Computational Methods and Function Theory Volume 2 (2002), No. 2, 319–336
Local Behavior of Entire Functions of Exponential Type Dimiter P. Dryanov, Mohammed A. Qazi, and Qazi I. Rahman (Communicated by Lawrence Zalcman) Abstract. We consider entire functions of exponential type τ > 0, whose modulus is bounded by a constant M at the extrema of sin(τ z), and which vanish at the origin. Extending a result of L. H¨ormander, we show that if f is any such function, then |f (x)| ≤ M | sin(τ x)| for all x ∈ (−π/(2τ ), π/(2τ )), provided that f (x) = o(x) as x → ±∞; furthermore, equality holds at any point x ∈ (−π/(2τ ), 0) ∪ (0, π/(2τ )) if and only if f (z) ≡ eiγ sin(τ z) for some γ ∈ R. This also generalizes a result due to R. P. Boas Jr. about trigonometric polynomials. Besides, we prove some other results for entire functions of order 1 and type τ > 0, one being an analogue of a result of M. Riesz about trigonometric polynomials whose zeros are all real and simple. Keywords. Entire functions, exponential type, local behaviour. 2000 MSC. 30A10, 30C15, 30D10, 30D15.
1. Introduction As in [2], we say that an entire function f is of exponential type τ if for every ε > 0 there exists a constant k(ε) such that |f (z)| < k(ε)e(τ +ε)|z| for all z ∈ C. Thus, any entire function of order ρ < 1 is of exponential type τ and so is any entire function of order 1 and type at most τ . The following result is due to L. H¨ormander (see [10, p. 26], the paragraph that precedes the statement of Theorem 4). Theorem A. Let f be a real entire function of exponential type τ > 0 such that |f (x)| ≤ M for all x ∈ R. Furthermore, let f (0) = M (and implicitly f 0 (0) = 0). Then π π f (x) ≥ M cos(τ x), − ≤x≤ . τ τ The estimate is best possible as the example f (z) := M cos(τ z) shows. Received March 5, 2003. c 2002 Heldermann Verlag ISSN 1617-9447/$ 2.50
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Here, it may be added that an entire function of exponential type 0 cannot be bounded on any line, unless it is a constant. It is probably not obvious, but Theorem A contains the following interesting result. Theorem B. Let f be an entire function of exponential type τ > 0 such that |f (x)| ≤ M for all x ∈ R. Furthermore, let f (0) = 0. Then π π (1) |f (x)| ≤ M | sin(τ x)|, − ≤x≤ . 2τ 2τ The estimate is best possible as the example f (z) := M sin(τ z) shows. In order to see how Theorem B follows from Theorem A, let us consider the entire function F (z) := M 2 − f (z)f¯(z), where f¯(z) := f (z). It is of exponential type 2τ and 0 ≤ F (x) ≤ M 2 . Using a well-known result (cf. [2, Theorem 7.5.1], [1, p. 154]), we may write ¯ M 2 − f (x)f¯(x) = F (x) = |B(x)|2 = B(x)B(x), P∞ where B(z) := ν=0 bν z ν is of exponential type τ and |B(x)| ≤ M for all x ∈ R with P∞ B(0)ν = M . Setting βν := Re bν for 0 ≤ ν < ∞, we see that B∗ (z) := ν=0 βν z is a real entire function of exponential type τ such that |B∗ (x)| ≤ |B(x)| ≤ M for all x ∈ R. Besides, B∗ (0) = β0 = M and B∗0 (0) = 0. Hence, by Theorem A, we have π π B∗ (x) ≥ M cos(τ x), − ≤x≤ , τ τ which implies that π π ≤x≤ , F (x) ≥ M 2 cos2 (τ x), − 2τ 2τ and so (1) holds. Let g be an entire function of exponential type τ such that |g(x)| ≤ M for all x ∈ R. Then, for any given x0 ∈ R, Theorem B applies to the function f (z) := [g(x0 + z) − g(x0 − z)]/2 and leads to the inequality π , |g(x + δ) − g(x − δ)| ≤ 2M sin(τ δ), x ∈ R, 0 ≤ δ ≤ 2τ which is a classical result of S. Bernstein [2, Theorem 11.4.1]. It can be shown that Theorem B also implies the following result ([3], [6], [2, p. 215]). Theorem C. Let f be a real entire function of exponential type τ such that |f (x)| ≤ M for all x ∈ R. Then τ 2 f (x)2 + f 0 (x)2 ≤ M 2 τ 2 ,
x ∈ R.
