Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 22, 1997, 39–74

ZEROS OF DERIVATIVES OF STRICTLY NON-REAL MEROMORPHIC FUNCTIONS

A. Hinkkanen

University of Illinois at Urbana–Champaign, Department of Mathematics Urbana, Illinois 61801, U.S.A.; aimo@math.uiuc.edu

Abstract. Let f be a meromorphic non-entire function in the plane, and suppose that for every n ≥ 0, the derivative f^{(n)} has only real zeros. We have proved that then f is rational and of a special form, and that all possibilities can be listed. In this paper we prove that part of this result which is related to properties of strictly non-real functions.

1. Introduction and results

1.1. Let f be a function meromorphic in the complex plane C. We consider the question of under what circumstances all the derivatives of f, including f itself, can have only real zeros. We may and will assume that f is not a polynomial so that none of the derivatives f^{(n)} vanishes identically. We say that f is real if f(z) is real or f(z) = ∞ whenever z is real. If f is not a constant multiple of a real function, then f is called strictly non-real. We have proved the following result. The proof is given partly in this paper and partly in the two companion papers [3], [4].

Theorem A. Let f be a non-entire meromorphic function in the complex plane, and suppose that for every integer k ≥ 0, the derivative f^{(k)} has only real zeros. Then there are real numbers a and b where a ≠ 0, and a polynomial P with only real zeros, such that

(i) f(az +b) = P (z)/Q(z), where Q(z) = z^{n} or Q(z) = (z^{2} +1)^{n}, n is a positive integer, and deg P ≤ deg Q + 1; or

(ii) f(az + b) = C(z − i)^{−n} where C is a non-zero complex constant; or (iii) f(az + b) = C(z − α)/(z − i), where α is a real number and C is a non-zero

complex constant.

Conversely, if f is as in (i) with deg P ≤ deg Q, or if f is as in (ii) or (iii), then f^{(k)} has only real zeros for all k ≥ 0. If f is as in (i) with deg P = deg Q+1 then f^{(k)} has only real zeros for all k ≥ 0 if, and only if, f′ (or, equivalently, zP′(z) − nP (z) or (z^{2} + 1)P′(z) − 2nzP (z)) has only real zeros.

1991 Mathematics Subject Classification: Primary 30D35.

Research partially supported by the Alfred P. Sloan Foundation and by the U.S. National Science Foundation grant DMS 94-00999.

If f is as in (i) of Theorem A and deg P = deg Q+1 then there are polynomials P for which f′ has only real zeros, and other polynomials P for which f′ has at least two non-real zeros (cf. [3]).

The complete proof of Theorem A is long, and is divided into three papers (this paper and [3], [4]). The structure of the proof of Theorem A, showing how the different pieces are put together, is explained in [3]. In this paper we prove the following result that basically deals with strictly non-real functions (referred to in [3] as [3, Theorem 2]).

Theorem 1. Let f be given by

(1.1)  f(z) = \frac{g(z)}{(z + i)^{m}(z − i)^{n}}

where g is an entire function with only real zeros with g(i)g(−i) ≠ 0 and m and n are integers, not necessarily equal, with m ≥ 0 and n ≥ 1. Suppose that f′ and f′′ have only real zeros.

If m = 0 then there is a non-zero complex constant C such that

(i) g(z) ≡ C; or (ii) g(z) ≡ C(z − α) for some real α, and then n = 1; or (iii) g(z) ≡ Ce^{inz}.

If g is as in (i) or (ii) then f^{(k)} has only real zeros for all k ≥ 0. If g is as in (iii) then f′′′ has at least one non-real zero.

If m ≥ 1 then m = n and g is a constant multiple of a real entire function.

We define a class of functions. We say that f ∈ U_{0} if f is of the form

f(z) = g(z) exp{−az^{2}}

where a ≥ 0 and g is a constant multiple of a real entire function with genus not exceeding 1 and with only real zeros. The class U_{0} is the so-called Laguerre–Pólya class. We have f ∈ U_{0} if, and only if, there are real polynomials P_{n} with only real zeros such that P_{n} → f locally uniformly in C. Also, f ∈ U_{0} if, and only if, we may write

f(z) = cz^{m}e^{−az^{2}+bz} ∏_{n=1}^{∞}(1 − \frac{z}{z_{n}} )e^{z/z_{n}}

where c is a non-zero complex constant, m is a non-negative integer, a ≥ 0, b is a real number, z_{n} ∈ R \ {0} for all n ≥ 1, and ∑_{n=1}^{∞} z_{n}^{−2} < ∞. Here R denotes the real axis. Thus if f ∈ U_{0} then f^{(n)} ∈ U_{0} and so f^{(n)} has only real zeros for all n ≥ 0.

Hellerstein, Shen and Williamson [2] developed a method for dealing with the reality of zeros of the first two derivatives of a strictly non-real meromorphic function with only real poles. We will follow their approach to the extent possible. Differences start to appear after the first few formulas have been obtained. They are due to the presence of non-real poles, even if there are at most two poles. One of the principal changes is that in some places, functions such as cos(az + b) + i sin(az + b) are replaced by functions of the form cos(az + b) + z sin(az + b). The latter functions can no longer be expressed as exponentials and have infinitely many zeros. Also we cannot take fractional powers of such functions and obtain single-valued functions in the plane. Therefore certain arguments that take up a few pages in [2] are replaced by much longer ones that amount to a careful analysis of what can be said about the new functions that we end up with. The case m ≥ 1 of Theorem 1 is still reasonably close to [2] but in the case m = 0 the situation changes more radically.

2. Proof of Theorem 1 when m ≥ 1

2.1. Suppose that f is as in (1.1) where g is an entire function with only real zeros with g(i)g(−i) ≠ 0 and m and n are positive integers. We first consider the case m ≠ n. Then f is strictly non-real since m ≠ n (for if f were a constant multiple of a real function then f would have a pole of the same order at i and at −i). Write

(2.1)  f(z) = \overline{f(\overline{z})}H(z),
(2.2)  f′(z) = \overline{f′(\overline{z})}G(z).

Then H and G are meromorphic in the plane and have modulus 1 on R. Suppose that f, f′, and f′′ have only real zeros. Then H and G have no zeros or poles at points other than ±i.

We denote various non-zero complex constants by A_{1}, A_{2}, …. Close to z = i we have

f(z) ∼ \frac{A_{1}}{(z − i)^{n}} , \overline{f(\overline{z})} ∼ \frac{A_{2}}{(z − i)^{m}}

so that H(z) ∼ A_{3}(z − i)^{m−n}. Similarly, G(z) ∼ A_{4}(z − i)^{m−n} as z → i. As z → −i we similarly get H(z) ∼ A_{5}(z + i)^{n−m} and G(z) ∼ A_{6}(z + i)^{n−m}. Hence there is an entire function h such that

H(z) = (\frac{z − i}{z + i})^{m−n}e^{ih(z)}

where |e^{ih(z)}| = 1 when z is real. Thus h is a real entire function. Similarly,

G(z) = (\frac{z − i}{z + i})^{m−n}e^{ig_{0}(z)}

where g_{0} is a real entire function. From (2.1) we get

(2.3)  \frac{f′}{f} (z) = \overline{\frac{f′}{f} (\overline{z})} + \frac{H′}{H} (z) = \overline{\frac{f′}{f} (\overline{z})} + ih′(z) + \frac{2i(m − n)}{z^{2} + 1}.

Following Hellerstein, Shen and Williamson [2, p. 321], we recall the functions from the Tsuji value distribution theory in a half plane defined by T_{0}(r, f) = m_{0}(r, f) + N_{0}(r, f),

m_{0}(r, f) = \frac{1}{2π} ∫_{arcsin(r^{−1})}^{π−arcsin(r^{−1})} log^{+}|f(r(sin θ)e^{iθ})| \frac{dθ}{r sin^{2} θ}

and

N_{0}(r, f) = ∫_{1}^{r} \frac{n(t, ∞)}{t^{2}} dt = ∑_{1≤r_{k}<r sin θ_{k}}(\frac{sin θ_{k}}{r_{k}} − \frac{1}{r}).

Here the points z_{k} = r_{k}e^{iθ_{k}} are the poles of f with Im z_{k} > 0, with due count of multiplicity, and n(t, ∞) denotes the number of poles of f in \{z : |z − \frac{1}{2}it| ≤ \frac{1}{2}t, |z| ≥ 1\}, also with due count of multiplicity. As in [2, Lemma D], we get

T_{0}(r, h′) = m_{0}(r, h′)

≤ m_{0}(r, f′/f) + m_{0}(r, \overline{(f′/f)(\overline{z})}) + m_{0}(r, 2(m − n)/(z^{2} + 1)) + log 3

= O(log r).

The same with z replaced by −z gives

T_{0}(r, h′(−z)) = m_{0}(r, h′(−z)) = O(log r).

As in [2, (2.11–12)], this gives T (r, h′) = O(r log r) so that the order of h′, and hence of h, is at most 1. Write

h_{1}(z) = h′(z) + \frac{2(m − n)}{z^{2} + 1}.

Then h_{1} is meromorphic with poles at ±i only and the order of h_{1} is the same as that of h, hence at most 1.

Since

(2.4)  \frac{f′}{f} (z) − \overline{\frac{f′}{f} (\overline{z})} = ih_{1}(z)

and all the zeros of f′, hence all the zeros of f′/f, are real, it follows that all the zeros of f′/f are also zeros of h_{1} . Let π_{1} be the canonical product of the zeros of f′/f. Then the exponent of convergence of π_{1} does not exceed 1 and

(2.5)  γ(z) ≡ π_{1}(z)\frac{f}{f′} (z)

is entire. Since the order of the real entire function π_{1} does not exceed 1, we see as in [2, p. 323] that

T_{0}(r, π_{1}) = m_{0}(r, π_{1}) = O(r^{ε}) for each ε > 0.

Hence

\begin{aligned}m_{0}(r, γ) & = T_{0}(r, γ) ≤ T_{0}(r, f/f′) + T_{0}(r, π_{1}) + log 2 \\ & ≤ T_{0}(r, f′/f) + T_{0}(r, π_{1}) + O(1) = O(log r) + O(r^{ε}) = O(r^{ε}) \end{aligned}

and similarly m_{0}(r, γ(−z)) = T_{0}(r, γ(−z)) = O(r^{ε}) for all ε > 0. (To see that T_{0}(r, f′/f) = O(log r), we apply [2, Lemma D, p. 322], noting that ff′′ has only real zeros and f has only finitely many non-real poles.) Thus the order of γ is at most 1. Let π_{2} be the canonical product of the zeros of γ. Then

(2.6)  γ(z) = π_{2}(z)e^{−(pz+q)}

for some complex constants p and q. Substituting (2.6) into (2.5) we get

(2.7)  \frac{f′}{f} (z) = \frac{π_{1}}{π_{2}} (z)e^{pz+q}.

All the zeros of f′/f, hence all of π_{1} , are real. The zeros of π_{2} are the poles of f′/f, that is, the zeros and poles of f, which are all real, apart from ±i, which are simple zeros of π_{2} . Hence in any case π_{2} is real if m ≥ 1. Thus

(2.8)  \frac{π_{1}}{π_{2}} (z) = \overline{\frac{π_{1}}{π_{2}} (\overline{z})}  if m ≥ 1.

Solving for π_{1}/π_{2} using (2.4), (2.7), and (2.8) and substituting into (2.7) we get

(2.9)  \frac{f′}{f} (z) = \frac{ih_{1}(z)e^{pz+q}}{e^{pz+q} − e^{\overline{p}z+\overline{q}}} = \frac{h_{1}(z)e^{i(az+b)}}{2 sin(az + b)}  if m ≥ 1

where a = Im p, b = Im q. Note that sin(az + b) ≢ 0 since h_{1} ≢ 0 (since h_{1}(±i) = ∞). So (2.9) makes sense.

Replacing f and f′ by f′ and f′′ above, we get a similar conclusion provided that T_{0}(r, f′′/f′) = O(log r). The conclusion is that

(2.10)  \frac{f′′}{f′} (z) = \frac{g_{1}(z)e^{i(Az+B)}}{2 sin(Az + B)}  if m ≥ 1

where

\frac{G′}{G} (z) = ig_{1}(z) = i(g_{0}(z) + \frac{2(m − n)}{z^{2} + 1}).

To prove that indeed T_{0}(r, f′′/f′) = O(log r), take the logarithmic derivative of (2.9) to get

(2.11)  \frac{f′′}{f′} (z) = \frac{f′}{f} (z) + \frac{h′_{1}}{h_{1}} (z) + ia − a cot(az + b).

