YET ANOTHER CHARACTERIZATION OF THE SINE FUNCTION

In this expository paper, it is shown that if an entire function of exponential type vanishes at least once in the complex plane and if it has exactly the same number of zeros (counting multiplicities) as its second derivative, then this function must take the form Asin(Bz

get stronger results in uch less space, the reader could consider this artlcle as an Invtatlon to the use of Nevanllnna theory in the study of differential equations.
In the set of entire functions, it is customary to classlfy functions according to the growth of their modulus.In this spirit, we give the follown8 deflntlon: an entire function f is of exponential type if there exist two real positive constants C and such that where designates the complex plane.
If f is a function of exponential type, then for every z E such that zl r > o, we my .rite in order to get the estimate "Ce T(').. C'.err.
'The last inequality is obtained va the preceding definition of a function of exponential type, and we may deduce from this inequality that the derivative of a function of exponential type is itself a function of exponential type.
The theorem we are about to establish may be formulated in the followng way: THEOREM A. Let f be an entire function of exponential type, possessing at least one zero.If f is such that z is a zero of multlpllclty m of f if and only if z is also a zero of multiplicity m of f", the second derivative of f, then f necessarily has the form where A, B and C are three complex constants.
To facilltate the exposition, we introduce the claas $ of entire funetns-f of exponential type that have at least one zero in the complex plane and that have the followlng propertT: z is a zero of f if and only if z is a zero of f", counting multipl.ic.itie.s.For convenience, we shall eliminate the function constantly 0 from S. are examples of memSers of S.More generally, f(z) A sin (Bz + C) is a function in S, and the preceding theorem asserts that every element of S is of this form.
We turn now to the proof of theorem k.
Let f be a function in S. Then the function Z" (z) is an entire function wthout zeros, and we shall show that, in this case, it must take the form (z) e h(z) for some entire function h.We observe that @'/@ is itself an entire function that must be the derivative of an entire function #, i.e. ' #'/.Consider now the new function (z) (z) e -(z).
Here, C is a constant.Hence each element f S satisfies the differential equation f"Cz) fCz)e h(z) for some entire function h.
We show now that in fact the function h must be a polynonal of degree at most one.To do this, we shall use Jensen's formula (cf.Convay [1], p. 283): xo Io1 ./log If(rete)J de : log (it|)" Here it is supposed Chat f is holomorphtc in [zJ r, and that a 1, ..., a n are the zeros of f contained in [z[ < r, repeated as many times as I2w log+(CeTr) d6 0 < Clr for a constant C 1. Finally, using the triangle inequality, we deduce the result 0 0 0 s 2Clr.
In the same fashion, we could demonstrate that, for another constant C 2, 2= d8 S 2C2r 0 on supposing also that If"(0) 1.
Eeturnln8 to the function h of the identity (2) and writing h(z) u(z) + Iv(z) we may use equation ( 2 where the constant D, independent of r, satisfies the inequality OO2r 02 2re iO + re i$ re i rei$ Since h is an entire function that grows no faster than a constant multiple of the independent variable, we may use a direct consequence of Liouvtlle's theorem to conclude that h is a polynomial of degree at most 1.
Thus, if we summarize the present situation, we have, for every f e S such that the identity and If"(0 Az+B P'(z) (z) e or equivalently " (z) ceZ(z) (4) for two possibly complex constants A and C. We now show directly that the wo hypotheses in (3) only onstitute a simple normalization.In the first place, if we had f(O) 0 (and hence f"(O) 0 since f S), the trouble would be that If(O) < 1 or [f"(O)l < I, so we could take fl(z) af(z) where 1 m I'(o)l The function fl belongs to S and satisfies (3).If fl takes the form indicated in theorem A, then f also does.In case f(O) 0 (and consequently f"(0) 0) then we perform the translatlon where is a constant chosen so that f2(0) O. Now one proceeds to show that f2 has the required form, and hence that f does.
In the sequel, we shall simplify the exposition by supposing, wthout loss of generality, that f e S, and that f(0)[ > 1 and [f"(O)[ > 1, so that f satisfies ().
Our aim, at this point, is to show that the constant A in (4) must be zero, so let us suppose otherwise.For s/mpltcity, we shall suppose A 1 in (4) since otherwise we could consider the function z(z) () 'which also belongs to S and satisfies the dlfferentlal equation F"(z) C' eZ(z) where C' is a constant.
Let f (z) Z a z n n-0 be the Taylor series of f.We may estimate the coefficients as follows: (n)

I%! ,co)
Ce TE for all r > 0 and n > O. r n Let us choose r n and use Stirling's formula to deduce the estimate 1 1 1 c" e 'r )E C' for all n z 0 where we suppose the elementary fact that limn I/n i. Consequently we find 1 sup ]ann]] n p < ", which sinlfies that the series ..ann'.
converges uniformly for Iwl > p' > 0 and thus defines a function that is holomor- phlc in a neighborhood of infinity and that vanishes at .Now consider, for Re(w) > max {0,T'}, the integral n-0z an f0 tn e-we dt a n n

Z n+l n=0 w
The interchange of the integration and the summation is Justified by (5) and its consequences.By the remark of the preceding paragraph, the function thus defined is holomorphic in a right half-plane.On the other hand, we have remarked that the derivative of an entire function of exponentlal type is again of exponentlal type.We may apply this to do the following integration by parts C-) f C) e -vt: cSt e- 'dr. w w2 0 Finally, since f" must satisfy the equation (4) with A I, the function satisfies the following relation: (.) -(o) '(o) -c (.-) where C is the same constant as in (4).Now we have remarked above that since is holomorphlc in a right half-plane, (and reca11ing that C O because we have ruled out f _= O), the last Inequallty allows us to continue aualytlcally to the whole complex plane, as follows.We know that is holomorphlc for Re(w) > B > max {p, T}, and the preceding equation allows us to continue analytically to Re(w) > B 1, then to Re(w) > B 2, and so on, untll the whole complex plane is covered, moving to the left by a band of wdth 1 each time.But we know also that is holomorphlc in a neighborhood of infinity.Hence the analytlc continuation of'@ is holomorphic on the whole Itlemann sphere, and must therefore be a constant.This constant is actually zero, since (=) 0. Now since we have n ann" f(z) Z a z and (w)-Z wn+l n=0 n n=0 where the coefficients an, n 0, I, 2, ..., that appear in the two developments are the same, and since we have shown that O, we have f O, which contradicts our excluslon of 0 from S. Hence the constant A of equation (4) must be zero.
Before ending, we remark that once it is established that a function f of the class S must satisfy a differential equation of the form (4), there are alternative elementary proofs at hand.The one we have chosen has the advantage of remaining in the field of functions of a complex variable, but one could alternatively proceed directly from the solution of (4) obtained by the classical methods of the theory of differential equations and a detailed examination of the solution to derive the conclusion of theorem A.
) to write logl f"(ree) logl f(reis) + u(rele) This leads us (tak/ng account of the preceding inequalities) C 3. Finally, we write the representation of h as a coplex Polsson ntegral (cf.Rudin[7],p.228) their multiplicity indicates.From this inequality, we deduce that