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P Any trigonometric polynomial t(x)P := nν=−n aν eiνx is the restriction to the real axis of the entire function t(z) := nν=−n aν eiνz , which is of exponential type n. Hence, Theorem B says in particular that if t is a trigonometric polynomial of degree at most n such that t(0) = 0 and |t(x)| ≤ M for 0 < x < 2π then |t(x)| ≤ M | sin(nx)| for |x| ≤ π/(2n). It was observed by Boas [4, Theorem 2] that the following stronger result holds. Theorem D. Let t be a trigonometric polynomial of degree n with (2ν + 1)π ≤ M for ν = 0, 1, . . . , 2n − 1. t(0) = 0 and t 2n Then π |t(x)| < M | sin(nx)|, 0 < |x| < , 2n unless t(x) ≡ M eiγ sin(nx) for some γ ∈ R. The question arises whether Theorem B can be similarly extended, and whether it would be sufficient to suppose that |f (x)| ≤ M for all x in an appropriately discrete set of points. The answer is yes. We shall not only obtain such an extension of Theorem B, but also that of Theorem A. In fact, we shall do considerably more. In order to place our first and probably the main result in perspective, we recall the following result of Duffin and Schaeffer [7, Theorem 1]. Theorem E. Let f be a real entire function of exponential type τ > 0 such that |f (x)| ≤ 1 for all real x and let Cτ (z) := cos(τ z). Then f − Cτ has only real zeros, or vanishes identically. Moreover, all the zeros are simple, except perhaps at points on the real axis where f (x) = ±1. More generally, H¨ormander [10, Theorem 1] proved that if f is as above and √ Cτ,α (z) := cos τ 2 z 2 + α2 , for α ≥ 0, then f − Cτ,α vanishes identically or it has a zero in every interval of the real axis where Cτ,α (x) varies between −1 and +1. Besides those there are at most 2k zeros, where k is the smallest integer larger than or equal to α/π. We prove the following generalization of Theorem E of Duffin and Schaeffer. We shall use it to obtain our extensions of Theorems A, B and D. Theorem 1. Let f be a real entire function of exponential type τ > 0 such that νπ (−1)ν f ≤1 for all ν ∈ Z τ and f (x) = o(x) as x → ±∞. Furthermore, let Cτ (z) := cos(τ z). Then f − Cτ has only real zeros, or vanishes identically. Besides, there can be at most one zero in (νπ/τ, (ν + 1)π/τ ) for any ν ∈ Z and then it can only be a simple zero. A zero at any point of the form νπ/τ , ν ∈ Z may be simple or double but not of higher multiplicity. If f − Cτ has a double zero at some point nπ/τ , n ∈ Z, then it cannot have any zeros in ((n − 1)π/τ, nπ/τ ) ∪ (nπ/τ, (n + 1)π/τ ).
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Remark 1. The function f (z) := [1 + cos(τ z)]/2 satisfies the conditions of Theorem 1 and is, in fact, bounded on the real axis. Nevertheless, the function f (z) − cos(τ z) = [1 − cos(τ z)]/2 = sin2 (τ z/2) has only double zeros, all lying at the points where f (x) = 1. This example shows that even in Theorem E the possibility of f − Cτ having multiple zeros is quite real. Remark 2. We wish to emphasize that the restriction on the growth of f cannot be relaxed any further. By this we mean that the result may fail if f (x) = O(x) as x → ±∞ but f (x) 6= o(x) as x → ±∞. This can happen in two different ways. First, we note that if f (z) := (π − 2τ z) sin(τ z) then f −Cτ has at least three zeros in (0, π/τ ). This function satisfies all the conditions of Theorem 1 except that f (x) 6= o(x) as x → ±∞ although f (x) = O(x). Next, we consider the function sin(τ z) − τ z sin(τ z) f (z) := τz which satisfies all the conditions of Theorem 1 except that f (x) is O(x) but not o(x) as x → ±∞. We have f (0) = Cτ (0), f 0 (0) = Cτ0 (0), f 00 (0) = −7τ 2 /3, Cτ00 (0) = −τ 2 and f 00 (0) − Cτ00 (0) < 0. Consequently, the graph of f (x) touches that of Cτ (x) at 0 but remains below it immediately to the left and immediately to the right of 0. However, at ±π/τ it lies above. Hence, the graph of f (x) cuts that of Cτ (x) at least once in each of the two intervals (−π/τ, 0) and (0, π/τ ), that is, f − Cτ has at least four zeros in (−π/τ, π/τ ). Corollary 1. Let f be a real entire function of exponential type τ > 0 such that f (x) = o(x) as x → ±∞ and suppose that for some M > 0, (2ν + 1)π ν (−1) f ≤M for all ν ∈ Z. 2τ Furthermore, let f (ξ) = M Sτ (ξ) := M sin(τ ξ) for some ξ ∈ (−π/(2τ ), π/(2τ )). Then either f (z) ≡ M sin(τ z) or p f 0 (ξ) < M Sτ0 (ξ) = τ M 2 − f 2 (ξ). Theorem 1 helps us obtain the following extension of Theorem A. Theorem 2. Let f be a real entire function of exponential type τ > 0 such that f (0) = M and f 0 (0) = 0. Furthermore, let νπ (−1)νf ≤M for all ν ∈ Z\{0} τ and f (x) = o(x) as x → ±∞. Then π f (x) > M cos(τ x), 0 < |x| < , τ except in the case where f (z) ≡ M cos(τ z).
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Remark 3. Clearly, f (z) := M cos(τ z) − z sin(τ z) is an entire function of exponential type τ such that (−1)ν f (νπ/τ ) = M for all ν ∈ Z. Furthermore, f 0 (0) = 0. However, f (x) < M cos(τ x) for 0 < |x| < π/(2τ ). This shows that the restriction on the growth of f (x) as x → ±∞ cannot be weakened to f (x) = O(x) as x → ±∞. Besides, the example n z sin(τ z) f (z) := M cos(τ z) + , |n| z − nπ τ where n ∈ Z\{0}, shows that the conclusion of Theorem 2 may not hold if the condition (−1)ν f (νπ/τ ) ≤ M is not satisfied even for one single ν ∈ Z\{0}. We find it desirable to state Theorem 2 in the following equivalent form. Theorem 20 . Let f be a real entire function of exponential type τ > 0 such that (2ν + 1)π ν ≤M for all ν ∈ Z (−1) f 2τ and f (x) = o(x) as x → ±∞. Furthermore, let f (π/(2τ )) = M and let f 0 (π/(2τ )) = 0. Then, either f (z) ≡ M sin(τ z) or π π f (x) > M sin(τ x) for 0 < x − < . 2τ τ As a supplement to Theorem 20 , we prove the following result.
Theorem 3. Let f be a real entire function of exponential type τ > 0 such that f (x) = o(x) as x → ±∞, and suppose that for some M > 0, (2ν + 1)π ν (−1) f ≤ M, for all ν ∈ Z. 2τ Furthermore, let f (ξ) = M sin(τ ξ) for some ξ ∈ (−π/(2τ ), π/(2τ )). Then π f (x) > M sin(τ x) for − < x < ξ, 2τ π f (x) < M sin(τ x) for ξ 0 such that (2ν + 1)π f ≤M for all ν ∈ Z 2τ
and f (x) = o(x) as x → ±∞. Furthermore, let f (0) = 0. Then π |f (x)| < M | sin(τ x)|, 0 < |x| < , 2τ except in the case where f (z) ≡ M eiγ sin(τ z) for some γ ∈ R.
