Extremal functions with vanishing condition
aa r X i v : . [ m a t h . C A ] A ug EXTREMAL FUNCTIONS WITH VANISHING CONDITION
FRIEDRICH LITTMANN AND MARK SPANIER
Abstract.
For a Hermite-Biehler function E of mean type τ we determinethe optimal (with respect to the de Branges measure of E ) majorant M + E andminorant M − E of exponential type τ for the truncation of x ( x + a ) − .We prove that Z R (cid:16) M + E ( x ) − M − E ( x ) (cid:17) | E ( x ) | − dx = 1 a K (0 , K is the reproducing kernel for the de Branges space H ( E ). As anapplication we determine the optimal majorant and minorant for the Heavisidefunction that vanish at a fixed point α = ia on the imaginary axis. Weshow that the difference of majorant and minorant has integral value ( πa − tanh( πa )) − πa . Results
Let δ ≥
0. We say that an entire function F is of exponential type δ ≥ ε > C ε > | F ( z ) | ≤ C ε e ( δ + ε ) | z | for all z ∈ C , and wewrite A ( δ ) for the class of entire functions of exponential type δ .Let µ be a Borel measure on R . We write A p ( δ, µ ) for the space of functions F ∈ A ( δ ) such that Z R | F ( x ) | p dµ ( x ) < ∞ , and we write A p ( δ ) if µ is Lebesgue measure.Let f : R → R be measurable and of polynomial growth, and let α ∈ C \ R .Consider the class F + = F + ( f, α ) ⊆ A ( δ ) of entire functions F satisfying(i) F ( x ) ≥ f ( x ) for all real x ,(ii) F ( α ) = 0,(iii) || F − f || L ( R ) < ∞ .Assume that F + = ∅ . We seek to find F + ∈ F + such that the inequality Z R ( F ( x ) − F + ( x )) dx ≥ F ∈ F + . If F + exists, we call it an extremal majorant of f (of type δ ) with vanishing at α . The class F − ( f, α ) is defined analogously by reversing theinequalities in (i) and (1.1); the corresponding optimal functions are called extremalminorants with vanishing at α . We note that condition (ii) implies F ( α ) = 0. Date : June 11, 2018.2000
Mathematics Subject Classification.
Primary 41A30, 41A52. Secondary 41A05, 41A44,42A82.
Key words and phrases.
Exponential type, bandlimited functions, best one-sided approxima-tion, de Branges space, Hermite-Biehler entire function.
This extremal problem was first considered in [14] for the function f ( x ) = χ [ β,γ ] ( x )with β < γ . M. Kelly found non-optimal functions in F + ( χ [ β,γ ] , α ) and obtainedbounds for || F ± − χ [ β,γ ] || as a function of δ . This choice of f is quite natural, sincethe extremal functions satisfying (i) and (1.1) have found frequent applications inanalytic number theory [10, 23, 26] and signal processing [9].The condition (ii) may allow applications to certain explicit formulas for L -functions. We briefly describe the setup. An L -function s L ( χ, s ) is given by L ( χ, s ) = ∞ X n =1 χ ( n ) n s where χ is a Dirichlet character. The generalized Riemann hypothesis states thatthe zeros of L in 0 < ℜ s < ℜ s = 1 /
2. An explicit formula isan identity that relates a series involving values L ( χ, ρ ) h ( ρ ) where L ( χ, ρ ) = 0 and h is a smooth function to values of the Fourier transform of h . Under assumptionof GRH and writing f ( t ) = L ( χ, + it ) the series involves only real values t , and h is frequently chosen to be a solution to the above extremal problem. Withoutassumption of GRH the series will involve non-real values of t , and an extremalfunction satisfying (ii) may be used to eliminate the contribution of such a zero.The results in this paper are an attempt to understand how the vanishing con-dition affects the construction of extremal functions. For this purpose we consider α = ia with a > x defined by x = ( x ≥ , F ± ∈ F ± ( x , ia ) of type 2 π . Theresults, given in Theorem 2 below, are obtained by changing measure and functionin such a way that condition (ii) may be dropped. This is possible since F + and F − are necessarily of the form F ± ( z ) = G ± ( z )( z + a )for some entire functions G ± ∈ A ( δ ). Hence, defining t a by t a ( x ) = ( ( x + a ) − if x ≥ , , we seek to find G + , G − ∈ A ( δ ) such that G − ( x ) ≤ t a ( x ) ≤ G + ( x ) for all real x ,and the integral Z ∞−∞ (cid:0) G + ( x ) − G − ( x ) (cid:1) ( x + a ) dx is minimal. This type of problem has been frequently investigated. An early resultis the solution by Beurling [2] and independently by Selberg [23, Chapter 20] forthe signum function, cf. the account in Vaaler [26]. Further references to recentdevelopments can be found in [8]. For the related subject of one-sided polynomialapproximation to the signum function see Bustamante et.al. [3], while the bestone-sided approximations to characteristic functions by trigonometric polynomialswere found recently by Babenko et.al. [1]. For the best approximation withoutconstraints we refer to Ganzburg [11]. For weighted polynomial approximation see XTREMAL FUNCTIONS WITH VANISHING CONDITION 3
