \Biggl|\sum _{k=1}^{n}\frac{\sin ^2(t-2^k)}{k^{1/2}2^k(t-2^k)^2} \Biggr|^2\,\mathrm {d}t \\ &\leq C\sum_{k=1}^n\int_{|t|>R_n} \frac{\,\mathrm {d}t}{(t-2^k)^4} \\ &\leq C\frac{n}{(R_n-2^n)^3} \leq C\frac{n}{2^{3n}}, \end{aligned}$$

which is smaller than \(C/R_{n}^{2}\) whenever n is large enough. We turn to estimating I 2. To this end, let w be the Fourier transform of (sint/t)2. By Parseval’s identity and the Cauchy-Schwarz inequality, we have

$$\begin{aligned} I_2 &\leq\int_{{\mathbb{R}}} \Biggl|\sum _{k=n+1}^{\infty}\frac {\sin ^2(t-2^k)}{k^{1/2}2^k(t-2^k)} \Biggr|^2dt\\ & = \int_{{\mathbb {R}}} \Biggl|\sum_{k=n+1}^{\infty} \frac{1}{k^{1/2}2^k}e^{2\pi \mathrm {i}2^k\xi}w(\xi ) \Biggr|^2\,\mathrm {d}\xi \\ & \leq \Biggl( \sum_{k=n+1}^{\infty} \frac{1}{k^{1/2}2^k} \Biggr)^2\int_{{\mathbb{R}}} \bigl|w(\xi)\bigr|^2\,\mathrm {d}\xi \leq C\frac{1}{n2^{2n}}, \end{aligned}$$

which is smaller than \(C/R_{n}^{2}\) whenever n is large enough. It follows that the function g satisfies condition (1), while condition (3) is not fulfilled. □

Remark 1

In fact, the function \(g + \hat{g}\) is an example of a function that does not satisfy condition (3), but for which both integrals in (1) diverge.

To better understand the relation between Theorem 1 and the classical Balian-Low theorem, we consider the symmetric case L=R. In this case, \(g \in L^{2}({\mathbb{R}})\) satisfies condition (3) of Theorem 1 if and only if

$$ \liminf_{R\rightarrow\infty}R^2 \biggl(\int _{|t|>R}\bigl|g(t)\bigr|^2\,\mathrm {d}t + \int_{|\xi|>R}\bigl| \hat{g}(\xi)\bigr|^2\,\mathrm {d}\xi \biggr)>C. $$
(22)

The classical Balian-Low theorem is related to a similar statement, with lim sup in place of lim inf. Indeed, the calculation (21) implies that if a function \(g \in L^{2}({\mathbb{R}})\) satisfies

$$ \limsup_{R\rightarrow\infty}R^2 \biggl(\int _{|t|>R}\bigl|g(t)\bigr|^2\,\mathrm {d}t + \int_{|\xi|>R}\bigl| \hat{g}(\xi)\bigr|^2\,\mathrm {d}\xi \biggr)>C, $$
(23)

then it also satisfies condition (1). Comparing conditions (22) and (23), the connection between Theorem 1 and the classical Balian-Low theorem becomes clearer. In fact, this was the main idea behind the construction of the function g in Proposition 1.

To conclude this discussion, we point out that condition (23) is only “slightly stronger” than condition (1). This follows from

Proposition 2

If a function \(g\in L^{2}({\mathbb{R}})\) satisfies ∫|x|2|g(x)|2 dx=∞ then, for every ϵ>0, we have

$$ \limsup_{R\rightarrow\infty}R^{2+\epsilon}\int _{|x|>R}\bigl|g(x)\bigr|^2\,\mathrm{d}x=\infty. $$
(24)

This claim is sharp in the sense that the ϵ in (24) can not be removed.

