On the Gabor frame set for compactly supported continuous functions

Transcription

1 Christensen et al. Journal of Inequalities and Applications :9 DOI /s R E S E A R C H Open Access On the Gaor frame set for compactly supported continuous functions Ole Christensen 1,HongOhKim and Rae Young Kim 3* * Correspondence: rykim@ynu.ac.kr 3 Department of Mathematics, Yeungnam University, 80 Daehak-Ro, Gyeongsan, Gyeonguk 3851, Repulic of Korea Full list of author information is availale at the end of the article Astract We identify a class of continuous compactly supported functions for which the known part of the Gaor frame set can e extended. At least for functions with support on an interval of length two, the curve determining the set touches the known ostructions. Easy verifiale sufficient conditions for a function to elong to the class are derived, and it is shown that the B-splines B,, and certain continuous and truncated versions of several classical functions e.g., the Gaussian and the two-sided exponential function elong to the class. The sufficient conditions for the frame property guarantees the existence of a dual window with a prescried size of the support. MSC: C15; C0 Keywords: Gaor frames; frame set; B-splines 1 Introduction Frames is a functional analytic tool to otain representations of the elements in a Hilert space as a typically infinite superposition of uilding locks. Frames indeed lead to decompositions that are similar to those otained via orthonormal ases, ut with much greater flexiility, due to the fact that the definition is significantly less restrictive. For example, in contrast to the case for a asis, the elements in a frame are not necessarily linearly independent, that is, frames can e redundant. One of the main manifestations of frame theory is within Gaor analysis, where the aim is to otain efficient representations of signals in a way that reflects the time-frequency distriution. For any a, > 0, consider the translation operator T a and the modulation operator E, oth acting on the particular Hilert space L R, given y T a f x = f x a ande f x=e πix f x, respectively. Given g L R, the collection of functions {E m T na g} m,n Z is called a Gaor frame if there exist constants A, B >0suchthat A f m,n Z f, Em T na g B f, f L R. If at least the upper condition is satisfied, {E m T na g} m,n Z is called a Bessel sequence. It is known that for every frame {E m T na g} m,n Z,thereexistsadualframe{E m T na h} m,n Z such 016 Christensen et al. This article is distriuted under the terms of the Creative Commons Attriution.0 International License which permits unrestricted use, distriution, and reproduction in any medium, provided you give appropriate credit to the original authors and the source, provide a link to the Creative Commons license, and indicate if changes were made.

2 Christensenetal. Journal of Inequalities and Applications :9 Page of 17 that each f L R has the decomposition f = f, E m T na h E m T na h. 1.1 m,n Z The prolem of determining g L R and parameters a, >0suchthat{E m T na g} m,n Z is a framehas attracted a lot of attention over the past 5 years. The frame set for a function g L Risdefinedastheset F g := { a, R + {E mt na g} m,n Z is a frame for L R }. Clearly, the size of the set F g reflects the flexiility of the function g in regard of otaining expansions of type 1.1. In particular, it is known that a 1isnecessaryfor {E m T na g} m,n Z to e a frame and that the numer a 1 is a measure of the redundance of the frame; the smaller the numer, the more redundant the frame. Thus, a reasonale function g should lead to a frame {E m T na g} m,n Z for values a 1 that are reasonaly close to one. We remark that F g is known to e open if g elongs to the Feichtinger algera; see 1, ]. Until recently, the exact frame set was only known for very few functions: the Gaussian gx =e x 3 5], the hyperolic secant 6], and the functions hx =e x, kx = e x χ 0, x 7, 8]. In 9] a characterization was otained for the class of totally positive functions of finite type, and ased on 10], the frame set for functions χ 0,c], c >0,was characterized in 11]. For applications of Gaor frames, it is essential that the window g is a continuous function with compact support. Most of the related literature deals with special types of functions like truncated trigonometric functions or various types of splines; see 1 15].Various classes of functions have also een considered, for example, functions yielding a partition of unity 16, 17], functions with short support or a finite numer of sign-changes 18 0], or functions that are ounded away from zero on a specified part of the support 1]. The case of B-spline generated Gaor systems has attracted special attention; see, for example, 1, 0 3]. Interesting results in the rational case are otained in ]. To the est of our knowledge, the frame set has not een characterized for any function g C c R \{0}. We will, among others, consider a class of functions for which we canextendtheknownsetofparametersa, yielding a Gaor frame. The class of functions contains the B-splines B,, and certain continuous and compactly supported variants of the mentioned functions g, h and other classical functions. Furthermore, the results guarantee the existence of dual windows with a support size given in terms of the translation parameter. In the rest of this introduction, we will descrie the relevant class of windows and their frame properties. Proofs of the frame properties are in Section,andeasyverifialeconditions for a function to elong to the class are derived in Section 3. Let us first collect some of the known results concerning frame properties for continuous compactly supported functions; i is classical, and we refer to 5] for a proof. Proposition 1.1 Let >0,and assume that g : R C is a continuous function with supp g, ]. Then the following holds: i If {E m T na g} m,n Z is a frame, then a <1and a <.