Lubinsky [19, 20]. For general facts regarding best approximation we refer to thebooks of Korneichuk [16] and Pinkus [21].In Theorem 1 we solve the extremal problem for t a with respect to a large classof measures. In order to state this result we require some terminology from thetheory of de Branges Hilbert spaces of entire functions. We refer to [4] and [13] forproofs of the following facts.For any entire function E we use the notation E ∗ ( z ) = E ( z ). An entire function E satisfying the inequality | E ( z ) | > | E ∗ ( z ) | (1.2)for all z with ℑ z > Hermite-Biehler (HB) function. An analyticfunction F defined in the upper half plane C + is said to have bounded type, if F is the quotient of two bounded functions. The number v ( F ) defined by v ( F ) = lim sup y →∞ y − log | F ( iy ) | is called the mean type of F . If F has bounded type in C + , then its mean type isfinite.We denote by H ( E ) the vector space of entire functions F such that Z ∞−∞ | F ( x ) /E ( x ) | dx < ∞ and the functions F/E and F ∗ /E have bounded type and nonpositive mean typein C + . It is a Hilbert space with scalar product h F, G i E = Z ∞−∞ F ( x ) G ∗ ( x ) | E ( x ) | − dx. We define A = (1 / E + E ∗ ) and B = ( i/ E − E ∗ ). A fundamental result ofde Branges from the 1960’s is the recognition that this space is a reproducing kernelHilbert space; we briefly sketch the argument. It follows from (1.2) that E has nozeros in the open upper half plane, and it follows from the definition of H ( E ) that F/E and F ∗ /E have no zeros in an open set containing the closed upper half plane.The condition that F/E and F ∗ /E have non-positive mean type implies that theCauchy integral formula for the upper half plane holds for F/E and F ∗ /E (e.g.,[4, Theorem 12], note also that Cauchy formulas for F/E ∗ and F ∗ /E ∗ in the lowerhalf plane hold), and it follows with an elementary calculation that H ( E ) has thereproducing kernel K given by K ( w, z ) = B ( z ) A ( ¯ w ) − A ( z ) B ( ¯ w ) π ( z − ¯ w ) (1.3)for z = ¯ w .The prototypical example of a de Branges space is the classical Paley-Wienerspace of square integrable functions of finite exponential type at most τ . In thiscase E ( z ) = e − iτz , and the condition that F/E and F ∗ /E have non-positive meantype implies that F and F ∗ grow no faster than E , i.e., this condition is essentiallyequivalent to the statement that F has exponential type τ . The reproducing kernelfor this space is the familiar sinc function( z, w ) sin τ ( z − ¯ w ) π ( z − ¯ w ) . FRIEDRICH LITTMANN AND MARK SPANIER
The classical proof that this is a reproducing kernel uses Fourier analysis, butthis can also be shown using a contour integration argument. It is the latter proofthat generalizes to the setting of de Branges spaces.We require the following conditions for E :(I) E is of bounded type in C + with mean type τ ,(II) E has no real zeros,(III) E ∗ ( z ) = E ( − z ) for all z ,(IV) B / ∈ H ( E ).These are technical assumptions; we return to these properties in Section 3,where we obtain the following result. Theorem 1.
Let E be a HB-function satisfying conditions (I) – (IV) . If F + , F − ∈A (2 τ ) with F − ( x ) ≤ t a ( x ) ≤ F + ( x ) (1.4) for all real x , then Z R ( F + ( x ) − F − ( x )) dx | E ( x ) | ≥ a K (0 , , (1.5) and there exist functions T ± a ∈ A (2 τ ) satisfying (1.4) such that there is equality in (1.5) for F + = T + a and F − = T − a . We consider now the problem of finding the extremal functions with vanishing at α = ia with respect to Lebesgue measure. In order to apply Theorem 1 we requirea Hermite-Biehler function E a satisfying (I) - (IV) such that Z ∞−∞ (cid:12)(cid:12)(cid:12)(cid:12) F ( x ) E a ( x ) (cid:12)(cid:12)(cid:12)(cid:12) dx = Z ∞−∞ | F ( x ) | ( x + a ) dx for all F ∈ A ( π, ( x + a ) dx ). We shall prove in Theorem 13 that E a defined by E a ( z ) = (cid:18) πa ) (cid:19) sin π ( z + ia ) z + ia . (1.6)has this property. Theorem 1 becomes applicable and we obtain the following resultfor extremal functions with vanishing at α = ia . Theorem 2.
Let a > . If S, T ∈ A (2 π ) with S ( ia ) = T ( ia ) = 0 and S ( x ) ≤ x ≤ T ( x ) for all real x , then Z ∞−∞ ( S ( x ) − T ( x )) dx ≥ πaπa − tanh( πa ) , (1.7) and there exist S + a , S − a ∈ A (2 π ) such that there is equality in (1.7) for S = S − a and T = S + a . XTREMAL FUNCTIONS WITH VANISHING CONDITION 5
It is instructive to consider the corresponding integral for the extremal functionsof type 2 πδ . To find these values, we temporarily set a = bδ . Then the extremalfunctions F ± ∈ A (2 π ) with F ( ibδ ) = 0 satisfy Z ∞−∞ ( F + ( x ) − F − ( x )) dx = πbδπbδ − tanh( πbδ ) . We note that the functions F ± δ given by F ± δ ( z ) = F ± ( δz ) are the extremalfunctions of type 2 πδ with F ± δ ( ib ) = 0, and a change of variable gives Z ∞−∞ ( F + δ ( x ) − F − δ ( x )) dx = πbπbδ − tanh( πbδ ) . This implies that for fixed b and δ → ∞ the integral is ∼ δ − , while for δ →
0+ the integral is ∼ πb ) − δ − . In [26, Theorem 8] the corresponding extremalproblem for the signum function without the vanishing condition is treated. Theintegral value is shown to be equal to δ − . This shows that the prescribed vanishingat α = ib substantially affects the integral value for small values of δ , but thevanishing condition leads only to a small change if δ becomes large.This paper is structured as follows. In Section 2 we give a general method toconstruct entire functions M ± that interpolate t a at the zeros of a given Laguerre-P´olya entire function F , and we prove that x F ( x ) (cid:0) M ± ( x ) − t a ( x ) (cid:1) is of one sign for all real x . These functions serve as candidates for the extremalfunctions in the approximation results of the subsequent sections. We prove The-orem 1 in Section 3. In Section 4 we investigate the Hermite-Biehler function E a from (1.6) and prove Theorem 2. Acknowledgement.