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Remark 4. The example f (z) := z cos(τ z) shows that in Corollary 2 it is not sufficient to assume f (x) = O(x) as x → ±∞. Remark 5. One may also ask if we really need |f (x)| to be bounded by M at all the points (2ν + 1)π/(2τ ), ν ∈ Z. The answer is yes. For this, let us leave out just one point xk := (2k + 1)π/(2τ ) and note that z cos(τ z) z − xk is an entire function of exponential type τ which vanishes at the origin and satisfies |f ((ν + 1/2)π/τ )| = 0 < M for all ν ∈ Z except ν = k and any M > 0. Besides, f (x) is uniformly bounded on the real axis. However, it is not true that |f (x)| < M | sin(τ x)| for 0 < |x| < π/(2τ ) and every M > 0. f (z) :=
2. An analogue of a result of M. Riesz Let t be a trigonometric polynomial of degree n, whose zeros are all real and simple. Let θ1 < θ2 < . . . < θ2n < θ2n+1 := θ1 + 2π be 2n + 1 of its zeros. The graph of |t(θ)| on [θ1 , θ1 + 2π] consists of 2n arches. Let hν be the height of the arch that stands on the interval [θν , θν+1 ]. Thus hν :=
max
θν ≤θ≤θν+1
|t(θ)|,
ν = 1, . . . , 2n.
Any arch whose height is equal to min{h1 , . . . , h2n } will be called “shortest”. It was noted by M. Riesz (see [8, p. 61]) that the shortest arch cannot be wider than π/n. We have not been able to find out if Riesz ever published his result. It has been asked if something similar holds also for non-periodic entire functions of order 1 and finite type. Whereas, a trigonometric polynomial of degree n has exactly 2n zeros in any strip of the form a ≤ Re z < a + 2π, the distribution of the zeros of a nonperiodic entire function f of order 1 and finite type can be complicated (see [2, Section 2.10]). So, we have to be quite careful in formulating an analogue of Riesz’s result for entire functions of order 1 and finite type. Let f be a canonical product of order 1 and type τ with only real and simple zeros. Then the graph of |f (x)| on (−∞, ∞) consists of an infinite number of arches. As the example f (z) := z −1 sin(τ z) shows, there may not be a shortest arch, but we shall suppose that there is one. Then, without loss of generality, we may also suppose that the point (0, h) ∈ R2 is a peak of the shortest arch and that f (0) = h. By considering f (z) := h cos(τ z), a function which has this property, we see that the shortest arch can be at least as wide as π/τ . In view of this example and the result of Riesz it would not be far-fetched that if f is a canonical product of order 1 and type τ with only real and simple zeros, such that the graph of |f (x)| on (−∞, ∞) has a shortest arch, then that arch cannot be wider than π/τ . However, at this stage, this is nothing more than a speculation.
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In order to get some insight as to why functions with non-real zeros need to be excluded, let us take τ = π and with ∞ Y 9 cos πz z2 ∗ 2 =− ϕ (z) := z − 1− 2 , 16 an z 2 − 14 z 2 − 94 n=1 where a1 := 3/4 and an := (2n + 1)/2 for n ≥ 2. Let us consider the function 1 ∗ 2 f (z) := − z + ϕ∗ (z). 16
2
1.5
1
0.5
-7.5
-5
-2.5
2.5
5
7.5
Figure 1. Graph of |f ∗ | on [−8, 8]. We see, by logarithmic differentiation, that −ϕ∗ (x) decreases from 1 to 0 when x increases from 0 to the first positive zero 3/4 of f ∗ . Hence, |ϕ∗ (x)| ≤ |ϕ∗ (0)| = 1 on [−3/4, 3/4], and so the function f ∗ has no zeros on (−3/4, 3/4) and 5 3 3 |f ∗ (x)| < , − ≤x≤ . 8 4 4 On the other hand, (−1)k f ∗ (k) =
k 2 + ( 14 )2 k 2 − ( 34 )2 > 1, k 2 − ( 12 )2 k 2 − ( 32 )2
k = ±2, ±3, . . . ,
which implies that all the other arches constituting the graph of |f ∗ (x)| are of height more than 1. Note, that the shortest arch stands over an interval of length 3/2, which is larger than π divided by the type of the function f ∗ . The entire function f ∗ (τ z/π), which is of order 1 and type τ , has an analogous property. This ‘explains’ why the zeros of f are required to be all real. We shall not explicitly require f to have only real zeros. Instead, we shall suppose that (−1)ν f (νπ/τ ) ≥ h for ν ∈ Z\{0}, where h is a positive number.