3 Christensenetal. Journal of Inequalities and Applications :9 Page 3 of 17 ii 1] Assume that 0<a <, 0< +a, and inf x a, a ] gx >0. Then {E m T na g} m,n Z is a frame, and there is a unique dual window h L R such that supp h = a, a ]. iii 0] Assume that a < and 0< < 1 a. If gx>0, x ],, then {E m T na g} m,n Z is a frame. We will now introduce the window class that will e used in the current paper; it is a suset ofthe set of functions g considered in Proposition 1.1iii. The definition is inspired y certain explicit estimates for B-splines given y Treels and Steidl 6]; this point will e clear in Proposition 3.1.First,fix >0and0<a <. Consider the first-order difference a f and the second-order difference af given y a f x=f x f x a, af x=fx fx a+f x a. We define the window class as the set of functions { V,a := f CR supp f =, ] }, f is real-valued and satisfies A1-A3, where A1 f is symmetric around the origin; A f is strictly increasing on,0]; A3 If a < 3,then a f x 0, x, + 3a ];ifa 3,then a f x 0, x,0] { + 3a }. ote that y the symmetry condition A1 a function f V,a is completely determined y its ehavior for x,0].ifa 3, then the point + 3a considered in A3 is not contained in, 0]; however, if desired, the symmetry condition allows us to formulate the condition a f + 3a 0 alternatively as f 3a f a 0 1. ecause the argument x a of the last term in the second-order difference is less than. The definition of V,a is technical, ut we will derive easy verifiale conditions for a function g to elong to this set in Proposition 3.1 and also provide several natural examples of such functions. Our main result extends the range of > 0 yielding a frame, compared with Proposition 1.1ii: Theorem 1. For >0,let 0<a < and +a < +3a. Assume that g V,a. Then the Gaor system {E m T na g} m,n Z is a frame for L R, and there is a unique dual window h L R such that supp h 3a, 3a ]. For an illustration of Proposition 1.1 and Theorem 1.,seeFigure1. Memership of a function g in a set V,a for some a ]0, only gives information aout the frame properties of {E m T na g} m,n Z for this specific value of the translation parameter a. In order to get an impression of the frame properties of {E m T na g} m,n Z in a region in the a, -plane, we need to consider a function g that elongs to V,a for an in-