The authors thank the anonymous referees for their valuablesuggestions. 2.
Interpolation at zeros of LP-functions
For certain discrete subsets
T ⊂ R we show how to construct entire functions F ∈ A (2 τ ) which interpolate t a at the elements of T (with prescribed multiplicity) so that F − t a has no sign changes between consecutive elements of T . This is thebasis for the construction of the extremal functions in Theorems 1 and 2.The interpolation theorems of this section are based on the representation az + a = Z ∞ e − zλ sin( aλ ) dλ, valid for all z with ℜ z >
0. The analogues of the crucial inequalities in Theorems 9and 10 below for the function x e − xλ χ [0 , ∞ ) ( x ) are known for every λ > aλ ) dλ is a signed measure introduces additional difficulties. Definition 3.
The Laguerre-P´olya class LP consists of all entire functions of theform F ( z ) = Ce − γz + bz z κ ∞ Y k =1 (cid:18) − za k (cid:19) e z/a k (2.1) FRIEDRICH LITTMANN AND MARK SPANIER where γ ≥ , κ ∈ N , b, C, a k ( k ∈ N ) are real, and ∞ X k =1 a k < ∞ . (2.2)We denote by T F the set of zeros of F . If the set of zeros is bounded above, weinclude ∞ ∈ T F , if the set is bounded below, we include −∞ ∈ T F .For c ∈ R \ T F we define g c ( t ) = 12 πi Z c + i ∞ c − i ∞ e st F ( s ) ds, (2.3)if this integral converges absolutely. (This is the case if γ > F has at least twozeros.) Let τ and τ be the consecutive elements from T F that satisfy τ < c < τ .A Fourier inversion shows that1 F ( z ) = Z ∞−∞ e − zt g c ( t ) dt (2.4)in the strip τ < ℜ z < τ . An application of the residue theorem shows that g c = g d for c, d ∈ ( τ , τ ).As part of a series of papers on total positivity, I.J. Schoenberg [24] gave an in-trinsic characterization of the functions g that may occur as Laplace inverse trans-formations of LP functions. We require estimates that can be found in [12]. Lemma 4.
Let F ∈ LP have at least n + 2 zeros counted with multiplicity, andlet τ , τ ∈ T F with ∈ ( τ , τ ) . Then g ( k )0 exists for k ≤ n and has at most k signchanges on the real line.Proof. This is [12, Chapter IV, Theorem 5.1]. (cid:3)
Lemma 5.
Let F ∈ LP have at least two zeros. Let τ < τ be two consecutiveelements in T F , and let c ∈ ( τ , τ ) . Then there exist polynomials P n and Q n suchthat | g ( n ) c ( t ) | ≤ | P n ( t ) | e τ t as t → ∞ , | g ( n ) c ( t ) | ≤ | Q n ( t ) | e τ t as t → −∞ . (2.5) Proof.
Let t > ℜ s = d with d < τ using the residue theorem, see, e.g., [12, Chapter V, Theorem2.1]. For t < ℜ s = d with d > τ . (cid:3) Lemma 6.
Let F be an even LP -function with a double zero at the origin and atleast one positive zero τ . Assume that F is positive in (0 , τ ) , and define g = g τ/ by (2.3) . (1) The derivatives g ′ and g ′′ exist and are nonnegative on the real line. (2) The function g ′′ is even. (3) If − τ < ℑ ζ < τ , then Z ∞−∞ g ′′ ( λ ) cos( ζλ ) dλ = − ζ F ( iζ ) . (2.6) In particular, F is real-valued on the imaginary axis. XTREMAL FUNCTIONS WITH VANISHING CONDITION 7 (4) If − τ < ℑ ζ < τ and w ∈ R , then Z ∞−∞ g ′′ ( λ + w ) sin( ζλ ) dλ = ζ sin( ζw ) F ( iζ ) . (2.7) Proof.
