Meromorphic extendibility and the argument principle

Publ. Mat. 52 (2008), 171–188 MEROMORPHIC EXTENDIBILITY AND THE ARGUMENT PRINCIPLE Josip Globevnik Abstract Let ∆ be the open unit disc in C. Given a continuous function ϕ:b∆→C\{0}denote by W(ϕ) the winding number of ϕaround the origin. We prove that a continuous function f:b∆→Cextends meromorphically through ∆ if and only if there is a number N∈N∪ {0}such that W(P f +Q)≥ −Nfor every pair P,Q of polynomials such that P f +Q6= 0 on b∆. If this is the case then the meromorphic extension has at most Npoles in ∆. 1. Introduction and the main result Let ∆ be the open unit disc in Cand let f:b∆→Cbe a continuous function. We say that fextends holomorphically through ∆ if fadmits a continous extension ˜ fto ∆ which is holomorphic on ∆. If this is the case then we say that f(or ˜ f) belongs to the disc algebra. Denote by C= C∪ {∞} the Riemann sphere. We say that fextends meromorphically through ∆ if there is a finite set A⊂∆ such that fhas a continuous extension to ∆\Awhich is holomorphic on ∆ \Aand has a pole at each point of A. Equivalently, fextends meromorphically through ∆ if it has a continous extension ˜ f:∆→Cwhich, as a function to C, is holomorphic on ∆. Given a continuous function ϕ:b∆→C\{0}we denote by W(ϕ) the winding number of ϕ(around the origin). So W(ϕ) equals 1/(2π) times the change of argument of ϕ(z) as zruns once around b∆ counterclockwise. In the present paper we show that meromorphic extendibility can be characterized in terms of the argument principle. For holomorphic extendibility this is already known: 2000 Mathematics Subject Classification. 30E25. Key words. Argument principle, meromorphic extensions. 172 J. Globevnik Theorem 1.0 ([G2]).A continuous function f:b∆→Cextends holomorphically through b∆if and only if W(f+Q)≥0for every polynomial Qsuch that f+Q6= 0 on b∆. If a continuous function f:b∆→C\ {0}extends meromorphically through ∆ then W(f)≥ −Nwhere Nis the number of poles of the meromorphic extension ˜ f(counted with multiplicity). Indeed, by the argument principle, W(f) = ν0(˜ f)−νp(˜ f) = ν0(˜ f)−N≥ −N where ν0(˜ f) is the number of zeros of ˜ fon ∆ and νp(˜ f) is the number of poles of ˜ fon ∆. Let f:b∆→Cbe a continuous function on b∆ which extends meromorphically through ∆ and whose meromorphic extension ˜ fhas Npoles on ∆. Then W(Pf +Q)≥ −Nfor all polynomials P,Qsuch that Pf +Q6= 0 on b∆. Indeed, P˜ f+Q, the meromorphic extension of Pf +Q, has no other poles than ˜ fand therefore, by the argument principle, W(Pf +Q)≥ −N. The following theorem, our main result, tells that this property characterizes meromorphic extendibility. Theorem 1.1. A continuous function f:b∆→Cextends meromorphically through ∆if and only if there is an N∈N∪ {0}such that (1.1) W(Pf +Q)≥ −N for all polynomials P,Qsuch that Pf +Q6= 0 on b∆. If this is the case then the meromorphic extension of fhas at most Npoles in ∆, counting multiplicity. 