4 Christensenetal. Journal of Inequalities and Applications :9 Page of 17 Figure 1 The Gaor frame set. The figure shows the following regions: B as in Proposition 1.1iii, and D as in Theorem 1.. The region A corresponds to the case where the frame operator is a multiplication operator, and {E m T na g} m,n Z is a frame if inf x 0,a] n Z gx na > 0. The region C is a part of the region determined y Proposition 1.1ii, corresponding to the new findings in the paper 1]. terval of a-values, preferaly for all a ]0,. Fortunately, several natural functions have this property. The following list collects some of the results we will otain in Section 3. Considering any \{1}, TheB-splineB of order elongs to 0<a< V,a; Thefunctionf x:=cos πx χ, ]x elongs to 0<a< V,a; Thefunctionh x:=e x e χ, ]x elongs to 0<a< V,a; Thefunctiong x:=e x e χ, ]x elongs to 3 7 a< V,a. In particular, Proposition 1.1 and Theorem 1. imply that for \{1}, the functions B, f,andh generate frames whenever 0 < a < and 0 < +3a,andg generates a frame whenever 3 7 a < and 0 < +3a. ote that the limit curve = in Theorem 1. touches the known ostructions for +3a Gaor frames. In fact, for =,weotainthat asa 0. Since it is known that the B-spline B does not generate a frame for =1, 16], we cannot go eyond this. We also know that at least for some functions g 0<a< V,a, parts of the region determined y the inequalities <,a <,a < 1 do not elong to the frame set. Considering, for example, the B-spline B 1] shows that the point a, = 7, 7 does not elong to the frame set. For a = 7,Theorem1. guarantees the frame property for < 7, which is close 5 to the ostruction. These considerations indicate that the frame region in Theorem 1. in a quite accurate way descries the maximally possile frame set elow =thatisvalid for all the functions in V,a,atleastfor =. Frame properties for functions g V,a The purpose of this section is to prove Theorem 1.. Since the functions g V,a are ounded and have compact support, they generate Bessel sequences {E m T na g} m,n Z for all a, > 0. By the duality conditions 7, 8] two ounded functions g, h with compact support generate dual frames {E m T na g} m,n Z and {E m T na h} m,n Z for some fixed a, >0 if and only if gx l/ + mahx + ma=δ l,0, m Z a.e.x a, a ] ;

5 Christensenetal. Journal of Inequalities and Applications :9 Page 5 of 17 in particular, a function g V,a and a ounded real-valued function h with support on 3a, 3a ] generate dual Gaor frames {E mt na g} m,n Z and {E m T na h} m,n Z for L R for some if and only if, for l =0,±1, +3a 1 gx l/ + mahx + ma=δ l,0, m= 1 a.e.x a, a ]..1 Given g V,a, we therefore consider the 3 3 matrix-valued function G on a, a ]defined y Gx:= gx l + ma 1 l,m 1 = In terms of the Gx, condition.1 simply meansthat gx + 1 a gx + 1 gx a gx a gx gx + a. gx 1 a gx 1 gx 1 + a hx a 0 Gx hx =, hx + a 0 a.e.x a, a ].. We will show that the matrix Gx is invertile for all x a, a ]; this will ultimately give us a ounded and compactly supported function h satisfying.1 andhereyprove Theorem 1.. The invertiility of Gx will e derived as a consequence of a series of lemmas,wherewefirstconsiderx a, 0]. ote that the proof of the first result does not use property A3. Lemma.1 For >0,let 0<a < and +a < +3a. Assume that g V,a and let x a,0].then the following hold: a gx a gx + 1 <gx + 1 a 0; gx 1 a gx 1 <gx 1 + a 0; c gx>gx a and gx gx + a with equality only for x = a. Proof For a, let x a,0].using and a <,wehave +3a x + 1 a 1 3a +3a It follows that x + 1 a 1 3a, 1 a ] 3a >. ],..3 Using that +3a < 1 a, it follows that x + a a 1 <0;thus, x 1 + a < x Since gx >0forx ],, we have y.3 thatgx + 1 a 0. By A we know that g is strictly decreasing on 0, ]. If x + 1 a 0, then we have y.3,., and the

6 Christensenetal. Journal of Inequalities and Applications :9 Page 6 of 17 symmetry of g that g x + 1 a = g x 1 + a > g x + 1 g x a ; if x+ 1 a >0,thenwehavegx+ 1 a>gx+ 1 gx+ 1 +a. Hence, a holds. Similarly, and c hold. Wenowshowthatifg V,a and a /3, then condition A3 automatically holds on alargerinterval. Lemma. For >0,let 0<a <. Assume that g V,a. Then agx 0 for all x, + 3a ]. Proof It suffices to show that, for a 3, a gx 0, x 0, + 3a ]. We first note that A1 and A imply that, for x 0, + 3a ], gx g + 3a and gx a g a..5 For a 3,wehave 5a < ; due to the compact support of g, f 5a =0;thus using A1 again, we have a g + 3a = g + 3a g a = g 3a g a. Together with.5, thisshowsthat 0 a g + 3a a gx, x 0, + 3a ], as desired. Let G ij x denotetheijth minor of Gx, the determinant of the sumatrix otained y removing the ith row and the jth column from Gx. Lemma.3 For >0,let 0<a < and +a < +3a. Assume that g V,a and let x a,0].then the following hold: a G 1 x 0, and equality holds iff gx + 1 =gx a=0; G 3 x 0, and equality holds iff gx 1 a=gx 1 =0; c G x G 1 x+g 3 x.