Since F has at least four zeros counted with multiplicity, (2.3) may bedifferentiated twice under the integral sign, which shows that g ′ and g ′′ exist. Twointegration by parts show that z j F ( z ) = Z ∞−∞ e − zt g ( j ) ( t ) dt ( j ∈ { , , } ) (2.8)for all z with 0 < ℜ z < τ . Since z − j F ( z ) is in LP for j ∈ { , , } , Lemma 4 with k = 0 implies that g ( j ) has no sign changes on the real line. Evaluation of (2.8)at z = τ / z − F ( z ) has no zero at the origin, (2.8) extends to − τ < ℜ z < τ for j = 2. Since z − F ( z ) is an even function of z , (2.3) with c = 0 gives g ′′ ( t ) = 12 π Z ∞−∞ ( iu ) e iut F ( iu ) du which implies that g ′′ is even.Let ζ such that − τ < ℑ ζ < τ . Since (2.8) extends to − τ < ℜ z < τ , we have Z ∞−∞ g ′′ ( λ ) cos( ζλ ) dλ = 12 (cid:18) ( iζ ) F ( iζ ) + ( − iζ ) F ( − iζ ) (cid:19) , and (2.6) follows since F is even. Since λ g ′′ ( λ ) sin ζλ is odd, the trigonometricidentity sin ζ ( λ − w ) = cos ζw sin ζλ − cos ζλ sin ζw implies Z ∞−∞ g ′′ ( λ ) sin ζ ( λ − w ) dλ = − sin ζw Z ∞−∞ g ′′ ( λ ) cos ζλdλ. An application of (2.6) gives (2.7). (cid:3)
Define for positive a the function h a : R → R by h a ( w ) = − Z −∞ g ( λ + w ) sin aλdλ, (2.9)and note that (2.5) with τ = τ implies that g ( t ) and g ′ ( t ) decay exponentially as t → −∞ . Hence h a ( w ) is finite for every w ∈ R , and an integration by parts givesfor all real w h a ( w ) = 1 a Z −∞ g ′ ( λ + w )(1 − cos aλ ) dλ. (2.10)We require bounds for the derivatives of h a and evaluations for special values. Lemma 7.
Let a > . Let F ∈ LP be even with a double zero at the origin, atleast one positive zero τ , and at least five zeros (counted with multiplicity). Assumethat F is positive in (0 , τ ) , and define g = g τ/ by (2.3) . (1) The inequalities ≤ h ( n ) a ( w ) ≤ a g ( n ) ( w ) (2.11) hold for n ∈ { , } and all real w , and for n = 2 and w ≤ . FRIEDRICH LITTMANN AND MARK SPANIER (2)
We have the representation h ′ a (0) = g ′ (0) a + a F ( ia ) . (2.12)(3) For all real w we have h ′′ a ( w ) − h ′′ a ( − w ) = − sin( aw ) a F ( ia ) . (2.13) Proof.
Equation (2.10) implies for all n and all real wh ( n ) a ( w ) = Z −∞ g ( n +1) ( λ + w ) 1 − cos( aλ ) a dλ. The functions g , g ′ , and g ′′ are nonnegative on R by Lemma 6, and g ′′′ hasexactly one change of sign on R by Lemma 4 applied to g ′′ . Since g ′′ is even, thesign change is located at the origin. It follows that for all real w and n ∈ { , } , aswell as for n = 2 and w ≤ ≤ h ( n ) a ( w ) ≤ Z −∞ g ( n +1) ( λ + w ) 2 a dλ = 2 a g ( n ) ( w ) , which implies inequality (2.11).To prove (2.12) we differentiate (2.9) and set w = 0 to get h ′ a (0) = − Z −∞ g ′ ( λ ) sin( aλ ) dλ. We perform an integration by parts, apply that g ′′ is even, and use (2.6) toobtain h ′ a (0) = g ′ (0) a − a Z −∞ g ′′ ( λ ) cos( aλ ) dλ = g ′ (0) a − a Z ∞−∞ g ′′ ( λ ) cos( aλ ) dλ = g ′ (0) a + a F ( ia )which finishes the proof of (2.12). Equations (2.9) and (2.7) give h ′′ a ( w ) = − Z ∞−∞ g ′′ ( λ + w ) sin( aλ ) dλ + Z ∞ g ′′ ( λ + w ) sin( aλ ) dλ = − sin( aw ) a F ( ia ) + h ′′ a ( − w ) , which is (2.13). (cid:3) The next proposition investigates two interpolations of t a in ℜ z ≥ ℜ z ≤ τ ,respectively, and shows that they are representations of a single entire function z A ( F, a, z ) which interpolates x t a ( x − ) at the zeros of F . See [7] for asimilar construction for the cutoff of an exponential function. XTREMAL FUNCTIONS WITH VANISHING CONDITION 9
Proposition 8.