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We emphasize that any point of the form (νπ/τ, (−1)ν f (νπ/τ )), ν 6= 0, may or may not be a peak of an arch of the graph of |f (x)|. In other words, f 0 (νπ/τ ) may be different from zero for any ν 6= 0. If xν and x0ν are the two zeros of f closest to νπ/τ such that xν < νπ/τ < x0ν then the arch of the graph of |f (x)| that is based on [xν , x0ν ] is of height at least h. Thus, to each point νπ/τ there corresponds an arch and none of them is shorter than the one that corresponds to 0. We then prove the following result which can be seen as an analogue of the above mentioned result of M. Riesz. Theorem 4. Let f be a real entire function of order 1 and type τ > 0 such that f (0) = h and f 0 (0) = 0. Furthermore, let νπ (−1)ν f ≥h>0 for all ν ∈ Z\{0} τ and f (x) = o(x) as x → ±∞. Then π f (x) < h cos(τ x), 0 < |x| < , τ except in the case where f (z) ≡ h cos(τ z). In particular, f has a zero in each of the intervals (−π/(2τ ), 0) and (0, π/(2τ )). Remark 6. Theorem 4 shows that the shortest arch of |f (x)| stands over an interval of length less than π/τ unless f (z) ≡ h cos(τ z). It may be noted that the conclusion of Theorem 4 does not hold if (−1)ν f (νπ/τ ) 6≥ h even for a single ν ∈ Z\{0}, say for ν = n. For example, the function n z f (z) := h cos(τ z) − h sin(τ z) |n| z − n πτ satisfies all the conditions of Theorem 4 except that (−1)n f (nπ/τ ) < h. Clearly, f (x) > h cos(τ x) for all x ∈ (−π/τ, 0) ∪ (0, π/τ ). The conclusion also fails if f (x) = O(x) but not o(x) as x → ±∞. This is illustrated by the function f (z) := h cos(τ z) + z sin(τ z).
3. Auxiliary results For sake of clarity and to keep the paper self-contained, we shall start with some lemmas. The first lemma, which is a result of Valiron ( see [12, p. 213]; also see [2, p. 161]), is the most crucial of them all. Lemma 1. Let f be an entire function of exponential type π such that zf (z)e−π|z| tends to zero uniformly as |z| → ∞ and vanishes at least once in each of the intervals [n, n + 1) for all n ∈ Z. Then f (z) ≡ 0. The following result of E. Lindel¨of [9, Theorem 18.3.5] plays an important role in the proof of Lemma 3.
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Lemma 2. Let Ω be holomorphic and bounded in the upper half-plane. Suppose that Ω is continuous at all finite points of the real axis. Furthermore, let Ω(x) → a as x → +∞. Then limz→∞ Ω(z) = a uniformly in any sector 0 ≤ arg z ≤ π − δ, δ > 0. Lemma 3. Let f be an entire function of exponential type τ > 0 such that f (x) = o(x) as x → ±∞ and let y := Im z. Then (2)
|f (x + iy)|e−τ |y| = o(|z|)
as
|z| → ∞.
Proof. Let ω(z) := [f (z) − f (0)]/z. Then Ω(z) := ω(z)eiτ z is of exponential type with π log |Ω(iy)| hΩ := lim sup ≤ 0. 2 y y→∞ Besides, it is bounded on the real axis. Hence, |Ω(z)| is bounded in the upper half-plane [2, Theorem 6.2.4]. Since Ω(x) → 0 as x → +∞, Lemma 2 implies that |Ω(z)| → 0 as z → ∞ uniformly in 0 ≤ arg z ≤ π/2. Thus |ω(z)|e−τ y → 0 as z → ∞ uniformly in 0 ≤ arg z ≤ π/2. The same argument may be applied to the function ω(z) to conclude that |ω(z)|e−τ |y| → 0 as z → ∞ uniformly in −π/2 ≤ arg z ≤ 0 and so uniformly in −π/2 ≤ arg z ≤ π/2. Since the function ω(−z) has the same properties as ω, we must also have ω(z) = o(eτ |y| ) as z → ∞ uniformly in π/2 ≤ arg z ≤ 3π/2. Hence, ω(z) = o(eτ |y| ) as |z| → ∞ uniformly in C. This readily implies (2). The formula for f 0 (0) appearing in the next lemma is known [2, p. 211] at least under the condition that f (x) is uniformly bounded on the real axis. Even in the form stated here it may be not new. The proof may still be of some interest, and noteworthy for its simplicity. Lemma 4. Let f be an entire function of exponential type τ > 0 such that f (x) = o(x) as x → ±∞. Then ∞ 4τ X (−1)ν 2ν + 1 0 (3) f (0) = 2 f π . π ν=−∞ (2ν + 1)2 2τ Proof. First, let τ = π and denote by Γn the square contour with corners at the four points (±1 ± i)n. Let us integrate the function Φ(z) :=
1 f (z) . z 2 cos(πz)
around Γn . The residue of Φ at 0 is f 0 (0) and its residue at (2ν + 1)/2 for any ν ∈ Z is 1 2ν + 1 ν+1 4 f . (−1) π (2ν + 1)2 2
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Hence, by the Residue Theorem Z n−1 1 4 X 1 2ν + 1 0 ν (4) Φ(z) dz = f (0) − (−1) f . 2πi Γn π ν=−n (2ν + 1)2 2 Simple calculations show that | cos(πz)| > 3−1 eπ|y| on Γn and so, in view of (2), we have 1 |Φ(z)| = o , z ∈ Γn . |z| R Consequently, Γn Φ(z) dz → 0 as n → ∞ and letting n tend to infinity in (4) we obtain the desired formula for f 0 (0) in the case where τ = π. If τ 6= π then we may consider the function f (πz/τ ). The next lemma is a consequence of the preceding one. Lemma 5. Let f be an entire function of exponential type τ > 0 such that f (x) = o(x) as x → ±∞. Suppose, in addition, that f (0) = 0 and that (2ν + 1)π ν (−1) f ≤M for all ν ∈ Z 2τ and some M ∈ R. Then f 0 (0) < M τ unless f (z) ≡ M sin(τ z). Proof. It is clear from (3) that f 0 (0) < M τ except in the case where we have f ((2ν + 1)π/(2τ )) = (−1)ν M for all ν ∈ Z. However, in this exceptional case (2ν + 1)π : ν∈Z . f (z) − M sin(τ z) = 0 for all z ∈ 2τ Thus, f (z) − M sin(τ z) cos(τ z) is an entire function such that φ(z) = o(z) on the square Γn introduced in the proof of Lemma 4 and so, by a well known generalization of Liouville’s Theorem, φ is a constant. Since φ(0) = 0, that constant must be zero. Hence, f (z) ≡ M sin(τ z). φ(z) :=
Applying the above argument to the function [f (z) − M sin(τ z)]/ cos(τ z) we can also prove the following lemma, which will come in handy. Lemma 6. Let f be an entire function of exponential type τ > 0 such that f (x) = o(x) as x → ±∞. Suppose, in addition, that f (0) = 0 and that (2ν + 1)π = (−1)ν M for all ν ∈ Z, f 2τ where M is some real number. Then f (z) ≡ M sin(τ z).