7 Christensenetal. Journal of Inequalities and Applications :9 Page 7 of 17 Proof Since g 0, a and follow from Lemma.1a,. For c, we note that a gx= gx gx a =g x g x + a = a g x + a ythesymmetryofg. ow a direct calculation shows that G x G 1 x G 3 x = a g x + 1 a g x 1 + a + a g x a a g x 1 = a g x 1 + a a g x 1 + a a g x 1 a g x 1. Hence, it suffices to show that a g x 1 + a a g x 1, x a ],0 and a g x 1 + a a g x 1, x a ],0 or, equivalently, that a gx 1 + a 0and a g x 1 + a 0, oth for x a,0].since 1 + a, 1 + a] 1 + a, 1 + 3a ]= 1 + a, 1 + 3a ], this means precisely that a gx 0, x 1 + a, 1 + 3a ]..6 We note that, for +a < +3a, 1 + a, 1 + 3a + 3a. Together with Lemma.,thisimplies that.6holds,asdesired. After this preparation, we can now show that Gx is indeed invertile for x a, a ] under the assumptions in Theorem 1.. Corollary. For >0,let 0<a < and +a < +3a. Assume that g V,a. Then det Gx 0for x a, a ]. Proof First, consider x a, 0]. By Lemma.3c we have det Gx = gx ag 1 x+gxg x gx + ag 3 x gx ag 1 x+gx G 1 x+g 3 x gx + ag 3 x = gx gx a G 1 x+ gx gx + a G 3 x =: A x. Using Lemma.1c and Lemma.3a,, we have gx gx a G1 x 0 and gx gx + a G 3 x 0.

8 Christensenetal. Journal of Inequalities and Applications :9 Page 8 of 17 Thus, A x 0, x a,0].if A x > 0 for all x a, 0], then the proof is completed; thus, the rest of the proof is focused on the case where A x 0 =0forsomex 0 a,0]. In this case, Lemma.1c shows that either G 1 x 0 =0, G 3 x 0 =0.7 or G 1 x 0 =0, x 0 = a..8 Thecase.8 actually cannot occur. Indeed, if G 1 a = 0, then y Lemma.3a we have g a + 1 = 0; y the symmetry of g this would imply that g a 1 = 0, which contradicts Lemma.1 with x = a. Thus, we only have to deal with the case.7. If G 1x 0 = G 3 x 0 = 0, then y Lemma.3a, we have g x = g x a = g x 0 1 a = g x 0 1 =0. Inserting this information into the entries of the matrix Gx 0 and applying Lemma.1 yield that det Gx 0 =g x a gx 0 g x a >0, as desired. This completes the proof that Gx >0forx a,0]. Since g is symmetric around the origin, we have det G x=det g x gx 1 l,m 1 l + ma = det + l ma = det gx gx 1 l,m 1 l ma = det l + ma = det Gx. 1 l,m 1 1 l,m 1 Thus, Gxisalsoinvertileforx ]0, a ]. We are now ready to prove Theorem 1.. Proof of Theorem 1. By Corollary. and the continuity of g, inf x a, a ] det Gx >0. We define h on 3a, 3a ]y hx a 0 hx = G 1 x, hx + a 0 x a, a ], which is a ounded function. On R \ 3a, 3a ], put hx = 0. It follows immediately y definition of h that then g and h are dual windows. Remark.5 We note that this approach is tailored to the region of parameters a, in Theorem 1.. For example, it does not apply to the region considered in Proposition 1.1ii.