2. Fourier series In this section we recall some well known facts. Let fbe a continuous function on b∆. For each integer nlet ˆ f(n) = 1 2πZπ −π e−inθf(eiθ)dθ so that ∞ X n=−∞ ˆ f(n)einθ is the Fourier series of f. We have (2.1) ∞ X n=−∞ |ˆ f(n)|2<∞. Meromorphic Extendibility 173 Define the functions Fand Gby F(z) = ˆ f(0) + ˆ f(1)z+ˆ f(2)z2+··· (z∈∆) G(z) = ˆ f(−1)z+ˆ f(−2)z2+··· (z∈∆). The functions Fand Gare holomorphic on ∆ and by (2.1) they belong to the space H2[R]. The function fbelongs to the disc algebra if and only if ˆ f(n) = 0 for all n < 0 or, equivalently, if and only if G≡0. Suppose now that fis smooth. Then the Fourier series converges uniformly to f. The functions Fand Gbelong to the disc algebra and have smooth boundary values. We have f(z) = F(z) + G(z) (z∈b∆). We shall need the following Proposition 2.1. Let Φ: b∆→Cbe a continuous function. Given N∈Nthere is a nonzero polynomial Pof degree not exceeding Nsuch that [ (PΦ)(j) = 0 (−N≤j≤ −1). Proof: PΦ is continuous on ∆ and a direct computation shows that for each integer jwe have [ (PΦ)(j) = ˆ P(0)ˆ Φ(j) + ˆ P(1)ˆ Φ(j−1) + ···+ˆ P(N)ˆ Φ(j−N) so \ (PF )(j) = 0 (−N≤j≤ −1) gives the homogeneous system ˆ P(0)ˆ Φ(−N) + ˆ P(1)ˆ Φ(−N−1) + ···+ˆ P(N)ˆ Φ(−2N) = 0 ˆ P(0)ˆ Φ(−N+ 1) + ˆ P(1)ˆ Φ(−N) + ···+ˆ P(N)ˆ Φ(−2N+ 1) = 0 . . . ˆ P(0)ˆ Φ(−1) + ˆ P(1)ˆ Φ(−2) + ···+ˆ P(N)ˆ Φ(−N−1) = 0 of Nlinear equations with N+ 1 unknowns ˆ P(0),ˆ P(1),..., ˆ P(N) which always has a nontrivial solution. This completes the proof. 174 J. Globevnik 3. The smooth case Theorem 3.1. Suppose that f:b∆→Cis of the form f(z) = G(z) + H(z) (z∈b∆) where the functions Gand Hbelong to the disc algebra and Hhas smooth boundary values. Assume that N∈N∪ {0}and that (3.1) (W(Pf +Q)≥ −Nwhenever P, Q, are polynomials, deg(P)≤N, such that Pf +Q6= 0 on b∆.. Then fextends meromorphically through ∆and the meromorphic extension has at most Npoles in ∆, counting multiplicity. Remark 3.2.To prove Theorem 1.1 we shall later use Theorem 3.1 only in the special case when His a rational function holomorphic in a neighbourhood of ∆ so with this in mind, we may assume as much smoothness as we want. In the proof of Theorem 3.1 below it is enough to assume that H|b∆ belongs to the Lipschitz class Cαwith α > 1/2. Before proceeding observe that if fis as in Theorem 3.1 and Pis a polynomial then P f has the same form. Indeed, we have P f =PG+PH on b∆ where the function z7→ P(z)H(z) is smooth on b∆ so on b∆ we have PH=F2+G1where F2,G1belong to the disc algebra and have smooth boundary values. So on b∆ we have Pf =P G +F2+G1= F1+G1where F1,G1are in the disc algebra and G1has smooth boundary values. Proof of Theorem 3.1: Assume that fis as in Theorem 3.1 and that (3.1) holds for some N∈N∪ {0}. If N= 0 then it is known that fextends holomorphically through ∆ [G2]. Assume that N≥1. By Proposition 2.1 there is a polynomial P, deg(P)≤N, such that (3.2) [ (Pf)(−1) = [ (Pf)(−2) = ···=[ (Pf)(−N) = 0. Now, Pf =F1+G1on b∆ where F1,G1are in the disc algebra and G1has smooth boundary values. With no loss of generality assume that G1(0) = 0. By (3.2) G1=zN+1G2where G2is again in the disc algebra and has smooth boundary values so that P(z)f(z)−F1(z) = zN+1G2(z) (z∈b∆). Suppose for a moment that G26≡ 0. We show that there is a constant α∈Csuch that zN+1G2(z) + α6= 0 (z∈b∆) and (3.3) W(zN+1G2(z) + α)≤ −N−1, Meromorphic Extendibility 175 that is, W(zN+1G2(z) + α)≥N+ 1. The function Φ(z) = zN+1G2(z) belongs to the disc algebra and has smooth boundary values. It has zero of order at least N+ 1 at the origin. If Φ(z)6= 0 (z∈b∆) then put α= 0. In this case W(Φ) equals the number of zeros of Φ in ∆ so W(Φ) ≥N+ 1. Suppose now that Φ(b∆) contains