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The following classical result of Montel [9, p. 170] is needed for the proof of Lemma 8. Lemma 7. Let f be holomorphic and bounded in the half-strip S defined by −η0 < y < η0 , x > 0. Suppose, in addition, that f (x) tends to a limit ` as x → ∞. Then f (x + iy) tends to ` when x → ∞ on every line y = η in S and indeed f (x + iy) → ` as x → ∞ uniformly for −η0 + δ ≤ y ≤ η0 − δ. Lemma 8. Let f be an entire function of exponential type τ > 0 such that f (x) = o(x) as x → ±∞. Then f 0 (x) = o(x) as x → ±∞. Proof. The entire function ω(z) := [f (z) − f (0)]/z is also of exponential type τ . Besides, there exists a finite number µ such that |ω(x)| ≤ µ for all real x. Hence, by a well-known result [2, Theorem 6.2.4], |ω(x + iy)| ≤ µeτ |y| ,
y ∈ R.
In particular, ω(z) is bounded in the half-strip S of Lemma 7. Furthermore, ω(x) → 0 as x → ∞. Now, let ε be an arbitrary positive number. By Lemma 7 there exists Xε > 0 such that |ω(x + iy)| < ε if x > Xε and |y| ≤ η0 /2. Applying Cauchy’s Integral Formula for the derivative of a holomorphic function with ρ := η0 /2 we conclude that Z ε 1 ω(ζ) η0 0 < = 2ε , dζ x > Xε + . |ω (x)| = 2 2πi (ζ − x) ρ η0 2 |ζ−x|=ρ
The number η0 being fixed and ε > 0 being arbitrary this shows that ω 0 (x) → 0 as x → ∞ and so f 0 (x) = xω 0 (x)+ω(x) = o(x) as x → ∞. Similarly, f 0 (x) = o(x) as x → −∞.
4. Proofs of the main results Proof of Theorem 1. Without loss of generality we may take τ = π. First of all we shall show that f − Cπ cannot have non-real zeros. Suppose it does and let f (z) − cos(πz) = (z − ζ)k (z − ζ)k ϕ(z), where k ∈ N, Im ζ > 0 and ϕ(ζ) 6= 0. Consider the function Eε (z) := f (z) − cos(πz) − ε cos(πz),
ε > 0.
It is easily checked that Eε (ν)Eε (ν + 1) < 0 for any ν ∈ Z. Hence, Eε has at least one zero in each of the open intervals (ν, ν + 1), ν ∈ Z. Now, choose ρ > 0 such that f (z) − cos(πz) 6= 0 for 0 < |z − ζ| ≤ ρ and let m := min|z−ζ|=ρ |f (z) − cos(πz)| > 0. There exists ε0 > 0 such that | − ε cos(πz)| < m for |z − ζ| < ρ for 0 < ε < ε0 . By Rouch´e’s Theorem Eε has exactly k zeros in |z − ζ| < ρ for 0 < ε < ε0 . Thus for any ε ∈ (0, ε0 ) we have Eε (z) = (z − ζ∗ (ε)) z − ζ∗ (ε) ϕε (z),
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where |ζ∗ (ε) − ζ| < ρ and ϕε has at least one zero in each of the open intervals (ν, ν + 1), ν ∈ Z. Applying Lemma 1 to the function ϕε we see that f (z) ≡ (1 + ε) cos(πz) for any ε ∈ (0, ε0 ) which is a contradiction. Next we show that f − Cπ cannot have more than one zero in any interval of the form (ν, ν + 1), ν ∈ Z, and that the zero, if any, must be simple. Let us take the interval (µ, µ + 1) for some µ ∈ Z. Since Eε (µ)Eε (µ + 1) < 0, the function Eε , which must have at least one zero in (µ, µ + 1), can only have an odd number of zeros in it. It is understood that each zero is counted as often as its multiplicity. If there is not just one simple zero in this interval then there are at least three, say ξµ1 , ξµ2 and ξµ3 , though not necessarily distinct. Applying Lemma 1 to the function Eε (z)/[(z − ξµ1 )(z − ξµ2 )] we see that f (z) ≡ (1 + ε) cos(πz) which is again a contradiction. Hence, Eε has one and only one (simple) zero in the open interval (µ, µ + 1) and no zeros at the end-points µ and µ + 1. Thus, for any ε > 0, the function Eε has one and only one zero in each of the open intervals (ν, ν + 1), ν ∈ Z, and no other zeros. In view of the distribution of the zeros of Eε just described a well-known theorem of Hurwitz [9, Theorem 14.3.4] implies that the limiting function f −Cπ can have at most one (simple) zero in (ν, ν + 1), ν ∈ Z, and at any point ν ∈ Z it can have a zero but not of multiplicity larger than 2. The last statement of Theorem 1 needs no further explanation. Remark 7. Let us denote by Zf −C ([a, b)) the number of zeros of the function f (z) − cos(πz) on [a, b), and by Zf −C ([a, b]) the number of zeros it has on [a, b]. For simplicity, let us take a function f satisfying the conditions of Theorem 1 with τ = π and M = 1. It is clear from the above discussion that for any n ∈ Z and any k ∈ N we have k − 1 ≤ Zf −C ([n, n + k)) ≤ k + 1, k ≤ Zf −C ([n, n + k]) ≤ k + 2. Are these lower and upper bounds for Zf −C ([n, n + k)) and Zf −C ([n, n + k]) attained? The answer is yes. Indeed, if !2 sin π(z−k) 2 f (z) := , where k = 2m, m ∈ N, π(z−k) 2
then Zf −C ([0, k)) = k − 1. Besides, if f (z) := [1 + cos(πz)]/2, where m ∈ N, then Zf −C ([0, 2m − 1)) = 2m and Zf −C ([−1, 2m − 1]) = 2m, whereas Zf −C ([0, 2m]) = 2m + 2. Proof of Corollary 1. Without loss of generality we may take τ = π and M = 1. Let f (z) 6≡ sin(πz). Applying Theorem 1 to the function f (z + 1/2) we see that f 0 (ξ) cannot be equal to Sπ0 (ξ), otherwise the function f (z+1/2)−cos(πz) would have a double zero at ξ − 1/2 without being identically zero.