Let a > . Let F ∈ LP be even with a double zero at the originand at least one positive zero τ . Assume that F is positive in (0 , τ ) , and define g = g τ/ by (2.3) and h a by (2.9) . Define A ( F, a, z ) = F ( z ) a Z −∞ h a ( w ) e − zw dw for ℜ z < τA ( F, a, z ) = 1 z + a − F ( z ) a Z ∞ h a ( w ) e − zw dw for ℜ z > . (2.14) Then z A ( F, a, z ) is analytic in ℜ z < τ , z A ( F, a, z ) is analytic in ℜ z > ,and these functions are restrictions of an entire function z A ( F, a, z ) . Moreover,there exists a constant c > so that | A ( F, a, z ) | ≤ c (1 + | F ( z ) | ) (2.15) for all z ∈ C .Proof. Inequality (2.5) with τ = τ implies that g ′ ( t ) decays exponentially as t →−∞ . Hence it follows from (2.10) that0 ≤ h a ( w ) ≤ a g ( w ) (2.16)for all real w , and (2.5) with τ = τ applied to g for t → −∞ implies that the integraldefining A ( F, a, z ) converges absolutely in ℜ z < τ . Inequality (2.5) implies with τ = 0 that g has polynomial growth on the positive real axis, hence the integralin the definition of A ( F, a, z ) converges absolutely for ℜ z >
0. It follows that A and A are analytic functions in ℜ z > τ and 0 < ℜ z , respectively.To prove that A and A are analytic continuations of each other it suffices toprove that they are equal in the strip 0 < ℜ z < τ . Starting point is the identity az + a = Z ∞ e − zλ sin aλdλ, (2.17)valid for ℜ z >
0. Combining this with (2.4) gives for 0 < ℜ z < τ z + a = F ( z ) a Z ∞−∞ Z ∞ e − z ( w + λ ) g ( w ) sin( aλ ) dλ dw = F ( z ) a Z ∞−∞ Z ∞ e − zw g ( w − λ ) sin( aλ ) dλ dw = F ( z ) a Z ∞−∞ h a ( w ) e − zw dw. Inserting this in (2.14) shows that A ( F, a, z ) = A ( F, a, z ) for 0 < ℜ z < τ . Toprove (2.15) we note that inequality (2.16) implies in ℜ z ≤ τ / | A ( F, a, z ) | ≤ | F ( z ) | a Z −∞ | h a ( w ) e − zw | dw ≤ | F ( z ) | a Z −∞ g ( w ) e − τw/ dw, (2.18)and an analogous calculation holds in ℜ z ≥ τ / (cid:3) Starting with the function A ( F, a, z ), we construct interpolations M ± of t a thatinterpolate t a at the zeros of F (with correct multiplicity) so that M ± − t a has nosign changes between two consecutive zeros of F . This is accomplished by selectingthe value at the origin appropriately. We assume that a >
0, and that F ∈ LP and τ > z M + ( F, a, z ) and z M − ( F, a, z ) by M − ( F, a, z ) = A ( F, a, z ) + h a (0) a F ( z ) z , (2.19) M + ( F, a, z ) = A ( F, a, z ) + h a (0) a F ( z ) z + 2 g ′ (0) a F ( z ) z (2.20)where A ( F, a, z ) is defined in (2.14). Evidently M + and M − are entire functions.Recall that T F ∩ R is the zero set of F . It is evident from the definitions that M ± ( F, a, ξ ) = t a ( ξ ) for all real ξ ∈ T F \{ } . Since F has a double zero at theorigin, we see that M − ( F, a,
0) = 0.Since g ′′ is nonnegative and integrable on R , (2.8) implies2 F ′′ (0) = Z ∞−∞ g ′′ ( w ) dw. As g ′′ is even and g ′ ( w ) decays exponentially as w → −∞ , we also have that R ∞−∞ g ′′ ( w ) dw = 2 g ′ (0). Hence, z − F ( z ) → / (2 g ′ (0)) as z → M + ( F, a,
0) =1 /a . This means that M ± ( F, a, ξ ) = t a ( ξ ± ) (2.21)for all real ξ ∈ T F , where t a ( ξ ± ) denotes the one sided limits at ξ . Theorem 9.
Let a > . Let F ∈ LP be even with a double zero at the origin andat least one positive zero τ . Assume that F is positive in (0 , τ ) , and define g = g τ/ by (2.3) and h a by (2.9) . If F ( ia ) < , then F ( x ) (cid:8) M + ( F, a, x ) − t a ( x ) (cid:9) ≥ for all real x .Proof. Consider first x <
0. An expansion of the second term in (2.20) in a Laplacetransform together with (2.14) gives M + ( F, a, x ) − t a ( x ) = F ( x ) a Z −∞ ( h a ( w ) − h a (0)) e − xw dw + F ( x ) x g ′ (0) a . (2.23)Two integration by parts and (2.12) lead to Z −∞ ( h a ( w ) − h a (0)) e − xw dw = 1 x Z −∞ h ′′ a ( w ) e − wx dx − h ′ a (0) x = 1 x Z −∞ h ′′ a ( w ) e − xw dw − g ′ (0) ax − a x F ( ia ) , and inserting this in (2.23) gives M + ( F, a, x ) − t a ( x ) = F ( x ) x (cid:18) g ′ (0) a − F ( ia ) + 1 a Z −∞ h ′′ a ( w ) e − xw dw (cid:19) . (2.24)By assumption − F ( ia ) >
0, and (2.11) implies h ′′ a ( w ) ≥
0. Since by assumption z − F ( z ) is a LP-function that is positive in (0 , τ ), it follows from Lemma 6 that g ′ (0) >
0. Hence (2.22) is shown for x < XTREMAL FUNCTIONS WITH VANISHING CONDITION 11
Let x >
0. From (2.14), (2.9), and (2.12) we get M + ( F, a, x ) − t a ( x ) = − F ( x ) a Z ∞ h a ( w ) e − xw dw + F ( x ) a h a (0) x + F ( x ) x g ′ (0) a = − F ( x ) a Z ∞ ( h a ( w ) − h a (0)) e − xw dw + F ( x ) x g ′ (0) a , and, analogously to (2.24), we obtain the representation M + ( F, a, x ) − t a ( x ) = F ( x ) x (cid:18) g ′ (0) a − F ( ia ) − a Z ∞ h ′′ a ( w ) e − xw dw (cid:19) (2.25)for x >
0. In order to investigate the sign of the right hand side, we multiply (2.13)by e − xw and integrate w over [0 , ∞ ) to get with (2.17) Z ∞ ( h ′′ a ( w ) − h ′′ a ( − w )) e − xw dw = − a F ( ia ) Z ∞ e − xw sin( aw ) dw = − a F ( ia ) ax + a . Hence − a Z ∞ h ′′ a ( w ) e − xw dw = a F ( ia ) 1 x + a − a Z ∞ h ′′ a ( − w ) e − xw dw. (2.26)Since h ′′ a ( − w ) ≥ w ≥
0, we have from (2.12) Z ∞ h ′′ a ( − w ) e − xw dw ≤ Z ∞ h ′′ a ( − w ) dw = g ′ (0) a + a F ( ia ) . (2.27)We multiply (2.27) by − a − and insert the resulting inequality into (2.26) to get − a Z ∞ h ′′ a ( w ) e − xw dw ≥ a F ( ia ) 1 x + a − g ′ (0) a − F ( ia ) . Inserting this into (2.25) gives M + ( F, a, x ) − t a ( x ) F ( x ) ≥ F ( ia ) 1 x (cid:18) x/a ) + 1 − (cid:19) , which is nonnegative since F ( ia ) <