the origin. Since Φ has smooth boundary values it follows that Φ(b∆) is nowhere dense. So there are α, arbitrarily close to the origin such that Φ(z)+α6= 0 (z∈b∆). Let ν≥N+1 be the multiplicity of the zero of Φ at the origin. A standard use of the argument principle on a sufficiently small disc Dcentered at the origin shows that for any α sufficiently close to the origin, α6= 0, the function z7→ Φ(z) + αhas exactly νzeros on Dwhich are arbitrarily close to the origin provided that αis sufficiently close to the origin. Thus, if α6= 0 is sufficiently close to 0 and Φ(z)+α6= 0 (z∈b∆) then Φ+αhas νzeros in a neighbourhood of the origin so the argument principle, now applied to the function Φ−α on ∆, implies that W(Φ + α)≥ν≥N+ 1 so that (3.3) holds. It follows that W(Pf −F1+α)≤ −N−1. A sufficiently good polynomial approximation Qof −F1+αthen satisfies W(Pf +Q)≤ −N−1, contradicting (3.1). It follows that G2≡0 so Pf =F1on b∆, that is, Pf belongs to the disc algebra. We need Proposition 3.3 ([G3, Proposition 5.1, p. 223]).Let Ψbe in the disc algebra, let a∈b∆and assume that the function z7→ Ψ(z)/(z−a) (z∈b∆\ {a})extends continuously to b∆. Then there is a function Ψ1 from the disc algebra such that Ψ1(z) = Ψ(z)/(z−a) (z∈∆\ {a}). Proof of Theorem 3.1 (continued): Writing P(z) = p0(z−a1)(z−a2)··· (z−aM) we have M≤Nand (z−a1)···(z−aM)f(z) = H1(z) (z∈b∆) where H1belongs to the disc algebra. Let α1,...,αJbe those of a1,...,aMwhich are contained in ∆. By Proposition 3.3 we may write (z−α1)···(z−αJ)f(z) = H(z) (z∈b∆) where Hbelongs to the disc algebra and J≤N. This completes the proof of Theorem 3.1. Remark 3.4.The preceding proof does not work without a smoothness assumption as it is known that there are functions hin the disc algebra such that h(b∆) = h(∆) [G1]. 176 J. Globevnik 4. The general case Lemma 4.1. Let N∈Nand let f:b∆→Cbe a continuous function such that the Fourier series ei(N+1)θf(eiθ)∼ ∞ X n=−∞ Aneinθ of the function eiθ 7→ ei(N+1)θf(eiθ)is such that A06= 0 and A1=A2=···=AN= 0, A−1=A−2=···=A−N= 0. There is a polynomial Qsuch that f+Q6= 0 on b∆and W(f+Q)≤ −N−1. Proof: With no loss of generality we may assume that A0= 1. Since zN+1fis continuous Fej´ers theorem implies that zN+1fis the uniform limit of the Ces`aro means of its Fourier series [Ho]. So, if Sk(eiθ) = k X j=−k Ajeijθ k= 0,1,2,... are the partial sums of the Fourier series then ei(N+1)θf(eiθ) is the uniform limit, as m→ ∞, of Cm(eiθ) = 1 m+ 1S0(eiθ) + S1(eiθ) + ···+Sm(eiθ). However, each partial sum Snand therefore each Ces`aro mean Cmhas the same coefficients vanishing property as the one which we have assumed for the Fourier series of zN+1f: [ (Cm)(0) = 1,[ (Cm)(j) = 0 (−N≤j≤N, j 6= 0) so that Cm(z) = 1 + zN+1Rm(z) + zN+1Tm(z) (z∈b∆) where Rm,Tmare polynomials. Choose mso large that (4.1) |Cm(z)−zN+1f(z)| ≤ 1 2(z∈b∆). We have Cm(z)−zN+1Rm(z)−zN+1Tm(z)∈1 + iR(z∈b∆) which, by (4.1) implies that zN+1f(z)−zN+1Rm(z)−zN+1Tm(z)∈[1/2,3/2] + iR(z∈b∆). Meromorphic Extendibility 177 It follows that zN+1f−zN+1Rm−zN+1Tm6= 0 on b∆ and W(zN+1(f− Rm−Tm)) = 0. Thus W(f−Rm−Tm) = −N−1 so putting Q=−Rm−Tmcompletes the proof. Remark 4.2.Note that the assumption in Lemma 4.1 is equivalent to saying that ˆ f(−N−1) 6= 0 and ˆ f(−1) = ˆ f(−2) = ···=ˆ f(−N) = 0 and ˆ f(−N−2) = ˆ f(−N−3) = ···=ˆ f(−2N−1) = 0. We now turn to