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Now, let us suppose that f 0 (ξ) > Sπ0 (ξ). First, let ξ = 0 and choose ε > 0 sufficiently small such that (1 − ε)f 0 (ξ) remains larger than Sπ0 (ξ). Then, setting g(z) := (1 − ε)f (z + 1/2) − cos(πz), we see that g(−1) > 0 and g(0) < 0. In addition, we observe that g(x) must be negative immediately to the left of the point −1/2 and positive immediately to its right since g(−1/2) = 0 and g 0 (−1/2) > 0. Hence, g must have at least one zero in each of the two intervals (−1, −1/2) and (−1/2, 0), that is, g has at least three zeros in (−1, 0). Thus, the function (1 − ε)f (z + 1/2) satisfies the conditions of Theorem 1 but its graph cuts that of cos(πz) at least thrice in (−1, 0) which is a contradiction. Next, let ξ ∈ (0, 1/2). Since f 0 (ξ) − Sπ0 (ξ) > 0, there exists δ > 0 such that f 0 (x) − Sπ0 (x) > 0 on [ξ − δ, ξ + δ]. Then f (x) − Sπ (x) must be increasing on [ξ − δ, ξ + δ]. Now, choose ε > 0 so small that (1 − ε)f 0 (x) − Sπ0 (x) > 0,
δ ξ≤x≤ξ+ , 2
and (1 − ε)f (ξ + δ/2) − Sπ (ξ + δ/2) > 0. Since (1 − ε)f (ξ) − Sπ (ξ) = −εf (ξ) < 0, the function (1−ε)f −Sπ must have at least one zero in (ξ, ξ +δ/2), say ξ + . Also, it has at least one zero in (−1/2, ξ) as (1 − ε)f (−1/2) − Sπ (−1/2) > 0. Besides, (1 − ε)f 0 (ξ + ) − Sπ0 (ξ + ) > 0, so that (1 − ε)f (x) − Sπ (x) is positive immediately to the right of ξ + . However, (1 − ε)f (1/2) − Sπ (1/2) < 0 and so (1 − ε)f − Sπ has at least three zeros in (−1/2, 1/2), which is not possible by Theorem 1. We can similarly treat the remaining case where ξ lies in (−1/2, 0). Remark 8. The above reasoning with little modification gives the following result which says more than what Corollary 1 does. Corollary 10 . Let f be a real entire function of exponential type τ > 0 such that f (x) = o(x) as x → ±∞. Also, let (2ν + 1)π ν (−1) f ≤M for all ν ∈ Z, 2τ where M > 0. Furthermore, let f (ξ) = M Sτ (ξ) := M sin(τ ξ) for some ξ in the open interval ((2µ − 1)π/(2τ ), (2µ + 1)π/(2τ )), µ ∈ Z. Then p (−1)µ f 0 (ξ) < (−1)µ M Sτ0 (ξ) = τ M 2 − f 2 (ξ), unless f (z) ≡ M sin(τ z). Proof of Theorem 2. By considering the function f (πz/τ )/M instead of f we may assume that τ = π and M = 1. Let f (z) 6≡ Cπ (z) := cos(πz). Then, Lemma 4 applied to the function f (z + (2ν + 1)/2) implies ∞ 4 X (−1)µ ν 0 2ν + 1 =− f (µ) ≥ −π, ν ∈ Z. (−1) f 2 π µ=−∞ (2µ + 1)2
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Since f 0 is of exponential type τ > 0 and since, by Lemma 8, f 0 (x) = o(x) as x → ±∞, we may apply Lemma 4 to obtain ∞ 4 X (−1)ν 0 2ν + 1 00 f (0) = f > −π 2 , π ν=∞ (2ν + 1)2 2 unless (−1)ν f 0 ((2ν + 1)/2) = −π for all ν ∈ Z. However, assume ν 0 2ν + 1 (−1) f = −π, ν ∈ Z, 2 then the function −f 0 /π satisfies all the conditions of Lemma 6 with τ = π and M = 1. Hence, f 0 (z) ≡ −π sin(πz). Since f (0) = 1, this implies that f (z) ≡ cos(πz), a contradiction. Thus, we have proved that f 00 (0) > −π 2 = C 00 (0). By hypothesis, the function f − Cπ has a double zero at 0. So, by Theorem 1, it is either strictly positive on (−1, 0) ∪ (0, 1) or strictly negative on it. However, the latter possibility is excluded because f 00 (0) > C 00 (0). Proof of Theorem 3. Suppose that f (z) 6≡ M sin(τ z). By Theorem 1, the graph of the function f (x) intersects that of the function sin(τ x) only once in (−π/(2τ ), π/(2τ )). Since, by Corollary 1, sin(τ x) > f (x) immediately to the right of ξ, it remains so for all x ∈ (ξ, π/(2τ )). Furthermore, f (x) must be smaller than sin(τ x) in (−π/(2τ ), ξ) since ξ is necessarily a simple zero of the function f (z) − sin(τ z). Proof of Corollary 2. Without loss of generality we may suppose that τ = π and that M = 1. Note that the result is only a special case of Theorem 3 if f (x) is real for real values of x. P n Now, let us suppose that the coefficients of f (z) := ∞ ν=0 an z are not all real. iγ Take anyPpoint x0 in (0, 1/2) and let f (x0 ) = |f (x0 )|e . Consider the function ν −iγ u(z) := ∞ ). Then u(x) is equal to Re(e−iγ f (x)) ν=0 αν z where αν := Re(aν e for all real x and u(x0 ) = |f (x0 )| > 0. It is clear that u is an entire function of exponential type π and |u(x)| = | Re(e−iγ f (x))| ≤ |f (x)| = o(x)
as x → ±∞.