0. This proves (2.22) for x > (cid:3) Theorem 10.
Let a > . Let F ∈ LP be even with a double zero at the origin andat least one positive zero τ . Assume that F is positive in (0 , τ ) , and define g = g τ/ by (2.3) and h a by (2.9) . Then F ( x ) (cid:8) M − ( F, a, x ) − t a ( x ) (cid:9) ≤ holds for all real x .Proof. We note that the integral representations for M + are valid even if F ( ia )is not negative. From the definition of M − and (2.23) we obtain for x < M − ( F, a, x ) − t a ( x ) = F ( x ) a Z −∞ ( h a ( w ) − h a (0)) e − xw dw, (2.29) and since h ′ a ( w ) ≥ w , it follows that h a ( w ) − h a (0) ≤ w ≤ x <
0. Analogously, for x > M − ( F, a, x ) − t a ( x ) = − F ( x ) a Z −∞ ( h a ( w ) − h a (0)) e − xw dw, (2.30)which gives (2.28) in this range. (cid:3) Proposition 11.
The functions M + and M − from Theorems 9 and 10 satisfy | M ± ( F, a, x ) − t a ( x ) | = O (cid:18) F ( x )1 + x (cid:19) (2.31) for all real x .Proof. Inequalities (2.23) and (2.25) yield (2.31) for M + , while (2.29) and (2.30)imply (2.31) for M − . (cid:3) Extremal functions for t a in de Branges spaces In this section we prove Theorem 1. Recall that E is an HB function that satisfies(I) E is of bounded type in C + with mean type τ , (II) E has no real zeros, (III) E ∗ ( z ) = E ( − z ) for all z , (IV) B / ∈ H ( E ). These conditions imply that E hascertain properties that we collect in the following lemma. We define the positiveBorel measure µ E by µ E ( A ) = Z A dx | E ( x ) | . (3.1) Lemma 12. If E is an HB function satisfying (I) - (IV), then (1) E has exponential type τ , (2) Every nonnegative F ∈ A (2 τ, µ E ) can be factored as F = U U ∗ with U ∈H ( E ) , (3) A = (1 / E + E ∗ ) is even and B = ( i/ E − E ∗ ) is odd, (4) For every U ∈ H ( E ) the identity Z ∞−∞ (cid:12)(cid:12)(cid:12)(cid:12) U ( x ) E ( x ) (cid:12)(cid:12)(cid:12)(cid:12) dx = X ξ ∈T B | U ( ξ ) | K ( ξ, ξ ) (3.2) is valid. (Recall that T B is the set of zeros of B .)Proof. Since E is of bounded type with mean type τ in C + , it follows from (1.2) that E ∗ is of bounded type with mean type ≤ τ in C + . It follows from Krein’s theorem[22, Theorems 6.17 and 6.18] that E has exponential type τ . The second propertyfollows from [17, Appendix V] together with the observation that F ∈ A (2 τ, µ E )implies Z ∞−∞ log + | F ( x ) | x dx < ∞ . This follows from Jensen’s inequality, see, e.g., the proof of [13, Lemma 12]. Thethird property is evident, and (3.2) is [4, Theorem 22]. (cid:3)
Proof of Theorem 1.