9 Christensenetal. Journal of Inequalities and Applications :9 Page 9 of 17 In fact, if 0 < +a, then the first row of Gx forx 1 + a, a ]isthezerovector, and the third row of Gx forx a, 1 a ]isthezerovector.hence,wehave inf x a, a ] det Gx=0. The conditions for g V,a are technical. We will now give an example, showing that the conclusion aout the size of the support of the dual window in Theorem 1. may reak down if g / V,a. Example.6 Let us consider the parameters =5,a = 1. Consider a symmetric and continuous function g on R with supp g = 5, 5 ]; assume further that g is increasing on,0]= 5,0]andthat g0 = 1, g 1 = 10, g =5, g 7 = Then a g 3 =g 3 g 7 3 +g 10 3 = 1<0.Hence,g does not satisfy A3 at x = 3, that is, g / V,a. We will show that, for = 3 +a = 3, there does not exist a ounded real-valued function 7 h L R withsupp h 3a, 3a ] such that the duality conditions.1 hold. In order to otain a contradiction, let us assume that such a dual window indeed exists. Let x 0 =0. Then x 0 a = x 0 a = 1, x a = x a = 3, x 0 1 = x 0 1 = 7 3, x 0 1 a = x 0 1 a = 10 3, and, consequently, gx a gx gx a Gx 0 = gx 0 a gx 0 gx 0 + a = gx 0 1 a gx 0 1 gx a By the continuity of g there exist continuous functions ɛ ij x, 1 i, j 3, such that 5+ɛ 11 x 3+ɛ 1 x ɛ 13 x Gx= 10 + ɛ 1 x 1+ɛ x 10+ɛ 3 x ɛ 31 x 3+ɛ 3 x 5+ɛ 33 x and ɛ ij x 0 asx x 0. Then.impliesthat 5+ɛ 11 x 3+ɛ 1 x ɛ 13 x hx a ɛ 1 x 1+ɛ x 10+ɛ 3 x hx = ɛ 31 x 3+ɛ 3 x 5+ɛ 33 x hx + a 0

10 Christensenetal.Journal of Inequalities and Applications :9 Page 10 of 17 for a.e. x a, a ]. By elementary row operations this leads to 5+ɛ 11 x 3+ɛ 1 x ɛ 13 x hx a 0 η 1 x η x η 3 x hx =, ɛ 31 x 3+ɛ 3 x 5+ɛ 33 x hx + a 0 a.e.x a, a ], where η i x:=ɛ i x ɛ 1i x ɛ 3i xfor1 i 3. Since h is a ounded function, ignoring a possile set of measure zero, this implies that = η 1 xhx a+η xhx+η 3 xhx + a 0 as x x 0.Thisisacontradiction. On the other hand, the condition g V,a is not necessary for {E m T na g} m,n Z to e a frame in the considered region. For example, let =1,a = 1, = and take g 1x := e x e 1 χ 1, 1 ]x. Then elementary calculations show that A3 does not hold for x 1 10, 1 16 ], + 3a ]. But since det Gx >0forx a, a ], we can prove that {E m T na g 1 } m,n Z is a frame y following the steps in the proof of Theorem ThesetV,a In this section, we give easily verifiale sufficient conditions for a function g to elong to V,a. Recall that a continuous function f : R C with supp f =, ]ispiecewise continuously differentiale if there exist finitely many x 0 = < x 1 < < x n = such that 1 f is continuously differentiale on ] x i 1, x i for every i {1,...,n}; the one-sided limits lim x x + i 1 f x and lim x x i f x exist for every i {1,...,n}. ote that if g is a continuous and piecewise continuously differentiale function, then the fundamental theorem of calculus yields that x a gx= g t dt for all x R, a > x a In order to avoid a tedious presentation, we will forego to mention the points where a piecewise continuously differentiale function is not differentiale, for example, in conditions c and d in the following Proposition 3.1. The result is inspired y explicit calculations for B-splines in Treels and Steidl 6], Lemma 1. Proposition 3.1 Let >0and assume that a continuous and piecewise continuously differentiale function g : R R with supp g =, ] satisfies the following conditions: a g is symmetric around the origin; g is strictly increasing on,0]; c g is increasing on ], ]; d g x g x for x,0. Then g 0<a< V,a. Proof ote that conditions a and are exactly the same as A1 and A. Thus, we will prove A3. In the entire argument, we will assume that g is differentiale; an elementary