Since u(0) = 0 and u(x) is real for real x we may apply Theorem 3 to conclude that either |f (x0 )| = u(x0 ) < sin(πx0 ) or u(z) is identically equal to sin(πz). It remains to show in the latter case that f (z) ≡ eiγ sin(πz). We shall do this by proving that if f is an entire function of exponential type π such that f (0) = 0 and |f ((2ν + 1)/2)| ≤ 1 for all ν ∈ Z while f (x) = o(x) as x → ±∞ and Re(e−iγ f (x)) = sin(πx) for all real x then e−iγ f (z) = sin(πz) for all z ∈ C. To this end we define v(z) := e−iγ f (z)−sin(πz) which is clearly an entire function of exponential type π vanishing at the origin. Besides, Re v(x) = 0 for all real x.
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In particular, (5)
2ν + 1 Re v = 0, 2
ν ∈ Z.
That is 2ν + 1 −iγ Re e f = (−1)ν , 2
ν ∈ Z.
Since |e−iγ f (x)| ≤ 1 at any point x of the form (2ν + 1)/2, ν ∈ Z, it follows that Im(e−iγ f (x)) = 0 at every such point and therefore 2ν + 1 2ν + 1 −iγ (6) Im v = Im e f = 0, ν ∈ Z. 2 2 From (5) and (6) it follows that v((2ν + 1)/2) = 0 for all ν ∈ Z. Since v(0) = 0 and v(x) = o(x) as x → ±∞, Lemma 6 applies with M := 0. Thus, v(z) ≡ 0, that is, f (z) ≡ eiγ sin(πz). Proof of Theorem 4. By considering the function f (πz/τ )/h instead of f , we may assume that τ = π and h = 1. Let f (z) 6≡ Cπ (z) := cos(πz). Then Lemma 4, applied to the function f (z + (2ν + 1)/2), shows that ν 0 2ν + 1 (−1) f ≤ −π, ν ∈ Z. 2 Since f 0 is of exponential type τ > 0 and since, by Lemma 8, f 0 (x) = o(x) as x → ±∞, we may apply Lemma 4 once again to conclude that f 00 (0) < −π 2 , unless (−1)ν f 0 (2ν + 1)/2) = −π for all ν ∈ Z. However, assume that ν 0 2ν + 1 (−1) f = −π, ν ∈ Z, 2 then the function −f 0 /π satisfies all the conditions of Lemma 6 with τ = π and M = 1. Hence, f 0 (z) ≡ −π sin(πz). Since f (0) = 1, this implies f (z) ≡ cos(πz), a contradiction. Thus, we have proved that f 00 (0) < −π 2 = C 00 (0). This means that the graph of f (x), which touches that of cos(πx) at 0, remains below it in (−δ, 0) ∪ (0, δ) for some δ > 0. Let us denote by ∆ the supremum of all δ such that the graph of f (x) remains below that of cos(πx) in (0, δ). We claim that ∆ ≥ 1. Suppose not, and let m := max0≤x≤∆ {cos(πx) − f (x)}. Denote by x1 the smallest number in (0, ∆) such that cos(πx1 ) − f (x1 ) = m and define F (η, z) := f (z) + η
sin(πz) , πz
η > 0.
It is clear that F (η, 0) = 1 + η > 1 and that (−1)ν F (η, ν) = (−1)ν f (ν) ≥ 1,
ν ∈ Z\{0}.
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Note that if G(η, z) := F (η, z) − cos(πz) then G(η, 0) > 0 and G(η, x1 ) < η − m < 0 for η > 0 and η < m. Hence, G(η, ·) has at least one zero in (0, x1 ) for any η ∈ (0, m). Besides, G(η, ∆) > 0 for any such η. Fix η and let H(ε, η, z) := G(η, z) + ε cos(πz) = F (η, z) − (1 − ε) cos(πz),
ε > 0.