Since T_{0}(r, f′/f) = O(log r), we obtain T_{0}(r, f′′/f′) = O(log r) provided that T_{0}(r, h′_{1}/h_{1}) = O(log r) and T_{0}(r, α′/α) = O(log r) where α(z) = sin(az + b). By [2, Lemma A] we have

m_{0}(r, α′/α) = O(log r + log T_{0}(r, α))

for all r outside a set of finite linear measure, while T_{0}(r, α) = O(r^{ε}) for all ε > 0 (in the same way as for T_{0}(r, γ) above) since α has order 1. Thus m_{0}(r, α′/α) = O(log r). Now T_{0}(r, α′/α) = m_{0}(r, α′/α) = O(log r) since α has only real zeros. (If a = 0, we need not consider α at all.) The same applies to h_{1} if we can show that h_{1} has only finitely many non-real zeros and poles (recall that the order of h_{1} is at most 1). From (2.9) we see that if h_{1}(z) = 0 then sin(az + b) = 0 or (f′/f)(z) = 0. All such zeros are real, so all zeros of h_{1} are real and hence T_{0}(r, h′_{1}/h_{1}) = O(log r) as required. This completes the proof of (2.10). Note that the function g_{1} in (2.10) has only real zeros, has poles at ±i only, both of them simple poles, and that g_{1} is real and of order at most 1.

Now by (2.10) and (2.11) we get

(2.12)  \begin{matrix}\frac{h′_{1}}{h_{1}} (z)+ & (\frac{h_{1}(z)}{2} − a) cot(az + b) + i(\frac{h_{1}(z)}{2} + a) \\ = \frac{g_{1}(z)}{2} cot(Az + B) + i\frac{g_{1}(z)}{2}. \end{matrix}

For z ∈ R, equate the real and imaginary parts in (2.12) to get

(2.13)  h_{1}(z) + 2a = g_{1}(z),

(2.14)  \frac{h′_{1}}{h_{1}} (z) + (\frac{h_{1}(z)}{2} − a) cot(az + b) = \frac{g_{1}(z)}{2} cot(Az + B),

which remain valid for all z ∈ C by analytic continuation. Using (2.13) in (2.14) to eliminate g_{1} we obtain

(2.15) \frac{h′_{1}}{h_{1}} (z) + \frac{h_{1}(z)}{2}(cot(az + b) − cot(Az + B)) = a(cot(az + b) + cot(Az + B)).

Now f′/f has residue −m at −i and −n at i. For each of g_{1} and h_{1} these residues are i(m−n) and −i(m−n). For f′′/f′ they are −(m+1) and −(n+1). So by (2.9),

(2.16)  −m = \frac{i(m − n)e^{i(−ai+b)}}{2 sin(−ai + b)} ,
(2.17)  −n = \frac{−i(m − n)e^{i(ai+b)}}{2 sin(ai + b)} ,

and by (2.10),

(2.18)  −(m + 1) = \frac{i(m − n)e^{i(−Ai+B)}}{2 sin(−Ai + B)} ,
(2.19)  −(n + 1) = \frac{−i(m − n)e^{i(Ai+B)}}{2 sin(Ai + B)} .

Dividing (2.16) by (2.17) and (2.18) by (2.19) we obtain

(2.20)  \frac{m}{n} = −\frac{sin(ai + b)}{sin(−ai + b)}e^{2a},  \frac{m + 1}{n + 1} = −\frac{sin(Ai + B)}{sin(−Ai + B)}e^{2A}

so that by taking absolute values we get

(2.21)  2a = log \frac{m}{n},  2A = log \frac{m + 1}{n + 1}.

We have a ≠ 0 ≠ A since m ≠ n. Furthermore, it follows that

sin(ai + b) = − sin(−ai + b) = sin(ai − b),

hence cos ai sin b = 0, so sin b = 0 and hence b = νπ for some integer ν.

If ν is even then sin(az + b) = sin az and e^{i(az+b)} = e^{iaz} so that we may take ν = 0. If ν is odd then

\frac{e^{i(az+b)}}{sin(az + b)} = \frac{−e^{iaz}}{− sin az} = \frac{e^{iaz}}{sin az}

so that we may always take b = 0 and similarly B = 0. Now (2.15) reads

(2.22)  \frac{h′_{1}}{h_{1}} (z) + \frac{h_{1}(z)}{2}(cot az − cot Az) = a(cot az + cot Az).

Dividing (2.22) by −h_{1} we get

(\frac{1}{h_{1}} )′ + a(cot az + cot Az)\frac{1}{h_{1}} = \frac{1}{2}(cot az − cot Az).

Thus for 0 < ε ≤ Az < π we have

\frac{1}{h_{1}} = e^{−(log sin az+(a/A) log sin Az)} ∫ ^{z}(sin at sin^{a/A}(At))\frac{1}{2}(cot at − cot At) dt

with a suitable implied constant of integration, so that if

h_{2}(z) ≡ \frac{sin az sin^{a/A} Az}{h_{1}(z)} for 0 < ε ≤ Az < π

then

h_{2}(z) − h_{2}(ε/A) = \frac{1}{2} ∫_{ε/A}^{z} sin^{(a/A)−1} At sin(A − a)t dt.

Since m ≠ n, we have

β ≡ \frac{a}{A} = \frac{log(m/n)}{log ((m + 1)/(n + 1))} > 0

so that the integral on the right has a finite limit as ε → 0+ or Az → π− and so h_{2} extends continuously to [0, Aπ] or [Aπ, 0]. Write h_{3}(z) = h_{2}(z/A). Then for 0 ≤ z ≤ π,

h_{3}(z) − h_{3}(ε) = \frac{1}{2A} ∫_{ε}^{z} sin^{β−1} t sin(1 − β)t dt

while h_{3}(z) = (sin βz sin^{β} z)/h_{1}(z/A).

If β is not an integer, then, since h_{1} is entire and h_{3} is continuous on [0, π], we see that h_{3}(0) = h_{3}(π) = 0. Hence

0 = ∫_{0}^{π} sin(β − 1)t sin^{β−1} t dt.

Proceeding in the same way with h̃_{1}(z) = h_{1}(z/A) we get

\frac{1}{2A}(cot βz − cot z) = \frac{1}{h̃_{1}(z)}β(cot βz + cot z) + (\frac{1}{h̃_{1}(z)})′.

Set ζ = z − π, where π < z < 2π, and write ĥ_{1}(ζ) = h̃_{1}(ζ + π). Then

(\frac{1}{ĥ′_{1}} )′ + \frac{1}{ĥ_{1}} β[cot(βζ + βπ) + cot ζ] = \frac{1}{2A}[cot(βζ + βπ) − cot ζ].

Now 0 < ζ < π and with

h̃_{2}(ζ) = \frac{sin(βζ + βπ) sin^{β}(ζ)}{ĥ_{1}(ζ)}

we obtain

h̃_{2}(ζ) − h̃_{2}(ε) = \frac{1}{2A} ∫_{ε}^{ζ} sin^{β−1} ξ sin((β − 1)ξ + βπ) dξ for 0 < ε < ζ < π.

As above we find that if β is not an integer then

0 = ∫_{0}^{π} sin^{β−1} t sin((β − 1)t + βπ) dt.

As in [2, pp. 326–327] we now see that β must be a positive integer, and then, as in [2, Lemma 5] we see that β ∈ {1, 2, 3}. Each of the choices β = 1, 2, 3 leads, as in [2, pp. 328–330], to an explicitly given function h_{1} . But none of these functions h_{1} has any non-real poles, which gives a contradiction since our h_{1} must have a pole at ±i.

We conclude that if f is of the form (1.1) with m ≥ 1, n ≥ 1, and m ≠ n then at least one of f, f′, and f′′ has at least one non-real zero. This completes the proof of Theorem 1 when m ≥ 1 and m ≠ n.

Suppose now that m = n ≥ 1 and that f (or, equivalently, g) is strictly non-real. Then we may proceed as above, starting at the beginning of §2.1 and continuing up to and including (2.9), apart from the possibility that h_{1} ≡ 0. Note that (2.11) is still valid. But now h_{1} = h′, and if h_{1} ≡ 0 then by (2.3), we have H′/H ≡ 0 so that H is constant. By (2.1) this means that f is a constant multiple of a real function, which is a contradiction. Hence h_{1} ≡ h′ ≢ 0, and so sin(az + b) ≢ 0 in (2.9).

If now f′ is a constant multiple of a real function then f′′/f′ is real and by (2.11), the function f′/f + ia is real. By (2.3), H′/H = −2ia so that H = e^{−2iaz} = e^{ih(z)} and again by (2.9), f′/f = \frac{1}{2}(−2a)(i + cot(az + b)) so that f(z) = e^{−aiz+C}/ sin(az + b), which is a contradiction both if a = 0 or if a ≠ 0, since then f has no poles at ±i. It follows that f′ is strictly non-real and we see that the previous argument is valid up to (2.15). Now (2.16)–(2.21) are still valid so that we get a = A = 0. Then if sin b ≠ 0, by (2.20) we obtain

1 = \frac{m}{n} = −\frac{sin b}{sin b} = −1,

which is a contradiction. Thus a = sin b = 0 so that sin(az + b) ≡ 0, which is also a contradiction. We conclude that if m = n ≥ 1 then f is a constant multiple of a real function, as asserted. This completes the proof of Theorem 1 when m ≥ 1. We remark that it has been proved in [3, Theorems 4, 5] that when m = n ≥ 1 (and f and g are real), the functions g and f are of finite order.

3. Preliminaries for the proof of Theorem 1 when m = 0

3.1. Suppose that f is as in Theorem 1 with m = 0 and n ≥ 1, and with g(i) ≠ 0. Then f is strictly non-real. Again we denote various non-zero complex constants by A_{1}, A_{2}, …. Close to z = i we have

f(z) ∼ \frac{A_{1}}{(z − i)^{n}} ,  f′(z) ∼ \frac{A_{2}}{(z − i)^{n+1}} ,  \overline{f(\overline{z})} ∼ A_{3},  \overline{f′(\overline{z})} ∼ A_{4}.

We define H and G as in (2.1) and (2.2). Hence

H(z) ∼ A_{5}(z − i)^{−n}  and  G(z) ∼ A_{7}(z − i)^{−(n+1)}  as z → −i,

while

H(z) ∼ A_{6}(z + i)^{n}  and  G(z) ∼ A_{8}(z + i)^{n+1}  as z → −i.

Thus there are real entire functions h and g_{0} such that

(3.1)  H(z) = (\frac{z + i}{z − i})^{n}e^{ih(z)},

G(z) = (\frac{z + i}{z − i})^{n+1}e^{ig_{0}(z)}.

From (2.1) and (3.1) we get

(3.2)  \frac{f′}{f} (z) − \overline{\frac{f′}{f} (\overline{z})} = \frac{H′}{H} (z) = ih′(z) − \frac{2in}{z^{2} + 1} ≡ ih_{1}(z).

Again h_{1} is real, and h′ and h_{1} have the same order, which does not exceed 1. Define the functions π_{1} , γ and π_{2} as before. Thus (2.5) holds, the functions π_{1} and γ have order at most 1, and (2.6) and (2.7) hold.

Now the zeros of π_{2} are real apart from a simple zero at z = i. Thus we may choose

π_{2}(z) = (z − i)π_{3}(z)

where π_{3} is real. Hence

(3.3)  \frac{π_{1}}{π_{2}} (z) = \overline{\frac{π_{1}}{π_{2}} (\overline{z})}\frac{z + i}{z − i}.

Substituting (2.7) and (3.3) into (2.4) (which still holds) we get

\frac{π_{1}}{π_{2}} (z)e^{pz+q} = \frac{π_{1}}{π_{2}} (z)e^{\overline{p}z+\overline{q}} \frac{z − i}{z + i} + ih_{1}(z).

Solving for π_{1}/π_{2} and substituting into (2.7) we obtain, with a = Im p and b = Im q that

(3.4)  \begin{aligned}\frac{f′}{f} (z) & = \frac{ih_{1}(z)e^{i(az+b)}}{e^{i(az+b)} − e^{−i(az+b)}(z − i)/(z + i)} \\ & = \frac{ih_{1}(z)(z + i)e^{i(az+b)}}{(z + i)e^{i(az+b)} − (z − i)e^{−i(az+b)}} \\ & = \frac{h_{1}(z)(z + i)e^{i(az+b)}}{2(cos(az + b) + z sin(az + b))}. \end{aligned}

The counterpart of (2.11) is

\frac{f′′}{f′} (z) = \frac{f′}{f} (z) + \frac{h′_{1}}{h_{1}} (z) + \frac{1}{z + i} + ia

− \frac{sin(az + b) + a(z cos(az + b) − sin(az + b))}{cos(az + b) + z sin(az + b)}.

If a = 0 then clearly T_{0}(r, f′′/f′) = O(log r). Suppose that a ≠ 0. We can deduce that T_{0}(r, f′′/f′) = O(log r) provided that cos(az + b) + z sin(az + b) has only finitely many non-real zeros. This last function vanishes at a point z if and only if tan(az + b) = −1/z, that is, if tan w = 1/(cw + d) for suitable real c, d with c ≠ 0, where w = az + b. It is easily seen that this last equation has only finitely many (in fact, at most two) non-real solutions (using the method of solution to [7, Problem V.172, pp. 65, 244]). Thus T_{0}(r, f′′/f′) = O(log r) as required.