11 Christensenetal.Journal of Inequalities and Applications :9 Page 11 of 17 consideration then extends the result to the case of piecewise differentiale functions. Let us first consider x. Then, y the mean value theorem, a gx=a gx gx a a gx a gx a = a g ξ g η a for some η x a, x a], ξ x a, x]. Since g is increasing up to x =,thisproves that a gx 0wheneverx. We now consider x, min + 3a,0].Then x a min a, a and a gx = g t g t a x dt + g t g t a dt = g x a x a g t g t a dt + x g a gx a+gx a g x gx a+g g t g t a dt + g a = g gx a g a ] gx a 3. + g gx a g x g a ]. 3.3 We now consider the terms 3.and3.3separately.For3., y the mean value theorem, g gx a g a ] gx a = g x + a gx a x + a g a gx a x + a = g x + a ξ g η for some η x a, a ], ξ x a, ]. Since a x a,wehaveη ξ ; thus, y assumption c, g η g ξ. Recalling that x + 3a < + a, wehave x + a > 0; thus, we conclude that the term in 3. indeed is nonnegative. For 3.3, we consider two cases. If x a x, then exactly the same argument as for 3.works.Ifx a < x, then we perform the same argument after a rearrangement of the terms. Indeed, g gx a g x g a ] = g g x gx a g a ]

12 Christensenetal.Journal of Inequalities and Applications :9 Page 1 of 17 = x + g g x x + g ξ g η = x + gx a g a x + for some η a, x a], ξ x, ]. As efore, this implies that 3.3 is nonnegative. We now consider x = + 3a ; according to 1., we must prove that g 3a g a First, if a > 3 i.e., + 3a >, then we have g 3a g a = a = a g 3a g a g a a g ξ g η 3 g + a for some ξ a, 3a ], η 3 + a, a ]; thus, 3. holds y assumption c. ow assume that < a 3 ;then + 3a,and a 3a 0, a < + a <0. Thus, g 3a g g a g a g a g a ; this can again e expressed in terms of the difference g ξ g η withξ a, ], η a, a ] and is hence positive. Finally,weassumethat 3 a.then + 3a ;since a < a,assumption implies that g 3a g a g 3a g a 3a a g t dt =:. Since a < 3a 0, d implies that 3a g t + a dt = g t dt =:. a 3 + 3a

13 Christensenetal.Journal of Inequalities and Applications :9 Page 13 of 17 Since + a and g is increasing on ], ], shifting the integration interval y + 3a 0totheleftyieldsthat a g t dt = g a ; thus, 3.holds,asdesired. Proposition 3.1 immediately leads to the following simple criterion for a function to elong to 0<a< V,a. Corollary 3. Let >0and assume that a continuous and piecewise continuously differentiale function g : R R with supp g =, ] satisfies the following conditions: a g is symmetric around the origin; g is positive and increasing on ],0. Then g 0<a< V,a. We will now descrie several functions elonging to V,a, either for all a from ]0, or from a suinterval hereof. Example 3.3 Consider the B-splines B,,definedrecursivelyy B 1 = χ 1/,1/], B +1 = B B 1. In 6], Lemma 1, it is proved that, for \{1}, i B is increasing on ], + 1 ]; ii B x B x for x,0. Thus, Proposition 3.1 implies that B 0<a< V,a for \{1}. Example 3. Let \{1} and define πx f x:=cos χ, ]x. Direct calculations show that, for x ], ], f π πx πx πx x= cos 3sin cos 0 and, for x,0, f x f x π = sin πx πx πx sin cos 0. By Proposition 3.1, f 0<a< V,a. In the following examples we consider continuous and compactly supported variants of the two-sided exponential function, the Gaussian, and other classical functions.