the proof of Theorem 1.1. We have already proved the only if part in Section 1. To prove the if part suppose that f:b∆→Cis a continuous function which satisfies (1.1) for all polynomials P,Qsuch that Pf +Q6= 0 on b∆. If N= 0 then we already know that fextends holomorphically through ∆ so assume that N≥1. Lemma 4.3. Let F:b∆→Cbe a continuous function. Assume that for some N∈Nwe have (4.2) W(PF +Q)≥ −N whenever P,Qare polynomials such that Pf +Q6= 0 on b∆. Then F(z) = G(z) + H(z) (z∈b∆) where Gbelongs to the disc algebra and His a rational function holomorphic in a neighbourhood of ∆. Assume for a moment that Lemma 4.3 holds. Since our rational function His smooth on b∆ the if part of Theorem 1.1 is now an immediate consequence of Lemma 4.3 and Theorem 3.1. It remains to prove Lemma 4.3. Given an infinite row A= (a1, a2,...) and J∈Nwe denote by A(J) the row containing the first Jentries of A, that is, A(J) = (a1, a2,...,aJ). Assume that F∈C(b∆) satisfies (4.2) whenever P,Qare polynomials such that P F +Q6= 0 on b∆. Lemma 4.1 implies that if Pis a polynomial such that \ (PF )(−N−2) = \ (PF )(−N−3) = ···=\ (PF )(−2N−1) = 0 \ (PF )(−1) = \ (PF )(−2) = ···=\ (PF )(−N) = 0 178 J. Globevnik then \ (PF )(−N−1) = 0. If P(z) = D0+D1z+···+DMzMthen \ (PF )(−1)=D0ˆ F(−1)+D1ˆ F(−2) + ···+DMˆ F(−M−1) . . . \ (PF )(−N+ 1) = D0ˆ F(−N+ 1) + D1ˆ F(−N) + ··· ···+DMˆ F(−M−N+ 1) \ (PF )(−N)=D0ˆ F(−N)+D1ˆ F(−N−1) + ···+DMˆ F(−M−N) \ (PF )(−N−1) = D0ˆ F(−N−1) + D1ˆ F(−N−2) + ··· ···+DMˆ F(−M−N−1) \ (PF )(−N−2) = D0ˆ F(−N−2) + D1ˆ F(−N−3) + ··· ···+DMˆ F(−M−N−2) . . . \ (PF )(−2N−1) = D0ˆ F(−2N−1) + D1ˆ F(−2N−2) + ··· ···+DMˆ F(−2N−M−1). Consider the infinite rows X−1= ( ˆ F(−1),ˆ F(−2),ˆ F(−3),...) X−2= ( ˆ F(−2),ˆ F(−3),ˆ F(−4),...) . . . X−N= ( ˆ F(−N),ˆ F(−N−1),ˆ F(−N−2),...) X−N−1= ( ˆ F(−N−1),ˆ F(−N−2),ˆ F(−N−3),...) X−N−2= ( ˆ F(−N−2),ˆ F(−N−3),ˆ F(−N−4),...) . . . X−2N−1= ( ˆ F(−2N−1),ˆ F(−2N−2),ˆ F(−2N−3),...). Meromorphic Extendibility 179 The preceding discussion shows that for every M∈Nthe following holds: if a row (D0, D1···DM) is orthogonal to the rows X−1(M+ 1), X−2(M+ 1),...,X−N+1(M+ 1), X−N(M+ 1) X−N−2(M+ 1), X−N−3(M+ 1),...,X−2N−1(M+ 1) (4.3) then it is orthogonal to X−N−1(M+1). This implies that for every M∈N the row X−N−1(M+ 1) is a linear combination of 2Mrows (4.3). It follows that there are numbers λj,−2N−1≤j≤ −1, j6=−N−1, such that (4.4) X−N−1=X −2N−1≤j≤−1, j6=−N−1 λjXj. Consider the function Ψ(z) = ˆ F(−N−1)z+ˆ F(−N−2)z2+··· . The function Ψ is holomorphic on ∆ and since −N−1 X n=−∞ |ˆ F(n)|2≤ ∞ X n=−∞ |ˆ F(n)|2<∞ it follows that Ψ belongs to the space H2[R]. We use (4.4) to show that Ψ is a rational function. Note that (4.4) implies that Ψ(z) = ˆ F(−N−1)z+ˆ F(−N−2)z2+··· =λ−1ˆ F(−1)z+ˆ F(−2)z2+··· +λ−2ˆ F(−2)z+ˆ F(−3)z2+··· . . . +λ−Nˆ F(−N)z+ˆ F(−N−1)z2+··· +λ−N−2ˆ F(−N−2)z+ˆ F(−N−3)z2+··· . . . +λ−2N−1ˆ F(−2N−1)z+ˆ F(−2N−2)z2+···. 