It now also follows from (3.4) that every zero of h_{1} is real, apart from the at most two non-real zeros of cos(az + b) + z sin(az + b), at which h_{1} must vanish. These non-real zeros are distinct from the points ±i since

cos(az + b) + z sin(az + b) = e^{i(ai±b)} ≠ 0  at  z = ±i.

The same analysis for f′ and f′′ instead of f and f′ gives

(3.5)  \frac{f′′}{f′} (z) = \frac{g_{1}(z)(z + i)e^{i(Az+B)}}{2(cos(Az + B) + z sin(Az + B))}.

Let us write from now on

(3.6)

c = cos(az + b),  s = sin(az + b),
C = cos(Az + B),  and  S = sin(Az + B).

The counterpart of (2.12) is

(3.7) \frac{g_{1}(z + i)(C + iS)}{2(C + zS)} = \frac{h_{1}(z + i)(c + is)}{2(c + zs)} + \frac{h′_{1}}{h_{1}} + \frac{1}{z + i} + ia − \frac{s + a(zc − s)}{c + zs}.

Considering the real and imaginary parts of (3.7) for z ∈ R and noting that h_{1} and g_{1} are real, we get

(3.8)  \frac{g_{1}}{2} = \frac{h_{1}}{2} + a − \frac{1}{z^{2} + 1}

from the imaginary parts and then, after a calculation,

(3.9)  \begin{aligned}\frac{g_{1}}{2} \frac{zC − S}{zS + C} & = \frac{h_{1}}{2} \frac{zc − s}{zs + c} + \frac{h′_{1}}{h_{1}} − \frac{zc − s}{zs + c}(a − \frac{1}{z^{2} + 1}) \\ & = \frac{h_{1}}{2} \frac{zc − s}{zs + c} + \frac{h′_{1}}{h_{1}} + \frac{z}{z^{2} + 1} − \frac{(1 − a)s + azc}{c + zs} \end{aligned}

from the real parts. Substituting (3.8) into (3.9) we obtain

(3.10) \{\frac{h_{1}}{2} + a − \frac{1}{z^{2} + 1}\}\frac{zC − S}{zS + C} = \frac{h′_{1}}{h_{1}} + \frac{z}{z^{2} + 1} + \frac{h_{1}}{2} \frac{zc − s}{c + zs} − \frac{(1 − a)s + azc}{c + zs}.

Note that by (3.8), the formulas (3.4) and (3.5) read

(3.11)  \frac{f′}{f} = \frac{h_{1}}{2} \frac{(z + i)(c + is)}{c + zs},
(3.12)  \frac{f′′}{f′} = (\frac{h_{1}}{2} + a − \frac{1}{z^{2} + 1})\frac{(z + i)(C + iS)}{C + zS}.

We have now developed the formalism along the lines of Hellerstein, Shen and Williamson [2]. The rest of the proof can be characterized as a struggle to find a contradiction, showing that no function f satisfying the assumptions of Theorem 1 can arise after all, except in three very special cases.

We define

(3.13)  h_{2} = \frac{f}{f′} \frac{e^{i(az+b)}(C + zS)}{z − i} = \frac{2}{h_{1}} \frac{(c + zs)(C + zS)}{z^{2} + 1}

and

(3.14)  h_{3} = \frac{h_{2}}{C + zS} = \frac{f}{f′} \frac{e^{i(az+b)}}{z − i} = \frac{2}{h_{1}} \frac{(c + zs)}{z^{2} + 1}.

We write

S_{0} = Sc−Cs = sin((A−a)z+(B−b)), C_{0} = Cc+Ss = cos((A−a)z+(B−b)).

We note that

(3.15)  h′_{2} = S_{0} + (A − a)\frac{zC − S}{C + zS} h_{2}

and

(3.16)  \frac{h′_{2}}{h_{2}} = \frac{S_{0}}{h_{2}} + (A − a)\frac{zC − S}{C + zS} .

Consequently, since h_{3} = h_{2}/(C + zS), we have

h′_{3} = \frac{S_{0}}{C + zS} − \frac{(1 − a)S + azC}{C + zS} h_{3},

that is,

(3.17)  (C + zS)h′_{3} = S_{0} − ((1 − a)S + azC)h_{3}.

We prove (3.15). Dividing by −h_{1}/2 we obtain from (3.10) that

(3.18)  \begin{aligned}(\frac{2}{h_{1}} )′ & + \frac{2}{h_{1}} \{(a − \frac{1}{z^{2} + 1})\frac{zC − S}{zS + C} − \frac{z}{z^{2} + 1} + \frac{(1 − a)s + azc}{c + zs}\} \\ & = \frac{−zC + S}{C + zS} + \frac{zc − s}{c + zs}. \end{aligned}

Since

\frac{−zC + S}{C + zS} + \frac{zc − s}{c + zs} = \frac{(z^{2} + 1)S_{0}}{(C + zS)(c + zs)}

and

\frac{1}{z^{2} + 1} \frac{zC − S}{C + zS} = \frac{z}{z^{2} + 1} − \frac{S}{C + zS} ,

we can write (3.18) as

\begin{aligned}(\frac{2}{h_{1}} )′ & + \frac{2}{h_{1}} \{(\frac{(C + zS)(c + zs)}{z^{2} + 1})′ / (\frac{(C + zS)(c + zs)}{z^{2} + 1}) + (a − A)\frac{zC − S}{C + zS} \} \\ & = \frac{(z^{2} + 1)S_{0}}{(C + zS)(c + zs)}. \end{aligned}

A calculation based on the definition (3.13) of h_{2} in terms of h_{1} now yields (3.15). Note that in all cases, neither one of the meromorphic functions h_{2} and h_{3} can vanish identically. Note further that each of h_{2} and h_{3} is real and is of order at most 1. Since the pole of f at z = i is cancelled out in h_{2} and h_{3} , and since f and f′ have only real zeros and no poles other than z = i, it follows that h_{3} has only real zeros and poles, and h_{2} has only real poles and only real zeros with the possible exception of the at most two non-real zeros of C + zS. (We saw earlier that c + zs has at most two non-real zeros, and the same applies to C + zS.)

We consider a number of cases:

Case I: A = a (so that S_{0} is constant) and S_{0} ≠ 0;

Case II: A = a and S_{0} = 0;

Case III: a ≠ A = 0; Case IV: A ≠ a = 0; Case V: A ≠ a and aA ≠ 0.

We shall need later on the fact that a function that differs from a function in U_{0} by a polynomial factor, behaves in certain ways like a function in U_{0} when |x| is large enough. We formulate this result as a lemma, which will be proved in the last Section 11 of this paper.

Lemma 1. Let P be a non-constant real rational function, and suppose that the real entire function Φ belongs to U_{0} and has no common zeros with P. Define Ψ = Φ/P. Then there is a positive number R such that on each of the intervals (−∞, −R) and (R, ∞), the zeros of Ψ and Ψ′ are interlaced. That is, if x_{1} and x_{2} are consecutive zeros of Ψ with R < x_{1} < x_{2} or with x_{1} < x_{2} < −R, then Ψ′ has exactly one zero on (x_{1}, x_{2}), and this point is a simple zero of Ψ′. Further, if x_{1} and x_{2} are consecutive zeros of Ψ′ with R < x_{1} < x_{2} or with x_{1} < x_{2} < −R, then Ψ has exactly one zero on (x_{1}, x_{2}), and this point is a simple zero of Ψ, unless it is the case that exactly one of x_{1} and x_{2} is also a zero of Ψ, in which case Ψ has no zeros on (x_{1}, x_{2}).

4. Case I

Here A = a and the constant S_{0} = sin(B − b) ≠ 0. Now (3.15) reads h′_{2} = S_{0} so that h_{2}(z) = S_{0}z + α for some real α. By (3.13),

(4.1)  \frac{f′}{f} = \frac{e^{i(az+b)}(C + zS)}{(z − i)(S_{0}z + α)},
\frac{h_{1}}{2} = \frac{(C + zS)(c + zs)}{(z^{2} + 1)(S_{0}z + α)}.

Considering residues of both sides of (4.1) at z = i, we obtain

−n = \frac{e^{−2A}e^{i(b+B)}}{S_{0}i + α} = \frac{e^{−2A}e^{i(b+B)}(−S_{0}i + α)}{S_{0}^{2} + α^{2}},

which gives after a calculation that α = S_{0} cot(b + B), so that

h_{2} = (z + cot(b + B))S_{0},  and  − nS_{0} = e^{−2A} sin(b + B) ≠ 0.

Thus

\frac{f′}{f} = \frac{e^{i(az+b)}(C + zS)}{(z − i)(z + cot(b + B))S_{0}} .

Since f′/f has only real zeros, also C + zS has only real zeros. If A = 0 then C + zS is a linear real polynomial, which has only real zeros. If A ≠ 0 then it follows that

tan(Az + B − \frac{1}{2}π) − z = tan(Az + B − \frac{1}{2}π)(Az + B − \frac{1}{2}π)/A + (B − \frac{1}{2}π)/A has only real zeros. By [5], this is possible only if 1/A ≥ 1, that is, 0 < A ≤ 1 (further, B must satisfy a more complicated condition given in [5], but we shall not make use of that).

Suppose first that a = A = 0 ≠ S_{0} . Then C = cos B and S = sin B are constants and

\frac{f′}{f} = \frac{e^{ib}(C + zS)}{(z − i)(z + cot(b + B))S_{0}} = \frac{−n}{z − i} + \frac{β}{z + cot(b + B)}

for some non-negative integer β while −nS_{0} = sin(b + B) ≠ 0. A calculation shows that

β = −\frac{e^{−iB} sin b}{sin(B − b)}.

Since β is real, we have sin B = 0 so that B = kπ and C = cos B = (−1)^{k} for some integer k. This gives β = 1 and n = 1. Hence f is a non-zero complex constant multiple of (z + cot(b + B))/(z − i) and is of the form (ii) in Theorem 1. The remaining statements in Theorem 1 concerning part (ii) are clear.

Suppose then that 0 < A ≤ 1. By (3.13), we have

\frac{h_{1}}{2} = \frac{(C + zS)(c + zs)}{(z^{2} + 1)(z + cot(b + B))S_{0}} .

The equation (3.5) still holds. Since f′′ and C +zS have only real zeros, it follows from (3.5) and (3.8) that g_{1} and hence

X(z) ≡ (z^{2} + 1)(z + cot(b + B))S_{0}g_{1}/2

= (z^{2} + 1)(z + cot(b + B))S_{0}h_{1}/2 + (a(z^{2} + 1) − 1)(z + cot(b + B))S_{0}

= (C + zS)(c + zs) + a(z^{2} + 1)(z + cot(b + B))S_{0} − (z + cot(b + B))S_{0}

have only real zeros. Hence X ∈ U_{0} . We have a ≠ 0 since a = A ∈ (0, 1]. Since (C +zS)(c+zs) = O(z^{2}) as z → ±∞ along the real axis, it follows that X(z) ≠ 0 for all real z such that |z| is sufficiently large. Hence X has only finitely many zeros so that X(z) = P (z)e^{γz}, where γ ∈ R and P is a polynomial with only real zeros. Differentiating four times we find that ((C + zS)(c + zs))^{(4)} = P_{1}(z)e^{γz}, where P_{1} is a polynomial with only real zeros. But since (C + zS)(c + zs) is a real function with infinitely many real zeros, it follows from Rolle’s theorem that ((C + zS)(c + zs))^{(4)} has infinitely many zeros, which gives a contradiction. Thus no function f satisfying the assumptions of Theorem 1 arises in this case. This completes our treatment of Case I.

5. Case II

Here A = a and the constant S_{0} = sin(B − b) = 0. Then B = b + kπ for some integer k, and C = (−1)^{k}c, S = (−1)^{k}s (recall the notation (3.6)). Now (3.15) reads h′_{2} = 0 so that h_{2}(z) = α for some non-zero real α. By (3.13),

\frac{f′}{f} = \frac{(−1)^{k}e^{i(Az+B)}(C + zS)}{(z − i)α} = \frac{(−1)^{k}e^{i(az+b)}(c + zs)}{(z − i)α},

so that C + zS and c + zs have only real zeros. Considering residues at z = i we get −n = (−1)^{k}e^{−2A}e^{2iB}/α so that α = (−1)^{k+1}e^{−2A}e^{2iB}/n and

(5.1)  \frac{f′}{f} = \frac{−ne^{2A}e^{−2iB}(c + is)(c + zs)}{z − i}.

Hence by (3.11),

(5.2)  \frac{h_{1}}{2} = \frac{f′}{f} \frac{c + zs}{(c + is)(z + i)} = \frac{−ne^{2A}e^{−2iB}(c + zs)^{2}}{z^{2} + 1} .