14 Christensenetal.Journal of Inequalities and Applications :9 Page 1 of 17 Example 3.5 Let >0anddefine h x:= e x e χ, ]x. Then h 0<a< V,a y Corollary 3.. Example 3.6 Let >0anddefine 1 k x:= 1+ x 1 1+ χ, ]x. Then k 0<a< V,a y Corollary 3.. In the following examples simple sufficient conditions in Proposition 3.1 and Corollary 3. are not satisfied. We will use the definition directly to show that the considered functions elong to V,a for certain ranges of the parameter a. Example 3.7 Let > 0 and consider 1 p x:= 1+x 1 1+ χ, ]x. We will show that a p 3 7 a< V,a; p 3 a< 3 V,a if 7 It is clear that A1 and A hold, so for the considered values of a,wenowchecka3.in fact, we will prove more, namely that a p x 0, x, + 3a ]. 3.5 Considering any a,, we have 3 < + 3a <, a < + 3a < a +, + 3a <a ; so the fact that p >0on],, p a>0on]a, a +, and p a>0on ]a,a + immediately shows that 1 p x 0, x ], + 3a ]; p x a=0, x, a ]; p x a 0, x ]a, + 3a ]; 3 p x a=0, x, + 3a ]. Thus, a p x= { p x, x, + a], p x p x a, x ] + a, + 3a ]. Since p x 0, x, p x p x a 0, + a], it now suffices to prove that x ] + a, + 3a ]. 3.6

15 Christensenetal.Journal of Inequalities and Applications :9 Page 15 of 17 Let x ] + a, + 3a ]. We see that p x p x a=p x p x a p x a 1 = 1+x 1 1+x a 1 1+x a 1 1+ = x a x 1 + x 1 + x a x a 1 + x a Inordertoprovea,wenowassumethat 3 7 0, we have a <.Sincex and x a x where p x p x a = hx:=x a x x a x x a x a 1 + x a 1 + hx x a, =x a a. ote that the quadratic function h is symmetric around x =a. Since + 3a 3 7 a <,wehave Thus, <a and hx h + 3a = 1 a7a p x p x a 0. Therefore, 3.5holds, that is, the proof of ais completed. 1 In order to prove, assume now that 5 and 3 a < 3 7.Sincex a x = a a xand x a = x + a + x a, 3.7impliesthat p x p x a= 1 a a x x + a + x a 1+x a 1+x 1+. Using + a < x + 3a, it follows that x + a < and + x a a x + a ; thus, p x p x a = 1 a a x a x 1+x a 1+x 1+ a x 1+x a qx 1 + x 1 +,

16 Christensenetal.Journal of Inequalities and Applications :9 Page 16 of 17 where qx :=a x. Since 3 a < 3 7 have 6 < x 1 ;sox < 36.Thus, and + a < x + 3a,we qx = Thus, p x p x a 0. Therefore, 3.5holds,asdesired Example 3.8 Let > 0 and consider g x:= e x e χ, ]x. We will show that g 3 7 a< V,a. As in Example 3.7 see 3.6, it suffices to check that g x g x a 0, x ] + a, + 3a ]. Let x ] + a, + 3a ], and a 3 7,. We see that g x g x a=g x g x a g x a = e x e x a e x a e = e x a e x a x 1 e e x a 1. Since x a and x a x 0, we have g x g x a e e x a x 1 e e x a 1 = e e x a x e x a = e x a e hx 1, where hx:=x a x.from3.8wehavehx 0. Thus, g x g x a 0, as desired. Competing interests The authors declare that they have no competing interests. Authors contriutions All authors contriuted equally to this work. All authors read and approved the final manuscript. Author details 1 Department of Applied Mathematics and Computer Science, Technical University of Denmark, Building 303, Lyngy, 800, Denmark. Division of General Studies, UIST, 50 UIST-gil, Ulsan, 919, Repulic of Korea. 3 Department of Mathematics, Yeungnam University, 80 Daehak-Ro, Gyeongsan, Gyeonguk 3851, Repulic of Korea.