Assume that E satisfies (I) - (IV). Inequality (1.2) impliesthat A and B have only real zeros, and since E has no real zeros, it follows that A and B have no common zeros. By [4, Problem 48] there exists a continuousincreasing function ϕ such that E ( x ) exp( iϕ ( x )) is real valued for all real x . The XTREMAL FUNCTIONS WITH VANISHING CONDITION 13 zeros of A are the values x where ϕ ( x ) = π + kπ for some k ∈ Z , and the zerosof B are the values x where ϕ ( x ) = kπ for some k ∈ Z . It follows that the zerosinterlace. It follows from [4, Problem 47] applied to f = A/B that the zeros of B are simple (see also the proof of Theorem 22 in [4]). Similarly, the zeros of A aresimple.Since E has exponential type τ , it follows that A and B are also entire functionsof exponential type τ . Hence they, and their squares, are Laguerre-P´olya entirefunctions. Evidently, B is an even LP function that has a double zero at theorigin, and the results of Section 2 are applicable.We define the entire functions T + a and T − a by T + a ( z ) = M + ( B , a, z ) T − a ( z ) = M − ( B , a, z )with M − and M + as in (2.19) and (2.20).We note that E ∗ ( z ) = E ( − z ) implies in particular that B ( ix ) = − B ( ix ) for real x , i.e., ℜ B ( ix ) = 0 for real x . It follows then that (cid:0) B ( ix ) (cid:1) < x , hence F = B satisfies the assumptions of Theorem 9. Since B ≥ R , inequality (2.22) implies T + a ( x ) ≥ t a ( x )for all real x , and (2.21) implies that T + a ( ξ ) = t a ( ξ )for all ξ with B ( ξ ) = 0. Since B /E is bounded on R , it follows from (2.31) that Z ∞−∞ T + a ( x ) − t a ( x ) | E ( x ) | dx < ∞ . A similar argument gives the same statement for t a − T − a . Since T − a ≤ t a ≤ T + a we obtain that | T + a − T − a | is integrable with respect to µ E = | E ( x ) | − dx . It followsfrom Lemma 12 that there exists U ∈ H ( E ) such that T + a − T − a = U U ∗ . (3.3)We prove next the optimality of T + a . Let F be a function of type 2 τ with F ≥ t a on R . We may assume that Z ∞−∞ F ( x ) − t a ( x ) | E ( x ) | dx < ∞ , (since otherwise there is nothing to show). The inequality T − a ≤ t a ≤ F gives | F ( x ) − T − a ( x ) | ≤ ( F ( x ) − t a ( x )) + ( t a ( x ) − T − a ( x ))and hence F − T − a is an entire function of exponential type 2 τ that is integrablewith respect to µ E . Evidently, F − T − a ≥ F − t a ≥
0. Applying Lemma 12 againimplies that there exists V ∈ H ( E ) such that F − T − a = V V ∗ , (3.4)It follows from (3.3) and (3.4) that F − T + a = V V ∗ − U U ∗ . An application of Lemma 12 (4) to U and V together with T + a ( ξ ) = t a ( ξ ) for all ξ ∈ R with B ( ξ ) = 0 implies Z ∞−∞ F ( x ) − T + a ( x ) | E ( x ) | dx = X B ( ξ )=0 F ( ξ ) − t a ( ξ ) K ( ξ, ξ ) ≥ , hence T + a is extremal.An analogous calculation (which we omit) shows that T − a is an extremal mino-rant. It remains to prove that Z ∞−∞ ( T + a ( x ) − T − a ( x )) dx | E ( x ) | = 1 a K (0 , . It follows from (3.3) and Lemma 12 (4) that Z ∞−∞ ( T + a ( x ) − T − a ( x )) dx | E ( x ) | = X B ( ξ )=0 T + a ( ξ ) − T − a ( ξ ) K ( ξ, ξ ) . The only non-zero summand is the term for ξ = 0. Since T + a (0) − T − a (0) = t a (0) − t a (0 − ) = 1 /a , the proof is complete. (cid:3) De Branges space and optimal functions for the vanishingcondition
Define the Borel measure µ a by µ a ( B ) = Z B ( x + a ) dx. Let a > E a is given by E a ( z ) = s πa ) sin π ( z + ia )( z + ia ) . We prove that E a is a Hermite-Biehler function whose associated de Brangesspace is isometrically equal to A ( π, µ a ). Theorem 13.
Let a > . The function E a satisfies (1.2) and properties (I) - (IV).Moreover, the space A ( π, µ a ) is isometrically equal to H ( E a ) .Proof. Since z sin πz is LP and hence of P´olya class, we have that z sin( π ( z + ia )) is also of P´olya class. By [4, Section 7, Lemma 1] it follows that E a is of P´olyaclass. This implies | E a ( z ) | ≥ | E ∗ a ( z ) | for all z with ℑ z >
0. Since E a has no zeros in the upper half plane, the function E ∗ a /E a is analytic in the upper half plane and has modulus bounded by 1. Sincethis quotient is not constant, the modulus is never equal to 1 by the maximumprinciple, hence E a satisfies (1.2).It can be checked directly that E a has bounded type π (or apply the reversedirection of Krein’s theorem). Evidently E a has no real zeros and E ∗ a ( z ) = E a ( − z ) XTREMAL FUNCTIONS WITH VANISHING CONDITION 15 for all z . A direct calculation gives A a ( z ) = s πa ) z cosh( πa ) sin( πz ) + a sinh( πa ) cos( πz ) z + a ,B a ( z ) = s πa ) a cosh( πa ) sin( πz ) − z sinh( πa ) cos( πz ) z + a , and in particular B a / ∈ H ( E a ). Hence E a satisfies (I) - (IV).Taking limits in (1.3) leads to the representation K a ( x, x ) = π ( a + x ) − a coth(2 πa ) + a cos(2 πx )csch(2 πa ) π ( a + x ) (4.1)for all real x . Recall that µ E a ( A ) = R A | E a ( x ) | − dx . It is straightforward to checkthat L ( R , µ a ) and L ( R , µ E a ) are equal as sets with equivalent norms. It followsthat A ( π, µ a ) and H ( E a ) are equal as sets and have equivalent norms. The mainstatement to prove is the fact that the two norms are equal on the smaller spaces.We note first that ( z + ia )( z − ia )sin( π ( z + ia )) sin( π ( z − ia )) = 2( z + a )cosh(2 πa ) − cos(2 πz ) (4.2)holds, in particular, the right hand side is 1-periodic after division by z + a .Furthermore, Z πa ) − cos(2 πx ) dx = 1sinh(2 πa ) . This means that p a defined by p a ( x ) = sinh(2 πa )(cosh(2 πa ) − cos(2 πx )) − − a >