Since f′′ and C + zS have only real zeros, it follows from (3.5) and (3.8) that g_{1} and hence

\frac{g_{1}}{2} = \frac{h_{1}}{2} + a − \frac{1}{z^{2} + 1} = \frac{−ne^{2A}e^{−2iB}(c + zs)^{2}}{z^{2} + 1} + a − \frac{1}{z^{2} + 1}

have only real zeros. Also, since g_{1} is real, we have e^{−2iB} = e^{−2ib} = ±1, and the function

(5.3)  X(z) = K(c + zs)^{2} + a(z^{2} + 1) − 1

has only real zeros, where K = −e^{−2ib}ne^{2a}. Hence X ∈ U_{0} . Since e^{−2ib} = ±1, the number b is an integral multiple of π/2. Since replacing b and B by b + lπ and B + lπ, respectively, for some integer l does not change the problem, we may assume that b ∈ {0, π/2}.

Since c+zs has only real zeros, we have c+zs ∈ U_{0} , so that (1−a)s+azc = (c + zs)′ ∈ U_{0} . Suppose that a ≠ 1 and a ≠ 0. Then

tan(az + b) + (1 − a)^{−1}(az + b) − b/(1 − a)

has only real zeros. By [5], this implies that −(1 − a)^{−1} ≥ 1, that is, 1 < a ≤ 2. Since c + zs has only real zeros, the function cos z + (sin z)(z − b)/a, which is c+zs evaluated at (z−b)/a, has only real zeros, and therefore so does the function tan z + a/(z − b). If b = 0, then the equation tan z = −a/z, where 1 < a ≤ 2, has only real solutions. Taking z = it, where t ∈ R \ {0}, we see that then tanh t = a/t has no real solutions, which is not true. Thus b ≠ 0, and so b = π/2. Hence e^{2ib} = −1 and so K = ne^{2a}.

Since K > 0 and a > 0, it follows that X(x) > 0 when x is real and |x| is large enough. Hence X has only finitely many zeros so that X(z) = P (z)e^{γz}, where γ ∈ R and P is a polynomial with only real zeros. Differentiating three times we find that ((c + zs)^{2})′′′ = X′′′ = P_{1}(z)e^{γz}, where P_{1} is a polynomial with only real zeros. But since c + zs ∈ U_{0} and c + zs has infinitely many zeros, it follows that ((c + zs)^{2})′′′ has infinitely many zeros, which gives a contradiction. This rules out the case a ∉ {0, 1}.

Suppose then that a = 1. Then the function cos z+(z−b) sin z, which is c+zs evaluated at z−b, has only real zeros, and therefore so does tan z+1/(z−b), where b ∈ {0, π/2}. If b = 0, we get a contradiction as above, considering tanh t = 1/t. Hence b = π/2 so that e^{2ib} = −1 and K = ne^{2a} = ne^{2} > 0. Now we get a final contradiction as in the previous paragraph, considering X′′′ for X defined by (5.3). This completes our treatment of the case when A = a ≠ 0 and S_{0} = 0.

Suppose that A = a = S_{0} = 0. Then by (5.1),

\frac{f′}{f} = \frac{−ne^{−2ib}e^{ib}(c + zs)}{z − i} = \frac{−ne^{−ib}(c + zs)}{z − i} = \frac{−n}{z − i} − nse^{−ib}.

Since h_{1}/2 = −ne^{−2ib}(c+zs)^{2}/(z^{2} +1), as given by (5.2), is real, we have e^{−2ib} = ±1. Again, we may assume that b ∈ {0, π/2}. If b = π/2 then c = 0 and −se^{−ib} = i. Hence

\frac{f′}{f} = \frac{−n}{z − i} + in,

which gives f(z) = A_{0}e^{inz}/(z − i)^{n} for some non-zero complex constant A_{0} . Thus f is as in (iii) of Theorem 1. Now f′ = A_{0}e^{inz}inz/(z − i)^{n} and f′′(z) = −A_{0}ne^{inz}(nz^{2} − 1)/(z − i)^{n+2} so that f, f′, and f′′ have only real zeros. But f′′′(z) = −iA_{0}n^{3}e^{inz}(z^{3} − (3/n)z − 2in^{−2})/(z − i)^{n+3} . Since z^{3} − (3/n)z − 2in^{−2} is not a constant multiple of a real polynomial, it follows that f′′′ has at least one non-real zero. In fact, all the three zeros of f′′′ are non-real.

If b = 0 then s = 0 and f′/f = −n/(z − i). Hence f(z) = A_{0}/(z − i)^{n} for some non-zero complex constant A_{0} . Thus f is as in (i) of Theorem 1. In this case f^{(k)} has no zeros and hence has only real zeros for all k ≥ 0. This completes our treatment of Case II.

6. Sign analysis: Case IV

We write sgn x = x/|x| when x ≠ 0 and sgn 0 = 0. As in [2, p. 325], we note that (h′_{1}/h_{1})(iy) = o(y) as y → ∞. Further, since (z^{2} + 1)h_{1} has order at most 1 and belongs to U_{0} , we see that |h_{1}(iy)| ≥ By for some fixed positive constant B as y → ∞ provided that h_{1} has at least three zeros with due count of multiplicity.

Suppose that A ≠ 0. As y → ∞, we have (S/C)(iy) = i sgn A + O(e^{−2|A|y}). Now we may write (3.10) for z = iy as

(6.1)  (i/2)h_{1}(iy)(sgn A + ic/s)(1 + o(1)) = o(y)  if a = 0 ≠ s,
(6.2)  (i/2)h_{1}(iy)(sgn A − y)(1 + o(1)) = o(y)  if a = s = 0,
(6.3)  (i/2)h_{1}(iy)(sgn A − sgn a)(1 + o(1)) = o(y)  if a ≠ 0.

Consider first the case a ≠ 0 so that (6.3) holds. This gives a contradiction as y → ∞ unless sgn A − sgn a = 0 or unless h_{1}(z) = e^{αz}P (z)/(z^{2} + 1) where α ∈ R and P is a polynomial of degree at most 2. Now if sgn A − sgn a = 0 then a/A > 0. Suppose then that sgn A − sgn a ≠ 0, so that sgn A = − sgn a, and that

(6.4)  h_{1}(z) = e^{αz} \frac{P (z)}{(z^{2} + 1)}.

As y → ∞, we have (s/c)(iy) = i sgn a + O(e^{−2|a|y}) → i sgn a = −i sgn A ≠ 0. We can write (3.10) as

(6.5) \begin{matrix}\{ & a − \frac{1}{z^{2} + 1} + \frac{e^{αz}P}{2(z^{2} + 1)}\} \frac{(C/S) − (1/z)}{1 + (C/S)(1/z)} \\ = α + \frac{P′}{P} − \frac{z}{z^{2} + 1} + \frac{e^{αz}P}{2(z^{2} + 1)} \frac{(c/s) − (1/z)}{1 + (c/s)(1/z)} − \frac{a(c/s) + ((1 − a)/z)}{1 + (c/s)(1/z)}. \end{matrix}

If α ≠ 0, we consider (6.5) for z = x where sgn(αx) = 1 and |x| → ∞ in a sequence such that 1/M ≤ |c/s| ≤ M and 1/M ≤ |C/S| ≤ M at all these x for some fixed positive number M. Clearly such a sequence can be found. Since the terms involving e^{αx} dominate, we deduce that (C/S) − (1/x) ∼ (c/s) − (1/x) and hence C/S ∼ c/s as |x| → ∞ in any such sequence. Thus Cs ∼ cS, that is, Cs = cS(1 + o(1)). Hence sin(az + b − Az − B) = Cs − cS = o(1) in any such sequence. If a ≠ A, we get a contradiction, for then we can choose the sequence of x = x_{k} so that for some fixed small ε > 0 we have |ax + b − (qπ/2)| > ε, |Ax + B − (qπ/2)| > ε and |ax + b − (Ax + B) − (qπ/2)| > ε for all integers q and all x = x_{k}. It follows that a = A, which contradicts our assumption that sgn A ≠ sgn a. We conclude that α = 0.

Since (C/S)(iy) = i sgn a + O(e^{−2|A|y}) and (c/s)(iy) = −i sgn a + O(e^{−2|a|y}) as y → ∞, a calculation shows that we may now write (3.10) or (6.5) as

\{a − \frac{1}{z^{2} + 1} + \frac{P}{2(z^{2} + 1)}\}\frac{i sgn a − (1/z)}{1 + (i sgn a)(1/z)}

= \frac{P′}{P} − \frac{z}{z^{2} + 1} + \frac{P}{2(z^{2} + 1)} \frac{(−i sgn a) − (1/z)}{1 − (i sgn a)(1/z)}

− \frac{−ai sgn a + ((1 − a)/z)}{1 − (i sgn a)(1/z)} + O(e^{−Ky})

and hence, as a longer calculation shows, as

(6.6)  \frac{−(z^{2} + 1)P′ + 2zP + P^{2}i sgn a}{(z^{2} + 1)P} = O(e^{−Ky})

valid for z = iy as y → ∞, with some constant K > 0. Now (6.6) can be valid only if −(z^{2} + 1)P′ + 2zP + P^{2}i sgn a ≡ 0. This can be written as ((z^{2} + 1)/P )′ = −i sgn a, which is impossible since (z^{2} + 1)/P is real. This contradiction shows that the assumption sgn a ≠ sgn A cannot be satisfied. This proves that when aA ≠ 0, we have sgn a = sgn A, that is, a/A > 0.

Suppose then that A ≠ 0 and a = s = 0, so that (6.2) holds. Now (6.2) implies that h_{1}(iy) = o(1) as y → ∞. It follows that h_{1} is of the form (6.4) where deg P ≤ 1. Now (3.10) reads

(6.7)  \{\frac{e^{αz}P (z)}{2(z^{2} + 1)} − \frac{1}{z^{2} + 1}\}\frac{zC − S}{zS + C} = α + \frac{P′}{P} − \frac{z}{z^{2} + 1} + \frac{e^{αz}zP (z)}{2(z^{2} + 1)}.

If α ≠ 0, we consider (6.7) in a sequence of x for which sgn(αx) = 1 and S(x) stays bounded away from zero, and find that C/S ∼ x, which is impossible. Thus α = 0. Hence (zC − S)/(C + zS) is a rational function, which is impossible. So this case cannot occur.

Suppose then that A ≠ 0 and a = 0 ≠ s, so that (6.1) holds. Now (6.1) implies that h_{1}(iy) = o(y) as y → ∞. It follows that h_{1} is of the form (6.4) where deg P ≤ 2. Now (3.10) reads

(6.8) \{\frac{e^{αz}P (z)}{2(z^{2} + 1)} − \frac{1}{z^{2} + 1}\}\frac{zC − S}{zS + C} = α+ \frac{P′}{P} − \frac{z}{z^{2} + 1} + \frac{e^{αz}P (z)}{2(z^{2} + 1)} \frac{zc − s}{c + zs} − \frac{s}{c + zs}.

If α ≠ 0, we consider (6.8) in a sequence of x for which sgn(αx) = 1 and S(x) stays bounded away from zero, and find that C/S ∼ c/s, which is impossible since c/s is a fixed constant while the sequence of values of x can be chosen so that C/S has no limit. Thus α = 0. Hence again (zC − S)/(C + zS) is a rational function, which is impossible. So this case cannot occur.

We conclude that if A ≠ 0 then a ≠ 0 and a/A > 0. In particular, Case IV, which means that A ≠ 0 = a, does not occur. The conclusion that if A ≠ 0 then a ≠ 0 and a/A > 0, will also be used when considering Case V later on.

7. Sign analysis: Case III

Consider Case III, so that a ≠ 0 = A. We have (s/c)(iy) → i sgn a as y → ∞. As before, we note that (h′_{1}/h_{1})(iy) = o(y) as y → ∞. Hence (3.10) gives, as y → ∞,

(7.1)  \frac{1}{2}h_{1}(iy)(iy + i sgn a + o(1)) + iay = o(y)  if S = 0,

and

(7.2)  \frac{1}{2}h_{1}(iy)(C/S + i sgn a + o(1)) = o(y)  if S ≠ 0.

Suppose that S = 0 so that (7.1) holds. Then h_{1}(iy) = −2a(1 + o(1)) as y → ∞. Hence h_{1}(z) = P (z)/(z^{2} + 1) where P is a real polynomial of degree 2 with leading coefficient −2a. Now (3.10) reads (c + zs)R_{1} = R_{2}c + R_{3}s where R_{1} , R_{2} , and R_{3} are certain real rational functions. This implies that R_{1} = R_{2} and zR_{1} = R_{3} . Now

\begin{aligned}R_{1} & = \{\frac{h_{1}}{2} + a − \frac{1}{z^{2} + 1}\}\frac{zC − S}{zS + C} − \frac{h′_{1}}{h_{1}} − \frac{z}{z^{2} + 1}, \\ R_{2} & = \frac{h_{1}}{2} z − az,  and  R_{3} = −\frac{h_{1}}{2} − (1 − a). \end{aligned}

We have zR_{2} = zR_{1} = R_{3} , which yields

(z + 1)h_{1}/2 = −1 + a(1 + z^{2}).

Thus

P (z) = (z^{2} + 1)h_{1}(z) = 2(z^{2} + 1)(−1 + a(1 + z^{2}))/(z + 1),

which is not a polynomial of degree 2 for any non-zero value of a. This contradiction shows that the case S = 0 is impossible.