186 J. Globevnik By the preceding discussion there are numbers dj,ℓ, 0 ≤ℓ≤pj, 1 ≤j≤ k, such that if for each j, 1 ≤j≤kthe function Φ satisfies Φ(z) = dj,0+dj,1(z−αj) + ···+dj,pj(z−αj)pj+··· then (6.3) holds for each j, 1 ≤j≤k. It is an easy application of the Weierstrass factorization theorem to construct an entire function Φ with this property [R, Theorem 15.13, p. 304]. The function Q=eΦ−Ψ (z−α1)p1···(z−αk)pk will have the required properties. This completes the proof. The following question is open: Question 6.2. Let f:b∆→Cbe a continuous function. Suppose that for some N∈Nwe have W(f+Q)≥ −Nfor all polynomials Qsuch that f+Q6= 0 on b∆. Must fextend meromorphically through ∆? In other words, we are asking whether in Theorem 1.1 it is enough to assume that P≡1 or, equivalently, whether the precise analogue of Theorem 1.0 holds for meromorphic extendibility. We do not know the answer even in the case when fis smooth. We conclude with a remark about holomorphic extendibility. It is an obvious question whether Theorem 1.0 holds for a smaller class of polynomials Q. That linear polynomials do not suffice was shown in [W], that polynomials of uniformly bounded degree do not suffice was shown in [G2]. One may ask, for instance, whether the polynomials Qsatisfying Q(0) = 0 suffice. The answer is no as shown by the example f(z) = z/(z−1/2) (z∈b∆). Indeed, writing Q(z) = zQ1(z) where Q1is a polynomial the argument principle implies that W(f+Q) = Wz z−1/2+Q=W z1 + (z−1/2)Q1 z−1/2!≥0 for all polynomials Qsuch that Q(0) = 0 and such that f+Q6= 0 on b∆, yet fdoes not extend holomorphically through ∆. However, there is no such example if fhas a meromorphic extension through ∆ which does not vanish at 0 by the following Proposition 6.3. Let Sbe a polynomial with all its zeros contained in ∆. Suppose that f:b∆→Cis a continuous function such that W(f+SQ)≥0for each polynomial Qsuch that f+SQ 6= 0 on b∆. If fextends meromorphically through ∆and the meromorphic extension has no common zeros with Sthen fextends holomorphically through ∆. Meromorphic Extendibility 187 Proof: Suppose that f(z) = H(z) (z−a1)···(z−aN)(z∈b∆) where aj∈∆ (1 ≤j≤N) and where His in the disc algebra, H(aj)6= 0 (1 ≤j≤N), such that Hhas no common zero with S. By Lemma 6.1 there is a polynomial Qsuch that H+ (z−a1)···(z−aN)SQ has no zero on ∆ It follows that 0≤ W(f+SQ) = WH+ (z−a1)···(z−aN)SQ (z−a1)···(z−aN)=−N which implies that N= 0, so fextends holomorphically through ∆. This completes the proof. Acknowledgements. The author is grateful to Sergey Ivashkovich for explaining the simple facts about meromorphic extendibility mentioned in the beginning of Section 5. This work was supported in part by the Ministry of Higher Education, Science and Technology of Slovenia through the research program Analysis and Geometry, Contract No. P1-0291. References [AW] H. Alexander and J. Wermer, Linking numbers and boundaries of varieties, Ann. of Math. (2) 151(1) (2000), 125–150. [D] P. 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Rudin,“Real and complex analysis”, Third edition, McGrawHill Book Co., New York, 1987. [S] E. L. Stout, Boundary values and mapping degree, Michigan Math. J. 47(2) (2000), 353–368. [W] J. Wermer, The argument principle and boundaries of analytic varieties, in: “Recent advances in operator theory and related topics” (Szeged, 1999), Oper. Theory Adv. Appl. 127, Birkh¨auser, Basel, 2001, pp. 639–659. [Z] A. Zygmund,“Trigonometric series”, 2nd ed., Vols. I, II, Cambridge University Press, New York, 1959. Institute of Mathematics, Physics and Mechanics University of Ljubljana Ljubljana Slovenia E-mail address:[email protected] Primera versi´o rebuda el 2 de febrer de 2007, darrera versi´o rebuda el 13 d’abril de 2007.