Proof

Assume that for some δ>0 we have

$$ \limsup_{R\rightarrow\infty} R^{2+\epsilon}\int_{|x|>R}\bigl|g(x)\bigr|^2\,\mathrm{d}x \leq\delta. $$

We use the formula of summation by parts given by

$$ \sum_{n=1}^N a_n (B_n-B_{n-1}) = a_N B_N - a_1 B_0 - \sum_{n=1}^{N-1} (a_{n+1}-a_n) B_n. $$

With the choices

$$ a_n=\int_{n<|x|<\infty}\bigl|g(x)\bigr|^2\,\mathrm{d}x \quad \text{and}\quad B_n=n^{2}, $$

this yields the identity

$$\begin{aligned} &\sum_{n=1}^N\bigl[n^{2}-(n-1)^{2} \bigr]\int_{n < |x| < \infty}\bigl|g(x)\bigr|^2 \,\mathrm {d}x \\ &\quad {}= N^2\int_{N<|x|<\infty}\bigl|g(x)\bigr|^2\,\mathrm {d}x - \sum _{n=1}^{N-1}n^{2}\int _{n<|x|<n+1}\bigl|g(x)\bigr|^2\,\mathrm {d}x. \end{aligned}$$

By our assumption, the first term on the right-hand side has limit equal to zero when N tends to infinity. Also, by assumption, the left-hand side satisfies

$$ \sum_{n=1}^N(2n-1)\int _{n<|x|<\infty} \bigl|g(x)\bigr|^2\,\mathrm{d}x\leq C\delta \sum _{n=1}^N\frac{1}{n^{1+\epsilon}}, $$

and therefore has also a finite limit. This implies that the second term on the right-hand side has a finite limit, whence ∫|x|2|g(x)|2 dx<∞.

To see that this claim is sharp put g(x)=1/(1+x 3/2log|x|). □

5.2 The (p,q) Balian-Low Theorem

Fix 1≤p≤2 and let q be such that 1/p+1/q=1. The (p,q) Balian-Low theorem states that if \(g\in L^{2}({\mathbb{R}})\) generates a Gabor Riesz basis on the integer lattice, then

$$ \int|t|^p\bigl|g(t)\bigr|^2\,\mathrm {d}t=\infty\quad \text{or}\quad\int|\xi |^q\bigl|\hat{g}(\xi)\bigr|^2\,\mathrm {d}\xi= \infty. $$
(25)

For p=1, this should be interpreted as

$$ \int|t|\bigl|g(t)\bigr|^2\,\mathrm {d}t=\infty\quad\text{or}\quad \text{$\hat{g}$ does not have compact support.} $$
(26)

For exponents p+ϵ and q+ϵ in place of p and q, the original proof follows by combining [7, Theorem 4.4] with [10, Theorem 1]. This was improved to (25) in [9] using methods from the theory of VMO functions.

The following proposition implies that the (p,q) Balian-Low theorem follows from Theorem 1.

Proposition 3

If a function in \(L^{2}({\mathbb{R}})\) satisfies condition (3), then it also satisfies condition (25). Moreover, there exists a function \(g\in L^{2}({\mathbb{R}})\) that satisfies condition (25), but does not satisfy condition (3).

Proof

Assume that \(g\in L^{2}({\mathbb{R}})\) satisfies condition (3). We first give the proof for 1<p≤2. Given R>0 and L=R p−1, (3) implies that

$$ C\leq\int_{|t|>R}|t|^{p}\bigl|g(t)\bigr|^2\,\mathrm {d}t+ \int_{|\xi|>R^{p-1}}|\xi |^{q}\bigl|\hat{g}(\xi)\bigr|^2 \,\mathrm {d}\xi, $$

which means that at least one of the integrals in (25) diverges. We turn to the case p=1. Assume that \(\hat{g}\) is supported on some interval [−L,L]. For R>0, (3) implies that

$$ C\leq RL\int_{|t|>R}\bigl|g(t)\bigr|^2\,\mathrm {d}t \leq L\int _{|t|>R}|t|\bigl|g(t)\bigr|^2\,\mathrm {d}t, $$

which means that the integral in (26) diverges. The second part of the statement follows from Remark 1. □

5.3 The Amalgam Balian-Low Theorem

The Amalgam Balian-Low Theorem [12, Corollary 7.5.3] states that if \(g\in L^{2}({\mathbb{R}})\) generates a Gabor Riesz Basis on the integer lattice, then

$$ \sum_n\bigl\|g(t)\bigr\|_{L^{\infty}[n,n+1]}= \infty\quad \text{or}\quad \sum_n\bigl\| \hat{g}(\xi)\bigr\|_{L^{\infty }[n,n+1]}=\infty. $$
(27)

This theorem is known to be independent of the classical Balian-Low theorem [3, Examples 3.3 and 3.4]. A straight-forward calculation shows that the same examples also yield that the Amalgam Balian-Low theorem is independent of Theorem 1:

Proposition 4

There exists a function \(g\in L^{2}({\mathbb{R}})\) that satisfies condition (3) but does not satisfy condition (27). On the other hand, there exists a function \(g\in L^{2}({\mathbb{R}})\) that satisfies condition (27), but does not satisfy condition (3).