Suppose that S ≠ 0 so that (7.2) holds. Then necessarily h_{1}(iy) = o(y), and thus h_{1}(z)/2 = e^{αz}P (z)/(z^{2} + 1) where α is real and P is a real polynomial of degree at most 2. Now (3.10) reads

(7.3)  (e^{αz}R_{4} + R_{5})(c + zs) = (e^{αz}R_{6} + R_{7})c + (e^{αz}R_{8} + R_{9})s,

where the R_{j} are certain real rational functions for 4 ≤ j ≤ 9.

If α ≠ 0, we consider both sides of (7.3) for z = x_{k} for various sequences x_{k} of real numbers such that sgn(αx_{k}) = 1, |x_{k}| → ∞ and either s(x_{k}) = 0 for all k or c(x_{k}) = 0 for all k. This shows that R_{4} = R_{6} , R_{5} = R_{7} , zR_{4} = R_{8} , and zR_{5} = R_{9} . Calculating the expressions for the R_{j} we deduce that

0 = zR_{6} − R_{8} = \frac{z^{2}P (z)}{z^{2} + 1} + \frac{P (z)}{z^{2} + 1},

hence P ≡ 0, which is impossible.

Thus α = 0, and we may adjust notation so that R_{4} = R_{6} = R_{8} = 0. Now R_{5} = R_{7} and zR_{5} = R_{9} . Thus

0 = zR_{7} − R_{9} = \frac{z^{2}P (z)}{z^{2} + 1} − az^{2} + \frac{P (z)}{z^{2} + 1} + (1 − a) = P (z) + 1 − a(z^{2} + 1).

This yields h_{1}/2 = a − (z^{2} + 1)^{−1} . Next,

(7.4) \begin{aligned}0 & = R_{5} − R_{7} = 2\{a − \frac{1}{z^{2} + 1}\}\frac{zC − S}{zS + C} − (\frac{−2az}{1 − a(z^{2} + 1)} − \frac{2z}{z^{2} + 1}) \\ & − \frac{z}{z^{2} + 1} − \frac{z(a(z^{2} + 1) − 1)}{z^{2} + 1} + az \\ & = 2\{a − \frac{1}{z^{2} + 1}\}\frac{zC − S}{zS + C} + \frac{−2az}{1 − a(z^{2} + 1)} + \frac{2z}{z^{2} + 1}. \end{aligned}

Suppose that a ≠ 1. Since a ≠ 0, the function 1 − a(z^{2} + 1) has two distinct zeros, neither one of which is ±i. At least one of these zeros is different from the zero of zS + C. At this zero of 1 − a(z^{2} + 1), the right hand side of (7.4) has a pole, which gives a contradiction. Thus a = 1, and (7.4) reads

(7.5)  2\{1 − \frac{1}{z^{2} + 1}\}\frac{zC − S}{zS + C} + \frac{2}{z} + \frac{2z}{z^{2} + 1} = 0.

Hence C = 0, for otherwise the origin is a pole of the left hand side of (7.5). Thus

−2\{1 − \frac{1}{z^{2} + 1}\} + 2 + \frac{2z^{2}}{z^{2} + 1} = 0.

This reads 2 = 0, which is a contradiction. This completes our treatment of the Case III, showing that Case III cannot occur at all.

8. Case V

Consider Case V so that A ≠ a and aA ≠ 0. We shall show that this leads to a contradiction so that in this case no function f satisfying the assumptions of Theorem 1 arises. We have already seen in Section 6 that now a/A > 0.

Consider the real meromorphic function h_{3} of order at most 1 defined by (3.14) and satisfying (3.17). By (3.14), h_{3} has only real zeros and poles since f/f′ does apart from the point z = i, at which h_{3} ≠ 0, ∞. Also the poles of h_{3} occur exactly at those zeros of f′ that are not zeros of f. We first show that h_{3} can have only finitely many poles, if any. Let x be a pole of h_{3} so that f′(x) = 0 ≠ f(x) and so (C + zS)(x) = 0, by (3.12). Let x be a zero of f′ of order ν ≥ 1, and hence a pole of h_{3} of order ν. Suppose that h_{2} = (C + zS)h_{3} ≠ 0, ∞ at x. Considering (3.15), we deduce that zC − S = 0. Since C = −zS at x so that S ≠ 0 since C^{2} + S^{2} = 1, we get x^{2} + 1 = 0, hence x = ±i. This is impossible since x is real. If h_{2}(x) = 0 then by (3.14), C + zS has a multiple zero at x. Since C + zS = 0 = (C + zS)′ = (1 − A)S + AzC at z = x, this can happen only if A ≠ 0 and x^{2} = 1 − A^{−1} , hence at no more than two points x. Suppose then that h_{2}(x) = ∞. Considering (3.15) at x, we see that x is a simple zero of C + zS (since (zC − S)(x) ≠ 0). Thus by (3.14), x is a pole of order ν − 1 of h_{2} and hence we must have ν ≥ 2. Comparing residues on both sides of (3.16) at x, we deduce that

−(ν − 1) = (A − a)\frac{−(x^{2} + 1)}{1 − A(x^{2} + 1)},

that is,

x^{2} + 1 = \frac{ν − 1}{νA − a}.

If A ≠ 0, the right hand side is bounded as ν varies over integers not less than 2 so that |x| is bounded. Thus there can be only finitely many poles x of h_{2} and hence of h_{3} . If A = 0 then C + zS is a polynomial of degree at most 1 so that by (3.12), f′ has at most one zero (ignoring multiplicity). Hence in all cases, all zeros of f′ apart from finitely many at most, occur at zeros of f. Also there is a real polynomial P with only real zeros such that if h_{4} = P h_{3} then h_{4} has no poles and h_{4} has the same zeros as h_{3} . Thus the order of h_{4} is at most 1 and h_{4} ∈ U_{0} .

9. Proof that a/A is an integer

We continue with Case V. Recall that a/A > 0. We shall next show that a/A is an integer. Note that C + zS has infinitely many real zeros so that there are infinitely many intervals I = (x_{1}, x_{2}) between successive simple real zeros x_{1} , x_{2} of C + zS such that C + zS > 0 on (x_{1}, x_{2}). For x ∈ I, define

(9.1) h_{5}(x) = h_{2}(x)[(C + zS)(x)]^{(a/A)−1} exp\{−(\frac{a}{A} − 1) ∫_{α}^{x} \frac{S(t) dt}{(C + zS)(t)}\}

for some α ∈ I. A calculation shows that by (3.15), we have

(9.2) h′_{5}(x) = S_{0}(x)[(C + zS)(x)]^{(a/A)−1} exp\{−(\frac{a}{A} − 1) ∫_{α}^{x} \frac{S(t) dt}{(C + zS)(t)}\}.

As t → x_{1} , we have

\frac{S(t)}{(C + zS)(t)} = \frac{1}{1 − A(1 + x_{1}^{2})} \frac{1}{t − x_{1}} + O(1).

An analogous equation holds with x_{1} replaced by x_{2} when t → x_{2} . When x_{1} is large enough, we have

(9.3)  \frac{a}{A} − \frac{(a/A) − 1}{1 − A(1 + x_{1}^{2})} > 0,

and similarly for x_{2} . Since h_{3} has only finitely many poles, so that by (3.14), h_{2} vanishes (at least as fast as z − x_{1} ) at all but finitely many zeros x_{1} of C + zS, it follows from (9.1) and (9.3) that h_{5} can be extended continuously from the open interval I to x_{1} and x_{2} by setting h_{5}(x_{1}) = h_{5}(x_{2}) = 0. Also we see that h′_{5} is integrable over [x_{1}, x_{2}] with

(9.4) ∫_{x_{1}}^{x_{2}} S_{0}(x)[(C + zS)(x)]^{(a/A)−1} exp\{−(\frac{a}{A} − 1) ∫_{α}^{x} \frac{S(t) dt}{(C + zS)(t)}\} = 0.

Without affecting the validity of (9.4), we may and will choose α to be the unique point in I with S(α) = 1. The factors in the integrand in (9.4) other than S_{0} are positive on (x_{1}, x_{2}). When |x_{1}| is large enough then, since S(x_{1}) = −C(x_{1})/x_{1} → 0, there is an integer k such that |(Ax_{1} + B) − kπ| ≤ 1/|x_{1}|, thus Ax_{1} + B − kπ is close to zero. Since x_{2} is the next larger zero of C + zS, we have |(Ax_{2} + B) − (k + 1)π| ≤ 1/|x_{2}| if A > 0 (if A < 0 we have the same with k + 1 replaced by k − 1). Let us assume, for simplicity, that A > 0. In fact, when |k| is large enough, there is a unique zero x of C + zS for which Ax + B is closer to kπ than to lπ for any other integer l, and each zero of C + zS that has a sufficiently large modulus corresponds to a unique integer k in this way. Let us denote such a zero by y_{k}.

Now the sign of S_{0}(t) can only change at a point t = t_{1} such that (A−a)t_{1} + (B − b) = lπ for some integer l. There must be at least one such point t_{1} on (x_{1}, x_{2}).

We proceed to show that a/A is a positive integer, hence a/A ≥ 2 since a ≠ A. Now x_{1} above is equal to a certain y_{l}. For simplicity, let us assume that l > 0 and that x_{1} > 0 and x_{1} is large. We can write

\frac{S(t)}{(C + zS)(t)} − \frac{1}{1 − A(1 + y_{l}^{2})} \frac{(C + zS)′(t)}{(C + zS)(t)} = −A\frac{y_{l}(y_{l}S(t) + C(t)) + (t − y_{l})C(t)}{(1 − A(1 + y_{l}^{2}))(C + zS)(t)}

= −A\{\frac{y_{l}}{1 − A(1 + y_{l}^{2})} + \frac{t − y_{l}}{(C + zS)(t)} \frac{C(t) − y_{l}S(t)}{1 − A(1 + y_{l}^{2})}\}

≡ −Aψ(t) ≡ −Aψ_{l}(t),

say, for x_{1} < t < x_{2} . We next prove that |ψ(t)| ≤ L/|y_{l}| for a positive absolute constant L provided that |y_{l}| is large enough and x_{1} < t ≤ α. For the first term (y_{l}/(1 − A(1 + y_{l}^{2})) in ψ(t), this is clear. Since |t − y_{l}| < 2π/A, say, the same holds for the second term in ψ(t) for those t for which |(C + zS)(t)| ≥ 1/3, say. Suppose then that |(C + zS)(t)| < 1/3. Then |tS(t)| < 4/3, hence

|y_{l}S(t)| = |tS(t) − (t − y_{l})S(t)| < 2π/A + 4/3,

and |C(t) − y_{l}S(t)| < 2π/A + 7/3. By the mean value theorem, there is a point t′ between y_{l} and t such that

\begin{aligned}\frac{(C + zS)(t)}{t − y_{l}} & = \frac{(C + zS)(t) − (C + zS)(y_{l})}{t − y_{l}} = (C + zS)′(t′) \\ & = ((1 − A)S + AzC)(t′). \end{aligned}

Since |(C + zS)(t)| < 1/3, we have |(C + zS)(t′)| < 1/3 for any such t′ (since y_{l} < t ≤ α), so that |t′S(t′)| < 4/3, |S(t′)| < 4/(3|t′|), |C(t′)| > 1/2, say, and so

\begin{aligned}|((1 − A)S + AzC)(t′)| & ≥ |A| |y_{l}| |C(t′)| − |1 − A| |S(t′)| \\ & > |A| |y_{l}|/2 − 4|1 − A|/(3|t′|) > 1, \end{aligned}

say. Hence

|\frac{t − y_{l}}{(C + zS)(t)} \frac{C(t) − y_{l}S(t)}{1 − A(1 + y_{l}^{2})}| = O(1/y_{l}^{2}).

This completes the proof of |ψ(t)| ≤ L/|y_{l}|. Similarly we see that |ψ_{l+1}(t)| ≤ L/|y_{l+1}| for α < t < y_{l+1} .

Now (9.4) implies, after multiplying both sides of (9.4) by α^{1−(a/A)} , that

0 =∫_{y_{l}}^{α}S_{0}(x)[\frac{(C + zS)(x)}{α}]^{((a/A)−1)(1−1/(1−A(1+y_{l}^{2})))}exp\{(\frac{a}{A} − 1) ∫_{α}^{x}Aψ_{l}(t)\} dx

+ ∫_{α}^{y_{l+1}} S_{0}(x)[\frac{(C + zS)(x)}{α}]^{((a/A)−1)(1−1/(1−A(1+y_{l+1}^{2})))}

× exp\{(\frac{a}{A} − 1) ∫_{α}^{x} Aψ_{l+1}(t)\} dx.

Note that by our choice of α, we have (C + zS)(α) = α.

Suppose that 0 < ε < 1. Clearly there is a large positive integer l_{0} such that if l ≥ l_{0} then, since the above integral equals zero, we have

(9.5)  |∫_{y_{l}}^{y_{l+1}} S_{0}(x)|S(x)|^{(a/A)−1} dx| < ε/|A|.