17 Christensenetal.Journal of Inequalities and Applications :9 Page 17 of 17 Acknowledgements The authors would like to thank the reviewers for many useful suggestions, which clearly improved the presentation. In particular, one reviewer suggested to formulate condition A3 for memership of V,a in terms of second-order differences, which is much more transparent than our original condition. He also suggested the approach in the current proof of Proposition 3.1, which is shorter than our original one. This research was supported y Basic Science Research Program through the ational Research Foundation of Korea RF funded y the Ministry of Education 013R1A1AA Received: 5 ovemer 015 Accepted: 16 Feruary 016 References 1. Feichtinger, HG, Kailinger, : Varying the time-frequency lattice of Gaor frames. Trans. Am. Math. Soc. 356, Ascensi, G, Feichtinger, HG, Kailinger, : Dilation of the Weyl symol and the Balian-Low theorem. Trans. Am. Math. Soc. 366, Lyuarskii, Y: Frames in the Bargmann space of entire functions. Adv. Sov. Math. 11, Seip, K: Density theorems for sampling and interpolation in the Bargmann-Fock space I. J. Reine Angew. Math. 9, Seip, K, Wallsten, R: Density theorems for sampling and interpolation in the Bargmann-Fock space II. J. Reine Angew. Math. 9, Janssen, AJEM, Strohmer, T: Hyperolic secants yield Gaor frames. Appl. Comput. Harmon. Anal. 1, Janssen, AJEM: Some Weyl-Heisenerg frame ound calculations. Indag. Math. 7, Janssen, AJEM: On generating tight Gaor frames at critical density. J. Fourier Anal. Appl. 9, Gröchenig, K, Stöckler, J: Gaor frames and totally positive functions. Duke Math. J. 16, Janssen, AJEM: Zak transforms with few zeros and the tie. In: Feichtinger, HG, Strohmer, T eds. Advances in Gaor Analysis. Appl. umer. Harmon. Anal., pp Birkhäuser, Boston Dai, XR, Sun, Q: The ac-prolem for Gaor systems. Mem. Am. Math. Soc. to appear 1. Del Prete, V: Estimates, decay properties, and computation of the dual function for Gaor frames. J. Fourier Anal. Appl. 5, Laugesen, RS: Gaor dual spline windows. Appl. Comput. Harmon. Anal. 7, Kloos, T, Stöckler, J: Zak transforms and Gaor frames of totally positive functions and exponential B-splines. J. Approx. Theory 18, Kim, I: Gaor frames with trigonometric spline dual windows. Asian-Eur. J. Math. 8, Gröchenig, K, Janssen, AJEM, Kailinger,, Pfander, G: ote on B-splines, wavelet scaling functions, and Gaor frames. IEEE Trans. Inf. Theory 9, Christensen, O, Kim, HO, Kim, RY: On entire functions restricted to intervals, partition of unities, and dual Gaor frames. Appl. Comput. Harmon. Anal. 38, Christensen, O, Kim, HO, Kim, RY: Gaor windows supported on 1, 1] and compactly supported dual windows. Appl. Comput. Harmon. Anal. 8, Christensen, O, Kim, HO, Kim, RY: Gaor windows supported on 1, 1] and dual windows with small support. Adv. Comput. Math. 36, Christensen, O, Kim, HO, Kim, RY: On Gaor frames generated y sign-changing windows and B-splines. Appl. Comput. Harmon. Anal. 39, Lemvig, J, ielsen, H: Counterexamples to the B-spline conjecture for Gaor frames. J. Fourier Anal. Appl. to appear. Prete, VD: Estimates, decay properties, and computation of the dual function for Gaor frames. J. Fourier Anal. Appl. 5, Gröchenig, K: Partitions of unity and new ostructions for Gaor frames. Preprint. Lyuarskii, Y, es, PG: Gaor frames with rational density. Appl. Comput. Harmon. Anal. 3, Christensen, O: An Introduction to Frames and Riesz Bases, nd expanded edn. Birkhäuser, Boston Treels, B, Steidl, G: Riesz ounds of Wilson ases generated y B-splines. J. Fourier Anal. Appl. 6, Ron, A, Shen, Z: Weyl-Heisenerg systems and Riesz ases in L R d. Duke Math. J. 89, Janssen, AJEM: The duality condition for Weyl-Heisenerg frames. In: Feichtinger, HG, Strohmer, T eds. Gaor Analysis and Algorithms: Theory and Applications, pp Birkhäuser, Boston 1998