We proceed to show that the Amalgam Balian-Low theorem and Theorem 1 are end-points of a more general Balian-Low type theorem.

Theorem 2

Let 1≤p≤2 and q be such that 1/p+1/q=1. If G(g) is a Riesz basis, then for R,L≥1 we have

$$ \frac{C}{(RL)^{p/q}}< \sum_{|k|>R} \|f \|_{L^q(k,k+1)}^p+\sum_{|k|>L} \| \hat{f}\|_{L^q(k,k+1)}^p , $$

where C is a positive constant depending only on the Riesz basis bounds.

Note that for p=q=2 we get Theorem 1, while for p=1 and q=∞ we get the Amalgam Balian-Low theorem.

Proof

Recall equations (4) and (5), which state that the Zak transform satisfies

$$ \|Zg\|_{L^2({[0,1]^2})}^2 = \|g\|_{L^2({\mathbb{R}})}^2= \sum _k \|g\|_{L^2(k,k+1)}^2, $$

and

$$ \|Zg\|_{L^\infty({[0,1]^2})} \leq\sum_k \|g \|_{L^\infty(k,k+1)}. $$

By the method of complex interpolation for vector valued sequence spaces (see, e.g., [4, Theorem 5.1.2]), this implies that

$$ \|Zg\|_{L^q({[0,1]^2})}^p \leq\sum _k \|g\|_{L^q(k,k+1)}^p. $$
(28)

We first prove the theorem for 1<p<2. Following the same technique as in the proof of Theorem 1, we use Lemma 3 to find that

$$\begin{aligned} \frac{C\delta^q}{RL} & \leq\iint_{[0,1]^2} \bigl|Z\hat{g} - Z(\hat {g} \ast \phi)\bigr|^q + \bigl|Zg - Z( g \ast\psi)\bigr|^q \,\mathrm {d}x \,\mathrm {d}y \\ &= \iint_{[0,1]^2} \bigl|Z \bigl( g(1- \hat{\phi}) \bigr)\bigr|^q + \bigl|Z \bigl( \hat{g}(1- \hat{\psi}) \bigr)\bigr|^q \,\mathrm {d}x \,\mathrm {d}y. \end{aligned}$$

By (28), this is smaller than

$$\begin{aligned} &\biggl(\sum_{k} \bigl\|g(1 - \hat{\phi}) \bigr\|_{L^q(k,k+1)}^p \biggr)^{q/p}+ \biggl(\sum _k \bigl\|\hat{g}(1- \hat{\psi })\bigr\| _{L^q(k,k+1)}^p \biggr)^{q/p} \\ &\quad {}\leq \biggl(\sum_{|k|>R} \|g \|_{L^q(k,k+1)}^p \biggr)^{q/p}+ \biggl(\sum_{|k|>L} \|\hat{g} \|_{L^q(k,k+1)}^p \biggr)^{q/p}. \end{aligned}$$

The desired inequality now follows from the fact that |x|β+|y|β≤C(|x|+|y|)β for all β≥0, with C>0 depending only on β.

The case p=1 is proved in much the same way. □

6 Open Problem

A much more general version of (2) was obtained by F. Nazarov [14]: There exist constants A,C>0 such that for \(g\in L^{2}({\mathbb{R}})\) and any two sets \(S,K\subset{\mathbb{R}}\) of finite measure the following inequality holds

$$ \int_{{\mathbb{R}}\setminus S} \bigl|g(x)\bigr|^2 \,\mathrm {d}x + \int _{{\mathbb{R}}\setminus K} \bigl|\hat{g}(\xi)\bigr|^2 \,\mathrm {d}\xi\geq C \mathrm{e}^{-A|S||K|}\|g\|^2_{L^2({\mathbb{R}})}. $$

Does this theorem have a version for functions g that generate Gabor Riesz bases on the integer lattice?