Write Ay_{l} + B = lπ + δ_{l}, where δ_{l} → 0 as |l| → ∞. (Then Aα + B = (l + 1/2)π.) Note that l is even (since C + zS > 0 on (y_{l}, y_{l+1}) and A > 0) and large but otherwise arbitrary. Changing variables in (9.5) we obtain

(9.6)  |∫_{δ_{l}}^{π+δ_{l+1}} sin(\frac{(A − a)(x + lπ − B)}{A} + B − b)| sin x|^{(a/A)−1} dx| < ε.

We find that (9.7)

lim_{{l→∞ \atop l even}} ∫_{0}^{π} sin((1 − (a/A))(x − B) + (1 − (a/A))lπ + B − b)| sin x|^{(a/A)−1} dx

= lim_{{l→∞ \atop l even}} ∫_{0}^{π}\{sin((1 − (a/A))(x − B) + B − b) cos((a/A)lπ)

− cos((1 − (a/A))(x − B) + B − b) sin((a/A)lπ)}(sin x)^{(a/A)−1} dx = 0.

We may deal with the intervals between successive zeros of C + zS on which C + zS < 0 in the same way. We find that (9.8)

lim_{{l→∞ \atop l odd}} ∫_{0}^{π} sin((1 − (a/A))(x − B) + (1 − (a/A))lπ + B − b)| sin x|^{(a/A)−1} dx

= lim_{{l→∞ \atop l odd}} ∫_{0}^{π}\{sin((1 − (a/A))(x − B) + B − b) cos((a/A)lπ)

− cos((1 − (a/A))(x − B) + B − b) sin((a/A)lπ)}(sin x)^{(a/A)−1} dx = 0.

If a/A is irrational then there is a subsequence of even integers l → ∞ along which cos((a/A)lπ) → 1 (so that sin((a/A)lπ) → 0) and another subsequence along which sin((a/A)lπ) → 1 (so that cos((a/A)lπ) → 0). This shows that

(9.9)

∫_{0}^{π} sin((1 − (a/A))(x − B) + B − b)(sin x)^{(a/A)−1} dx
= ∫_{0}^{π} cos((1 − (a/A))(x − B) + B − b)(sin x)^{(a/A)−1} dx = 0.

We now use the formula due to Cauchy ([1, pp. 41–89], [6, p. 158]), also quoted by Hellerstein, Shen and Williamson [2, (3.15), p. 326], which states that for γ > 0,

(9.10) ∫_{0}^{π}(sin^{γ−1} t) exp\{i[(γ − 1)t + δ]\} dt = \frac{π}{2^{γ−1}} exp\{i[π(γ − 1)/2 + δ]\} ≠ 0.

We apply this with γ = a/A and δ = −(B − b) − ((a/A) − 1)B and get a contradiction since the integral ought to vanish by (9.9) but it does not, by (9.10).

If a/A is rational but not an integer then there are only finitely many distinct pairs (cos((a/A)lπ), sin((a/A)lπ)). Thus (9.7) and (9.8) imply that (9.11)

cos((a/A)lπ) ∫_{0}^{π} sin((1 − (a/A))(x − B) + B − b)(sin x)^{(a/A)−1} dx

− sin((a/A)lπ) ∫_{0}^{π} cos((1 − (a/A))(x − B) + B − b)(sin x)^{(a/A)−1} dx = 0. for all l, both even and odd. Taking real and imaginary parts in (9.10) and combining the result with (9.11) we obtain (after dividing by π/2^{(a/A)−1} )

(9.12)

sin(π((a/A) − 1)/2 − (B − b) − ((a/A) − 1)B + (a/A)lπ)
= cos((a/A)lπ) sin(π((a/A) − 1)/2 − (B − b) − ((a/A) − 1)B)
+ sin((a/A)lπ) cos(π((a/A) − 1)/2 − (B − b) − ((a/A) − 1)B) = 0

for all l. This is possible only if (a/A)l is an integer for all l, which is not the case when a/A is not an integer. So it follows that a/A is an integer, as claimed. Write p = (a/A) − 1. Since a ≠ A, it follows that p is a positive integer. Furthermore, it now follows from (9.12) that

(9.13)  sin(πp/2 − (B − b) − pB) = 0.

We have derived these results assuming that A > 0. If A < 0 then a similar argument yields the same results. The only difference is that if x_{1} and x_{2} are successive simple real zeros of C + zS with x_{1} < x_{2} , corresponding to integers k and k + 1 via the fact that (Ax_{j} + B)/π is close to an integer for j = 1, 2, then k corresponds now to x_{2} , not x_{1} . We leave further details to the reader.

It follows from (9.13) that there is an integer q such that πp/2−(B−b)−pB = −qπ and hence

(9.14)  (B − b) + pB = πp/2 + qπ.

Thus

S_{0}(z) = sin((A − a)\frac{(Az + B) − B}{A} + B − b)
= sin(−p(Az + B) + (B − b) + pB)

= sin(−p(Az + B)) cos(πp/2 + qπ) + cos(−p(Az + B)) sin(πp/2 + qπ).

We obtain

(9.15)  S_{0}(z) = (−1)^{q+1+(p/2)} sin(p(Az + B))  if p is even

and

(9.16)  S_{0}(z) = (−1)^{q+((p−1)/2)} cos(p(Az + B))  if p is odd.

By (3.15), the function h_{6} = (C + zS)^{p}h_{2} , which has only finitely many poles, has at most two distinct non-real zeros (at the non-real zeros of C + zS, if any), and is real of order at most 1, satisfies

(9.17)  h′_{6} = \frac{ph_{6}S}{C + zS} + S_{0}(C + zS)^{p}.

10. Conclusion of the proof in Case V

We continue our study of Case V. Suppose first that p is even. For simplicity and definiteness, we assume that A > 0. An inspection of the arguments that follow shows that a similar reasoning works if A < 0. Let x_{1} and x_{2} be successive zeros of C + zS, with x_{1} < x_{2} , with |x_{1}|, and hence |x_{2}|, sufficiently large. Suppose that sgn(C + zS) = σ ∈ {1, −1} on (x_{1}, x_{2}). We denote a sufficiently small positive quantity by ε, not necessarily the same at every occurrence. Suppose that sgn h′_{6}(x) = τ on (x_{1}, x_{1} + ε). Then clearly sgn h_{6}(x) = τ. By (9.17), sgn S_{0}(x) = τ on this interval also, since |(h_{6}S)/(h′_{6}(C + zS))| is comparable to 1/x^{2} close to x = x_{1} .

At x = x_{1} , we have S = −C/x so that writing Ax_{1} + B = lπ + ε_{1} for some integer l we find that C(x_{1}) ≈ (−1)^{l} and Ax + B ≈ lπ − (1/x) close to x = x_{1} . If x_{1} > 0, we then have lπ − (1/x) < lπ at x_{1} . Let the zeros of S_{0} on (x_{1}, x_{2}) be y_{1} < y_{2} < ⋯ < y_{p} (all zeros of S_{0} are simple). At these points, p(Ax + B) = μπ where the integer μ increases (since A > 0) from pl to p(l + 1) − 1 for the integer l found above, and Ax + B = μπ/p. In this case, y_{1} − x_{1} is very small and tends to zero as x_{1} → ∞. If x_{1} < 0 then x_{2} − y_{p} > 0 is small and tends to zero as x_{1} → −∞. We shall go through the case x_{1} > 0 (and |x_{1}| large) in detail and then simply state what the corresponding results are for x_{1} < 0. As one can understand from the above remark, the principal difference is that certain action moves from the x_{1} -end of the interval (x_{1}, x_{2}) to the x_{2} -end when x_{1} is taken to be negative.

The zeros of h_{6} on [x_{1}, x_{2}] coincide with the zeros of the function h_{2} and hence with the zeros of h_{5} on this interval. Between consecutive zeros like that, there is a zero of h′_{6} and a zero of h′_{5} (these might occur at the same point, of course). By (9.2), these zeros of h′_{5} occur at the zeros of S_{0} , that is, at the points y_{j}. It follows that any interval (y_{j}, y_{j+1}) for 1 ≤ j ≤ p−1 contains at most one zero of h_{6} . Similarly, the intervals (x_{1}, y_{1}] and [y_{p}, x_{2}) contain no zeros of h_{6} since h_{5} and h_{6} vanish at x_{1} and at x_{2} . In particular, h_{6}(y_{1}) ≠ 0 ≠ h_{6}(y_{p}). If h_{6}(y_{j}) = 0 for some j with 2 ≤ j ≤ p − 1, then by (9.17), h′_{6}(y_{j}) = 0. If h_{6}(y_{j}) = 0 then inspection of multiplicities in (9.17) shows that y_{j} is a zero of h_{6} of order 2. Further, if h_{6}(y_{j}) = 0 then y_{j} is a zero of h_{5} of order 2, and if this happens for a certain j then neither h_{5} nor h_{6} can have any zeros on (y_{j−1}, y_{j}) ∪ (y_{j}, y_{j+1}). Also we cannot have h_{6}(y_{j}) = h_{6}(y_{j+1}) = 0 for any j with 1 ≤ j ≤ p − 1. Considering the various possibilities that can arise, one can see that h_{6} has at most p − 1 zeros on (x_{1}, x_{2}), with due count of multiplicity. We shall next show that in fact h_{6} has p − 1 zeros on (x_{1}, x_{2}).

Now suppose that x_{1} is large and positive. Claim: the function h_{6} has a zero on (x_{1}, y_{2}), or otherwise a double zero at y_{2} . If h_{6} ≠ 0 on (x_{1}, y_{2}), then sgn h_{6} = τ on (x_{1}, y_{2}). At x_{1} , we have (C + zS)′ = (1 − A(1 + x_{1}^{2}))S whose sign is σ since (C + zS)(x_{1}) = 0 and sgn(C + zS) = σ on (x_{1}, x_{2}). Thus, since A > 0 and |x_{1}| is large, we have sgn S(x_{1}) = −σ. Further, C(x_{1}) = −x_{1}S(x_{1}) so that sgn C(x_{1}) = σ, and |C(x_{1})| ≈ 1 since |S(x_{1})| = |C(x_{1})/x_{1}| ≈ 0. Incidentally, this shows that σ = (−1)^{l}. We have S(y_{1}) = 0 since Ay_{1} + B = (plπ)/p = lπ. Thus sgn S(x) = −σ on (x_{1}, y_{1}) and sgn S(x) = σ on (y_{1}, x_{2} + ε). By (9.17), we have h′_{6}(y_{1}) = 0 since S_{0}(y_{1}) = S(y_{1}) = 0. We may apply Lemma 1 to Ψ = h_{6} . This shows that y_{1} is a simple zero of h′_{6} when |x_{1}| is large enough.

If h_{6}(y_{2}) = 0 then h′_{6}(y_{2}) = 0 by (9.17) since also S_{0}(y_{2}) = 0. Considering multiplicities in (9.17) we see that y_{2} must be a zero of h_{6} of order 2, being a simple zero of S_{0} and not a zero of C + zS. In this case h_{6} cannot have any zeros on (x_{1}, y_{2}) since they would be zeros of h_{5} also and since y_{1} is the only zero of h′_{5} on (x_{1}, y_{2}).

Suppose that h_{6}(y_{2}) ≠ 0 (and that h_{6} has no zeros on (x_{1}, y_{2})). Since S_{0}(y_{2}) = 0, we have by (9.17)

(10.1)  \frac{h′_{6}}{h_{6}} (y_{2}) = \frac{pS(y_{2})}{(C + zS)(y_{2})} > 0.

Thus sgn h′_{6}(y_{2}) = τ = sgn h′_{6}(x_{1} + ε). So h′_{6} must have at least two zeros on (x_{1}, y_{2}), not only y_{1} but also some ζ_{1} ≠ y_{1} . But then by Lemma 1, h_{6} must have a zero between ζ_{1} and y_{1} , and hence on (x_{1}, y_{2}), as claimed. Considering h_{5} as above we see that h_{6} has exactly one zero on (x_{1}, y_{2}), which furthermore is a simple zero of h_{6} and lies on (y_{1}, y_{2}). This proves the Claim.

We prove by induction on j that for 1 ≤ j ≤ p − 1, the interval (y_{j}, y_{j+1}) contains points ξ_{j}, η_{j} with h′_{6}(ξ_{j}) = h_{6}(η_{j}) = 0 and ξ_{j} ≤ η_{j}. (If h_{6}(y_{j}) = 0 for some j, a simple modification is required, and we omit the details.) We have done this for j = 1. In this case only we consider the zero of h′_{6} at y_{1} to substitute for a zero on (y_{1}, y_{2}). If y_{2} is a double zero of h_{6} then if substitutes for a zero of h_{6} on (y_{1}, y_{2}) and on (y_{2}, y_{3}) and for a zero of h′_{6} on (y_{2}, y_{3}). In this case h_{6}(y_{2}) = 0, and therefore, we have proved our claim also for j = 2, and we have sgn h_{6}(x) = τ for x ∈ (y_{2}, y_{2} + ε).