0, this function is infinitelydifferentiable on the real line. It follows that the Fourier series of p a convergesabsolutely and uniformly, and that it represents the function, i.e., there exists asequence a n so that p a ( x ) = X n =0 a n e πinx (4.3)for all real x .Let H ∈ L ( R ) be an entire function of exponential type 2 π . Since H ∈ L ( R )by [6, Theorem 6.7.1], the Paley-Wiener theorem [25, Theorem 4.1] implies thatthe Fourier transform b H defined by b H ( t ) = Z ∞−∞ e − πixt H ( x ) dx satisfies b H ( t ) = 0 for | t | >
1. Since H ∈ L ( R ) it follows that b H is continuous,hence b H ( t ) = 0 for | t | ≥
1. This implies Z ∞−∞ H ( x ) X | n |≤ Nn =0 a n e πinx dx = X | n |≤ Nn =0 a n b H ( − n ) = 0 . Since the partial sums of the series in (4.3) converge uniformly, we obtain withan application of Lebesgue dominated convergence that Z ∞−∞ H ( x ) (cid:18) sinh(2 πa )cosh(2 πa ) − cos(2 πx ) − (cid:19) dx = 0 . (4.4)Let F, G ∈ A ( π, µ a ) and define H by H ( z ) = F ( z ) G ∗ ( z )( z + a ). It followsfrom (4.2) that h F, G i H ( E a ) − h F, G i L ( R ,µ a ) = Z ∞−∞ F ( x ) G ∗ ( x ) {| E a ( x ) | − − ( x + a ) } dx = Z ∞−∞ H ( x ) (cid:18) sinh(2 πa )cosh(2 πa ) − cos(2 πx ) − (cid:19) dx, and since H is a Lebesgue integrable entire function of exponential type 2 π , itfollows from (4.4) that h F, G i H ( E a ) = h F, G i L ( R ,µ a ) as claimed. (cid:3) Sketch of an alternate proof.
Define for a > W a by W a ( z ) = − e − πa a + iza − iz and note that W a is analytic and has modulus ≤ E a ( z ) + E ∗ a ( z ) W a ( z ) E a ( z ) − E ∗ a ( z ) W a ( z ) = coth(2 πa ) − e πiz csch(2 πa )is valid for z ∈ C , and for real x and y > yπ Z ∞−∞ ( t + a ) | E a ( t ) | dt ( x − t ) + y = coth(2 πa ) − e − πy cos(2 πx )csch(2 πa )holds. Theorem V.A of [5] with dµ ( t ) = ( t + a ) | E a ( t ) | dt implies Z ∞−∞ | f ( x ) | ( x + a ) dx = Z ∞−∞ (cid:12)(cid:12)(cid:12)(cid:12) f ( x ) E a ( x ) (cid:12)(cid:12)(cid:12)(cid:12) dx for every f ∈ H ( E a ). (cid:3) Proof of Theorem 2.
We define S + a and S − a by S + a ( z ) = M + ( B a , a, z )( z + a ) ,S − a ( z ) = M − ( B a , a, z )( z + a ) . Since B a is odd, we have B a (0) = 0. Since E a is Hermite-Biehler and is ofbounded type in the upper half plane, it follows that B a is of bounded type in theupper half plane. Since B a has only real zeros, [4, Problem 34] implies that B a isof P´olya class, and [4, Theorem 7] implies that B a ∈ LP . Hence F = B a satisfiesthe assumptions of Proposition 8. Furthermore, it can be checked directly that B a ( ia ) <
0. For the remainder of the proof we set M ± ( z ) = M ± ( B a , a, z ). Since B a ∈ A (2 π ) we obtain that M + , M − ∈ A (2 π ). Theorems 9 and 10 imply M − ( x ) ≤ t a ( x ) ≤ M + ( x ) XTREMAL FUNCTIONS WITH VANISHING CONDITION 17 for all real x . The proof of Theorem 1 shows that M + = T + a and M − = T − a withrespect to the measure dµ E a . It follows from (1.5) that Z ∞−∞ ( M + ( x ) − M − ( x )) dx | E a ( x ) | = 1 a K a (0 ,
0) = πaπa − tanh( πa ) . (4.5)By definition of S ± a we have Z ∞−∞ ( S + a ( z ) − S − a ( x )) dx = Z ∞−∞ ( M + ( x ) − M − ( x ))( x + a ) dx. (4.6)Since M + − M − ∈ A (2 π, µ a ), it follows that M + − M − ∈ A (2 π, µ E a ). Lemma12 (2) implies that M + − M − = U U ∗ with U ∈ H ( E a ). Theorem 13, (4.5), and(4.6) imply Z ∞−∞ ( S + a ( x ) − S − a ( x )) dx = πaπa − tanh( πa ) , which gives the case of equality in (1.7).Let now S, T ∈ A (2 π ) such that S ( ia ) = T ( ia ) = 0 and S ( x ) ≤ t a ( x ) ≤ T ( x ) onthe real line. We may assume that S − M − and T − M + are integrable with respectto ( x + a ) dx . Since S and T are real entire, it follows that S ( − ia ) = T ( − ia ) = 0,hence S ( z ) = ( z + a ) σ ( z )and T ( z ) = ( z + a ) τ ( z )where σ, τ are entire and have exponential type 2 π . Furthermore, σ − t a and τ − t a are integrable and σ ( x ) ≤ t a ( x ) ≤ τ ( x )for all real x . It follows from Theorem 1 that Z ∞−∞ ( σ ( x ) − τ ( x ))( x + a ) dx ≥ a K a (0 , , which is (1.7). (cid:3) References [1] A.G. Babenko, Yu.V. Kryakin, V.A. Yudin,
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