Furthermore, let ε be so small that H(ε, η, ∆) remains positive and H(ε, η, x1 ) remains negative. Then, H(ε, η, ·) must have at least two zeros in (0, ∆). Call them ξ10 and ξ20 . There is an additional zero ξ30 in (∆, 1) since H(ε, η, ∆) is positive and H(ε, η, 1) is negative. Since (−1)ν H(ε, η, ν) = (−1)ν [F (η, ν) − (1 − ε) cos(πν)] > 0,
ν ∈ Z,
the function H(ε, η, ·) has at least one zero in each of the intervals (ν, ν + 1), ν ∈ Z\{0}. Applying Lemma 1 to the function H(ε, η, z)/[(z − ξ10 )(z − ξ20 )], we conclude that F (η, z) ≡ (1 − ε) cos(πz) for all sufficiently small ε > 0. Since the function F (η, ·) is independent of ε, we conclude that F (η, z) ≡ cos(πz), that is sin(πz) + cos(πz) πz for all sufficiently small η > 0 and so f (z) ≡ cos(πz). This contradiction proves that f (x) < cos(πx) for all x ∈ (0, 1), unless f (z) ≡ cos(πz). By symmetry, the same holds for all x ∈ (−1, 0). f (z) ≡ −η
Example. We shall now present an example of an entire function f of order 1 and type τ > 0 which is not bounded on the real axis but to which Theorem 4 applies. Let α ∈ (0, 1) and d := (1 − α)/2. The exponent of convergence of the zeros of the canonical product ∞ Y z λ(z) := 1+ (n + d)2 n=0 being 1/2 [2, Theorem 2.6.5] it defines an entire function of order 1/2. Its zeros are all negative and λ(0) = 1. Besides, if nλ (r) denotes the number of zeros of λ in |z| ≤ r then nλ (r) ∼ r1/2 as r → ∞ and therefore, by [2, Theorem 4.1.1], log λ(r) ∼ πr1/2 as r → ∞. This implies that the infinite product ∞ Y z2 (7) f (z) := 1− (n + d)2 n=0 defines an entire function of order 1 and type π such that f (0) − 1 = f 0 (0) = 0. The function f can be expressed in terms of Euler’s gamma function Γ. For this we may use either the Weierstrass’ product expansion ([5, pp. 215–216], [13, p. 236]), ∞ h Y 1 z −z/n i γz = ze 1+ e Γ(z) n n=1
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or the Euler’s limit formula ([5, pp. 208–211], [13, p. 237]) n!nz , n→∞ z(z + 1) · · · (z + n)
Γ(z) = lim
where the convergence is uniform on any compact subset of C not containing any of the singularities of Γ, to conclude that f (z) =
Γ(d)2 . Γ(z + d)Γ(−z + d)
In view of the functional equation Γ(z)Γ(1 − z) = π cosec(πz), which holds ([5, p. 211], [13, p. 239]) for all z 6∈ Z, we may write f (x) = −
Γ(d)2 Γ(x + 1 − d) sin(π(x − d)). π Γ(x + d)
Hence, using the asymptotic expansion ([5, pp. 222–223], [13, p. 253]) of Γ(z), we find that Γ(d)2 1−2d x | sin(π(x − d))| π 2 1 1−α 1−α α = Γ x sin π x − π 2 2
|f (x)| ∼
as x − (1 − α)/2 → ∞ through non-integral values. Since the function f is even, we do not need to discuss the behavior of f (x) as x → −∞. In addition, we see that for any ν ∈ N, Γ(d)2 Γ(ν + 1 − d) sin(πd) πΓ(ν + d) ν 1Y µ−d = Γ(d)Γ(1 − d) sin(πd) π µ=1 µ − (1 − d)
(−1)ν f (ν) =
=
ν Y
µ−d > 1. µ − (1 − d) µ=1
Once again, since f is even, (−1)ν f (ν) > 1 for all ν ∈ Z\{0}. The function defined in (7) satisfies all the conditions of Theorem 4 with τ = π and h = 1. It is not bounded on the real axis. However, f (x) = O(|x|α )
as x → ±∞,
where α ∈ (0, 1), so that f (x) = o(x) as x → ±∞. Theorem 4 tells us that the graph of f (x) lies below that of cos(πx) for x ∈ (−1, 0) ∪ (0, 1).
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Remark 9. The observation made in the above example can be seen as a property of the gamma function. We have shown that if, as usual, Γ denotes the Euler’s gamma function then Γ(d)2 1 < cos(πx), x ∈ (−1, 0) ∪ (0, 1), 0 < d < . Γ(x + d)Γ(−x + d) 2 Although we have not seen it mentioned anywhere, this inequality may or may not be new. It can be easily seen by inspection for 0 < x < 1/2, but is more delicate to be established directly for x > 1/2.
References 1. N. I. Achieser, Theory of Approximation, Frederick Ungar Publishing Co., New York, 1956. 2. R. P. Boas, Jr., Entire Functions, Academic Press, New York, 1954. 3. , Some theorems on Fourier transforms and conjugate trigonometric integrals, Trans. Amer. Math. Soc. 40 (1936), 287–308. 4. , Inequalities for polynomials with a prescribed zero, in: Studies in Mathematical Analysis and Related Topics (Essays in honour of George P´ olya), Stanford University Press, Stanford, California, 1962, 42–47. 5. E. T. Copson, Theory of Functions of a Complex Variable, Oxford University Press, Oxford, 1935. 6. R. J. Duffin and A. C. Schaeffer, Some inequalities concerning functions of exponential type, Bull. Amer. Math. Soc. 43 (1937), 554–556. 7. , Some properties of functions of exponential type, Bull. Amer. Math. Soc. 44 (1938), 236–240. 8. P. Erd˝ os, On the uniform distribution of the roots of certain polynomials, Ann. of Math. 43 (1942), 59–64. 9. E. Hille, Analytic Function Theory, Vol. II, Ginn and Company, Boston, 1962. 10. L. H¨ ormander, Some inequalities for functions of exponential type, Math. Scand. 3 (1955), 21–27. 11. E. C. Titchmarsh, The Theory of Functions (2nd ed.), Oxford University Press, 1939. 12. G. Valiron, Sur la formule d’interpolation de Lagrange, Bull. Sci. Math. 49 (1925), 181–192, 203–224. 13. E. T. Whittaker and G. N. Watson, A Course of Modern Analysis (4th ed.), Cambridge University Press, Cambridge, 1927. Dimiter P. Dryanov E-mail:
[email protected] Address: D´epartament de Math´ematique et de Statistique, Universit´e de Montr´eal, Montr´eal (QC) H3C 3J7, Canada Mohammed A. Qazi E-mail:
[email protected] Address: Department of Mathematics, Tuskegee University, Tuskegee, Alabama 36088, U.S.A. Qazi I. Rahman E-mail:
[email protected] Address: D´epartament de Math´ematique et de Statistique, Universit´e de Montr´eal, Montr´eal (QC) H3C 3J7, Canada