If h′_{6} has two distinct zeros, say z_{1} < z_{2} , on (y_{j}, y_{j+1}), then by (9.17) and the fact that each of S, S_{0} , and C + zS is non-zero and retains its sign on (y_{j}, y_{j+1}), we see that h_{6}(z_{1}) and h_{6}(z_{2}) are non-zero and of the same sign. Choose z_{1} and z_{2} to be consecutive (distinct) zeros of h′_{6} . By Rolle’s theorem there is at most one zero of h_{6} on (z_{1}, z_{2}) with due count of multiplicity. Thus we find that h_{6} has no zeros on (z_{1}, z_{2}). This contradicts Lemma 1 applied to Ψ = h_{6} when |x_{1}| is large enough. Also by Lemma 1, any zero z_{0} of h′_{6} on (y_{j}, y_{j+1}) is a simple zero of h′_{6} when h_{6}(z_{0}) ≠ 0, and if h_{6}(z_{0}) = h′_{6}(z_{0}) = 0 ≠ (S_{0}(C +zS)^{p})(z_{0}) then (9.17) gives a contradiction.

By this and by earlier analysis, (y_{j}, y_{j+1}) contains at most one zero of each of h_{6} and h′_{6} , and so there is exactly one zero. To get to the induction step, we now assume that our claim has been proved up to a certain j with 1 ≤ j ≤ p − 2 (and we assume that p ≥ 4 for otherwise there is nothing more to prove). Thus sgn h_{6}(x) = sgn h′_{6}(x) = (−1)^{j}τ for x ∈ (y_{j+1}, y_{j+1}+ε). Analogously to (10.1), we have (h′_{6}/h_{6})(y_{m}) > 0 for all m ≥ 2. Therefore either both h_{6} and h′_{6} have a zero on (y_{j+1}, y_{j+2}) or neither function has any zeros there. To get a contradiction, suppose that neither h_{6} nor h′_{6} vanishes on (y_{j+1}, y_{j+2}) so that then sgn h_{6}(x) = sgn h′_{6}(x) = (−1)^{j}τ for x ∈ (y_{j+1}, y_{j+2}). By the induction assumption, there are points t_{1}, t_{2} with h′_{6}(t_{1}) = h_{6}(t_{2}) = 0 and with y_{j} ≤ t_{1} ≤ t_{2} < y_{j+1} . We have sgn S_{0} = (−1)^{j}τ on (y_{j}, y_{j+1}). Thus there is a unique point w ∈ (y_{j+1}, y_{j+2}) such that S_{0}(w) = (−1)^{j+1}τ. So by (9.17), at z = w we have

sgn(ph_{6}S + (−1)^{j+1}τ(C + zS)^{p+1}) = (−1)^{j}στ.

Hence (−1)^{j}στph_{6}S − σ(C + zS)^{p+1} > 0 at z = w. Note that sgn S(w) = σ. By the mean value theorem there is a point t ∈ (t_{2}, w) with

(10.2)  (−1)^{j}τh′_{6}(t) = \frac{h_{6}(w)(−1)^{j}τ}{w − t_{2}} .

On the other hand, by (9.17),

(10.3)  h′_{6}(t) = \frac{pS(t)h_{6}(t)}{(C + zS)(t)} + S_{0}(t)(C + zS)^{p}(t).

The function (−1)^{j}τh_{6} has positive derivative and hence is increasing on (t_{1}, y_{j+2}). Furthermore, this function is positive on (t_{2}, y_{j+2}).

Suppose that t ∈ [y_{j+1}, w) so that (−1)^{j}τS_{0}(t) ≤ 0. Then 0 < (−1)^{j}τh_{6}(t) < (−1)^{j}τh_{6}(w), and so combining (10.3) and (10.2) we obtain

1 < \frac{(w − t_{2})pS(t)}{(C + zS)(t)}.

Considering the relationship between C + zS, S, and S_{0} , we see that there is a positive constant ε_{0} depending on p only such that σS(t) ≥ ε_{0} and σ(C+zS)(t) ≥ ε_{0}t for any t ∈ [y_{2}, y_{p}] (recall that x_{1} > 0). Also

w − t_{2} ≤ w − y_{j} ≤ y_{p} − y_{1} ≤ (p − 1)2π/(pA) < 2π/A.

Thus

ε_{0}x_{1} ≤ ε_{0}t ≤ σ(C + zS)(t) ≤ pσS(t)(w − t_{2}) ≤ p(w − t_{2}) < (2πp)/A,

which gives a contradiction when x_{1} is large enough.

Hence we must have t ∈ (t_{2}, y_{j+1}) and so (−1)^{j}τS_{0}(t) > 0. Then, since we still have 0 < (−1)^{j}τh_{6}(t) < (−1)^{j}τh_{6}(w), we obtain by (10.2) and (9.17) that

\frac{1}{2} \frac{(−1)^{j}τh_{6}(w)}{(2π/A)} < \frac{1}{2} \frac{(−1)^{j}τh_{6}(w)}{w − y_{j}} < \frac{1}{2} \frac{(−1)^{j}τh_{6}(w)}{w − t_{2}}

< (−1)^{j}τh_{6}(w)(\frac{1}{w − t_{2}} − \frac{pS(t)}{(C + zS)(t)})

= (−1)^{j}τ(h′_{6}(t) − \frac{pS(t)h_{6}(w)}{(C + zS)(t)})

< (−1)^{j}τ(h′_{6}(t) − \frac{pS(t)h_{6}(t)}{(C + zS)(t)})

= (−1)^{j}τS_{0}(t)(C + zS)^{p}(t) < (−1)^{j}τKt^{p}S_{0}(t)

where we note, as above, that pS(t)/(C + zS)(t) is comparable to 1/x_{1} . Here K > 0 is a constant depending only on p. Now we get

(−1)^{j}τh_{6}(w) < (−1)^{j}τK_{1}t^{p}S_{0}(t) < K_{1}w^{p}

where K_{1} > 0 depends only on p and A. Furthermore, by (9.17),

(−1)^{j}τh′_{6}(w) = \frac{(−1)^{j}τpS(w)h_{6}(w)}{(C + zS)(w)} + (−1)^{j}τS_{0}(w)(C + zS)^{p}(w)

so that

0 < (−1)^{j}τσ(C + zS)(w)h′_{6}(w)
= (−1)^{j}τσpS(w)h_{6}(w) + (−1)^{j}τσS_{0}(w)(C + zS)^{p+1}(w)
< K_{1}pw^{p} − |(C + zS)(w)|^{p+1}.

But |(C + zS)(w)| > ε_{0}w. Hence (−1)^{j}τh′_{6}(w) < 0, which is a contradiction. This proves our claim that each of h_{6} and h′_{6} has a zero on (y_{j+1}, y_{j+2}). Hence h_{6} has exactly p − 1 zeros on (x_{1}, x_{2}), with due count of multiplicity.

These p−1 zeros of h_{6} are also zeros of c+zs, which has exactly p+1 zeros, all simple, on (x_{1}, x_{2}) when |x_{1}| is large enough (see (3.13)). These zeros occur at points where s = −c/z is close to zero. By (9.14),

s = sin(az + b) = sin((p + 1)(Az + B) − (p + 1)B + b)

= sin((p + 1)(Az + B) − (q + (p/2))π) = (−1)^{q+(p/2)} sin((p + 1)(Az + B)).

Thus the zeros of c + zs are close to the points where Az + B = l_{0}π/(p + 1) for some integer l_{0} . The points y_{j} are points where Az + B = μπ/p, and here μ varies from lp to (l + 1)p − 1 for some integer l. Thus at y_{1} , we have Az + B = ((p + 1)l)π/(p + 1) so that y_{1} is close to a zero of c + zs. As we have seen, we have Ax_{1} + B ≈ lπ − (1/x_{1}) so that

\begin{aligned}(c + zs)(x_{1}) & = (−1)^{q+(p/2)}((−1)^{l(p+1)} + x_{1}(−1)^{l(p+1)}(−(p + 1)/x_{1})) + o(1) \\ & = −p(−1)^{q+(p/2)+l(p+1)} + o(1) = −pτ(−1)^{l} + o(1) = −pτσ + o(1). \end{aligned}

Here we have used the fact that σ = (−1)^{l} and that τ = (−1)^{q+1+(p/2)+pl+1} , which follows since by (9.15), we have sgn S_{0}(x) = (−1)^{q+1+(p/2)} sgn sin(p(Az + B)) = τ on (x_{1}, x_{1} + ε). The o(1)-term tends to zero as |x_{1}| → ∞. We further have (c+zs)′(x_{1}) = ((1−a)s+azc)(x_{1}) = ax_{1}(−1)^{q+(p/2)} cos((p+1)(Ax_{1}+B))+o(1) = (−1)^{q+(p/2)+(p+1)l} +o(1) = τ(−1)^{l} +o(1) = τσ +o(1). We have a > 0 since A > 0 and a/A > 0. Thus sgn(c + zs)′(x_{1}) = − sgn(c + zs)(x_{1}). Hence the zero of c + zs which is close to both x_{1} and y_{1} is slightly greater than x_{1} . We have (c + zs)(y_{1}) = cos(ay_{1} + b) = (−1)^{q+(p/2)+(p+1)l} = τσ = − sgn(c + zs)(x_{1}). We conclude that c + zs has a zero on (x_{1}, y_{1}). This cannot be a zero of h_{6} as we have seen above. The next p − 1 zeros of c + zs lie on the intervals (y_{j}, y_{j+1}) for 1 ≤ j ≤ p − 1 so that they must be zeros of h_{6} . When |x_{1}| is large enough, c + zs has exactly p + 1 zeros on (x_{1}, x_{2}). There is one more zero of c + zs on (x_{1}, x_{2}). It lies in (y_{p}, x_{2}) and is not a zero of h_{6} . We may express this by saying that if for each large l, the two zeros of c + zs immediately smaller than the value of x corresponding to Ax + B = lπ are not zeros of h_{6} while all other sufficiently large positive zeros of c + zs are also zeros of h_{6} . Let us denote the two zeros mentioned by t_{l} and z_{l}, where t_{l} < z_{l}. As l → ∞, we have

At_{l} + B = lπ + O(1/l)  and  Az_{l} + B = (l − (p + 1)^{−1})π + O(1/l).

Still assuming that A > 0 and that p is even, suppose that x_{1} < x_{2} < 0 and that |x_{2}| is large enough. Then an analysis similar to the above shows that at the zeros y_{1} < ⋯ < y_{p} of S_{0} on (x_{1}, x_{2}), we have Ay_{j} + B = (π/p)(lp + j) for some negative integer l, and we now have S = S_{0} = 0 at y_{p} (instead of y_{1} ), and for each sufficiently large negative integer l, the two zeros of c+zs immediately larger than the value of x corresponding to Ax + B = lπ are not zeros of h_{6} while all other sufficiently large negative zeros of c + zs are also zeros of h_{6} . Let us denote the two zeros mentioned by t_{l} and z_{l}, where z_{l} < t_{l}. (We have z_{l} < x_{1} < t_{l} < y_{1} .) As l → −∞, we have

Az_{l} + B = lπ + O(1/l)  and  At_{l} + B = (l + (p + 1)^{−1})π + O(1/l).

Thus

(10.4) z_{l} + z_{−l} = −2B/A + O(1/l)  and  t_{l} + t_{−l} = −2B/A + O(1/l)

as l → ∞.

Since h_{6} is a real meromorphic function with only finitely many poles and of order at most 1 and since h_{6} has the zeros described above, we conclude that there is a real number α and a real rational function R, not identically zero, such that

(10.5)  h_{6} = e^{αz}(C + zS)^{p+1}(c + zs)R/h_{7} ≡ φ/h_{7}

where

(10.6)  h_{7}(z) = ∏_{l=l_{0}}^{∞}((1 − \frac{z}{z_{l}} )(1 − \frac{z}{z_{−l}} )(1 − \frac{z}{t_{l}} )(1 − \frac{z}{t_{−l}} )).

It is easily seen that the infinite product for h_{7} converges and defines a real entire function of order 1 when the factors are grouped as above (it suffices to pair z_{l} and z_{−l} together, and separately t_{l} and t_{−l} together). Now (9.17) can be written as

\frac{φ′}{h_{7}} − \frac{φh′_{7}}{h_{7}^{2}} = S_{0}(C + zS)^{p} + \frac{φ}{h_{7}} \frac{pS}{C + zS} .

At a point where S_{0} = 0, this gives

(10.7) \begin{aligned}\frac{h′_{7}}{h_{7}} & = \frac{φ′}{φ} − \frac{pS}{C + zS} \\ & = −\frac{pS}{C + zS} + α + \frac{R′}{R} + (p + 1)\frac{(1 − A)S + AzC}{C + zS} + \frac{(1 − a)s + azc}{c + zs}. \end{aligned}

Let the notation be as before and consider (10.7) at z = y_{1} so that S_{0} = 0, as required, where y_{1} is large and positive. Thus Ay_{1} + B = l_{0}π for some large integer l_{0} . Now y_{1} is close to the point x_{1} at which C +zS = 0 and S = O(1/x_{1}), and close to the common zero z_{l} of h_{7} and c + zs. Also S(y_{1}) = 0. We obtain

\frac{h′_{7}}{h_{7}} (y_{1}) = (p + 1)Ay_{1} + \frac{1}{y_{1} − z_{l}} + O(1)

where the O(1)-term remains bounded as x_{1} (and hence y_{1} ) tends to infinity. On the other hand, the definition of h_{7} together with (10.4) allows us to estimate

\frac{h′_{7}}{h_{7}} (y_{1}) = \frac{1}{y_{1} − z_{l}} + \frac{1}{y_{1} − t_{l}} + O(1).

Now y_{1} → ∞ and y_{1} − t_{l} → π/(A(p + 1)) as x_{1} → ∞. Thus we clearly obtain a contradiction whenever x_{1} is sufficiently large. This contradiction shows that the case when p is even and A > 0 cannot occur. The case when p is even and A < 0 is dealt with in the same way, and we omit the details.

If p is odd then by (9.16), we have S_{0}(z) = (−1)^{q+((p−1)/2)} cos(p(Az + B)), s(z) = (−1)^{q+((p+1)/2)} cos((p + 1)(Az + B)), and c(z) = (−1)^{q+((p−1)/2)} sin((p + 1)(Az + B)) for some integer q. We argue as above on an interval (x_{1}, x_{2}) where C + zS retains its sign. The location of the zeros of S_{0} in this interval is now different but the number p of zeros is the same. In the particularly simple case when p = 1, the function h_{6} has no zeros on (x_{1}, x_{2}) and we may use h_{7} = c + zs in (10.5). Then S_{0} ≡ (−1)^{q}C, and (10.5) and (9.17) imply first that α = 0 in (10.5) and then yield a contradiction, considering sequences of real points tending to ±∞ at which S = 0 or C = 0.

Suppose that p is odd and p ≥ 3, A > 0. The argument is analogous to the one used when p is even, up to (10.6). However, the zeros y_{j} of S_{0} on (x_{1}, x_{2}) are now bounded away from x_{1} and x_{2} , and so are the zeros of c+zs and hence those of h_{6} . Therefore a consideration of (h′_{7}/h_{7})(y_{1}) as in the case when p is even does not seem to help. Instead, we may argue as follows. When |x_{1}| is large, the zeros of c + zs and hence those of h_{6} occur closer and closer to points where s = 0 since s = −c/z at these points. Indeed, since (c + zs)′ = −c(1 − a − az^{2})/z when c + zs = 0, the mean value theorem and the monotonicity of (c + zs)′ between a zero of c + zs and the nearest zero of s (where c + zs = c = ±1, (c + zs)′ = acz) imply that the distance between these zeros is O(1/z). When |x_{1}| is large enough, the zeros of c + zs that are zeros of h_{7} are close to those zeros of s at which Az + B = kπ ± π/(2(p + 1)) for all integers k with |k| sufficiently large. Write β = sin(π/(2(p + 1))). Comparing the Hadamard product representations of h_{7} and of sin(Az + B + π/(2(p + 1))) sin(Az + B − π/(2(p + 1))) = S^{2} − β^{2} and considering (10.5), we conclude that there are a real rational function R_{1} and a real number α_{1} such that if we define

(10.8)  h_{8} = \frac{R_{1}e^{α_{1}z}(C + zS)^{p+1}(c + zs)}{S^{2} − β^{2}},

then for any small ε > 0, we have

(10.9)  h_{6}(x)/h_{8}(x) → 1  and  (h′_{6}/h_{6})(x) − (h′_{8}/h_{8})(x) → 0

as x → ±∞ outside ε-neighbourhoods of the zeros of S^{2} − β^{2} .

We obtain from (9.17), (10.5), and (10.8)–(10.9) that

(10.10) \begin{aligned}\frac{h′_{8}}{h_{8}} − \frac{h′_{6}}{h_{6}} & = α_{1} + \frac{R′_{1}}{R_{1}} + (p + 1)\frac{(1 − A)S + AzC}{C + zS} + \frac{(1 − a)s + azc}{c + zs} \\ & − \frac{2ASC}{S^{2} − β^{2}} − \frac{pS}{C + zS} − \frac{h_{8}}{h_{6}} \frac{S_{0}(S^{2} − β^{2})}{R_{1}e^{α_{1}z}(C + zS)(c + zs)} → 0 \end{aligned}

as x → ±∞ outside ε-neighbourhoods of the zeros of S^{2}−β^{2} . Considering (10.10) at the points where C = 0, as we may, so that also S_{0} = c = 0, we see, as x → ∞, that α_{1} = 0. Next, we may take x = y_{j} so that S_{0} = 0, and noting that the functions C/S, s, c always take the same values at x = y_{j} (depending on j but independent of x_{1} ), we obtain, as x → ∞, that

(10.11)  \frac{(p + 1)AC}{S} + \frac{ac}{s} − \frac{2ASC}{S^{2} − β^{2}} = 0  at x = y_{j}

for all j with 1 ≤ j ≤ p. This is trivially true for j = (p + 1)/2 since then C = c = 0. But since p ≥ 3, there are other values of j to consider also. A calculation based on trigonometric identities and the fact that a = (p+1)A shows that when C ≠ 0, (10.11) is equivalent to

(10.12)  \frac{p + 1}{p} = \frac{sin^{2}(\frac{π}{2p}(2j − 1))}{β^{2}},  for 1 ≤ |j − \frac{p + 1}{2}| ≤ \frac{p − 1}{2}.

If p = 3 and j = 1, (10.12) reads 4/3 = (1/4)/ sin^{2}(π/8), which is false. If p ≥ 5, (10.12) cannot be satisfied since sin^{2}(π(2j − 1)/(2p)) will not take the same value for all j. This gives a contradiction, showing that the case when p is odd and A > 0 cannot occur. The case when p is odd and A < 0 is dealt with in the same way, and we omit the details. This completes our treatment of Case V.

This completes the proof of Theorem 1 when m = 0. Hence the proof of Theorem 1 is now complete.

11. Proof of Lemma 1

Let the assumptions of Lemma 1 be satisfied. If Φ has only finitely many zeros, then so does Ψ′, so that the conclusion of Lemma 1 is obtained trivially by taking R to be sufficiently large. Hence we may assume that Φ has infinitely many zeros.

We have

Ψ′ = \frac{P Φ′ − P′Φ}{P^{2}}.

Choose R_{0} > 1 so that |a| < R_{0} whenever P (a) = 0 or P (a) = ∞. Let x_{1} and x_{2} be two consecutive zeros of Ψ and hence of Φ such that x_{1} < x_{2} and |x_{j}| > R_{0} for j = 1, 2. By Rolle’s theorem, Ψ′ has at least one zero on (x_{1}, x_{2}). The zeros of Ψ′ on (x_{1}, x_{2}) coincide with those of

\frac{Φ′}{Φ} − \frac{P′}{P}

on (x_{1}, x_{2}). If there are at least two such zeros, with due count of multiplicity, then (x_{1}, x_{2}) contains a point x_{3} at which (Φ′/Φ)′ − (P′/P)′ vanishes. There is a positive constant K such that |(P′/P)′(x)| < K/|x|^{2} for |x| > R_{0} . Now Φ′/Φ has no zeros on (x_{1}, x_{2}) and tends to ∞ and −∞ at the endpoints. We can write

Φ(z) = Az^{m}e^{−az^{2}+bz} ∏_{j=1}^{∞}(1 − \frac{z}{a_{j}} )e^{z/a_{j}}

where A is a non-zero real constant, m is a non-negative integer, a ≥ 0, b is real, and the a_{j} are non-zero real numbers with ∑_{j} a_{j}^{−2} < ∞. Thus

(\frac{P′}{P} )′(x_{3}) = (\frac{Φ′}{Φ} )′(x_{3}) = \frac{−m}{x_{3}^{2}} − 2a − ∑_{j=1}^{∞} \frac{1}{(x_{3} − a_{j})^{2}} < − ∑_{j=1}^{∞} \frac{1}{(x_{3} − a_{j})^{2}} < 0.

We obtain a contradiction if we can show that

(11.1)  \frac{K}{x^{2}} < ∑_{j=1}^{∞} \frac{1}{(x − a_{j})^{2}}

whenever |x| is large enough and applying this to x = x_{3} .

Suppose that there are infinitely many positive numbers a_{j}. Then (11.1) holds for a given x > 0 provided that the number of zeros a_{j} on (0, x) is larger than K. We argue similarly when x < 0 if Φ has infinitely many negative zeros. Suppose then that Φ has infinitely many positive zeros but only finitely many negative zeros, if any, and that x < 0. To prove (11.1), we may assume that all the a_{j} are positive. Choose a positive integer N > K. For any j with 1 ≤ j ≤ N, there is ζ_{j} < 0 such that x^{−2} < (x − a_{2j−1})^{−2} + (x − a_{2j})^{−2} for all x < ζ_{j}. Thus (11.1) holds for all x < min{ζ_{j} : 1 ≤ j ≤ N}. A similar argument works if x > 0 and if Φ has infinitely many negative zeros but only finitely many positive zeros. We have now proved that (x_{1}, x_{2}) contains exactly one zero of Ψ′, with due count of multiplicity, provided that the consecutive zeros x_{1} and x_{2} of Φ are of the same sign and that |x_{1}| and |x_{2}| are large enough, say |x_{1}| > R and |x_{2}| > R, where R > R_{0} .

The same argument shows that if Φ has infinitely many negative zeros but only finitely many positive zeros, then Ψ′ can have at most one zero on (R_{1}, ∞) for a suitable R_{1} > 0, and hence no zeros on (R_{2}, ∞) for some R_{2} > 0. Choosing R > R_{2} in this case we do not have to consider the possible existence of zeros of Φ between consecutive positive zeros of Ψ′. An analogous conclusion is obtained if Φ has infinitely many positive zeros but only finitely many negative zeros.

Suppose then that x_{1} and x_{2} are consecutive zeros of Ψ′ with |x_{j}| > R for j = 1, 2. Suppose first that Φ vanishes at both x_{1} and x_{2} , or at neither x_{1} nor x_{2} . We claim that (x_{1}, x_{2}) contains exactly one zero of Φ, with due count of multiplicity, provided that |x_{1}| and |x_{2}| are large enough. If (x_{1}, x_{2}) contains at least two zeros of Φ, with due count of multiplicity, then it follows from Rolle’s theorem that Ψ′ has a zero on (x_{1}, x_{2}), which is against our assumption that x_{1} and x_{2} are consecutive zeros of Ψ′. Thus (x_{1}, x_{2}) contains at most one zero of Φ. If Φ does not vanish anywhere on (x_{1}, x_{2}), then let x_{3} be the largest zero of Φ with x_{3} ≤ x_{1} , and let x_{4} be the smallest zero of Φ with x_{2} ≤ x_{4} . (By the previous paragraph, we may assume that x_{3} and x_{4} exist with x_{4} < −R or x_{3} > R.) By what we have proved already, (x_{3}, x_{4}) contains exactly one zero of Ψ′, with due count of multiplicity. This gives an immediate contradiction if Φ(x_{1})Φ(x_{2}) ≠ 0 (so that x_{3} < x_{1} < x_{2} < x_{4} ). If Φ(x_{1}) = Φ(x_{2}) = 0 then x_{3} = x_{1} and x_{4} = x_{2} . But then Ψ′ has a zero on (x_{1}, x_{2}) by Rolle’s theorem, which gives a contradiction. Thus Ψ′ has exactly one zero on (x_{1}, x_{2}) with due count of multiplicity, as claimed. Suppose then that exactly one of x_{1} and x_{2} is a zero of Φ. If Φ has a zero on (x_{1}, x_{2}) then by Rolle’s theorem, Ψ′ has a zero on (x_{1}, x_{2}), which gives a contradiction. Hence Φ has no zeros on (x_{1}, x_{2}) in this case. This completes the proof of Lemma 1.

References

[1] Cauchy, A.L.: Oeuvres de Cauchy 2ième série, Tom XV. - Gauthier-Villars, Paris, 1974.

[2] Hellerstein, S., L.C. Shen, and J. Williamson: Reality of the zeros of an entire function and its derivatives. - Trans. Amer. Math. Soc. 275, 1983, 319–331.

[3] Hinkkanen, A.: Reality of zeros of derivatives of meromorphic functions. - Ann. Acad. Sci. Fenn. Math. (to appear).

[4] Hinkkanen, A.: Iteration, level sets, and zeros of derivatives of meromorphic functions. - Preprint.

[5] Hinkkanen, A., and J. Rossi: On a problem of Hellerstein, Shen and Williamson. - Proc. Amer. Math. Soc. 92, 1984, 72–74.

[6] Nielsen, N.: Handbuch der Theorie der Gammafunktion. - Teubner, Leipzig, 1906; reprinted in Die Gammafunktion, Chelsea, New York, 1964.

[7] Pólya, G., and G. Szegö: Problems and Theorems in Analysis II. - Springer-Verlag, Berlin, 1976.

Received 18 July 1995